Justin

Combinations – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Combinations.

He will start with defining a combination and move on to running through an example of how combinations are used in the statistics and probability portion of the exam.

This Fundamentals of Engineering Exam Review of Combinations is part of the global subject Probability and Statistics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Combinations.

When you need to count the number of groupings without regard to order then combinations is the way to go. Now in a previous tutorial, I went over permutations.

Permutations specifically count the number of ways a task can be arranged or ordered…that is the fundamental difference between the two.

A permutation is with regard to order and combinations is without regard to order.

So if you guys need a refresher on that you can go ahead and jump over to that tutorial once this one’s complete this one will be completely standalone so you can continue to watch and follow along.

So let’s start off our Fundamentals of Engineering Exam Review of Combinations by defining exactly what Combinations are.

A combination is an arrangement of our objects without

regard to order and without repetition

selected from n distinct objects is

called a combination so to put that in

into an equation a combination a

combination oops a combination of n

objects taken R at a time is equal to n

factorial divided by n minus R factorial

R factorial so the difference between

combinations and permutations again is

in combinations you are counting groups

order is not important and in

permutations you are counting different

ways to arrange items with regard to

order the N and R mean the same thing in

both permutation and combinations but

the formula differs and it differs by

this

our factorial in the denominator so

let’s define that one more time and

arrangement

of our objects

without regard and I will emphasize

without without regard to order and

without repetition

selected from selected from n distinct

objects

is called a combination

that’s it so without regard without

regard to order combinations with regard

to order permutation so let’s go through

an example to illustrate that let’s say

that we want to draw three marbles at

random from a bag containing three red

and five white marbles so we want to

take three marbles at random from a bag

that contains three red and five white

so this is the bag and the bag has eight

total marbles and we want to take three

so how many different draws or how many

different ways or how many different

ways can can three marbles come out of

this bag of eight this bag of eight

that’s broken up into three red and five

white this would be a combination

problem because the draw would be a

group of marbles without regard to order

it is like grabbing a handful of marbles

and just looking at them note that there

are no special conditions placed on the

marbles that we draw so this is

straightforward combination problem and

note that if we are putting these

marbles in any kind of order then we

would need to use permutations to solve

the problem but in this case order does

not matter so we’re going to use

combinations so let’s go ahead and

define N and R for a combination

equation n is the number of marbles we

can choose from in this bag once again

has eight total marbles so n is equal to

eight R is the number of marbles we will

be drawing at one time and we want to

choose three marbles so R is equal to

three and all we need to do is plug this

into our combination equation which is n

factorial divided by

n minus R factorial R factorial and

plugging that in n is equal to 8 and R

is equal to 3 and simplifying that we

got 8 factorial divided by 5 factorial

and 3 factorial so once again you can

simplify this further by recognizing

that 8 factorial is 8 times 7 times 6

times 5 times 4 times 3 times 2 all the

way down to 1 and then 5 factorial is 5

times 4 times 3 times 2 times 1 and 3

factorials 3 times 2 times 1 and notice

that we can just eliminate a couple of

these factors here because they just

divide out and our final answer is going

to be 8 times 7 times 6 divided by 3

times 2 times 1 and in this case there’s

it’s going to equal 56 and so there are

56 different draws that can come from

this bag of eight total marbles so let’s

move on with using this same bag of

marbles let’s move on to another example

to illustrate another scenario so let’s

say this time that we want to know how

many different draws would contain only

red marbles so remember there’s eight

total marbles and that three are red and

that five are white and out of these

eight marbles we’re still choosing three

at random

so once again this is a combination

problem because the draw would be a

group of the marbles and with without

regard to order and so on Part A we just

we determine how many different possible

draws there are which is 56 now we want

to know out of those 56 how many would

give us 3 red so our draw would look

like just three red marbles so let’s go

ahead and define N and R so in in this

case is n is the number of red marbles

we can choose from and in this case

there’s only three red marbles that are

in the bag so n is equal to 3 and R is

the number of red marbles that are drawn

at a time once again 3 R 3 are randomly

chosen but we want all three of those to

be red marbles so R is also going to

equal 3 so plugging that into our

combination equation n factorial divided

by n minus R factorial R factorial n is

equal to 3 factorial 3 minus 3 factorial

R factorial oops

R is equal to 3 this simplifies out the

3 factorial over zero factorial times 3

factorial and we can figure out that

that’s equal to 1 so out of the 56

possible draws only one draw would

contain 3 red marbles so let’s go one

more let’s go just a bit further to

complicated a little bit more but to

illustrate the use of combination in a

different scenario let’s use the same

bag of eight marbles so once again we

have eight marbles

three of which are red and five are

white so in this case how many different

drawers would contain one red marble and

two white marbles so we want one red

marble and we want two white marbles so

once again this is a combination problem

because this draw would be just a group

of marbles without regard to order and

like like we stated before in the first

part of this problem there’s 56 possible

drawers and of those drawers we want to

know how many of them would have one red

and two white so first we need to

determine n NR together that would make

up one draw we are going to have to use

the counting principle in this problem

to help us out note how one draw one

draw of three marbles is split into two

parts red and white we cannot combine

them together because we need a

particular number for each one so we

will figure out how many ways to get one

red marble and how many ways to get two

white marbles and use the counting

principle we will multiply these numbers

together to get our final possible

outcomes possible drawers so let’s start

out with one red so in in this case

would be the number of red marbles we

can choose from so in the bag there’s

three red marbles and our would be the

number of red marbles we are drawing at

a time and in this draw we want one red

marble so plugging that in to the

equations we get n is equal to one

and our sorry n is equal to 3 and R is

equal to 1 and so that gives us if we

calculate that out that gives us or

simplify that gives us 3 2 factorial

times 1 factorial

now before we calculate this out let’s

go ahead and just look at the white

scenario let’s say we want to whites so

in this case n is equal to 5 because

there’s five white marbles in the bag

and in this case we also want two of

those to the three that we choose to be

white so R is equal to two so plugging

that in to the equation we have n is

equal to five and we have R is equal to

two so to simplify that that’s five

factorial divided by 3 factorial times 2

factorial now to get the total number of

possible draws that would contain one

red and two white we use D we use the

counting rule and multiply these two two

possibilities together to get the final

value so that would give us 3 factorial

divided by 2 factorial times one

factorial for our red scenario right

here multiply that out by 5 factorial

divided by 3 factorial times 2 factorial

and so we can just use our calculator in

this case would probably be the most

simple way but you can simplify it by

you know simplifying each of these and

two simpler more simple terms so we got

3 over 1 multiplied by 5 times 4 divided

by 2 times 1 and that equals 30 so out

of the 56 possible draws that we

determine in the first part 30 of those

would contain one red and two white

marbles so that’s it that’s all for our Fundamentals of Engineering Exam Review of Combinations and what Combinations consist of I hope that

helped you guys out a little bit

refreshed it just a tiny bit if you guys

have any more questions don’t hesitate

to contact me over at engineering

training exam calm and be more than

happy to help you out in whatever way

you need so we’ll be talking too soon

too soon

—

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Permutations – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Permutations.

We will begin by defining a factorial, move on to defining a Permutation, and then run through an example of how permutations are calculated in the statistics and probability portion of the exam.

This Fundamentals of Engineering Exam Review of Permutations is part of the global subject Probability and Statistics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Permutations.

Permutations are an offshoot of the fundamental counting principle which I covered in a previous review. Permutations are used to specifically count the number of ways a task can be arranged or ordered.

So let’s start off our Fundamentals of Engineering Exam Review of Permutations by defining exactly factorials are.

