okay in this example we're gonna solve

the inequality X minus 2 less than 2x

plus 6 being divided by 4 so so here's

my little recipe on when I see this kind

of problem I think you know there's a

procedure to do it it's just like being

a baker you know there's an order in

what you do things and the same thing

for this type of inequality so the first

thing I'm thinking is well it would be

nice to get rid of that fraction I've

got this fraction on the right side you

know fractions can sometimes be a little

just tedious to deal with so maybe if we

can get rid of the fraction that'll be

that'll be helpful

the next thing I'm gonna do is I say

well you know I've got terms involving X

I'm gonna have constants floating around

let's get all the X's combined and let's

also combine all the constants the last

thing I want to do is get one X all by

itself and then I'll have my my solution

okay so the first thing we said was

let's get rid of this fraction so I've

got X minus 2 less than 2x plus 6

divided by 4 well since I'm dividing by

4 that's kind of the thing I want to go

away I wish that wasn't there right I

wish I just had 2 X plus 6 left over I

would feel a little bit better in that

I'm getting closer to solving the

problems since I'm dividing by 4 I'm

gonna multiply both sides by 4 so make

sure you add parentheses on the left

side because you want to use this

distributive property if not you'll

certainly make a mistake so 4 times X

will be 4 X 4 times negative 2 will be

negative 8 on the left side the fours

will just cancel out and we'll be left

with 2x plus 6 okay so now I've gotten

gotten rid of the fraction the next

thing I want to do is I want to combine

the terms involving X and I want to also

combine the just the constants so it

doesn't sort of matter the order in

which you do it first let's get the

constants together so I'm gonna put all

the X's on the left side and all the

constants on the right side just just to

pick an order I mean you could do an

opposite it doesn't matter

so you could put all the the terms

involving X on the right side and all

the constants on the left so okay so if

I want to keep the X's on the left that

means I'm going to move this constant

over

so since I'm subtracting eight I'm gonna

add 8 to both sides well I'll just be

left with 4 X on the left side Oh

that'll be less than 2x plus 14 and now

the same thing I want to combine all the

terms involving X well I said I'm gonna

put all the terms involving X on the

left side so let's let's move this

positive 2x by subtracting 2x from both

sides

well 4x minus 2x that's gonna leave me

with 2x 2x minus 2x well hey that's 0x

we still have our positive 14 on the

right side and now the last thing I want

to do is get X all by itself so since

I'm multiplying by positive 2 I'm gonna

divide both sides by positive 2 so the

twos will cancel we'll be left with X on

the left side that's less than well 14

divided by 2 is positive 7 and that's

our solution recall that if you multiply

or divide by a negative number you have

to flip the inequality nowhere did we do

that in this example but certainly

something I want you to keep in mind and

just to remember and to be aware of that

fact

GED 'Most Missed' Math Practice Problems , Code A.3.a,  #3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The GED Testing Service keeps a list of the most missed math topics and can be seen here: http://www.gedtestingservice.com/uploads/files/126de0283c94ff6323b3b5b3f310da5d.pdf
This video shows how to work a sample problem using their coding system that is shown on the left using the link above.


okay so here we're going to do some more

examples related to implicit

differentiation and these equations are

now implicit equations we kind of have

the X and the Y um on the same side of

the equal sign so that's kind of what I

look for what sort of Clues me in so the

main thing I remember when we're doing

implicit differentiation so uh let's

look at x^2 + y^2 = 25 mechanically what

you have to remember okay we're taking

the derivative with respect to X of both

sides whenever you take the derivative

of you know the variable X we just do it

like normal so the derivative of x^2 is

just 2x now we're going to take the

derivative of y^2 and we do it again

quote unquote like normal um we get 2 y

um but what we have to do anytime you

take the derivative of something

involving y we have to multiply that by

Dy

DX and on the right side when we take

the derivative with respect to X we'll

just get zero so normally we have the

dydx all by itself it's all isolated and

that's what we want to do in this case

so what I'm going to do is I can just

subtract 2x from both

sides so if we subtract 2x from both

sides we'll just get -2X and then we can

just divide both sides by 2

Y and then we'll have our solution it

says Dy DX is going to equal well we can

simplify um -2 over 2 is just going to

be -1 uh X over y and now we've got our

derivative so let's look at uh at least

one other example here uh maybe one

that's uh a touch more

complicated again hopefully nothing

nothing too

crazy so let's figure out uh let's do

the derivative of cosine of x^2 + S of Y

Cub = 6 we'll take the derivative of

this all right well we'll have to use

the chain rule so the derivative of

cosine of x^2 when we do the derivative

of cosine we'll get Negative s we'll

leave the inside alone and then we have

to take the derivative of the inside

which will just be

2x then we have to do the derivative of

the s y cubed so the derivative of s is

cosine we'll leave the inside alone and

now we're going to take the derivative

of something involving y so we get uh

when we take the derivative of y cubed

we just get 3 y^2 but this again is when

I have to remember to tack on the

dydx so I took the derivative of

something involving y um right here I

tack on the Dy

DX well on the right side uh 6 uh the

derivative of that's just going to be

zero and now uh again our goal is to

solve for dydx so this first term I'm

just going to add it over to the right

side so instead of having um you know

Nega sin x^2 * 2x I'm going to bring

that over and I'm going to rewrite it as

positive

2x sin x^2 so I'm just adding this term

to both sides and we'll be left with um

our second term and I'm going to rewrite

that as 3 y^ 2 * cosine of y 3 power *

dy over

DX and now just the same thing to get Dy

DX all by itself we can just divide so

dy over DX we'll divide both sides by

the 3 y^ 2 cosine y cubed so on the

right side we'll just be left with 2x *

sin of

x^2 again all over 3 y^ 2 * cine of Y

the 3 power and again now we've got our

derivative

❖ Implicit Differentiation for Calculus - More Examples,  #1 ❖

Implicit Differentiation for Calculus - More Examples.
In this video I look at using implicit differentiation in order to find the derivative of an implicit function.  This is a common technique used and taught in any calculus course.
Recall that implicit differentiation is really us just using the chain rule on a function defined implicitly by x.
In this video I show solutions for the first two examples.

