so here's a little proof for the

existence of God I always thought it was

interesting I was a I was initially a

philosophy major after being undecided

until my favorite philosophy Professor

told me I was crazy for being a

philosophy major major which uh I think

he was absolutely probably correct at

least for me so um anyway here's a

little proof and I thought it was

interesting the guy uh the guy was a you

know a philosopher mathematician and

astronomer but this is how it goes

so okay so this guy Guido he thought

that uh he had a little I don't know

that necessarily he even proved it or

came up with it but he thought that this

Justified the existence of God so um as

he said something had been created from

nothing

but I think he was a bit uh incorrect In

His approach so this is how it

goes nothing too crazy

so we start off with all right well 0 =

0 + 0 + 0 plus whatever um well we can

rewrite the zeros as 1 - one 1 - one 1 -

one well then let's get rid of the

parenthesis well let's regroup the

parentheses let's just kind of shift

them over a little bit well now we have

1 +- 1 + 1 uh plus 1 + 1 Etc and now

what we have is well 1 + 0 + 0 plus 0 +

0 and hey that's going to give us one

so

well probably figured out the just by

now what he has

shown

is Tada 0 equals 1 so again he's shown

that 0 equals one so

um if only life were so easy so maybe

you can figure out where the the the

problem is but again I just think it's

kind of interesting

um a lot of the old mathematicians were

definitely philosophers at the same time

and you know um I think now we kind of

see math as this sort of a dry boring

thing we're taught to memorize but um I

think there's something something much

more interesting you can do with it but

that's just my own opinion so again

maybe not necessarily prove the

existence of God but um it's not as dry

and as boring and as mundane I think

some times as we we think it is so

anyways I think it's an interesting

little argument um hopefully you find it

interesting um I certainly

do

A Proof for the Existence of God

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! A Proof for the Existence of God. Get upset, cause you will.  Ya, but the problem is, the math is wrong.  Just a little video showing the proof that a theologian / philosopher of the past gave. Before you get your panties in a knot, it is not my proof, nor do I endorse it. : )


all right here we're going to look at

one more example involving a little

division of decimal so

suppose patrick drives at an average

speed of 46.25

miles per hour we want to know how long

does it take him to travel 185

miles

well all we're going to have to do is

take 185

and divide it by

or equivalently we'll do 185 over 46.25

and now i'm going to do the same thing

as before i'm going to

multiply top and bottom of my fraction

by 100 to move the decimal

in the denominator we'll be left with uh

four six two five and the numerator will

be left with one eight five zero zero

and now i'm going to do my long division

and again these are kind of numbers that

uh

have the potential to be

um to make life a little miserable you

know if you're dividing this number by

this number by hand but okay well

that's that's how it goes so i'm gonna

check here

you know uh i kind of round things off

you know to me uh 46 that's somewhere

i kind of look basically at the very

first digit you know so four i'm going

to count i've rounded up to five um you

know if i think about this number um the

first number is being five

if i look at the first two numbers

inside of here that's somewhere around

18. okay so i'm kind of thinking you

know roughly what's 18 divided by five

well that's somewhere between three and

four

so i'm guessing maybe three or four

forty six point two five will go into

eighteen thousand five hundred

uh maybe you know four or five times

three or four or five times um

and again all i'm doing is i'm getting

rid of the first three digits underneath

and then i'm getting rid of basically

the first three digits of the number out

front and that's kind of what i'm

estimating i'm saying you know roughly

18 divided by four you know that's

somewhere between four and five

um

again depending on how you estimate

obviously you'll get slightly different

values but i know that 46 25 isn't going

to go into this say seven times

okay so again i'm just getting rid of

three digits getting rid of three digits

and i'm thinking about that division to

give me a guess

well i'm going to check 46 to 5 maybe

maybe let's try four

so let's see 4 times 5 is 20

we'll carry the 2 4 times 8 is 10 or

excuse me 4 times 2 is 8 plus 2 is 10.

carry the 1.

4 times 6 is 24 plus 1 is 25 we'll carry

the 2

4 times 4 16 plus 2 is 18 hey lo and

behold it goes in there exactly four

times so

things were set up nicely for us again

four times 46 25 will leave us with

eighteen five zero zero exactly

if we subtract we've got no remainder

so it says it'll take a good old patrick

that that wonderful guy

it'll take him uh four hours to make

that

to travel that distance

Dividing Decimals, Ex 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Dividing Decimals, Ex 3


all right here we've got an example

we're going to use the graph to find the

absolute and local maximum and minimum

value so absolute maximum and minimums

and also local maximum and minimums

oftentimes people just give the Y values

so we can kind of use both notations

here so if we want to figure out sort of

the local minimums and also sort of the

local maximum values

well again I just kind of imagine I'm

looking sort of at a mountain range off

in the distance you know obviously kind

of a weird mountain range because it

keeps going forever but if we're looking

for local minimums I'm kind of looking

at you know sort of imagining like you

know kind of looking at the valley of my

mountain range sort of a you know just

where it bottoms out so at this point

negative 5 comma 3 we would say we have

a local minimum and oftentimes people

will just specify the Y value so we

could say the Y value of 3 corresponds

to a local minimum well let's see when

we get to this point negative 2 comma 5

notice this is kind of like a little top

of the hill that would be considered a

local maximum so we could say we have a

local maximum at the Y value of 5 again

as we go down it kind of bottoms out

here at this point 1 comma negative 4 so

again that would be a local minimum so

we have a local minimum at the Y value

of negative 4 here at the point 4 comma

5 we kind of have the you know the

little top of the hill so again 5 would

correspond to a local maximum value but

we've already got it listed so we don't

have to list it again and again at this

point 6 comma 4 you're at the bottom so

we could say that y equals 4 there's a

local minimum if you actually want to

specify you know the points obviously

you just do the same thing so you could

say you have a local minimum at negative

5 comma 3 at the point 1 comma negative

4 and also at the point 6 comma 4

squeeze that guy back in there

we have local maximums

at the points negative 2 comma 5 and

also at the point 4 comma 5 okay so that

takes care of local maximums and

minimums when we talk about the absolute

maximum and the absolute minimum

well now we're just kind of thinking you

know I just think of all the y-values if

I were to list every single Y value on

this graph all infinitely many of them

I'm just thinking what would be the the

largest Y value that ever gets used if

there is one and what would be the

smallest Y value that ever gets used if

there is one well it looks like to me my

function since it keeps going up forever

and ever and ever on either side

well in that case there's really not an

absolute maximum because you know the Y

values always get bigger bigger bigger

bigger so they sort of know there's no

top no biggest Y value so I would say

there's no absolute maximum but when I

think about the absolute minimum well

this point down here at the bottom at 1

negative 4 to me that looks like the

smallest Y value that ever gets used so

we could say our absolute minimum is the

Y value of negative 4 and again you

could specify you know specify the point

like we did you could say the absolute

minimum is at the point 1 comma negative

📈 Local and Absolute Maximum and Minimum from a Graph 📈

Finding Local and Absolute Maximum and Minimum from a Graph

In this video, we delve into the world of calculus as we explore how to identify local (relative) maximums and minimums, as well as absolute maximums and minimums from a graph. Understanding these concepts is crucial for analyzing functions and making informed decisions based on graphical data.

What You’ll Learn:

Identifying Local Maximums and Minimums: Discover how to spot relative extrema on a graph and what they signify about the function's behavior.

Finding Absolute Maximums and Minimums: Learn the steps to determine the overall highest and lowest points on the graph.

Graphical Analysis Techniques: Gain insights into how to interpret graphs effectively and apply these techniques to various problems.

Why Watch This Video?

Perfect for Students: This tutorial is ideal for high school and college students tackling calculus or introductory math courses.
Clear and Engaging Explanations: Our step-by-step approach makes complex topics easy to understand and apply.
Real-World Applications: Understand how these concepts are used in fields like economics, physics, and engineering.

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okay so let's finish off this integral

from from part one so we said the area

between these two curves y Cube minus y

1 - y 4th we graphed them uh sorry I I

fudged there for a little bit I was

thinking um this graph for sure is of

course it makes sense um and then we had

to calculate this integral from- 1 to 1

1 - y 4 - y Cub - y um so let's do that

so I'm going to get rid of my nice

lovely graph here

and simply calculate this out so let's

write it a little bigger

um so from negative 1 to one looks like

my Pen's starting to die let's get rid

of the parentheses so 1 minus y 4th

distribute we'll have negative y cubed

um plus

y

Dy um I always I don't know just like to

have these in descending order so

negative y to 4th

um minus y

cubed +

y +

one Dy okay so now let's just calculate

this so there's really no way to combine

like terms we'll just do it a bit at a

time so we'll get y 5th over

5 we'll get - y 4 over

4 um we'll get + y^2 over

2 + y and then our limits of integration

just - 1:

1 so if we plug one in it looks like

we'll get -5

-4 + 12 um + one so there's our upper

limit of

integration let's subtract away our

lower limit well negative 1 to the 5th

is negative one5 but we'll get an extra

negative so it'll make it positive one5

-1 to 4th is still 1 so we'll be left

with Min

-4 um we'll also get a positive2 when we

square and then we'll get a

minus1 so it looks like we have -5 -4 +

12 + 1 -5

+4 um - 12 + 1 so what cancels out we've

got a negative 1/4 and a POS

1/4 um we've got a positive 1/2 and a

negative

1/2 um it looks like we have a negative

1 and a negative 1/5 which is negative

whoops two

fths um soga - 2 fths and then we have

plus two so get common denominators -2

fths um top and bottom by five so 10

fths that'll give us positive 8 fths

left over as the

area all right so I hope this video

helps and makes some sense if you have

any questions feel free to post comments

and hopefully um somebody can help you

out if not me

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Area Between Curves - Integrating with Respect to y - Part 2.
In this video, I actually do the integration and calculate the area.
For more free math videos, visit http://PatrickJMT.com


all right in this example we're gonna do

a couple a couple different problems

where we evaluate expressions just given

some different values so in Part A we

want to evaluate the expression X plus y

if x equals negative 5 and also we have

that y equals negative 7 so all we're

gonna do is substitute when I plug

things in I like to put things in

parenthesis so X is gonna be negative 5

and in this case you certainly don't

have to but in some cases it'll help

emphasize that you have to distribute

signs I think it's just a good policy so

you'll see me do this some times but

okay negative 5 plus negative 7 well

that's simply gonna give us negative 12

so that would be our solution here for

Part A for Part B again we'll just do

the same thing Part B we have that X

we're gonna evaluate x divided by y

again if X is negative 75 and Y is

positive 5 so we'll get negative 75

divided by Y which is positive 5 well we

have a negative over a positive which is

a negative let's see

75 divided by 5 that's gonna leave us

with 15 so the solution for Part B will

be negative 15

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Evaluating Algebraic Expressions for Given Values of X and Y.  Here we evaluate two basic algebraic expressions for given values of x and y.


okay here we're gonna look at another

couple examples of solving some

equations where we have to do a couple

steps and we can do both B and C in this

example I think in this video so we've

got 9x minus 4 equals 77 again to get

the the term involving X isolated I'm

going to start off by adding forward to

both sides so we'll be left with simply

9x on the Left negative 4 plus 4 will be

there's our equal sign 77 + 4 will give

us 81 and now 2 to get our solution

again we want to be left with just 1x

well since I'm multiplying by 9 I'm

going to divide both sides by 9 9

divided by 9 is 1 X + 81 divided by 9 is

also 9 so our solution for Part B here

will be x equals 9 and again notice you

can always check these 9 times 9 would

be 81 81 - 4 hey that is 77 so you can

always check them Part C here same idea

okay we've got a fraction no big deal so

I've got an N over 5 we've got negative

4 plus and over 5 equals negative 8 I

want to get the term involving the

variable in this case and I want to get

that by itself since I have a negative 4

I'm going to add 4 to both sides so

negative 4 plus 4 will just be 0 we can

drop down our 10 over 5 and negative 8

plus 4 is going to be negative 4 and to

get end by itself since I've got N and

I'm dividing by 5 I'm going to multiply

both sides by 5 so on the left side 5

divided by 5 is just one in a negative

and a positive as a negative 4 times 5

will be 20 and we have our solution for

Part C so

equal negative 20

Solving Two - Step Linear Equations ,  Another Example

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Two - Step Linear Equations.  Here we look at solving linear equations where we have to perform addition / subtraction as well as division / multiplication to solve.


all right in this video we're going to

look at finding some sums of geometric

series if they exist so in part A here

we've got the series uh from nals 0 to

Infinity of 1 over5 to the N so in this

case I'm going to rewrite it um Nal 0 to

Infinity I'm just going to write it as 1

over5 all raised to the N power again uh

if we take one to the N power in the

numerator 1 to the N is just one so

again in this case this is going to be a

convergent geometric series remember the

number that's getting raised to the

power that's our r value so in this case

r

is5 and again these

converge if our r value is in between

positive one and negative 1 and they

converge to sometimes people will write

it as a over 1 - R I like to think about

it as being the first term over 1 -

R and in this case well let's see if we

plug in n equals 0 That's How we'll get

our first term we'll have well just 1

raised to the 0o over 1 minus the R

value which again is 1/5 well 1/ to the

0er is just 1 1 minus a fth is going to

be four fths and 1 over four FS gives us

5 over 4 that will be the value of of

our convergent geometric series

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !!
Sum of an Infinite Geometric Series, Ex 1. In this video, I show how to find the sum of a convergent infinite series.


okay so one more example here of using

our half angle identities and here we're

simply going to evaluate s of the angle

a over 2 we're given that cosine of a is

3 over5 and we're also told that angle a

is between 0 and 90° so that means s of

a over 2 that would be somewhere in

quadrant

one so now I'm just going to go back and

use our

identity so it says for for sign again

it says either get positive or negative

well in this case again we're going to

use the positive one so s of a over 2

again since our angle was in quadrant

one we'll use the positive square

root and then it says we would simply

get one minus well cosine of

a over two and well now I'm going to

make use of the fact that we were given

so we were given the value of cosine of

a

cosine of a was given to us as just the

value 3 fths so now all I have to do is

just really just plug that in so it says

we'll get the square Ro T of 1 -

35ths over

2 well in the numerator uh you can write

one as 5 over 5 5 over 5 - 3 over 5 and

the numerator would leave us with two

fths or 2 over five the denominator okay

this is all still under the square root

we can write two as simply 2 over

one well so underneath the square root

I've got a fraction divided by a

fraction so we can write that as two

fths times well we could flip and

multiply by the denominator 1 over two

those would simply cancel out and we

would be left with the square root of 1

over5 uh we could also take the square

root of the numerator which is one the

square root of the denominator would

just beunk five we could also

rationalize

that the denominator by multiplying top

and bottom by square root of five and

that'll simply leave us withun 5 over

five again you can always uh plug this

into your calculator and get a decimal

approximation but to me this is now is a

nice exact answer

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Half Angle Identities to Evaluate Trigonometric Expressions, Example 3. In this video, I give some half angle identities and show how they can be used to evaluate trigonometric expressions.


all right here we're going to look at a

little question related to a sequence

and in this case we're going to ask

ourselves is the sequence negative 1 to

the N times 1 over N squared is an

increasing decreasing or neither also

you know is it bounded above below both

for neither so you know there's lots of

different ways to argue these or you

know address these types of questions I

think this one's kind of easy to just

plug in numbers and see what's going to

happen I mean you know this negative 1

to the N basically as n increases all

that's going to do it's going to make

the sequence just alternate and sign

well that to me makes me automatically

feel like it can't be strictly

increasing or decreasing because it

looks like it's going to bounce back and

forth between you know negative positive

negative positive so it's it's not

strictly getting bigger or smaller the 1

over N squared if you think about that

this just gets closer to 0 as n

increases you know so and you can even

plug in numbers if we plug in 1 you know

we would get negative 1 times 1 over 1

so we would get negative 1 if we plug in

2 we would get a positive 1 over 2

squared or 1/4 if we plug in 3 we would

get a negative 1 over 3 squared which

would give us negative 1 over 9 and then

positive again if you plug in 4 we'll

get 16 etc so it's you know it's clear

to see that this is neither increasing

or decreasing

you know and again you're starting like

at negative one and you're uh put forth

a negative nine at the sixteenth but if

you think about it okay so it certainly

looks like to me it is bounded it's

going to be bounded to me both above and

below and to me it looks like it's going

to be bound bounded below you could say

it's bounded below by negative one and

then it's bounded above it looks like

the biggest positive number you would

ever get out of our sequence would be

positive 1/4 so it's bounded below and

above by both of these numbers again if

a sequence is neither strictly

increasing or decreasing it is bounded

above and below you know in that case

you can't really say for sure that it

converges or not the question that nask

is the sequence converge but we could

always address it I think it's pretty

easy to see that it's going to converge

to zero and again in that case we could

use the absolute value theorem so

remember the absolute value theorem we

just take the absolute value of the

sequence in that case if we take the

absolute value the negative 1 to the N

would go away we would just be left with

1 over N squared if we simplify this and

certainly as n goes to infinity this

limit equals zero and therefore by the

absolute value theorem the original

sequence in this case and you know the

seeking particular sequence it would

also converge to zero so again lots of

different ways to kind of justify these

a lot of times to me with these you know

it's to get a solution a lot of times

you can just plug in values and get a

feel for what's going on

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Intro to Monotonic and Bounded Sequences, Ex 1.  Just a little question to decide if a sequence is monotonic and/or bounded.


okay in this video I'm gonna do another

example of simplifying radical

expressions involving variables this

time it'll be I guess a little more

complicated because we have a fraction

in here so we've got the square root of

20 U to the fourth B squared W to the

fifth over divided by 80 U to the sixth

B to the third W to the third again a

couple different ways you can start

about simplifying this I could rewrite

all of this to the one-half power but my

own personal preference is certainly

before I do anything I like to simplify

you know inside the inside the root I'm

gonna simplify this fraction a little

bit so I think you know 20 over 80

well 20 will go over go into 20 one time

and 20 will go into 80 four times so 20

over 80 simply reduces to the fraction

1/4 okay I have U to the fourth and U to

the sixth again the bigger exponents in

the denominator so the use will stay in

the denominator if you take the bigger

number 6-4 we'll get to the second power

I notice hey there's these on top B's on

the bottom again we can do this

cancellation because there's all

multiplication in the numerator and the

denominator just to point that out so

again the bigger exponents in the

denominator so 3 minus 2 will give us B

to the first power and then in the

numerator we have W to the fifth over W

to the third the bigger exponents on top

if you take 5 minus 3 that'll give us to

the second power

okay so now I am gonna do this step of

rewriting this all to the 1/2 power

because that's equivalent notation when

we have a square root we have 1 times W

squared but I'm just gonna simply write

that as W squared then we have 4 to the

first power u squared and V to the first

power in the denominator if you have an

exponent applied to a fraction

equivalently the numerator gets raised

to that to that exponent to the one-half

power and then we'll also take the

denominator for u squared V to the first

and we'll raise that also to the one

half power and now again I'm just going

to multiply the exponents since

everything's in parentheses so in the

numerator I've got two times 1/2 which

will be W to the first power in the

denominator I have for my exponent of 1

times 1/2 that'll give me 4 to the 1/2

power if I take my U squared and raise

that to the 1/2 again I multiply so 2

times 1/2 would just give us u to the

first power and then if I multiply on

the V's I would get V to the one-half

power so let me rewrite this one more

time so I've got W on top remember 4 to

the one-half power is the same thing as

the square root of 4 my use just hanging

out I can rewrite the V as V the square

root of V

and the last thing I would do is I think

you know the kind of the one

simplification I could make is I could

certainly take the square root of four

the square root of four is positive two

will leave the you alone will leave the

square root of V alone and this to me

would look like a nice simplified

version of this original problem

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Simplifying Radical Expressions Involving Variables - Example 2