Factorial written as in or any
number exclamation point that’s n
factorial and n factorial is the product
of whatever n is n times n minus 1 times
n minus 2 all the way down to 1 so let’s
say you want to get 7 factorial it would
first be written as 7 factorial and say
you want the product or what that equals
the answer to that would be 7 times 6
times 5 times 4 all the way down to 1
and so the answer to that if you just
put that into your calculators real
quick would be 5040 so real quick before
we go on factorials are pretty basic and
I know you guys all know them but it
just wanted to touch on those real quick
but one special definition of factorial
is we need to know that 0 factorial is
always equal to 1 so if N in feature
problems we see zero factorial it’s not
that we’re dividing by 0 that actually
is equal to 1 so moving on what is a
permutation now a permutation is an
order of arrangements of our objects
without repetition selected from n
distinct objects is called a permutation
of n objects
can are at a time and that is denoted as
in P R is equal to n factorial divided
by n minus R factorial so let’s define
that again and order and that’s
important so make sure you underline
that an order of arrangements so I
running out of room here an order of
arrangements of our objects selected or
taken from n distinct objects is called
a permutation of n objects taken R at a
time so that might look a little
confusing butting in equation form you
can just note that the permutation of n
objects taken R at a time is equal to n
factorial divided by n minus R factorial
so to simplify my complicated definition
here that might have confused you guys
sorry about that but to simplify when
you need to count the number of ways you
can arrange items where order I bring
that up again where order is important
then you can use permutations to count
so once again when you need to count the
number of ways you can arrange items
where order is important then you can
use permutation to count so let’s look
at it let’s illustrate this at using in
an example here so let’s say that we
have 8 pictures and we want to hang all
eight of those pictures on to the wall
so the question would be in how many
ways
can eight pictures be hung on the wall
so since we are arranging these pictures
on the wall that means order is
important and because order is important
we can use permutations to count how
many ways so first we need to we need to
define the number of objects or pictures
and so in like I said is the pictures in
this in this example so since there’s
eight pictures in is going to equal
eight now are on the other hand is the
number of pictures we are going to use
at a time so once again we want to hang
all eight pictures on the wall so all
eight pictures are going to be used at
the same time so R is going to equal 8
so taking our permutations equation that
we defined before we got n factorial
divided by n minus R factorial and all
we need to do is plug in our our numbers
or our values for this specific problem
so we have eight objects so n is going
to equal 8 and we want to hang all eight
of them on the wall at the same time so
R equals 8 so we can simplify that as 8
over 0 factorial and once again I said
zero vectorial is equal to 1 so we can
go ahead and just simplify this as 8
times 7 times 6 times 5 times 4 all the
way down to 1
and when we multiply that all out we
find that there it’s 40,320 so that’s a
very large number and that tells us that
there’s four 40,320 ways that eight
pictures can be hung on the wall all at
the same time so let’s move on to
another example let’s say that we have
an engineering group and that
engineering group has 20 members and out
of those 20 members that engineering
group needs to select a president it
needs to select a vice president and it
needs to select a speaker and assuming
that the same person cannot hold that to
position two positions at the same time
assuming that a member can hold only one
position at a time that’s our assumption
so one member can’t be vice president
and president so we can go move forward
and define our values in an AR and once
again since we are choosing positions
this is a way to rank members and rank
equals order so order is important in
this problem once again so we can use
the equations of permutations to
determine the number of ways that three
members can be chosen from the twenty
within the group so we first need to
define n so n is the number of members
we can choose from and in this case we
have 20 members of those 20 members we
want to choose 3 for this the positions
the president the vice president and the
speaker so since we’re choosing 3 at a
time
r is going to equal 3 so once again all
we need to do is take our standard
permutation equation
we got NP R is equal to n factorial
divided by n minus R factorial and we
know that n is equal to 20 and we know
that R is equal to 3 so to simplify that
we got 20 factorial over 17 factorial
now let’s just say you know most of us
are going to be using the factorial key
on our calculators to determine the what
what these equations are equally and out
to be but let’s just say you you don’t
have that just for conversations sake
here you can simplify this equation so
you don’t have to write it out as 20
times 19 times 18 times 17 etc and since
I already wrote it out like that let me
show you how you can simplify it so