#maths #calculus #calculusbc #implicitdifferentiation #implicit_differentiation #derivative #differentiation #implicitfunction #apcalculus #apcalc


all right in this video I'm going to do

another example related to the product

rule so here we're going to take the

derivative of G ofx = 3x to 4 + 2x -1 *

the quantity X 5th -

2x^2 so all right well if we take our

derivative here if I do the derivative

of the first

Factor okay so the four would come out

front and multiply by the three and that

would give us 12 x and then we take one

away from our EXP which will give us to

the thir power um the derivative of POS

2x is just going to be pos2 and then the

derivative of our constant negative-1

just turns into a zero um so we took the

derivative of the first Factor we'll

leave the second one

alone and then we'll take uh we'll leave

the first Factor alone so we'll have 3x

4th + 2x - 1 and then when we take the

derivative of uh X to 5th well the five

will come out front we'll take one away

from the exponent and then

uh when we take the derivative of -2X

^2 okay so the -2 comes along and then

uh the two will multiply by the two

which will give us four and then we'll

have X we take one away from the

exponent so that'll be the second

Factor um and now all right well uh we

could always clean this up a little bit

um probably what I usually start looking

for at this point are common factors uh

notice there's not anything there's not

really the same thing inside the

parentheses on any of these so I think

probably what I would do in this case is

just leave this one alone and that's our

derivative you could always multiply all

this stuff back out collect all your

like terms but um again just to

illustrate the product rule I think I'm

going to leave it there

Product Rule to Find a Derivative : Basic Example , No Simplifying

In this video I look at a quick example of finding the derivative of a function by using the product rule.  Notice that we could perform some algebra on this function before taking the derivative but I just jump into things in order to illustrate the procedure.

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all right in this video I just want to

do a couple examples about finding the

maximum and minimum values for uh

different signing cosine functions and

in this case we've got 6 s of 7x and

then -2 cosine of 3

pix okay a couple different ways uh I

guess you can go by thinking about this

to me you know when I look at this um

the first thing I see for example is the

sign

function so s if you think about you

know if you think about the graph of it

you know a couple different arguments I

guess here um Remember by multiplying by

the seven all that's really going to do

is just change the uh the period so I'm

forgetting about the six for a second

here um all that's going to do is change

the

period so okay maybe it's going to uh do

things a little

faster but still sign uh sign of some

value is always between is at most

positive one and uh the smallest that

ever is is negative 1 so really I'm just

thinking about the range of 7 sinx it's

always between negative 1 and positive 1

well what that means is if we multiply

the inside by six so it says 6 s of 7x

well I would have to multiply the left

side by six as well which is -6 I would

have to multiply the right side by six

as well which would give me positive six

and these are going to be your maximum

and minimum values the maximum value is

going to be uh just six and the minimum

value again is going to be

-6 so um again definitely taking this

this coefficient into account kind of

thinking about the amplitude here how

this coefficient changes the amplitude

instead of being a positive one and

negative 1 it's not going to go up to 6

and -6 when we graph

it

okay so the same thing for uh our next

one here -2 * cosine of 3 pix well again

the same

argument cosine of anything is in

between positive one and negative 1 and

again notice there's no restrictions on

the domains so you know we can let X be

whatever number we want from Infinity to

positive infinity and plug it inside of

this cosine function so uh

correspondingly cosine is going to hit

every value between negative 1 and

positive one well we have to be a little

careful if I do this little argument

that I did a second ago if I multiply

the inside by -

one2 remember if you multiply uh by a

negative number you have to flip the

inequalities so -1 * 1 12 would be

positive2 again we'll flip the

inequality because we're multiplying Now

by -2 -2 * 1 is going to give us just a

- one2 and again we've got our maximum

and our minimum it says our function -2

cosine of 3 pix is less than or equal to

1 12 but it's greater than or equal to

-2 so in this case our max value is just

positive2 and our minimum value will be

- one2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Maximum and Minimum Values of Sine and Cosine Functions, Ex 1. In this video, I talk about how to find the maximum and minimum values of sine and cosine functions with different coefficients thrown in!


all right so here we're going to look at

one more example related to the quotient

rule so um here we've got the function f

ofx = x^2 - x + 1 over 3x + 2 we want to

find the equation of the tangent line at

the point 0 comma 1 12 so using point

slope formula y - y1 um equals the slope

* x - x sub1 um we've used this a few

times now to find equations of tangent

lines well the y coordinate is just

going to be 1

the x coordinate is zero that's given to

us again the slope is what we have to

figure out but the slope comes from the

derivative and in this case we'll have

to plug zero into the derivative to get

the slope of our tangent line well to

take the derivative of this function we

can just use the quotient

rule so it says you get whatever's in

the the uh denominator 3x + 2 in this

case we'll multiply that by um the

derivative of the numerator

so the derivative of x to the second

power the two would come out front and

then we would take one away from the

exponent the derivative of -1x will just

be -1 and then um the derivative of the

constant is just zero so that'll be kind

of the first part of the product rule

excuse me the quotient

rule um and then we'll put a minus sign

in between so at the bottom times the

derivative of the top minus the stuff in

the top so x^2 - x + 1 and then we have

to multiply by the derivative of the

denominator so the derivative of 3x is

just three um the constant again just

turns into a zero when we take the

derivative and that'll give us the uh

the numerator of our quotient Rule and

then it says we take whatever is

downstairs 3x + 2 and we Square it all

right so again um in one of the quotient

rule problems I kind of simplified all

this but to me all we're trying to do is

plug in xal 0 so I don't know to you

could simplify it but um in terms of

getting the slope of the tangent line we

certainly don't need to do that here so

notice if we plug in xal 0 we'll get 3 *

0 or zero then we'll just be left with

two if we plug in zero for the X in our

second uh Factor we'll get 2 * 0 -1 or

just we'll be left with the

-1 um and the second part let's see if

we plug in 0er so there's our minus in

between we'll just be left with 0 - 0 +

1 which is 1 and then we'll multiply

that by 3 and then in the denominator

well 3 * 0 is 0 + 2^ 2ar so we've got 2^

2 left

over well let's see this is going to be

-2 - 3 which uh looks like -5 to me and

then 2^2 is 4 so now we've got our slope

of our our tangent line so it says the

slope of the line um excuse me the

equation of the line we can just write

that as y -

12 equals okay we now know the slope by

using the derivative so we've got -5

over4 and x - 0 we can just write that

as X and that's now the equation of our

tangent line of course you can also add

the 1/2 over you could make it y equals

negative -5 over

4x + 12

and again now we've got the equation of

our line either way so um again finding

slopes of lines what do you got to do

you got to take the derivative and

basically just plug the point into the

formula um so again in this case to take

the derivative well we just had to use

the quotient rule

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Calculus : Equation of the Tangent Line at Point using the Quotient Rule