all right so one more problem here

related to limits and this is actually

going to have a little bit to do with

continuity so we want to figure out what

value of a is this piecewise function

continuous on the entire real line so

we've got the peie wise function x^2 + 3

if x is less than 1 and then we have ax

+ 6 if x is greater than or equal to 1

so we've seen this result that

polinomial functions are continuous on

their domain x^2 + 3 is a polinomial

whatever value we pick for a when we

multiply that by x + 6 that's going to

be continuous as well so definitely it's

continuous for everything less than one

and everything greater than one again

where they kind of break at this x

coordinate of one they may or may not be

continuous depending on the value of a

but we're trying to again pick that

value of a so that it does happen so

again we're going to use our definition

what we want to occur is we want the

limit as X approaches one of the

function

we want that to equal F of one okay well

for the limit to exist the limit as X

approaches one from the left that's

going to have to equal the limit as X

approaches one from the right so my

little sideways equal sign there um well

if x is approaching one from the left

we're using values of X smaller than one

so we would be using the function x^2 +

3 when we do the limit as X approaches 1

from the right well now we're taking x

coordinates larger than 1 so we would

use the function ax +

6 well for the first one if we plug in x

= 1 we'll get 1^ 2 + 3 or we'll get the

value

four um if we plug one into the second

expression again we're plugging it in

for X we'll get a * 1 + 6 or we'll just

be left with a + 6

but again for the limit to exist the

left hand limit has to equal the right

hand limit so that would tell us that

that means that four would have to equal

well a + 6 and well we can just subtract

six from both sides and that would give

us -2 as our value for a so in this case

we've made the limit um equal okay CU if

we use the a value of -2 the left hand

limit and the right hand limit will both

equ Al 4 and um so the limit

exists and notice you know now if you

plug in the x coordinate of one again

we've now figured out that this is the

function we should be using -2x + 6

notice if you plug one into that again

we'll just get -2 + 6 which again will

give us the value four that we want so

it says if we make a equal to -2 the

limits will exist um and in a sense by

default for this particular function

it'll also give you the correct value so

really kind of the uh you know I think

uh in a sense to do the problem you just

need to set the left-and limit equal to

the right hand limit and solve

it

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Continuity : Making a Piecewise Function Continuous.
Here we have a piecewise function.  We have to find the value for a missing coefficient of one of the functions to make the whole thing into a continuous function.
These problems can be done rather quickly, but a typical college professor would want to see limits and the definition of continuity on an exam ( I would have, at least).


so suppose chris made 18 out of 20 free

throws

we want to know what percent did he make

so again if we can write a fraction in

the form n over 100

that's the same thing as in percent so

chris made 18 out of a total of

20. so our fraction is going to be 18

over 20.

well if we want to make the denominator

into a 100

we could just multiply the denominator

by five

and so the numerator by five well hey

we've now got that 100 in the

denominator

just like we wanted let's see five times

ten is fifty

five five times eight is going to be

forty 50 and 40 would give us 90.

90 over 100 is going to be the same

thing as

90 percent so if chris makes

18 out of 20 of his free throws we could

say he's a 90 percent

free throw shooter

Finding the Percent of Free Throws Made ( Word Problem using Basketball)

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all right so just a few more quick

questions about uh just kind of the

basics of the trigonometric functions so

here we've got a little angle and we've

got a point uh on there 3 comma 4 and we

want to figure out the value of Cent

Theta and also tangent Theta that'll be

the first thing we'll do here well again

you can really think about a little we

really have a little triangle right here

if we just drop a you know drop a line

down that's perpendicular the x

coordinate has value three

so since the x coordinate has value

three well that really means we moved

over three units the Y value is four so

that means this thing has a height of

four and the hypotenuse you may remember

this is just a 345 triangle we could use

Pythagorean theorem as well 3^ 2 + 4^ 2

maybe I'll call the missing side x s

well that's 9 + 16 = x^2 9 + 16 16 is 25

if we take the square root of both sides

normally we would get positive and Nega

five but here it's clearly got to be you

know a positive answer so again we're

just going to get uh our 345 triangle

and now it just basically comes back to

knowing the definition uh coent of theta

we said coent is going to be the

adjacent side divided by the length of

the opposite

side so coent so one more time adjacent

which is has l length three over the

opposite side which is length four so

cent of theta we would say just has

value

34s tangent of theta so again tangent of

theta we said that's opposite over

adjacent so same thing the opposite side

has length four the adjacent side has

length three and that's going to be a

relationship for tangent and cotangent

they're uh basically the reciprocal of

each other always have the same sign but

one is just going to be the flip of the

other one all right so uh just one more

uh actually two more quick questions

here so we've got our little right

triangle here uh 5 7 and square < TK of

74 you can check that this does in fact

satisfy Pythagorean theorem which does

in fact show it's a right triangle we

want to find secant of the angle p and

we also want to find cosecant of the

angle P okay so Shand for secant we just

write SEC Shand for cosecant we write

CSC and again now it's just a matter of

just knowing uh these definitions secant

of an angle we say that's the length of

the hypotenuse over the adjacent side

well the length of the hypotenuse is

square < TK of 74 and the adjacent side

is simply equal to five and now I've got

the value for secant of of P P the angle

p uh P for Patrick cosecant of the angle

P or Theta when we originally had had it

labeled cosecant of an angle is defined

to be the length of the hypotenuse over

the length of the uh opposite side so

here the length of the hypotenuse again

is square < TK of 74 the opposite side

has length seven so we would simply say

the value of cosecant of the angle p is

< TK of 74 / 7

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Trigonometric Functions: The Basics! Example 2. Just a few more examples of finding values for the trig functions given a right triangle.


all right so another example here

related to hyperbolas and in this case

what we're going to do is we're going to

find again the vertices ASM tootes and

the fosi of this hyperbola y^2 over 100

- x^2 36 = 1 so kind of the first thing

I usually do is I try to think you know

is it going to look like the one on the

left or the one on the

right well notice in this case we've got

y^ 2 over a number - x^2 over a number

so that's going to be more like the

second type here notice we we've got our

value A2 is going to be equal to um 100

it's the value underneath the

y^2 and if we take the square root of

both sides we would get positive and -

10 but again when I fill in my formulas

I just use the positive one okay the b^2

is going to be the value underneath the

x s term so b^2 equals 36 well again if

I take the square root I'm just going to

keep the the positive Sol solution to

this equation which would be just simply

B =

6 okay to get the vertices in this

case so let's see here give myself a

little more

room so to figure out the vertices in

this

example well it says the vertices occur

at 0 positive a and at 0 negative a well

since a is 10 then that means our

vertices will be at 0 pos1 and at 0

-10 okay if we want to find the fosi

again I need to somehow find this value

C but again C satisfies this

relationship that c^2 = A2 +

B2 so c^2 = A2 which in this case will

be

100 plus b^2 which in this case will be

36 so that says c^2

= 36 and again if we want to uh uh solve

for C here we could just take the square

root of both sides again I'm just going

to keep the positive solution here and

now we could uh start reducing this or

at least try to so definitely 136 uh is

even I think we could make that 2 *

68 uh so I'm going to break this down a

little further usually if I see an even

number I just start dividing by

two so let's see 68 we could write that

as 2 *

34 so I've got my two I've got my two 34

we could write that as 2 * 17 well I

think what we could do here so 2 * 2 is

4 we could pull that out as a two 2 and

17 don't really break down any further

so I'm going to go ahead and multiply

those back together and make that the

square otk of 34 so according to our

diagram here it says in this case the

vertices will be at 0 C and at 0 netive

c um excuse me not the vertices but the

fosi so the fosi will be at 0 positive C

and at 0 neg C so in this case our fosi

will be at

0 comma 2 < TK of 34 and at 0 comma -2 <

TK of

34 last but not least if we want to find

the ASM

tootes to get the as ASM tootes we just

let uh we take y = positive and netive A

over B X well again we said our a value

is 10 our B value is 6 so we would get

positive and - 10 over 6X and we could

simply reduce this to get y = positive

and 5/3 X and that would now be our Asm

tootes

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Conic Sections: Hyperbolas, An Introduction - In this video, I find the vertices and foci for a couple of different hyperbolas.