let’s expand on 20 factorial that’s 20
times 19 times 18 times 17 times 16 all
the way down to 117 factorial 17 times
16 times 15 times all the way down to 1
now all we need to need to notice here
is that we can cancel out every value
from the denominator that’s on the
numerator and starting at 17 all the way
down so our answer here will really be
20 times 19 times 18 and in this case
that equals 6 that 68 40 so running the
calculations this tells us that there
are six thousand eight hundred and forty
different ways to select three members
of the 20 member group for the three
positions so that’s it guys that’s all
there is to permutations I know it might
sound a bit complicated just the word in
itself but it’s really all not not that
big of a deal once again were concerns
with order in these problems so let me
just tell you once again that the
definition is an order of arrangements
of
are objects without repetition selected
from n distinct objects is called a
permutation so that’s it if you guys
have any more questions or need any more
guidance for preparing for the
engineering training exam don’t hesitate
to visit my site engineer and training
exam comm and I look forward to meeting
you guys and helping you in any way that
I’m able so for now take care and look
forward to talking to you soon

—

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Counting Principle – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Counting Principle.

He will begin by defining the Counting Principle and then run through an example in how we can use the counting principle in the probability and statistics portion of the exam.

This Fundamentals of Engineering Exam Review of Counting Principle is part of the global subject Probability and Statistics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Counting Principle.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Counting Principle by defining exactly what Law of Sines is.

The Counting Principle will allow us to
count the number of ways a task can
occur given a series of events so
basically you are multiplying the number
of possibilities
each event within a task can occur so
let’s say suppose that you have a task
say task a and let’s say task a has K a
sequence of K choices and let’s say that
n1 is the number of ways the first stage
or event can occur and n 2 is the number
of ways the second stage or event can
occur following the first stage and
continue on continuing on let’s say n K
is the number of ways the K the stage or
event can occur after the first K minus
1 events or stages have occurred so the
total number of different ways that this
certain task a can occur is simply n 1
multiplied by n 2 multiplied by all your
stages multiplied by n K so whatever
this equals is the total number of
possibilities of task a
so let’s illustrate this using a quick
example so let’s say that we walk up to
a restaurant and we see that they have a
special that includes a sandwich a soup
a dessert and a drink so we we can have
a sandwich which comes also with a soup
we get a dessert and we can get a drink
so let’s say that we walk up to the
counter and we can we first choose our
sandwich so that’s the first stage and
we can choose from a chicken turkey tuna
or roast beef sandwich after this we can
choose our soup which can be tomato clam
chowder or vegetable after we choose our
soup we can choose our dessert which can
be say ice cream or pie and finally we
write before we sit down we can choose
our drink which is tea coffee soda diet
soda or milk so the questionnaire here
is how many possible specials
so how many ways or how many possible
specials are there so there’s four
stages and one and two and three and
four so the total possible ways is going
to be N 1 times n 2 times n 3 times n 4
so let’s define each stage in their
possibilities so let’s say a stage 1 we
have we can choose our sandwich and like
I said we can choose from a chicken
turkey tuna or roast beef so there’s
four possibilities as stage 1 when we go
and choose our soup we can choose from
tomato clam chowder or vegetable so
that’s three possibilities at stage two
at stage three we can choose from ice
cream or pie so that’s two possibilities
and finally we can choose our drink at
stage 4 which is tea coffee soda diet
soda or milk which is five possibilities
so determine to do to determine the
number of possible specials we multiply
n 1 times n 2 times n 3 times n 4 which
is simply 4 times 3 times 2 times 5 and
that equals 120 so the number of
possible specials given are given the
number of stages and given the
circumstances that we’ve defined there’s
120 possible specials that can be chosen
so that’s it in a nutshell it’s it’s not
really all that difficult all we do is
define our task we define the stages
within our task and then we define the
number of possibilities for each stage
within that task and then we finally to
finish it off we just multiply the
number of possibilities for each stage
to get our overall possibilities for the
entire task so that’s it I hope this
shined a little light and helped you
guys refresh a little bit on the
counting principle if you guys need any
further direction or guidance for the
engineering training exam don’t hesitate
to visit my site at engineer in
trainings am calm and I’ll be happy to
help you guys and guide you as best that
I can so that’s it for now take care and
and be good bye