here we're going to find the S cosine

tangent cosecant secant and cent of

theta given that Theta is an angle in

standard position and the given point is

on the terminal side of theta so in this

example we're going to use -1 comma -2

our x value will be -1 our y value is -2

to get our r value we'll use the r^2 =

+ y^

2 well this will give us 1 + 4 which

will be

5 and if we take the square root of both

sides keeping the positive solution

we'll get that R equals the sare < TK

5 so let's see s of

theta cosine of

theta tangent of theta we can just

simply use our formulas s of theta is y

/ R again Y is -2 R is theare < TK of 5

and if we multiply the numerator and

denominator by < TK 5 we'll get -2 * the

> 5 /

5 cosine of theta is going to be x / R

so again X in this case is1 r is the

sare < TK of 5 multiply numerator and

denominator by the square < TK of 5

we'll get < TK 5 /

5 tangent is y / X the Y value was -2

the x value is -1 so that'll leave us

with

positive2 and now to get the other three

I'm just going to use reciprocals so

cosecant of theta that's 1/ sin Theta

and I use the value before we we

rationalize I just flip that I'll get

theun 5 over -2 I'm going to put the

Nega up front and the reason I use it uh

before we rationalize because if I use

the other form I'm going to actually end

up making a radical appear in the

denominator and then we'll have to

rationalize all over so to me it's

easier to do it this way secant of theta

well that's 1 over cosine

Theta again cosine of thet was- 1 overun

5 if we take the reciprocal we'll just

simply getun 5

and last but not least cotangent of

theta hey that's one over

tangent of theta well we had a positive2

or 2 over one if we take the reciprocal

of that we'll simply get

1/2 and we've now got all six of our

values so again I always just find S

cosine tangent and then I just take the

reciprocals of those to get the other

three

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Trigonometric Values Given a Point in the Plane.
 In this example, we find the sine, cosine, tangent, cotangent, secant and cosecent given a point in a plane and an angle in standard position.


all right here we're going to start uh

looking at some examples of finding the

radius and interval of convergence so in

part A here we've got the series from 1

to infinity x to the n * 2 the N Over N

factorial so again to do these we're

going to use the ratio test so we'll

take the limit as n goes to Infinity

everywhere there's an N I'm going to

replace it with n +

one so then we would have n + one

factorial in the

denominator okay then we have to flip

and multiply by the reciprocal so n

factorial over x n * 2 the N so let's

see if we simplify here we've got the

limit as n goes to

infinity x the n + 1 over X the N is

going to leave us with X to the

1st um 2 the n + 1 over 2 the N is going

to leave us with a

two and then n factorial Over N +1

factorial is going to leave us with n +

one you could pull the 2x out and keep

that an absolute value and then we have

the limit as n goes to

Infinity of 1/ n +

1 but this limit is simply going to

equal zero and again we

want we want to figure out what values

of X we can plug in so that at the end

of the day when we multiply we get

something less than one one well no

matter what you plug in for x and

multiply by zero you'll always get

something less than

one so that means the solution um the

interval of convergence is going to be

from negative Infinity to positive

Infinity so this will be our interval of

convergence well in this case if you

think about the uh you know the length

of this interval from negative Infinity

to positive Infinity that's just

Infinity so we would say the radius is

also Infinity so the interval of

convergence all real numbers the radius

uh again the radius is of a number or in

this case we would just say positive

Infinity

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Interval and Radius of Convergence for a Series, Ex 3. In this video, I show another example of finding the interval and radius of convergence for a series.


all right so here we're going to look at

some more examples of multiplying

fractions and again I'm going to use

this this idea of getting rid of common

factors immediately so we have 7 over 5

times 5 over 6 again we could write that

as 7 times 5 over 5 times 6 well we

could multiply immediately but then we

would just immediately have to get rid

of common factors so I notice you know

there's a multiplication by 5 in the

numerator and a multiplication by 5 in

the denominator I can cancel those out

again 5 divided by 5 is just 1 so we're

left with you know if you want to you

can put a 1 back in there so you have 7

times 1 in the numerator which is 7 and

then 1 times 6 in the denominator which

is 6 and again we have our solution

suppose we have 12 times 7 over 15 well

the first thing I do is usually I'll

write my numbers I'll write everything

as a fraction I can write 12 as just 12

over 1

okay so we're multiplying by again 7

over 15 well again I recognize that 12

and 15 both have a common factor of 3 so

I say well 12 divided by 3 that's going

to leave me with 4 15 divided by 3

that's going to leave me with a 5 so all

I'm really reducing all I'm really doing

is I'm reducing 12 over 15 and reducing

that to 4 over 5 and again we just

multiply what's left over so 4 times 7

in the numerator will leave us with 28

and in the denominator we have 1 times 5

which is 5 so again got our solution so

maybe do one more here so 6 over 45

times 9 again I'm going to write 9 as

just 9 over 1 but now I can do the same

thing as before

again just cancelling out factors okay

so 9 and 45 have a common factor namely

9 well let's see 9 divided by 9

you would leave me with 140 5/9 would

leave me with a 5 so now I've got a 6 in

the numerator a 1 in the numerator 5 in

the denominator and again just a 1 in

the denominator

there's no more common factors so I'm

just going to multiply 6 times 1 is 6 5

times 1 is 5 and again now we've got our

solution and it's nice and reduced

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


Alright: in this video I'm gonna do an  example of
rewriting complex fractions

and there's two different ways that you can
go about simplifying complex fractions