okay in this video I'm going to do an

introduction to solving inequalities so

I'm going to solve some pretty

straightforward uh inequalities here to

me the main rule to remember is

basically for inequalities uh you know

at least for linear inequalities which

is what I'm going to do here at the

beginning uh one of the main rules is

simply if you multiply or divide by a

negative number hey you're going to have

to flip around um the inequality it

flips direction is what happens so if

you don't do that your answers are going

to get all crazy okay okay but okay so

in this problem we have 3x - 6 greater

than 8x + 7 in my head I'm almost just

pretending like this is an equal sign

and I'm going to basically solve it the

same way um again just remembering to

flip the inequality If I multiply or

divide by negatives so I'm going to

start off by subtracting 3x from both

sides so on the left we would have -6 uh

greater than 8x - uh 3x is going to give

us 5x + 7 7 and now I'll simply just

subtract seven from both sides -6 - 7 is

-13 that's greater than

5x and now what I'm going to do is

divide both sides by five um the

inequality is going to keep facing the

same direction I am dividing but I'm not

I'm just dividing by a positive number

so if you multiply or divide by a

positive number the inequality stays

where it was and this says -13 over 5 is

greater than x equivalently sometimes

people don't like to see it this way we

could rewrite this equivalently we could

say x is less than ne-3 over 5 so I'm

writing the X first notice if you if you

think about this is kind of being an

arrow it's still pointing at The X and

hey it's still pointing at The X so this

is the same thing um if we were to write

this in interval notation so I'm going

to make a little graph here first so

here's -3 over5 since we don't include

that number I'm going to make a little

open circle that's usually the kind of

the convention that people use and then

I'm going to shade everything that works

which would be every number to the left

of it and we would go forever and ever

and ever out to negative Infinity we

would never get there but again this

just says goes just go as far left just

keep going the way that we can write

this in interval notation is we'll we'll

put a little parentheses always a

parentheses around negative infinity and

then we go up to -13

over5 um but we don't include it so

again we use parentheses uh to indicate

that we don't include that number so

this would be the interval notation that

would provide the solution to our very

original inequality

An Introduction To Solving Linear Inequalities - Example 1

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okay in this video I'm going to do one

more example of factoring perfect square

trinomials and I'm going to factor this

expression 2p plus T squared plus 6

times 2 P plus T plus 9 and I think at

first this is one of those that looks a

little confusing if you you know you ran

into it you might be inclined to try to

multiply everything out collect any like

terms and to me then you'd have a big

old mess and you'd be trying to factor

that maybe it would work but uh I think

the thing that we need to recognize here

again is really this is a perfect square

trinomial

okay and I'm going to try to rewrite it

in this you know in this form just to

illustrate it just to really emphasize

that so notice okay

2p plus T squared there that is we have

6 so I'm going to leave it as 6 the only

thing I'm really going to do here at the

first step is rewrite my 9 as well I can

rewrite that as positive 3 squared and

it says in this formula it says this

would be our a value so a would equal to

P plus T that's what's being squared B

would have the value of 3 and notice it

says in the middle we take 2 times a

which is 2 P plus T times B which is 3

notice if you multiply all those

together 2 times 3 is 6 and then we'll

get our 2 P plus T we do get this middle

term back so

Soviet says we do have this form a

squared plus 2 a B plus B squared so it

says well we can simply factor that now

it says we take a which is 2p plus T and

it says we add to that B which is

positive 3 and then it says we just

rewrite it again it's a perfect square

okay and in this case you could drop the

parentheses on the inside 2p plus T plus

three again you can think about there's

being a coefficient of positive one so

if you multiply it out it just gets rid

of it then we have two P plus T plus

three or again even we could rewrite

this as 2 P plus T plus 3 since we have

one and two of them being multiplied we

could rewrite that as that whole

quantity square so again you know it

looks a little tricky at the beginning

but just trying to really emphasize this

form this a squared plus 2 a B plus B

squared hey that's a perfect square

trinomial

and it actually turns out it factors a

little bit a little bit more nicely than

I think you might first believe

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okay in this example I want to do just a

couple more examples of applying rules

of exponents and this time we just have

a few fractions floating around and the

main rule you need to remember with

fractions so suppose I have x to the N

over y to the m all being raised to the

P power what it says is it says we can

basically sort of distribute the

exponent to the top part distribute the

exponent to the bottom part and then we

have the rule that we saw in a previous

video where you just multiply the two

exponents together so n * p and on the

bottom we'll have y raised to the m *

P so that's what we'll do in this case

um so here I've got x^2 over 2 again you

could make this to the first Power if

you want to see an exponent um

never hurts I think so here we would

have X2 being raised to the 4th

power over 2 to the 1st being raised to

the 4th

power okay so now we're at this step and

then it says again we just multiply the

exponents so 2 * 4 will be to the e88th

power um if we take two we have an

exponent of one and then a four we

multiply those that'll give us 2 to the

4th power there's not much you can do in

terms of simplifying the numerator we'll

just leave that X to 8 but we can

rewrite 2 4th that's 2 * 2 * 2 *

2 2 * 2 is 4 * another 2 is 8 * another

two will give us 16 so it says our final

solution is X to 8 over

16 okay so we can do the same thing here

with our second example we have 4 over n

the 6 all being squared I'm going to

stick an exponent in

there so again we can write that as 4

the 1st

squared um into the

6 squared I think when you get a little

more comfortable these let me go down

here so it doesn't run into my other

problem I think when you get more

comfortable you can actually kind of

skip this step I I definitely tend to

but I think at the beginning it's

certainly a good idea to write things

out always write things out when you're

unsure um I think it's a good it's just

good form it keeps you from making so

many mistakes so 1 time 2 that'll give

us to the second

power again on the bottom we have 6 * 2

which is to the

12th 4^ 2ar is 4 * 4 that'll give us 16

and then we have our n the 12th just

hanging out on the bottom okay so

that'll be uh our original 4 over n 6

all squared we can simplify that by

saying that 16 over n into the 12th

power

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all right in this video I'm going to

graph a basic rational function and the

idea is um the graph of 1 /x is the

graph here in the red and it's got a

horizontal ASM toote of yals 0 or

equivalently the Y AIS excuse me the x-

axis it has a a vertical ASM toote of uh

xal 0 which is the Y AIS okay and then

it's kind of sandwiched in you know the

bottom left corner the upper right

corner so I'm going to use some of our

sort of Reflections and shifting to help

us get some new graphs here so suppose

we want to graph 1x +

3 well the negative 1 overx all that

does is remember it's just going to take

our graph and reflected about the x-axis

so the part that had been in the top

right will now be in the in the um the

bottom

right and then the part that had been in

the bottom left will now go

up in the top left so this would be my

graph of f ofx =

-1x again the uh horizontal ASM toote is

still at the same place and the vertical

ASM toote is still at the same place

remember if we add three to a function

though all that does is it just I always

just imagine if the graph you know the X

and Y axis couldn't move but everything

else was rigid I just move it up three

units so this uh horizontal Al Asm toote

that had been at yal Z will now be moved

up here

to to y equal

3 okay and let's see a couple things

here um the the vertical ASM toote is

still in the same place so let's see

this is still our x axis and our y AIS

now the the portion that had been in the

uh the the the bottom right is going to

get moved up a little bit the portion

that had been in the bottom or excuse me

the the top left will get moved up so

I'm going to move that part right up

here and I know this part's also going

to get shifted up so it's going

to it's basically going to look like

that we could even find out this x

intercept if we wanted to um we could

find out the coordinates because again

to find X intercepts we just plug in Zer

for f ofx so we would get 0 =

1x + 3 we can add 1X to both

sides I'm going to write 3 is 3 over 1

if we cross multiply we would get 3x = 1

or x = 1/3 so it's actually crossing the

x axis here at this point um uh at the x

coordinate of 1/3 and now we've got our

basic graph just again kind of reflect

it the 1 overx graph and move it up so

um you know again you could all always

plot points plot points plot points and

get this General shape but this basic

graph 1/x is a useful graph to know so

that you can kind of do all these

shifting and uh uh U moving around of it

to help you get new graphs

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working with fractions in we're gonna do
in this case is just you

a little division involving fractions
and the main thing

you need to remember when dividing
fractions

is that we turned division into

multiplication so typically the ideas

we have one fraction say a/b divided
by another fraction

c/d what we do is we leave that the
first alone

that one we're dividing by we we
actually flipped end over

so the "d" goes on top, the "c" goes on the
bottom

and that's it you're the good thing
about multiplication and as

when you multiply fractions you just
multiply whatever's on top

so "a" times "d" and then you just multiply
whatever's on the bottom

"b" times "c" so remember if you are adding or
subtracting fractions you have to have

common denominators but

multiplications actually a little bit
easier because you just multiply across

the top and across the bottom in

if you can reduce you simplify you do
but I am

added net that's that's it so a case on
the first one we have six divided by two

over seven

the first thing I always do if I have
any

member that's not you know not it's not
written as a fraction we can always

write

any number as a fraction by just dividing
it by one

6 divided by one is the same thing as
six

okay so we were dividing by to over
seven

and now all I'm gonna do you just apply
my little rule here and I'm flip the

second fraction

and make this six over one times

7 over two you and we can

do this we can multiply either cancel
first but

maybe we just multiply six times seven
is 42

divided by two and forty-two divided
by two is

21 notice the other thing we could have done is we could have
 canceled we could have said

you know so there's a 6 on top and two on the
bottom

well 2 goes into 2 one time when times

and two will go into  six 3 times so we simplify
we will get  three times seven

which would be twenty-one and then on the
bottom we have one times one so you can

get 21 over one war

21 so typically I tried to cancel things
out before and multiply I found

in my own experience and it makes things
a little quick and a little easier but

as long as you simplify at the end it doesn't  really
matter okay so what's your other example

here

we have -7/4 divided by 4/7 same idea

we just leave the first fraction 11 so
negative 7 number 4

and then i turn my division into
multiplication

in my second fraction I just flip it
over so now the 7 goes on top

and the four goes on the bottom and again now I just
 multiply the numbers on top so I have -7