—

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Law of Sines – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Law of Sines.

He will start by defining the Law of Sines and the run through a couple of instances in which we can use the law of sines to solve for unknown characteristics of a triangle.

This Fundamentals of Engineering Exam Review of Law of Sines is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Law of Sines.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Law of Sines by defining exactly what Law of Sines is.

The law of cosines and the law of sines
allows us to solve triangles that are
not right-angled in our called oblique
triangles it states the following the
sides of a triangle are to one another
in the same ratio as the signs of their
opposite angles so to illustrate that
let’s look at the triangle here let’s
say this is side a this is side B side C
this is angle a this is angle C and this
is angle B so the algebraic statement of
the law of sines states that sine of a
over a is equal to sine a B over B which
is equal to the sine of C over C so
let’s take into take and look at example
here let’s look at the triangle let’s
say that we know this angle right here
65 degrees we got this angle 75 degrees
and this angle is 40 degrees let’s say
this side is 10 and let’s say we do not
know this side which we’ll call X and we
want to find that side so all we need to
do is employ the law of sines so again
its sine of a over a is equal to sine of
B over B which is equal to sine of C
over C so in this case we can we can
just take the angle for T in the ankle
75 and the side of 10 and the side 10 to
determine X so let’s go ahead and take
that ratio let’s say the sine of 75 over
10 is going to be equal to the sine of
40
over X now just rearranging and solving
that we get 10 sine of 40 over the sine
of 75 is equal to X and using our
calculators we find that X is equal to
six point six five so what if we are
given a triangle that has an obtuse
angle how do how do we go about solving
that well no differently we do the same
thing we just take the ratios of the
sides and the angles and a soft form so
let’s look at an example there example
let’s say that we have in triangle that
looks something like this and let’s say
this angle is our obtuse angle and it’s
105 and we got this angle which is 25 in
this angle let’s say 50 and let’s say
that we have this side is 10 and this is
our unknown side X so all we need to do
once again is look at our definition of
the law of sines and know that all we
need to do is have plug in a couple few
couple knowns to get that unknown X so
let’s say we got sine of 25 which is the
opposite of our known sine of 10 divided
by 10 is equal to sine of 105 which is
opposite our unknown X and once again
just rearranging that we got X is equal
to 10 sine of 105 divided by sine of 25
solving that using our calculators we
find that X is equal to 22 point eight
so no different just solve it in the
same way just use the ratios so what if
we want to find an unknown angle and all
these all these previous examples we’ve
we kind of just used the law of sines to
determine unknown sides but we can
so you use the law of sines to find the
unknown angle so let’s look at another
example here example let’s say we have a
triangle and let’s say we know this side
is five point five we know this side is
four point seven and we know this side
or this angle is 63 degrees so what we
want to do is find angle B so how do we
do that let’s say this is angle B since
we know side B and we know we know a
side and and that 63 is the angle
opposite to our our second known side
all we need to do again is take our sine
of 63 divided by the known side of
five-and-a-half and our unknowns angle B
divided by our known side of four point
seven so again all we need to do is
rearrange sine of B is equal to four
point seven sine of 63 divided by five
point five and we can take the inverse
sine of this all and let’s calculate
this out this is all getting equal point
seven six one taking that inverse sine
we find that B is equal to forty nine
point six degrees so we’re also able to
find an unknown angle using the law of
sines so that’s it for now guys I hope
you appreciated that quick review that
simple review I know it was probably
pretty basic to you guys but definitely
something that you need to brush off up
on because as easy as it may seem you
will run across some of these questions
on the exam and being able to crank
these out as quick as possible is
definitely going to be of benefit to you
so anyways if you guys have any more
questions about the engineer in training
exam go ahead and visit my site and
engineering training exam comm and also
take a minute and
or quick second and subscribe to this
channel I’d be honored to have you as a
subscriber
I’m definitely have plans to expand this
channel and get more content up for you
guys to check out so if you have also
any topics that you would like me to
cover in future tutorials
definitely don’t hesitate to contact me
I’m definitely open to all emails and
contacts and I’m just out there to help
you guys out so I’ll do the best that I
can for you so for now take care and I
look forward to talking to you guys soon
all right bye

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Law of Cosines – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Law of Cosines.