and again it's just a complex fraction
all that is is you know you got a big

fraction

and then within that you have even more
fractions is the basic idea so

again two ways to do 'em one way is to
look at all the denominators

and you can multiply the top and bottom
by the least common multiple

of the of the denominators and the ideas
that'll get rid of

everything except for it you know the
one big attraction

another thing is to simply write the
numerator as a single fraction

and the denominators a single fraction
and then we're gonna do you is just use

that

the trick that we can flip and multiply
and that's how I've always done so

that's how I'm gonna do

this example and the others as well
so we have 2 over "x" plus three divided

by 1 plus 3 divided by "x"

you can think about the wine is being
one divided by one

and again what I want to do is the
numerator is already a single fraction

I'm gonna write the denominator also is
a single fraction

so in this case I look at the denominators
we have 1 and "x"

so to get a common denominator I could
multiply the first term

a by "x" over "x" and in the numerator
will simply leave that alone

2 over x+3 and

then and the denominator we would have
x over x

plus three over x and again the

the whole reason we we did that is so
that we could write the denominators a

single fraction

so often times all you can skip this step
so

in the numerator of the denominator
we would have  x +3

which we can't really do much with
divided by our common denominator

of x so this is kinda

the first half but getting the numerator
single fraction

and now the denominator as a single
fraction and

remember the trick is if you have a
fraction divided by a fraction

we leave the top fraction alone and we
take the denominator

we flip it over but we turn it into
multiplication

okay so instead of divided by

X plus three over X we're multiplying by
x-over

X plus three and in this case on

you know if there is anything to you to
factor and cancel

I would certainly try to factor in
cancel

but in this case in the numerator we
have a 2 times an x

in the denominator we have x plus three in
x plus three

I would probably rewrite this simply is
2 x

and then X plus 3 quantity squared

again you can multiply denominator out
you could FOIL it out

x squared plus 6x plus 9 but I think

my own personal preference would again
just to be leave it

in this nice factored form

❖ Simplifying Complex Fractions - Ex 1 ❖

Simplifying Complex Fractions - Example 1
In this video, I break down the process of simplifying a complex fraction. Starting from the basics of finding a common denominator for the terms in the original denominator, I then demonstrate how to flip, multiply, and cancel to simplify the fraction effectively. This method is essential for handling more complicated expressions and is a foundational skill in algebra and calculus.

What You Will Learn:
Finding a Common Denominator: Learn the first step in simplifying complex fractions by finding a common denominator for the terms in the original denominator.
Rewriting the Fraction: Understand how to rewrite the complex fraction by flipping and multiplying, a key process in simplification.
Canceling Terms to Simplify: Watch how to cancel terms to reduce the fraction to its simplest form.
Step-by-Step Example: Follow a clear, guided example that covers each part of the process, making it easy to understand and apply to other problems.
If you find this video helpful, please subscribe for more tutorials on algebra, calculus, and other math topics! Don't forget to like, comment, and share the video. Support me on Patreon to access more exclusive content:
https://www.patreon.com/patrickjmt?ty=c

#complexfractions #algebra #fractionmultiplication #simplifyingfractions #commondenominator #patrickjmt #mathhelp #mathvideos #math #rationalexpressions #fractionreduction #algebratips #mathtutorial #fractionsimplification #mathpractice #mathstepbystep #maths


okay here we're gonna look at some anti

derivatives of some trig and some

exponential functions and the only rule

of anti derivative formula that we'll

need here for the exponential is just

the antiderivative of e to the X so

again we could have a different

exponential instead of e we could have 2

or 3 or 1/2 or some different number but

for now we'll just stick with e to the X

and again the antiderivative of e to the

X is just e to the X and again plus C so

we'll make use of this formula so when

we do the first integral here so the

integral with negative 5 e to the X plus

7 sine X well the negative 5 I can just

you know carry that along we just saw

the antiderivative of e to the X is e to

the X let's see then I think whose

derivative is sine X or equivalently

what's the antiderivative of sine X and

at first maybe you say hey cosine X but

recall the derivative of cosine X would

actually give us a negative sign so to

cancel out that negative we'll use a

negative cosine X and now I think we've

done everything all we have to do is

just include our plus C the last thing

I'm gonna do is just sort of clean it up

that's negative 5 e to the X we could

write you know positive 7 and negative

cosine is just negative 7 cosine X plus

C so that would be our antiderivative

all right let's look at a let's look at

the other one as well here I think you

can knock them both out ok so the first

part let's see we've got 2 e to the X I

say hey no problem there we've got a

negative 1 and hey no problem there but

then I see the sine times cosecant and I

think this is a product so I have to be

a little more careful when I go to you

know find the antiderivative of this

function and I don't think that's what I

know just immediately off the top of my

head you know again you can't do them

individually when they're being

multiplied but notice kind of

you know the trick on this I mean what

is cosecant X well cosecant X is the

same thing as 1 over sine X so really

just kind of a sneaky little problem

because really when we take positive

sign and divide it by sine X that's just

gonna give us a positive 1 well we can

just cancel out the negative 1 and the

positive 1 so now we're really just

finding the antiderivative of 2 e to the

X DX

well again the antiderivative of e to

the X is just e to the X and then we'll

be left with our plus C we have to tack

that on so that would be the

antiderivative now of our second

function here so again you know

sometimes people will give you these

sort of you know contrived problems and

again what they're just trying to

illustrate to is a you know make sure

you remember trig identities this is

important and just the idea that you do

a lot of times have to do some sort of

algebraic manipulation to make things

work out so rarely do we get things

where it just sort of immediately works

Calculus : Antiderivative / Integration Problems with Trig and Exponential Functions

In this video we find the antiderivatives / integrate two different functions, each of which involve an exponential function as well as a trigonometric function.
Notice that it is often good to look for any algebra simplifications before starting, such as in the second example.