times

positive 7 that will give me a negative 49

and then 4 times 4 is 16

and in this case nothing will cancel out or
 reduced so this from your

final answer okay so nothing too bad

again the main thing with dividing
fractions you have to remember

division to do that we turn it into
multiplication

and again when multiplying you just do the
cross the top across the bottom

and you know if you can I would say make
use of any canceling that's out there

Basic Math:  Dividing Fractions

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all right here we're going to start

looking at some videos uh related to

converting between fractions and

decimals and um kind of in general going

from fractions decimals and even

percentages is something that you know

to me that's uh that's useful math to be

able to to understand those things and

to be able to switch from one to the

other so what we're going to do though

is we're going to take decimals um and

we're going to write them as

fractions so kind of the basic procedure

and this to me is what I do in you know

any type of uh uh problem like this the

first thing I you know I do is just

always put everything over one I mean

you can write any number by dividing it

by one 0.4 divided by one is certainly

still

0.4 and now what we want to do is simply

get rid of decimals and to get rid of

decimals you'll either multiply by you

know 10 over 10 or 100 over 100 or a th

over a thousand Etc and again all that's

going to do is uh move the decimal

place so in this case we're going to

multiply by 10/

10 so again multiplying by 10 basically

has the effect of moving the decimal

place one to the

right well 0.4 * 10 then will be uh

simply equal to 4 1 * 10 is just 10 and

hey now we've got our decimal as a

fraction but of course we can reduce

this one a little bit they're both even

so we could write the numerator as 2 * 2

the denominator as 2 * 5 we could cancel

out those common factors of two and be

left with 2 over 5 or two fths so 0.4 is

the same fraction as 2 over5

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Writing a Decimal as a Fraction, Example 1.  Here I look at how to convert a decimal into a reduced fraction.


all right in this video I'm going to do

another example of solving some inqu an

inequality and again this is just a

linear inequality any linear

inequalities are the the kind of the

nice inequalities you'll see as you go

further that when you have um say for

example quadratic inequalities there's

actually a bit more of a process it gets

a a little more um a little more

cumbersome but so for now all I'm going

to do is again just kind of pretend this

is an equation and just solve it you

know using my same technique so I'm

going to distribute to 5 so we would get

5 * x 5 * 3 would be POS 15 on the right

if I distribute we'll get 2 * X and then

2 * -5 will give us

-10 um I like to keep my x's positive so

since the bigger one's on the left I'm

going to subtract 2x from both sides so

that'll leave me with

3x + 15 less than or equal to

-10 and again now I'm just going to

subtract the 15 from both sides that'll

give me 3x is less than or equal

to

25 and now I'm just going to divide both

sides by three divide by 3 and this says

X has to be less than or equal to

-25 over 3 again if we wanted to write

this in interval notation um first I I

always like to make a little graph I

almost think about it in my head so

here's - 255 over3 I'm just kind of

making a little number line arbitrarily

it can be less than or equal to that's

what the little line means that number

so now we're going to shade it in to

indicate that that number works and then

it says everything less than that also

works so we would just keep going going

going going going

forever negative Infinity uh so again

just to indicate that you go arbitrarily

far to the left um so if we write this

in interval notation we'll write a

little parentheses around our negative

infinity and then we have - 255 to 3

excuse me - 255 over 3 and since we can

use that number the notation is we put a

little bracket around that number to

indicate that we can use it I always you

know kind of thought about the little

bracket you know I always imagine those

were like little hooks or little

barriers and that would catch that

number and keep it from getting away so

that we include it so that was my little

weird way of kind of remembering the

difference in notation and if it's

rounded um you know if we didn't include

um if we didn't include -25

over3 we would use parentheses and those

to me are kind of more rounded and kind

of uh not as tight and the number

somehow sneaks away so all right

probably more information than you

wanted to need uh needed to know but

again just kind of a little way to maybe

I I've seen uh in my experience it seems

like people forget this notation um

quite frequently so just you know um

however you can remember it uh just try

to internalize it and remember it any

way that works for you

An Introduction To Solving Linear Inequalities - Example 2

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all right in this video I just want to

do a couple examples of simplifying

products of binomials involving trig

functions so just some binomial

expressions I'm just going to clean them

up here a little bit so in the first one

we have cosine X minus one times cosine

X plus one if we were to distribute this

out we would get cosine times cosine

which would be cosine squared X we would

get cosine x positive one which would be

positive cosine X then we get negative

one times cosine X X which would be

negative cosine X then we would get

negative one times positive one which

would be negative one well we've got a

positive positive cosine and a negative

cosine that's just canceled out so we'll

be left with cosine squared X minus one

but recall we've got again this identity

cosine squared X plus sine squared x

equals 1 so we can always I think sort

of manipulate this identity to make it

look like this so notice we could

subtract one from both sides that would

give us cosine squared X plus sine

squared X minus 1 equals 0 and then we

could also subtract the sine squared

from both sides and get cosine squared X

minus 1 equals negative

sine squared X so which one is

preferable I don't know probably the

second one just because there's you know

one fewer term so that's it so I would

say the first one just simply reduces to

negative sine squared X maybe let's look

at the second one real quick as well

we've got secant X plus tangent x times

secant X minus tangent X you're going to

do the same thing which is just simply

to distribute it out so we would have

secant times secant which would be

secant squared X we would have secant

times negative tangent which would be

negative secant X times tangent X on the

inside we would have a positive tangent

times secant but since it's

multiplication I can order it however I

want so I'm going to write secant x

times tangent X and then we would have a

positive tangent x times a negative

tangent X so that's going to give us a

negative tangent squared X so our

negative secant tangent and our positive

secant tangent will cancel

we'll have secant squared X minus

tangent squared X but again there's an

identity relating secant squared and

tangent squared so tangent squared X

plus 1 equals secant squared X so I

think if we subtract the tangent squared

from both sides we'll get that one

equals secant squared X minus

tangent squared X so that's what we're

left with over here on the left so

secant squared minus tangent squared

just by manipulating that one sort of

known identity we can see that secant

squared minus tangent squared is simply

going to equal one

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okay in this video we're going to talk a

little bit about the conjugate pair

theorem and what the conjugate pair

theorem says it says if you have a

complex number of the form for example

A+ bi if we know that that's a zero of

our polinomial then it says a minus bi

is also a zero conversely if uh a minus

bi is a zero then A+ bi is a zero so

this it's this idea that complex numbers

the zeros appear in pairs okay

so suppose

uh suppose we know that f of x is a

polom and it's got real coefficients and

it has the following zeros Square < TK

of 3 netive I 5 - 4 I and 1 + I / 8

suppose those are all zeros and we want

to know what three other zeros would be

well we can write um negative I as a

complex number by writing it as 0 - well

1 I and since 0 - 1 I is a0 that would

imply that 0 + 1 I which just equals

positive I is also a

zero okay so just flip the sign on and

is all it says since 5 - 4 I is a zero

again the conjugate all we do is just

change the sign in the middle so since 5

- 4 I is a zero that means 5

+ 4 I is also a

zero and 1 + I over 8 since that's a

zero again you can think about this as

you know you don't have to do this we

could write as 1/8 plus I over

8 and since that's a zero again we'll

just change the sign in the

middle we could make it 1/8 - I over 8

or equivalently 1 - i/ rate is also a

zero

The Conjugate Pair Theorem - Example 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Conjugate Pair Theorem - Example 1.  In this video, I show that if we know a complex root, we can use that to find another complex root using the conjugate pair theorem.  Just a basic idea of the theorem.


all right in this video I want to find

the Lao transform of a function and here

we're going to use uh we're going to

find the llao transform of 4 * cosine of

2T - 5T + 2 * e

3T so this uh won't be too bad at all so

if we take the llao transform of of all

of

this so we'll have the L

transform of 4 cosine of 2T

remember the lass transform is linear so

we can break all this up minus the L

transform of

5T plus the Lelo transform of 2 e to the

3T but also recall that for uh

coefficients we can simply Factor the

coefficients out

front so really we're just going to be

left with uh four * the LL

transform of cosine

2T okay minus 5 * L transform of

t plus 2 * LL transform of e to the

3T and now we're just going to use our

table much like you would use you know a

table of derivatives when you first

start doing calculus so let's see here

let's find our table so let's see um so

for cosine of 8 we said there you get s/

s^2 + a 2 so in this case our number

being squ is

2 so the luse transform we'll get

s/ s^2 plus again our a is going to be

two so we'll get 2^

2 minus 5 * the Lao transform of t uh

the Lao transform at just T where' It Go

um so the Lelo transform if it's t to

the first Power well you would get uh

one factorial which would just be well

one let's write it out we would get 1

factorial and then we would get S raised

to the 1 + 1 or we would get s^

squar um let's see

here notice uh well excuse me so last

but not least if we just take the Lael

transform of e to the 3T no no noticing

anything uh let's see so we said if you

have e to the a you get 1 / s minus the

a value so in this case we'll get 1/ s -

3 and now this is our lass transform um

you're pretty I mean you you found it at

this point um you can always just

multiply the

constants whoops let's make that a plus

4 instead of 2^ s um you know you can

always just multiply the constant back

in but that's all there is to it uh just

kind of knowing the table and then just

kind of taking the Lao transform of each

function uh individually

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Calculating the Laplace Transform of a Function Using Tables.  In this video, we find the Laplace transform of a function using linearity and tables.  This is a very basic example!


Okay, in this question it says if y = kx where k is a constant

and it says y = 24 when x = 6,

we wanna know what's the value of y when x = 5.

So this one, you know, hopefully won't be too bad,

again it's just using this information that y = 24

and x = 6 that's gonna allow us to solve for k.

Once we solve for k, we can substitute in also x = 5 and we'll get the value of y.

So let's use this info that y = 24 when x = 6..