We will begin by defining the Law of Cosines and then run through a couple of instances in which we can use the law of cosines to solve for any unknown characteristics of a triangle.

This review of Law of Cosines is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Law of Cosines.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Law of Cosines by defining exactly what Law of Cosines is.

We use the law of cosines and the law of sines to

solve triangles that are not right

angled such triangles are called oblique

triangles the law of cosines is use much

more frequently than the law of sines is

and more specifically when we know two

sides of a given triangle and the angle

in between those two sides then the law

of cosines enables us to determine the

third side of that triangle so let’s

illustrate the definition of the law of

cosines let’s look at the triangle here

say we have a triangle and this is side

a we got side B and we got C and this

angle right here side c so the log

cosine states that c squared is equal to

a squared plus B squared minus to a B

cosine of C so let’s look at a quick

example of how the law of cosines can be

used let’s look at the triangle again

say that we’re given that this side is

eight we’re given that this side right

here is 10 we don’t know this side which

is d let’s say that this angle D is 60

degrees and this is angle f and this is

angle e which are both unknowns but

since we have two sides eight and 10 and

the angle between those two sides we are

able to employ the law of cosines to

determine what the third side is so

let’s go ahead and just plug all of our

numbers into the into our equation d

squared is equal to e squared plus F

squared minus 2 e F cosine of D so

that’s our that’s our standard equation

we just want to plug in all that we know

so let’s just go ahead this is angle e

so opposite is going to be side e which

is going to be this first term which is

going to be a squared plus 10 squared

minus 2 times 8 times 10 cosine of 60

and so we can just use our calculators

and calculate this out it comes down to

d squared is equal to 80 for taking the

square root we get d is equal to 9.17 so

that is the length of our third unknown

side so how do we remember the equation

of the law of cosines are there is one

thing we can do is first thing we just

got to know that is it essentially is

the Pythagorean theorem with some extra

making it work for making it work for

all triangles so the Pythagorean theorem

you know is a squared plus B squared is

equal to C squared and we know that this

only works for right triangles

now the law of cosines is a squared plus

B squared minus to a B cosine of the

angle c which is equal to C squared and

this works for all triangles if you put

if you plugged in 90 degrees right here

to make this a right triangle this whole

term would disappear making it the

Pythagorean theorem so that’s basically

just how you can a quick way you can

remember what the law of cosines is so

when do you use the law of cosines

there’s there’s a two two instances in

which you could use the law of cosines

you can use it when the third side of a

triangle

I is needed needed and you know two

sides plus the angle between it also can

be used when you have when the angles of

a triangle are are needed and you know

all three sides so to use the law of

cosines essentially you need to know

there needs to be three things that you

need to know number one you need to know

either two sides and the angle between

or if you if you know all the side

you’re able to determine all the angles

within the triangle by plugging in all

your knowns so let’s again look at an

example let’s take the triangle here

let’s say we know all three sides let’s

say this side is five this side is nine

and this side is eight let’s say we have

an angle here that’s unknown a we have

an angle here B which is unknown and we

have an angle here C which is unknown so

what we want to do is determine what

let’s say angle C is so all we need to

do is start with our standard equation C

squared is equal to a squared plus B

squared minus to a B cosine of C and we

just start plugging in all of the known

sites all the known quantities that we

know so we got eight squared because

there’s a C and C is always opposite or

in this case we’re using a B and C where

before we used EF and G or yeah and so

the sides are just opposite their angles

so if this is angle C this is going to

be side see this is angle B so this is

going to be side B this is angle a so

this is going to be side a so just plug

in all your knowns eight squared is

equal to 9 squared plus 5 squared minus

two x 9 x 5 cosine of c and then all you

need to do now is calculate it out using

a calculator and you get down to the

point where it’s where it’s a see cosine

of c is equal to 42

/ 90 and then to determine what C is

just take the inverse cosine so c is

equal to cosine and negative 1 or

inverse cosine 42 / 90 and we’re going

to find that c is equal to 60 2.2

degrees so we could do that for each and

every one of the angles within that

triangle because we do know all three

sides and all three sides are defined so

using the law of cosines we can go ahead

and uh and determine more information

than what we’re given so that’s all I

got for you today guys that’s a real

quick tutorial and real basic tutorial I

hope it cleared some things up or at

least was a decent review of the subject

if you guys need any more information

regarding the engineering training exam

just go ahead and visit my site at

engineer in training exam com shoot me

an email if you have any more questions

or if you want to see any future video

tutorials on any subject it doesn’t

matter just shoot them to me and I look

forward to talking to you guys soon

alright so take care

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LHopital Rule – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of LHopital Rule.

We will begin by defining L’Hopital’s Rule and run through a couple of instances in which we can use this rule to determine the limit of a particular function.