Integrate e^x
Integrate sinx

#math, #algebra, #khanacademy, #khan, #geometry, #education, #montessori, #calculus, #homeschooling, #functions, #notation, #organicchemistry, , #calculus, #derivative, #powerrule, #limits,  #limit, #graph, #graphing, #conjugate, #infinity, #patrickjmt, #rational, #shortcut, #trick, #commondenominator, #fraction, #university, #admissions, #test #continuity #onesided #infinite #absolutevalue #trigonometry #trigonometric #integration #integrate


all right in this video I'm going to do

one more example of solving an equation

containing a radical so in this case we

have the square root of 3x plus 9 minus

x equals negative 3 so the first thing

that I do is i isolate the radical I get

that all by itself so in this case to do

that I'm going to add X to both sides so

we would be left with the square root of

3x plus 9 on the right side I could

write negative 3 plus X but I think I'm

just gonna go ahead and write that

equivalently as X minus 3 now again to

get rid of the square root we simply

square both sides the square root of 3x

plus 9 squared is just going to give us

3x plus 9x minus 3 squared we have to

foil that out or distribute everything

out so on the Left again we just have 3x

plus 9 on the right when I multiply it

out I'll get x times X which is x

squared I'm going to get a negative 3x

and another negative 3x which is

negative 6x negative 3 and negative 3 is

going to give us a positive 9 so now

again I've got a more familiar quadratic

equation notice we could simply subtract

9 from both sides and that would get rid

of the nines the other thing I'm going

to do is I'm going to subtract 3x from

both sides so that would leave me with 0

on the left and then I would have x

squared minus 9x

and again I like the quadratic equations

where there's no constant because

basically all you have to do in this

case is we have to factor out the X and

then in parentheses I think we would

have X minus 9 left over if we set the

first factor equal to zero we'll simply

get x equals zero as a solution if we

set X minus nine equal to zero we can

add 9 to both sides and then simply get

x equals positive 9 as our other

solution I should say potential

solutions again you always need to go

back and check notice if you plug 0 into

the original equation we'll just have

the square root of 9 which is 3 minus 0

that would give us 3 but 3 is not equal

to negative 3 slip solution doesn't work

notice if we plug 9 in we would get 3

times 9 which is 27 plus 9 is gonna be

what 36 so then we would have the square

root of 36 minus 9 and we're asking

ourselves does that give us negative 3

well the square root of 36 is 6 6 minus

9 does equal negative 3

yea that all works so it says x equals 9

is a solution to our original equation

involving the radical

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving an Equation Involving a Single Radical (Square Root) - Example 3


okay so one quick question here uh here

we want to find the X and Y intercepts

of this graph Ral 5 - 4 cosine Theta and

the idea is for these types of graphs

what they're going to turn out to be is

what's called a

lemon and if your a value is greater

than your B value so here our a value

would be five kind of forget about the

negative your B value would be four um

and if both of those are greater than

zero okay so a is 5 b is 4 it says then

to get the X intercepts we simply add a

plus b and we take the negative of it

and then so that'll be one of our X

intercepts and then we take a minus B

that'll be our other x intercept the Y

intercepts are just negative a and

positive a if your B value is bigger

than your a

value it's going to turn out that the X

and Y intercepts also includes zero but

we don't have that in this

case so simply um we we've got five is

greater than four is greater than zero

this is our a value this is our B value

again it says take negative of a + b and

then take a minus

B so a plus b would just be 5 + 4 which

would give us -9 that's going to be one

of our X intercepts if we do a minus B

that'll be 5 - 4 or pos1 so that'll be

our other x intercept and and then it

says the Y intercepts we just take

negative a and positive a so the Y

intercepts will be at5 and at posi five

and that's it now we're done so we've

got our X intercepts over here we've got

our two y intercepts here and that's it

so again uh you'll see polar curves here

and there uh if you end up taking a

Calculus class kind of sprinkled

throughout the course um uh these are

formulas that definitely I kind of

forget I am aware in fact that there are

formulas out there so you know when I do

start doing polar stuff again I say oh

yeah there's these formulas and I dig

out my pre-calculus book or my

trigonometry book and I look it up and I

say oh yeah that was it so um you know

definitely stuff that's probably easy to

forget but at least be aware that

they're out there because sometimes

graphing these polar curves can be a bit

of a pain so some of these little

shortcuts will definitely make uh make

producing a good graph a lot easier

Basic Info About a Limacon

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Graphing a Limacon - Just some info about how to find x and y intercepts for a limacon (I do not actually graph it, sorry!).


all right in this video I want to talk

about what's called chbby shev's theorem

and chbby shev's Theorem is pretty

simple and easy to use and what it says

it says the proportion of any

distribution that lies within K standard

deviations of the mean is at least 1 - 1

k s where K is any positive number

greater than one the big use is that

this theorem applies to all

distributions distributions of data um

you know so if you know your data is

normally distributed you can come up

with a stronger result than but often

times you don't know

that so for example

um this theorem says that within two

standard deviations of the mean you'll

always find at least well since in this

case k is going to equal two we'll get 1

- 1 over 2^ 2 which is 1 - a 4 or 34s

which is 75 or it says you're going to

get at least 75% of your

data let's do uh kind of a specific

example

here so suppose that the average score

on a math test is an 84 with a standard

deviation of Four Points according to

Chevy Chef's theorem at least what per

of the tests have a grade of a 72 and at

most a

96 well okay so the idea

again so here's our average 84 that's

our mean mu notice if we count over um

one standard deviation so if we go over

a standard deviation again we're

basically adding four points that would

put us at a grade of an 88 if we go over

another standard deviation or another

four points that's going to put us at a

92 and if we go over another standard

deviation that's going to put us at our

uh our grade of a 96 which we wanted

likewise notice if you go down three

standard deviations you'll be at an 80

if you subtract Four Points will be at a

76 if you subtract Four Points you'll be

at a

72 so what we're doing is we're looks to

me like we're going three standard

deviations away from the mean okay so in

this case our K value is simply going to

equal three so according to chbby shev's

uh theorem we'll get 1 - 1 over K which

is 3^ sared that's going to be 1 - 1 9th

well we can write that as 9 over 99 - 1

over9 do the arithmetic on the top we

get 8 9 8 9 which is going to be uh

roughly equal

to 89 if we round and 89 equals hey

89% so it says if uh in your class if if

the teacher told you you know hey the

the you know the the test scores were an

84 the average was an 84 and a standard

deviation of Four Points um it says that

89% of the class made a grade uh between

72 and 96

Chebyshev's Theorem

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Chebyshev's Theorem - In this video, I state Chebyshev's Theorem and use it in a 'real life' problem.