So I'm gonna put that on the left side of my equation,

so we've got 24 for y, again we're trying to solve for k.

So k multiplied by x, which is gonna be 6..

So, k * 6, you could right this as 6k but whatever.

So to solve for k I'm gonna divide both sides by 6..

24/6, that's just gonna give us 4..

On the right side, we have k..

So now, we know that our equation is of the form y = 4x,

again we just solved for k,

and now all I have to do is substitute in x = 5..

So not too bad.

When x = 5, we'll get that y = 4 * 5, or, y will equal 20.

So, answer choice C is the correct choice in this case.

SAT Math - Question 2, Section 2, 2016 Edition and Beyond

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Section 2 - A Calculator IS allowed!


all right so just a little word problem

here involving

fractions and division of fractions

suppose we've got

a board that's five and uh 1 4

feet long we're going to cut this up

into little

planks that are three-quarters of a foot

long and we want to know how many of

these little planks can we get

so you can probably see it through there

i've got my little picture here so

suppose we cut you know we've got our

board here

it's five and uh one-fourth feet long

and we're going to cut it up into these

little planks that are

uh you know three-fourths of a foot

long each and okay so i just put some

arbitrary number in there don't count

them up and

say oh that's not right because i doubt

it is so

but again that's the idea right we've

got a long piece of wood we want to chop

it up into

a bunch of smaller pieces and we want to

know how many of these smaller pieces

can we get well again what we're doing

is

we're taking five or uh

our piece of wood that's five and one uh

one-fourth feet long and we're going to

divide it

into pieces that are three-fourths

of a foot long each okay so that's what

we have to do we've got to take

the five and one-fourths and divide it

by three-fourths and

again sometimes if you have a problem

like this and maybe the numbers are kind

of throwing you off and you're like do

you have to multiply or divide or i

don't know

you know make it easier numbers suppose

it was 10 feet long

and suppose they were you know two feet

each

you know maybe this is a little bit

easier right you know if it's ten feet

long

and they're two feet each well okay um

yeah you can get five and

well how do you get that you take the

total length and you divide it by

you know the length of each little each

smaller piece

so sometimes i think about these

problems you know when i do these

problems

i think about them with easier numbers

just to sort of make sure that i'm doing

the right thing

in this case division so okay division

makes sense

and now i say oh i've got a mixed number

divided by you know a fraction

i'm going to make these both into

fractions 4 times 5 is 20 plus 1 will be

21 over 4

divided by 3 over 4.

well now i'm going to make my

division into multiplication by

multiplying by the reciprocal

but now i can cancel out some factors

immediately

okay so the 4 and the 4 we could just

say well

4 divided by 4 is 1 4 divided by 4

divided by 4 is also

1. we can do the same thing with the 21

over the 3.

i'm just going to be lazy and write it

though 21 times 1 is 21 over 3

well 21 divided by 3 is going to give us

7. so it says actually uh if we chop up

this piece of wood

this long piece of wood we should get

exactly one two three four five

six seven so um i've got too many

too many parts so it says you would get

exactly

uh seven little little uh planks of wood

that are three-fourths of a foot each

and again all we're doing

just a little bit of division again you

have to make your mixed number

into a fraction division by a fraction

turns into multiplication by the

reciprocal and then it's just a matter

of doing the arithmetic and

simplifying it down

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all right here we're going to do some

examples of finding mcloren pols so all

a Muran polinomial is it's just a

special case of a Taylor polinomial and

a mclen polinomial is by assumption

centered at a equals Zer so I just kind

of rewrote our new formula here

emphasizing the fact that a is equal to

zero all right so in this example we're

going to find the third degree mclen

polom for f ofx = e

4X so again I'm just going to write it

out generically because that's what I

like to do so we'll have F of

0 plus FR Prime of 0 over 1 factorial x

to the 1st F Prime of 0 over 2 factorial

x^2 and then F triple prime of 0 3

factorial * X cubed so let's see our

function

just e to 4X well these derivatives

aren't too bad we have to use the chain

rule so e to the 4X when we take the

derivative we'll get e to 4X but then we

have to take the derivative of the

exponent so we'll get uh an extra four

so we have 4 e

4X well when we take the second

derivative uh we'll get an e to 4X time

another four so that'll give us 16 * e

4X and then our third derivative

evaluated at zero we're going to have to

multiply by yet another four so that's

going to give us 64 * e

4X but when we plug in

zero f of0 will get e to the 0 which is

one F Prime of 0 we'll get well four * e

to the 0 are just four F Prime of 0

will'll be left with 16 * 1 or

16 and our F triple prime of 0 we'll

just be left with a

64 so now if I fill in you know our very

beginning

formula we will have our degree 3 muren

polinomial so F of0 we said is equal to

one F Prime of 0 is four so 4 over 1

factorial is Just 4 * X the

1 um f Prime of zero that's 16 divided

two which is 8 so 8 x^ SAR

let's see um f triple prime of 0 that's

64 over 3 factorial so 3 * 2 * 1 well

let's see two goes into 64 I guess 32

times so we'll be left with uh so F

triple prime so um 64 divided 6 we'll

reduce that to 32 over3 then we have our

x to the 3 power and this will now be

our degree 3 the chlorin

polom

📚 Finding the Third-Degree Taylor Polynomial for f(x) = e^(4x) 📚

In this video, I walk through the process of finding the third-degree Taylor polynomial for the function f(x) = e^(4x), centered at x = 0. This example shows how to apply the Taylor series formula to compute a polynomial approximation of an exponential function.

🚀 What’s covered:

Step-by-step derivation of the Taylor series for e^(4x).
Detailed explanation of how to compute the third-degree Taylor polynomial.
How to find higher-order derivatives and use them to build the polynomial approximation.
Understanding Taylor polynomials is essential in calculus for approximating functions, so this is a great exercise to strengthen your skills in series expansions and polynomial approximations.

👍 Don’t forget to like, subscribe, and hit the notification bell for more calculus tutorials and polynomial approximations! 🔔✨

#TaylorPolynomial #ExponentialFunction #TaylorSeries #MaclaurinSeries #Calculus #MathTutorial #Mathematics #SeriesExpansion #LearningMath #MathHelp #PolynomialApproximation #STEM


okay so the next question again is going

to involve this graph that we saw in the

previous problem and it says which of

the following represents the

relationship between H and C so again

I've labeled some points on here you

could always start substituting in

values right the first cordinate

represents H the second coordinate

represents

C so again the number of hours rented

comma the total cost

involved so there's two ways to do this

um

I mean if you wanted to the way I would

think about it is just think about it

like in terms of y = mx +

b right this is slope intercept

form slope

intercept well instead of Y the vertical

axis is labeled as capital

c instead of the variable X you know our

x axis is replaced by this variable H

the number of hours

but again the same thing m still

represents the

slope B still represents the you know

the normally we call it the Y intercept

but in this case it would be the C

intercept okay well the C intercept had

a value of five it's going through the

point 0 comma 5 so I know

that the plus b is going to be replaced

with + 5 well what's the slope that's

the rise over the run so how much do you

go up and how much do you go over to get

to the next Point well if we go from 1A

8 up to 2A 11 I'm going up 1 2 3 units

I'm going over one unit again my my

scale is definitely off here but you

could do slope change in again change in

y over change in X or in this case it

would be change in C over change in h

but let's think about it as before you

you do the difference of the second

coordinates over the difference of the

coordinates so you would have 11 - 8 so

I'm doing 11 - 8/ 2 -1 that's going to

give you 3 over 1 or three so that's

going to be the slope so the formula

here would be c = 3 H + 5 which would

also be answer Choice C and again once

you had a couple points on here you

could always start substituting these um

you know for example we know that 2

comma 11 is on there if we substitute in

2 for H and answer Choice a we would

have 5 * 2 which is 10 but we should be

getting the value of 11 so we know

answer Choice a is not correct you could

go through and do elimination and get

rid of B and D as well by doing that but

again you're just doing the slope

intercept method you know finding the

equation of a line it's just switching

out again instead of X and Y you've got

H and C is all your you know the only

kind of difference here

SAT Math - Question 16, Section 2, 2016 Edition and Beyond

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Section 2 - A Calculator IS allowed!


alright so another example here

simplifying one of our expressions so

here we're going to simplify cotangent

of PI over 3 minus X again we can't

quite use our cofunction identity

because again we need a PI over 2 what

I'm going to simply do those rewrite

cotangent I can write cotangent as 1

over tangent so we would have tangent of

PI over 3 minus X and again now we can

do the same thing we can just simply use

our identity for tangent so let's see if

we can squeeze it in here let's see so

it says so we're actually going to get

another little fraction in the

denominator so it says we would get

tangent PI over 3 since there's a minus

sign in between we keep the minus sign

so minus tangent X and that's all gonna

be divided by 1 and again if there's a

minus sign in between it says now we use

the plus sign so we'll get tangent of PI

over 3 times tangent of

X okay so on the right side if you have

one over a fraction all that's going to

do is flip your fraction so I'm just

going to go ahead and do that so we'll

get one plus tangent of PI over three

times tangent of X in our denominator we

would have tangent of PI over three

minus tangent of X okay well tangent of

PI over three to evaluate that again

tangent of PI over three that's just

going to be sine of PI over three over

cosine of PI over three

well sine of PI over three recall that's

the square root of three over two cosine

of PI over three is just going to be one

half and if we flip and multiply this

will simply reduce down to the square

root of three so now I'm just gonna plug

that in it says we would get one plus

square root of three times tangent X and

in our denominator again tangent of PI

over three is going to be square root of

three minus tangent X and to me you know

of course there's other things you can

do to it but I think I would just leave

it there and now we have just come up

with a new expression for that original

expression

Cofunction Identities, Example 2

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okay in this video I want to talk about