This Fundamentals of Engineering Exam Review of LHopital Rule is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of LHopital Rule.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of LHopital Rule by defining exactly what LHopital Rule.

let’s
consider the limit as X goes to a of f
of X over G of X if at a both f of X and
G of X are finite and G sub a is not
equal to zero then the typical process
of finding the limit would be plugging
in a wherever we see X in those
functions so our limit would simply be F
of a over G sub a but what happens if
both the numerator and the denominator
10 to 0 it is not clear then what the
limit would be in fact depending on what
functions f of X and G sub X are the
limit can really be anything so when we
get when we approach functions are asked
to find functions and we get an
indeterminate form of say 0/0 that’s not
really telling us much about a limit for
example let me show you a quick example
here let’s take the limit as X goes to 0
of X to the 4th divided by X to the
third so if we are asked to find this
limit we would first try to just plug in
0 for wherever we find X and in this
case doing that would give us the
indeterminate form of 0 over 0 so that’s
not telling us much about what the limit
of that function is so we need to use
l’hopital’s rule to do a couple steps to
help us determine what the real limit of
that function is so let’s take a look at
the first indeterminate form of 0/0 so
suppose that we have that same limit f
of X over G sub X as X goes to a
and when we take the limit in our
standard procedures we find that both f
of X and G of X are going to be 0/0 so
l’hopital’s rule what it does is it
gives us two two rules here number one
if we can take the derivative of both f
of X and G of X then the limit as X goes
to a of f prime of X divided by G prime
of X if that exists and is some number L
then the limit of our original function
f of X over G sub X is also L and number
two it tells us that if in on the other
hand F prime of X divided by G prime of
X if that tends to plus or minus
infinity then our original function also
tends to plus or minus infinity so let’s
take a quick look at an example here
let’s take the limit as X goes to 1 of 2
natural log of X divided by X minus 1 so
what we’d try to do here first is just
plug in 1 wherever we see X and we find
real quickly that we would get the
indeterminate form of 0 over 0 so what
we need to do is employ la PETA’s rule
and the first thing we need to do is
take the derivatives of both the
numerator and the denominator and once
that’s complete all we need to do is
plug that back into the limit and take
the limit so let’s take the limit as X
goes to 1 and take the derivatives of
both the numerator and the denominator
and we get 2 divided by X divided by X
oops sorry divided by 1 so plug in 1
wherever we see X and we find that the
that the limit is going to be 2 and la
PETA’s rule excuse me tells us that the
limit of
the original function is now – so what
if we have the indeterminate form or
another and rather another indeterminate
form we might encounter is infinity over
infinity so this isn’t telling us much
again about the limit so we have to
employ la PETA’s rule again and the same
rules apply if we can find a limit at F
prime of X G prime of X if we can find
this limit then that’s good that’s the
limit of the original function and in
the same way if we find that F prime of
X G prime of X tends to plus or minus
infinity then we know that our original
function also will tend to plus or minus
infinity so let’s take a look at an
example here let’s look at the example
limit as X goes to infinity of e sub X
divided by X so our our first shot here
will just plug in an infinity into
wherever we see X and we’ll see that we
have an indeterminate form of infinity
over infinity so once again not telling
us much about the limit of the function
so what we need to do is take the
derivatives of both the numerator and
the denominator and we will get limit as
X goes to infinity of e sub X over 1 so
now plugging in infinity we see that the
function will tend to plus infinity so
that tells us what our limit of the
original function does sometimes you
know we’ll encounter
equations in which we have to use
l’hopital’s rule of multiple times what
I mean by that is will will be given in
a function say our function is 1 minus
cosine of X divided by x squared and
we’ll be asked to take that limit
and let’s say it approaches zero so if
we plug in zero here we’ll find that
once again we’ll get the indeterminate
form zero over zero so now that we’ve
been through the process we know that we
need to take the derivative of both the
numerator and the denominator using
l’hopital’s rule to determine the limit
so what so let’s do that real quick
let’s take the limit as X goes to zero
of the derivatives and we’ll see that
that’s sine X divided by 2x so usually
right here well we can find a limit
usually it takes one step in using
l’hopital’s rule to find that limit but
in this instance we plug in zero and
once again we get 0/0 so what do you do
in this case it’s it’s not very
difficult all you need to do is employ
la PETA’s rule again take the derivative
of both functions and plug it back into
the limit X goes to 0 of cosine of X
divided by 2 and so now in this case we
plug in 0 and we find that the limit is
1/2 and that’s going to be the limit of
our original function so if you continue
to hit indeterminate forms as you work
your way through la PETA’s rule just
continue to employ until you find that
that definitive limit so that’s it for
now guys I appreciate you guys taking
the time out to check out this quick
tutorial head on over to engineer in
training exam comm for more resources as
you prepare for the engineering training
exam and if you have any questions just
contact me through the site or here on
YouTube
I appreciate have a good day guys but
you

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Quadratic Equations – Fundamentals of Engineering Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Quadratic Equations.

He will start by defining the standard form of the quadratic equation and run through a common way people go about solving such equations and follow it up with a way to simplify our work in complicated situations.

We will wrap this tutorial up with introducing the Discriminant and what it can tell us when using the quadratic equation.