all right in this video I'm gonna graph

the equation y equals 4x minus 1 by

plotting points ok it turns out down the

road you'll see this if you have a

problem just involving a Y to the first

power X to the first power and some

numbers you know kind of a few more

restrictions on that but this is going

to turn out to just make a straight line

when you graph it but you know assuming

we weren't aware of that at this point

let's just do the same thing of plots

and points

maybe we can plug in x equals negative 2

negative 1 0 positive 1 and positive 2

again you can use any numbers you want

typically I try to graph stuff around

the origin and just you know smaller

numbers to make the arithmetic easier so

if we use x equals negative 2

we'll get that y equals 4 times negative

2 minus 1 4 times negative 2 is negative

8 negative 8 minus 1 is going to give us

negative 9 so it says the Y value of

negative 9 goes with X the x value of

negative 2 let's see x equals negative 1

1 minus 1 so negative 4 minus 1 is going

to give us negative 5

if we plug 0 in we'll just be left with

negative one so that one's not too bad

if we plug in x equals positive one

that'll give y equals four times one

minus one well four minus one equals

three

and then lastly one more point if I plug

in two it looks like we get y equals

four times 2 minus 1 which is 8 minus 1

or 7 notice as we jump up as we increase

the x-values by one unit by one unit by

one unit on the left notice the

y-coordinates are jumping up by four

units by four units by four units by

four units and I think if you think

about it you can kind of convince

yourself of that that trend is going to

continue so now I'm just gonna plot

these points here again you know kind of

maybe a little rough by hand so negative

1 negative 2 negative 1 2 3 4 5 6 7 8

negative 9 so let's see if I can make

this look like the line by hand so

they're supposed to be negative 2

negative 9 here's negative 1 we go down

1 2 3 4 5 units there's another point it

says at 0 we're getting negative 1 out

it says when you plug one end we get 1 2

3 out and then it says when you plug 2

in we're up here at positive 7 and again

the idea is if you were to graph this

and then maybe if you had some graph

paper to produce a slightly better graph

than what I have

the idea is if you were to kind of play

connect the dots if you had a better

graph you would see that this line goes

through all the points and it makes a

straight line

okay so again eventually you're kind of

learned shortcuts to say well depending

on the type of equation you'll be able

so obviously I know this isn't a perfect

line it's a little a little broken there

but again it's supposed to just be a

straight line eventually you'll be able

to look at you know the exponents on the

X's and Y's and be able to say you know

you'll be able to recognize immediately

oh that's the equation of a line or

that's the equation of a circle that's

the equation of a parabola etc

eventually those are things that you'll

want to know just because graphing ends

up being a very useful it's a use of the

picture is worth a thousand words a good

graph a lot a lot of times give you a

lot of information about what's actually

how the equation is behaving

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Graph a Line by Plotting Points - Example 2
Here look at graphing an equation by plotting points.


okay so here we're going to do another

uh problem related to the equation of a

tangent line and this is kind of a very

common type of problem so uh you've got

some function in this case square otk of

x uh it turns out that the derivative of

square otk of X is 1 over 2 * the < TK

of X and what we want to do is we want

to find the point on the graph and maybe

there's more than one in this case

they're actually won't be but uh we're

going to find the point uh XY so that

the tangent line to the gra graph has a

slope equal to 1 18 so again we're

trying to figure out again you know

where on the original graph would the

tangent line have some particular slope

and to do this it's easy um in a sense

and once you have the derivative uh

because again the derivative basically

tells you for tells you a formula for

the slope of the tangent line and it

says well our formula for the slope of

the tangent line uh at any particular x

coordinate is given by 1 1 2 * the < TK

X and well we want that to equal 1 18 so

all we have to do in this case is just

solve the uh equation 1/ 2 < tkx = 1

18 so uh to do this uh to solve this

equation this is one that's not not too

bad um typically anytime I have a

fraction on both sides I just cross

multiply so uh we can do 1 * 18 that'll

give us 18 and then we can do well uh 1

* 2 < TK X that'll just give us well 2 <

TK

X we can divide both sides by two so by

two 18 over 2 is 9 on the right we still

are just left with square otk of X well

to solve for x all we would need to do

now is just Square both

sides so on the right side we're left

with x and 9 * 9 is going to be 81 so uh

if you graph the original function

square otk of X recall square otk of X

looks something like

that and what we've done is we've gone

over to the x coordinate of 81 so it

says at the x coordinate of 81 uh it

says the slope of that tangent line is

going to be uh 1 over

18 okay so we're almost there um the

question uh actually asks for we want

the the X and the y coordinate on the

original uh function function well again

if you if you're using the square root

function if you plug 81 in the square

root of 81 is just 9 so this point will

be 81 comma 9 so again all we're doing

is we're taking our x coordinate we're

not putting it into the derivative but

we're putting it into the original

function so again if you take the square

root of 81 again we get n so it says the

point that we're looking for would be

the point 81 comma 9 so again it's

common uh they'll you'll try to figure

out

where um you know you'll try to figure

out points on the graph where the

tangent line equals some number A lot of

times we'll we'll try to figure out

where the tangent line equals zero and

you'll see that um in a few sections

that turns out to kind of be an

important question but again that's all

there is to it in this case just take

the derivative set it equal to the slope

that you want that'll give you the x

coordinate plug the x coordinate back

into the original function to find the

corresponding y value

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Find Where a Tangent Line to a Curve Has a Desired Slope. In this video, we are given a curve and its derivative.  We have to determine at which point on the curve has a specified slope.