um just a different example of a type of

homogeneous second order linear

differential equation so um this is the

form of a homogeneous second order

linear differential equation P of x q

ofx r ofx g of X have to be continuous

functions in general these are hard to

solve we're going to look at an easier

case where PQ and R are constants

replaced with by values a b and c we

want a to be non zero the idea is we're

going to solve we're going to we're

going to

turn the um differential equation into

basically a quadratic equation we call

this the characteristic equation solve

the corresponding characteristic

equation based on the types of solutions

to the characteristic equation it tells

us the general solution to the original

differential equation um so obviously we

have a quadratic equation different

things can happen in regards to our

Roots one of the results is the

following so it says if your roots are

complex then

it says if the roots are of the form a

plus or minus um beta I it says then the

general solution of this differential

equation is going to be of the form e to

the alpha x time the quantity C1 cosine

beta X Plus C2 sin beta X so um here

we're going to solve y Prime - 6 y Prime

+ 13 y so again we have to look at the

uh equation R 2 - 6 r + 13 = 0 well I

don't think this is going to factor so

I'm just going to use the quadratic

formula B plus or minus the square < TK

of b^ 2 - 4 * a * C 4 * 1 * 13 is 52

over 2 * a okay so if we simplify this

down we'll get 6 plus or minus the

square < TK of -16 so there's our

complex Solutions cropping up

and we can just simply bust this up as 6

plus or minus < TK of 16 excuse me Nega

16 would be 4 I over 2 well let's just

go ahead and reduce this to 3 plus or

minus 2

I so we're basically there at this point

um Alpha is going to be the value three

beta is going to be the

value two don't worry about the plus or

minus um and now all we have to do is

just drop that into

our basically our little formula up here

so I'm going to fill in Alpha = 3 betaal

2 drop in Alpha Beta And beta and then

we'll have our general solution so it

says our solution will look like e to

the alpha which is

3x C1 cosine of beta Co beta being um

two in this case is what we found so

2x um plus C2 s of beta again which is 2

um time x and Tada we now have our

general solution so nothing too bad um

the only thing that's really tricky I

would say in this case is just

recognizing and remembering the formula

um so other than that it's just

basically just a solving a quadratic

equation so things don't get much better

than that and differential equations

class so if you have any questions or

comments please feel free to post them

um and hopefully me or somebody else can

help you out out there

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For more free math videos, visit http://PatrickJMT.com


okay suppose at the area of a circle is

740 square inches we want to know what's

the diameter of the circle to the

nearest inch all right so there's our

circle the area is equal to 740 that's

given to us now one thing we'll want to

use is the area formula for a circle and

that is pi times the radius squared

recall that the radius is the distance

from the center of the circle out to one

edge we're gonna approximate PI by using

the value 3.14 so the idea is we're

gonna be able to use this information

about the area to solve for the radius

and recall that the diameter the

diameter is actually the distance all

the way across the circle so once we get

the value for R will simply multiply

that by 2 and that's gonna give us the

diameter all right so we've got 740

again because that's the area that

equals 3 point 1 4 multiplied by R

squared so the first thing I'm gonna do

is I'm gonna divide both sides by 3.14

that's gonna leave us with r squared on

the right side so here I'm gonna use my

calculator so let's see 740 divided by

3.14 I'm getting that to be 235 point

six six eight so I'm gonna round that to

point six seven and now I have to get R

by itself since R is being squared we're

gonna take the square root of both sides

now normally for an algebra algebraic

equation when you take the square root

of both sides you have to include a

positive and negative on one side of the

equation but in this case you know so

taking the square root of r squared is

going to give us R the radius clearly

has to be positive so I'm only going to

keep the positive solution so let's see

here I'm gonna take the square root of

two thirty five point six seven I'm

getting that to be fifteen point three

five after

so that's the radius okay so we said

that the diameter that's gonna be twice

the radius we have to double that so the

diameter is going to be 2 multiplied by

15 0.35 that's gonna give us 30 point

seven zero after doing the arithmetic so

2 times 15 is 32 times 35 is 70 so

that's where I'm getting the 30 point 70

and since we're rounding that to the

nearest we said to the nearest inch the

diameter is roughly going to be equal to

31 inches

GED 'Most Missed' Math Practice Problems , Code Q.4.b,  #1

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This video shows how to work a sample problem using their coding system that is shown on the left using the link above.


All right, in this video I want to talk about solving some trigonometric equations, and the equations

we're going to see are going to be relatively,

you know, not too bad.

But the idea is typically you're going to have, you know, more complicated equations

that eventually reduce to, you know, finding solutions to these types of equations.

Start here with the basics. So suppose we want to find all solutions to sine x equals root 3 over 2.

This is one of those kinds of things, you know, that you want to have memorized.

So go back to the unit circle, and recall that on the unit circle, at the angle  Pi/3 radians,

the x coordinate is 1/2, but the y coordinate is root 3 over 2, so

definitely, we can say one of our solutions here is going to be x equals Pi/3.

Now there are other places on the unit circle where certainly sine equals root 3 over 2.

Likewise, if we go over to the angle

2 Pi/ 3. (There's my bad circle.) So now, at the angle 2Pi/3,

the x coordinate is now going to be negative 1/2, but the y coordinate again is still root 3 over 2.

So 2Pi/3 is going to be another solution. So x equals 2Pi/3, and

now, well, you know

certainly those aren't the only solutions. Those would be the only solutions between 0  and 2 Pi,

but we're not given any restrictions here, really, on the, you know, the interval for x.

So to me the idea is again since I'm at Pi/3, I can either add or subtract multiples of 2 Pi.

That'll put me at the same

equivalent place on the unit circle, so we can simply add multiples of 2 Pi by saying

plus 2 Pi times n, where n is an integer,

and

we'll have to do the same thing for our other solution 2 Pi/3.

Because, again, if you add or subtract multiples of 2 Pi, it's going to put you back at that same

equivalent angle, so the other solutions would simply be x equals 2 Pi/3

plus 2 Pi times n.   Again, n is

an integer.

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a Basic Trigonometric Equation, Example 1


all right in this example we're going to

find a position Vector for a vector that

starts at the point 3 comma 7 and ends

at4 comma 2 in Part B we're actually

going to find the length of the vector

as

well so to find this position ve Vector

um so again we start at 37 you don't

have to draw the picture at all

um but I'm going to um it starts at 3

comma 7 and goes to the

point4 comma 2 so this is what our our

Vector would look like if we were to

draw it to find this position

Vector all we have to do is okay so I'm

going to list the coordinates again this

is -4 comma 2 this was 3 comma

7 what we do is we take um our so I'm

going to look at the x coordinates we we

take the stopping point which was -4 and

we subtract away whatever the starting

point was which was the x coordinate of

three and then we do the same thing uh

for the Y coordinates so we started at

the uh we stopped at the y-coordinate of

two and we started at the y-coordinate

of

seven and that's now going to be our

position Vector so4 minus 3 that's -7 2

- 7 that's going to be -5 and again all

this to me really says it says if you're

sitting at some point it says what have

you done you've gone seven units to the

left and then you've gone five units

downwards that's all that's happened so

that's our position

Vector Part B we wanted to know the

length well to get the length all we do

is we just take the square root we take

each component and we Square it we take

-7 squared we take5 squared and then we

add those together

so let's see this is going to be the

square root of uh 49 +

25 so I guess uh 49 + 25 that's going to

give us the sare < TK of

74 um we could factor that I guess as 2

* 37 but 37 isn't going to factor any

further so I would simply uh stop here

and say the length of our Vector is the

square root of 74

❖ Position Vector and Magnitude / Length ❖

In this video I find the position vector and also the length / magnitude of that position vector.
In this example we are given the coordinates of where our vector starts and the coordinates of where the vector ends.
Once we find the components of our vector, we simply use the pythagorean theorem to find the magnitude/length.

#vector #magnitude #components


all right here we're going to look at

Absolute and conditional convergence and

just a few examples in trying to decide

whether they're uh absolutely convergent

conditionally convergent or simply

Divergent so part A here we' got the

series -1 to Infinity of1 to the n * 4

2N over 5n um I'm going to simplify this

a little

bit this is -1 to the N we could write

this as 4^2 raised to to the N but that

would just give us 16 to the N over 5 to

the N okay so now if you want to ask

yourself you know is this uh you know

absolutely

convergent well to look to decide if

it's absolutely convergent or

not we would put our

series we would put the formula inside

of absolute value and all that would do

is just basically get rid of the Nega

one to the n

but then we would be left with 16 over5

all raised to the N power this is a

geometric series but this is going to be

a Divergent geometric

series because the R value is not in

between negative 1 and positive one so

uh when we ask ourselves is it

absolutely convergent no definitely not

we could ask ourselves is it

conditionally convergent

well is it conditionally convergent no

um and basically for the same reason um

you know if we look at it the way it

stands this is a uh an alternating

series our B subn in this case again we

can forget about the alternating part

our B subn would be 16 over5 raised to

the N well again we have to ask

ourselves two questions is it

decreasing and does the limit go to zero

well neither one of these are going to

happen if you plug in Nal 1 2 3 4 five

uh you know 16 over five is a number

larger than one so if we start

multiplying it by itself it's definitely

not going to decrease it's going to

increase so neither one of these things

happen so it's likewise uh when we ask

ourselves is it conditionally convergent

we would say no well uh in this case our

series is simply going to be Divergent

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Absolute Convergence, Conditional Convergence, Another Example 1.  Here we looks at some more examples to determine whether a series is absolutely convergent, conditionally convergent or divergent.