This Fundamentals of Engineering Exam Review of Quadratic Equations is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Quadratic Equations.

We’ll start off this tutorial by defining a quadratic equation, we’ll move on to solving quadratic equations using factoring, then we’ll move on to the quadratic formula and finish off with defining the discriminant.

So let’s start off our Fundamentals of Engineering Exam Review of Quadratic Equations by defining exactly what a Quadratic Equation is.

So what is a Quadratic equation?

in its simplest form a
quadratic equation is ax squared plus BX
plus C is equal to 0 where a cannot
equal 0 so in other terms a quadratic
equation must have squared term as its
highest power so how do we solve
quadratic equations there are 4 methods
of solving these equations 1 we can
factor 2 completing the square factor by
grouping or using the quadratic formula
in this tutorial in this quick review
we’re going to talk about factoring and
the quadratic formula so a quadratic
equation can be solved by factoring if
after writing that equation in its
standard form the quadratic expression
factors completing this process can be
done in four steps step number one is
getting an equation we want to simplify
it we want to take away any parenthesis
division we want to add like terms
once simplified we want to get it that
equation into the standard form of ax
squared plus BX plus C is equal to 0
step number 3 factor
once we have factored we want to solve
so that’s all let’s look at a quick
example let’s take the equation x
squared plus 3x minus 10 is equal to 0
so the first step is to simplify that’s
completed step number two we want to get
in its standard form
once again convenient it’s completed
step number three we want to factor so
in this case this equation factors out
2x plus five multiplied by X minus 2 is
equal to 0 so we want this product to
equal to 0 so in step number four we
need to solve that and we can just
visualize and see that we’re going to
get X is equal to negative 5 and X is
equal to 2 so this quadratic quadratic
equation has two solutions X equal
negative 5 and X equal 2 so let’s talk
about the quadratic formula you can
solve any quadratic equation by using
the quadratic formula this is convenient
when a quadratic equation does not
factor or is difficult to factor so once
again there’s 4 steps or excuse me
there’s 5 steps in completing this
process step number one is given an
equation we want to simplify just like
that previous factoring we want to get
rid of any parenthesis division we want
to get rid of or combine like terms
step number two is the same as well we
want to get it into standard form so
once it’s in standard form of ax squared
plus BX plus C is equal to zero we can
move on to step number three which is we
need to identify what a is equal what B
is equal and what C is equal once we
identify what a B and C is we can take
the
podriatic formula which is X is equal to
negative b plus or minus the square root
of b squared minus 4ac divided by 2a so
we just take our values of a B and C and
we plug them into this equation and then
step number five would be whatever those
values all are if we can simplify and
make them complete whole numbers and
that’s great let’s do that so once again
let’s look at a quick example let’s take
the equation 2x squared minus 5x plus 1
is equal to zero
once again step number one of
simplifying is done step number two of
getting it in its standard form is
completed step number three in this case
a is equal to 2 B is equal to negative
five and C is equal to one step number
four we take the quadratic formula and
populate it so in this instance where
you go – of negative 5 plus or minus the
square root of negative five squared
minus four times two times one and
that’s all over two times two so let’s
move up here now that we have all of our
values for a B and C plugged into the
quadratic formula we can solve and we
get X is equal to 5 plus or minus the
square root of 25 minus eight divided by
four and this doesn’t simplify and do
whole numbers or anything but we can
simplify it a little bit as X is equal
to 5 plus or minus the square root of 17
divided by 4 so those are our two
solutions for this quadratic equation
so let’s talk about the discriminant
when we’re given an equation or a
quadratic equation x squared once again
BX plus C is equal to zero right off the
bat we can take a look at it and know if
we can divide define a B and C we can
define the discriminant and if we can
define the discriminant then we can use
a couple known factors to determine what
our solutions are going to look like to
these this equation so in other words
let me define it like this the
discriminant in the quadratic formula is
that under the square root of the
quadratic formula on the top which is b
squared minus 4ac so we have all these
defined in a given quadratic equation
and if we do we can determine what type
of solutions we should expect to see so
in other words we can put this into a
table say we define our discriminant and
that discriminant is less than 0 if it’s
less than zero then there are going to
be two distinct real number answers
solutions to that quadratic equation if
that discriminant is equal to zero then
there’s going to be one real number
solution if that discriminant is greater
than zero then there’s going to be two
distinct complex
imaginary numbers solutions so that’s it
so the discriminant once again the
quadratic equation minus B plus or minus
the square root of b squared minus 4ac
divided by 2a this right here that’s the
discriminant so once we get a quadratic
equation we can determine what the value
is of that discriminant take it to our
table and know what we know what to
expect when we determine our solutions
so that’s that that’s that quick
overview of quadratic equations I
appreciate you guys checking out the
YouTube channel don’t forget to head
back over to engineering training exam
comm and check out more free resources
for you as you prepare for the engineer
and training exam so good luck everybody
we’ll be talking soon.

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Functions – Fundamentals of Engineering Exam Review

December 13, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Functions.