all right here we're going to look at an

example related to addition of integers

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

going to write an addition statement for

each of these number lines and again all

the number lines are it's just kind of a

visual way of sort of a you know

thinking about adding positive numbers

and negative numbers and uh just kind of

a way to to to uh to see it you know

again on a number line so in part A here

you know we we're starting at zero and

then we're going uh four units to the

right so we're going off in the positive

direction positive 1 2 3 4 so our

addition statement for this would be

well positive 4 and then we're going to

add um something to it well it looks

like from four now what we're interested

in is the length of the arrow so be

careful so okay so we go from four back

to three but what we're really doing is

we're kind of going uh you know sort of

one unit to the left to the left in the

negative Direction you can think about

it that way so we we've got pos4 + -1

well and from the number line you can

see that if we go four units to the

right and then if we add to that

negative one which would give us take us

back one unit to the left this would be

our addition statement um and we could

say well hey that equals

positive3 uh we'll do the same thing for

part uh Part B here do both of these at

once okay so on the top part it looks

like we're going two units to the left

well two units to the left is going to

give us well -2 and then we're going to

add to that and again let's count so

we're starting here we're going over uh

we're going one unit two units three

units four units five units six units so

we're adding six units by moving to the

right so we have -2 plus a positive 6

and as you can see from the number line

if we go two units to the left and then

we add six unit to that that's going to

put us at the value of

Using a Number Line to Write an Addition of Integer Statement

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Using a Numberline to Write an Addition of Integer Statement.
Here we look at an example of using a number line and the notion of direction to help visualize what is happening when adding integers.


all right here I'm going to take a

fraction uh 24 over 32 and I'm going to

write it in lowest terms so what that

means is uh basically get rid of any

common

factors so one thing we could do is we

could write 24 over 32 I could say well

24 is the same thing as 2 *

12 and then in the denominator 32 that's

really the same thing as 2 * 16 there's

certainly different uh uh uh ways we can

do it besid besides mine the way I'm

doing it right now um we could have

factored for example a four out of the

numerator and denominator you could have

factor that an eight as well but a lot

of times when I look at these I say hey

they're both even I can certainly reduce

them okay so now the idea is we can

cancel out common factors and the idea

is really what we have is okay so 2 * 12

over 2 * 16 uh we could really write

that as 2 over2 * 12 over 16 but but 2

divided two is the same thing as 1 so 1

* 12 16 that'll just give us 12 over 16

okay so I've definitely reduced 24 over

32 I've reduced it to 12 over 16 but now

I ask myself does it reduce any anymore

well again it certainly could they're

still both even I could factor a two out

just like I did but I say oh I think

there's a bigger number that goes into

both of these so notice I can write 12

as 4 * 3 16 I can write that as 4 *

4 and again people typically we don't

typically write this step but that's you

know the reason why you can just cancel

these out okay that's one way of

thinking about it I'm just going to

cancel out my common factors of uh you

know the four and the four so we would

be left with 3 over four and now there's

nothing else we can't really Factor them

they're no more common factors other

than one we could write three as 3 * 1

we could write four as 2 * 2 there would

be nothing else that we could really

cancel out so we've now reduced our

fraction uh to lowest terms and again

the idea is if you make 24 over 32 into

a decimal you'll get the same thing as 3

over4 if you make that into a decimal

um

Reducing a Fraction to Lowest Terms

Reducing a Fraction to Lowest Terms
Learn how to simplify fractions to their lowest terms by factoring and canceling common factors. This video walks you through reducing fractions step-by-step, making it easy to understand and apply to any fraction problem.


all right in this video we're going to

do an example of uh finding an angle the

measure of an angle if we know a little

more information so in this case suppose

we know that cosine of theta is equal to

< TK 3/2 we want to find the angle

um what the measure is for our angle

here

Theta so remember cosine of an angle is

defined to be the adjacent side over the

hypotenuse and again for our angle Theta

this would be the adjacent so I'm simply

going to label the adjacent side as squ

< TK

3 I'm going to label the hypotenuse here

with length

two and now we would have to figure out

you know this missing side you may know

this uh you know if you uh from geometry

but we can always figure it out just

using Pythagorean

theorem so we'll take the < TK of 3 S

Plus this missing side we'll call it b^

2 that equals 2^

2 well uh the < TK of 3 S is just going

to be 3 plus b^ 2 uh 2^ s is going to

give us four and now if we subtract

three from both sides we'll just get

that b ^2 = 1 and when we take the

square root we'll get positive Nega 1

but again it's got to be the positive

one uh because we're talking about a

length here so this simply has a length

of

one so again what we're trying to do is

we're trying to figure out uh our our

our angle Theta so now uh we have to be

a little bit clever here and what we're

going to do is I'm going to kind of uh

take this triangle and I'm going to

reflect it Okay so we've got our

triangle here this has length

two um this has length two this has

length one this has length square > of

three but I don't really care about that

uh I'm going to imagine sort of uh just

reflecting this triangle over to the

other side okay so just imagine it sort

of getting reflected over um it's going

to go down

here so we reflect it over it's going to

come down

here okay so this was length two okay so

this is not great Artistry here but if

it's the same triangle this head length

two this head length one that means the

hypotenuse here would have length two

and this other side again would have

length one but if you think about it now

what we really have is we've got um an

equilateral

triangle because this is length two this

is length two this entire length adds up

to two so that means uh since all the

side lengths are the same that means all

the angles inside are the same and since

all the angles have to add up to

180° we we know that each one of those

angles has to be 60°

individually so okay so that means again

this angle is

60° but notice our little angle Theta

here that was kind of

sitting you know kind of sitting in this

uh this upper region right this whole

angle if we trace it out that's what's

going to be

60° so in this case Theta is only going

to be half of that so we would have to

take our 60°

and divide it by two and that would

simply give us that Theta has uh the

measure of uh 30° or we could put that

in radians and say that's pi/ 6 radians

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding an Angle Given the Value of a Trigonometric Function - Example 1.  In this video, I deduce the angle associated with a given trigonometric value.