okay here we're going to graph y equals

five times tangent of one-third X we'll

pick out vertical asymptotes and zeros

this is going to look like the graph of

tangent X except for it's been stretched

vertically by a factor of five and the

period is changed so the period to get

the period be careful for sine and

cosine we use 2pi but tangent has a

period of Pi so we take pi and divide it

by the absolute value of the B value

which in this case is 1/3 and that's

going to give us a value of 3pi

okay so for tangent if you take the

period and divide it by two so it'll be

well three PI over two if you move over

that many units from the y-axis to the

right and to the left those will be your

first vertical asymptotes so here's

going to be the line x equals 3 PI over

2 this will be the vertical asymptote x

equals negative 3 PI over 2 and we can

fill in one point if we want to at least

so normally at PI over 4 it goes through

1 but now at PI over 4 it's going to go

through 5 but it still has that same

general shape to it so again we said the

period in this case is equal to 3 PI so

again certainly not to scale right

here's PI over 4 my vertical asymptotes

already over here 3 PI over 2 so but if

we move over another 3 PI units so we'll

put it right there there's 3 pi 3 PI

over 2 plus 3 PI well that'll be 6 PI

over to our next vertical asymptote our

next positive vertical asymptote is

going to occur at x equals 9 PI over 2

and again it's got that same that same

shape to it so there's two two pieces of

the graph

the zeroes are going to occur the first

one occurs at zero and then it's at

multiples of three pi so those will

occur at the value three pi n where n is

an integer and the asymptotes the

vertical asymptotes the vertical

asymptotes are going to occur at x

equals 3 PI over 2 and again we're

adding 3 pi to get to the next one so

multiples of 3 PI again where n is an

integer

Graphing a Tangent Function - EX 2

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all right here we're going to look at

one more example of multiplying decimals

so we're going to do 51.5 *

0.014 so let's see

51.5 *

0.014 all right so let's see four time 5

is going to be 20 so I'm going to drop

down the zero and carry the two 4 * 1

will be 4 + 2 will be 6

and then 4 * 5 will be

20 okay let's see so I'm going to add my

zero and then 1 * 5 is five 1 * 1 is 1 1

Time 5 is

five two zeros and then if we multiply

we would just have 0 Time 5 one and five

which will be Z 0o and zero then we can

add in our next row this time we'll have

three zeros but again if you multiply by

zero you'll just get a bunch of zeros

okay so now we're going to add all this

stuff together the first column we'll

just get zero 6 + 5 is 11 so we'll keep

the one and carry the one 1 + 1 is 2 2 +

5 is 7 and then we were left with a

couple zeros and again now I just count

so to the right of the decimal sign we

have one digit two digits three digits

four digits total to the right of the

decimal sign so that's how many places

I've moved my decimal so one place place

two places three places four

places so it says if we multiply 51.5 *

uh

0.014 will'll be left with

0.721 Z as our solution

Multiplying Decimals, Ex 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Multiplying Decimals, Ex 2


all right in this video we're going to

start looking at the route tests to

determine whether or not a series

converges or diverges and again the

route test says we take our series the

formula and we look at the limit as n

goes to infinity of the nth root of the

absolute value of a sub n just like the

root test or excuse me just like the

ratio test the root test has similar

results of the the limits less than 1

the root test says the series converges

if it's greater than 1 it diverges if an

equals 1 it's inconclusive

so just a few different examples here

we'll decide whether it converges or

diverges so Part A here we've got from 1

to infinity of 3 of it and e to the

negative n so we'll take the limit as n

goes to infinity we can take the

absolute value of the stuff again we're

going to take the nth root but taking

the nth root is equivalent to raising it

to the 1 over N power well both of these

are going to be positive no matter what

value of n we plug in so we can just

simply drop you know get rid of the

absolute value well then we'll raise it

to the 1 over N power well we have to

multiply exponents so we'll get n times

1 over n that will give us 3 to the

first then we'll get negative n times 1

over N and that will give us e to the

negative first power this is just a

constant so the limit will equal 3 times

e to the negative 1 but that's going to

be 3 over e to the positive first power

and well this is greater than 1 because

E is smaller than 3 so that's going to

tell us that our series diverges and

again you could have recognized that at

the very beginning by rewriting 3 to the

N over e to the negative n as 3 over e

raised to the N power so you could have

also recognized the original series as

being a geometric series where the R

value is not between negative 1 and

positive 1 so in this case our series

will diverge

The Root Test - Another Example, #1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Root Test - Another Example, #1.  Just another example showing that a series converges or diverges using the root test.


okay this example is going to be about

estimating um a series using the

alternating series estimation theorem

and here we want to know how many terms

of this series we would need to sum up

to make sure our approximation is

accurate to two decimal places so I'm

just going to plug in Nal 1 2 3 4 Etc

into our Series so when we plug in Nal 1

we'll get 1 over

two when we plug in Nal

2 we'll get 2 the 2 which is 4 * 2^ 2

which is another four then we'll get

positive let's see we're plugging in

three so 2 the 3 is 8 3^2 is going to be

9 and then we'll get a negative so now

if we plug in four 2 to 4th is 16 4 SAR

is another 16 and maybe one more term

we'll plug in N = 5 we would get uh 2

5th which is 32 5^2 is 25 and then our

series is just going to keep going so I

plug these into a calculator so well2

that's just 0.5 5 um 1 over 16 is

0.0625 1 over 72 that's roughly

0138 then we have 1 over 16 sared which

is 1 over 256 that's

0.0039 and now I'm going to

stop because it says if we sum up the

first three terms it says the

error is at most

the absolute value of the size of the

first neglected term and we can even say

it's actually less than but it's going

to be

um less than the absolute value of

.39 which well is

00 or

0.0039 so um in this case we're going to

get accuracy uh correct to two decimal

places so we're basically looking for um

the first number where the first two

decimal the first two numbers after the

decimal place are zero so now we've got

two decimal places of

accuracy two decimal places of accuracy

so in this case it says we would just

have to sum up a total of three

terms and then our um our approximation

would be correct within the desired uh

error

Alternating Series - Error Estimation

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Alternating Series - Error Estimation.  In this example, I find the number of terms required so that we can estimate the value of our convergent alternating series correct to two decimal places.


all right in this video I'm going to do

another example of solving a rational

equation and these are going to be

pretty nicely set up simply because

notice both sides of the equation we

already have just a single fraction um

and the denominators aren't too crazy so

uh it's going to make life a little bit

easier on us so the first thing I'm

going to do in this example is a lot of

times if you can notice if if anything

factors and cancels at the beginning

it'll make your life a little bit easier

so notice the top left side this 3x -3

you know in my head I I think well we

could factor out a

three then we would need an xus one in

the parentheses but hey lo and behold we

have an xus one in the denominator so

that's good I'm going to be able to

cancel that stuff out and it'll just

kind of reduce you know reduce

everything that's present in the problem

so I'm going to cancel out my x -

1's and on the left side you know you

could think about there's being a one in

the denominator so we still have 3 over

one on the left side on the right side I

I think well I could factor a two out of

the numerator but that's not really

going to cancel out anything so I'm just

going to leave it

alone and now I'm going to do the same

thing as in some of my other examples

I'm just simply going to cross multiply

and then that that's going to turn at

least this problem into a nice little

linear equation that won't be hopefully

too bad to to solve so if we take 3 * 5x

that'll give us

15x we would have to take our positive 1

and multiply it by 2x + 4 and I'm going

to write this step out again just

emphasizing that you should put this in

parentheses and distribute everything

out but obviously if you you know

distribute out the one you just get hey

2x + four and now I want to try to put

all my x's on one side and all the

numbers on the other so in this case um

I'm going to subtract this 2x from both

sides and the reason I do that is well

then I'll have all my x's on one side

I'll just have my numbers or my number

um on the other side so 15x - 2x would

give us

13x and then we've got positive4 left

over on the right

side and now we've basically just got

one little step left left to do um 3x =

excuse me 13x = 4 so since we're

multiplying by 13 I just want to get

positive 1X all by itself so since we're

multiplying by 13 we'll do sort of the

opposite which is to divide by 13 on the

left side 13 over 13 is just 1 X on the

right side we get 4 over 13 and that'll

be our solution again as a rule of thumb

you should always kind of go back and

plug this in and make sure that that it

does work in general as long as it's not

making one of your fractions undefined

you should be okay but you know to be

perfectly sure you can always plug it in

and make sure um it is possible to get

extraneous solutions that actually don't

work at the beginning so just one other

little thing that you have to

unfortunately be worried about when you

solve rational equations

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a Basic Rational Equation - Ex 3.  In this video, I solve a rational equation.


okay here we're going to find the volume

of a box and for a nice uh a nice shape

like this basically what we're going to

do is we're going to

find the area of one of the bases and

then multiply it by the corresponding

height so what I'm going to do here is

I'm going to find the area of this front

face and then multiply it by the height

so if you can kind of imagine if the you

know the Box were on its side or again

you know for a rectangle the area is

basically the length time the width time

the height but again the length times

the width that's going to give us the

area of one of the

bases so we'll multiply 4 * 5 and then

just simply multiply that by 10 so 4 * 5

is 20 * 10 will be 200 the units will be

centime time cenm time cenm so we would

have cm cubed in this case so the answer

of our the answer for the volume will be

200 cm cubed

Volume of a Prism (A Box!)

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! For tons of free videos, visit:  http://PatrickJMT.com
 Volume of a Prism (A Box!).   Here we look at a quick example of finding the volume of a box.  All we need to do is multiply the length, width and height together!