The goal is to reinforce both your understanding of functions as well as function notation, two topics commonly encountered on the Engineer In Training Exam.

He will first start with defining a function and function notation, then discuss how to evaluate a function, discuss domain and range, and wrap it up with a discussion on function composition.

This Fundamentals of Engineering Exam Review of Functions is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s gong on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s quick tutorial we are going to present a Fundamentals of Engineering Exam Review of Functions.

We’ll start off by defining a function and function notation, discuss how to evaluate a function, define domain and range and then finish off with function composition.

So let’s start off our Fundamentals of Engineering Exam Review of Functions by defining exactly what a Function is.

An equation will be a function if for any X
in the domain of that equation the
equation will yield exactly one value of
y so the definition criteria for an
equation to be a function is for all X
in that domain there’s only one value of
y so if we can satisfy this with an
equation then that equation is a
function so let’s let’s illustrate that
look at y is equal to x squared plus 1
so is this a function quickly I would
say yes this is a function because given
any X there is only one way to square it
and then add one to the result so no
matter what value of x we plug in here
there’s only going to be one value of y
one value of Y satisfies our criteria so
this is a function let’s look at Y
squared is equal to X plus 1 for the
second equation the only difference is
that we move the exponent off the X and
onto the Y this small change is all that
is required in this case to change the
equation from a function to something
that isn’t a function we can see that
this isn’t a function fairly simply by
plugging in any X let’s just plug in 3
and solve for so Y squared is equal to 3
plus 1 which is equal to 4 let’s solve
for y and we see that it’s going to be
plus or minus 2 that’s two values of Y
it does not satisfy our definition
criteria for function so this is not
this equation is not a function so let’s
talk function notation
function notation is nothing more than
an elaborate way of writing y so from so
let’s look at the equation y is equal to
2x squared minus 5x plus 3 so if we want
to write this equation using function
notation we can substitute the Y as f of
X we can substitute it as G sub X we can
substitute as H of X etc we just plug in
what’s on the right side and that’s
function notation so recall that this is
not some letter say f multiplied by X
it’s just like I said an elaborate way
of writing y using function notation
allows us to represent the value of the
function at a given X within a domain in
a clean and compact way so how do you
evaluate a function let’s take that same
function f of X is equal to 2x squared
minus 5x plus 3 and so what to evaluate
this wherever we see an X on this left
side we plug in whatever X we decide to
plug into the function on the left side
so let’s take f of negative 3 and just
plug it in on the right side it’s that
simple
and then use a math we’ll just go ahead
and quickly solve that just trust me
that it equals 36 so let’s talk about
domain and range this is one of the more
important concepts of a function the
domain of a function is the set of all
values of x all values that can be
plugged into a function and have the
function exist and have a real number
for a value so the domain we need to
avoid anything of division by zero we
need to avoid square roots of negative
numbers logarithms of 0 etc or anything
that makes a function non existent the
range of a function is simply the set of
all possible values that a function can
so let’s quickly find the domain of a
function let’s say we have f of X is
equal to X minus 4 divided by x squared
minus 2x minus 15 so in this case we
want to avoid division by 0 so let’s
isolate x squared minus 2x minus 15 and
since we don’t want it to be a division
by 0 we have to determine what values of
x will make that equal to 0 so let’s set
this function equal to 0 factor it X
minus 5x plus 3 and we can solve that
and see that X is equal to negative 3
and X is equal to 5 so the domain for
this function the domain will be all
values of X except X equal negative 3
and X equal 5
so let’s talk about function composition
let’s say we have two functions f of X
and we have G sub X the function
composition of f of X and G sub X can be
written as f G sub X which is equal to F
with G sub X plugged into that function
so in other words compositions are
evaluated by plugging in the sex second
function which in this case is G sub X
into the first function which is f of X
so let’s look at an example here let’s
take that same composition of F G sub X
and let’s use the equation so let’s say
we have f of X is equal to 3x squared
minus X plus 10 and let’s say we have G
sub X is equal to 1 minus 20 X so all we
need to do is plug G sub X into f of X
so this will equal 3 120 x squared minus
120 X plus 10 so we can solve that out
just trust me that it’s going to come
out to 12,000 12,000 pourtant
interchanging the order will usually
result in a different answer and we can
illustrate that by using the same
functions here let’s use f of X is equal
to 3x squared minus X plus 10 let’s use
G sub X the same one as 1 minus 20x this
time let’s plug f of X and the G sub X
and see what we get
that will come out to be G sub 3 x
squared minus X plus 10 which is equal
to 1 minus 23 x squared minus X plus 10
which is equal to minus 60 x squared
plus 20x minus 199 so to compare that to
the previous example we see that by
interchanging the same functions that
we’ve got two different answers so
that’s all I got for you guys today I
hope you guys enjoyed that quick review
I know it was a fundamental and that you
probably right up on this stuff but I
just wanted to touch briefly on it so
check out my website at engineering
training exam com for more resources and
leadership I look forward to meeting
guys and getting getting to know you and
help you on your journey as you take on
and prepare for the engineering training
exam

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