all right so another example here using

the law of signs so suppose we've got

our a triangle here we've got angle of

35° 120° we don't know our angle beta um

the side opposite 120 is

10.45 we want to find the side length B

well kind of the same idea U we know

that the angles have to add up to 180 so

we've got

120° plus

35° well that would give us 155°

well if we take 180 and uh subtract our

155 that's simply going to leave us with

25 degrees so our angle beta must be uh

25° down here and now we can simply just

use our law of signs so I'm going to

take uh Little B over s of 25 Dees so

little B over s of 25° the only uh side

length we're given is the

10.45 so I'm going to use the 10.45 and

the

120° so we have

over s of

120° and now all we have to do uh is

simply solve for b we can multiply both

sides by S of

25° so that'll leave us with B =

10.45 * s of 20

25° all over s of

120° so again now I think this is

probably good place to start plugging

things into a

calculator so let's see um sign of 25

degrees so I'm getting s of 25° to be

um after

rounding and S of

120° um a sign of 120 degrees I'm

getting that to be

0.866 after rounding a little bit so now

we'll simply take

10.45 multiply that

by

423 I'm getting that to be 4

42035 again we're going to divide that

by

0.866 so when I divide

this so

0.866 I'm getting this to be uh

5.14 after rounding so uh We've now got

our our missing length it simply equals

5104 looks like like we were using

centimeters so those will still be our

our units um so again that's all there

is to it just figure out the missing

angle we know they add up to 180 set up

your uh your equation using the law of

signs and then it's just a matter of you

know to again to make sense out of it I

would definitely want a calculator to

get some decimal values

The Law of Sines, Example 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Law of Sines, Example 2.
  This video talks a bit about the Law of Sines.  I give the formula and proceed to do an example using it.


all right so here I'm gonna look at

another example of subtracting whole

numbers so in Part A we have four

hundred twelve thousand three hundred

forty seven minus two hundred sixteen

thousand five hundred seventeen I'm

gonna do that one first

so my first column there's no problem I

don't have to borrow a digit from the

next column or anything as long as the

number on top is greater than or equal

to the number on the bottom we can just

do our subtraction so seven minus seven

is gonna leave us with zero if I look at

my next column well four minus one is

gonna leave us with three well now this

is our first kind of against spot where

we have to be careful because now the

number on top is larger than the number

on the bottom so if you think about it

if you've got three in a sense I guess

somebody couldn't take away five well

you can you've got negative numbers of

course but again when we do subtraction

that's not how we do it so if the number

on top is smaller than the number on the

bottom what we do again is we borrow a

digit from the next the next column so

instead of a two I'm gonna bump this

down by one and well that would make it

into a 1 and instead of a three I'm

gonna now pretend this is a thirteen

okay so I'm kind of borrowing you know

one digit from that next column so 13

minus five well we can do that that's

gonna leave us with eight all right but

now we've really got the same problem

again because now I'm thinking about one

minus six well the number on top is

smaller so what I'm gonna have to do

again is borrow a digit so instead of a

one I'm gonna make this into one smaller

which again is a zero and again then we

just tack a one in front of our original

number so now we have 11 minus six which

is gonna give us five you can just drop

down or comma again we've got this

problem again because now we have zero

minus one well I'm gonna have to borrow

a digit so instead of a

or this is gonna become a three again we

just put a 1 out front 10 minus 1 is 9

and then 3 minus 2 is 1 so it says if we

do our subtraction it says our final

answer will be 190 5830

and I think I'll do Part B here just as

a separate example

Subtracting Whole Numbers , Example 3, Longer Examples

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  Here we look at two examples of subtracting whole numbers.


all right so another example here

figuring out whether a function is even

odd or neither and here we're going to

do x 4 * sinx * cine 2

x um I'm going to rewrite this just a

little bit first so G of X that's going

to be X 4th we've got s of X and cosine

2 x we can just pull that out so we'll

have

cosine of x all squared Okay so that all

says the same thing so now I'm going to

think about G

ofx well that's just X to

4th sin

ofx cine

ofx squared and now um I just have to

simplify things down so okay wellx to

4th uh if you take a negative one and

raise it to an even power that'll give

us positive one we be left with the X to

4th we said s ofx in terms of being even

and odd we said sign was odd therefore s

ofx =

sinx okay cosine ofx we said cosine ofx

is even so that's just going to equal

cosine X so this is just cosine

x

squared and now what I'm going to do

again is just pull the sign all the way

out front so we'll pull the negative um

out front we've got X 4th we would still

have our sinx we can put the square back

uh cine 2 x and now let's see the

original function we started with was x

4 sin x cine

SAR well that just means we have the

negative of that and again in terms of

being even or odd um this is just simply

going to be an odd function

so our very original

function we would say that's an odd

function

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some some more examples here using the

limit comparison test and this is kind

of similar to what we've been doing so

here we're going to use the limit

comparison test on the series N equals 1

to infinity of n cubed minus n plus 14

over N to the fifth plus n to the third

plus 1 by comparing it to the series 1

over N squared and again we know that

this is just a convergent P series so if

we do the limit as n goes to infinity so

we've got n cubed minus n plus 14 over N

to the fifth plus n cubed plus 1 again

we would divide by 1 over N squared or

equivalently multiplied by n squared

over 1 well notice in the numerator when

we distribute we're going to get an end

to the fifth and then the rest of it's

going to be smaller degree will have an

end to the fifth over N to the fifth

since the degree of the numerator equals

the degree of the denominator the limit

exists and the coefficients would just

be 1 so in this case the limit is going

to equal 1 again a positive finite

number says they both do the same thing

well since we're using a convergent p

series to compare it to that means our

original series again also converges

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Limit Comparison Test for Series - Another Example 6. In this video, I use the limit comparison test to determine whether or not a given series converges or diverges. I do not delve heavily into the theory in this video (but do in another video if that is what you need to see!).


okay here's another sample sat problem

taken from the prins and review it says

if 20% of X is q and 70% of X is

T it says then in terms of X what is the

value of T minus

q and the solutions they give are

2x 3x 4X 5x or

6X um so again on this type of problem I

always ask myself um is there a for any

sat problem you should always ask

yourself should I do it the mathy way or

should I do it a plug and chug way I

don't really see a good way to do this

one with plug and chug I mean this is

something we're just going to actually

have to have to work out here so let's

check here so it says if 20% of X is Q

so mathematically 20% of X is Q well

remember 20% is 02 so it's says 2 * x =

q and then it says

70% of X is T well that means 7 * x = t

it says in terms of X what is the value

of T minus Q well T minus Q is going to

equal we said t is

7x and Q is

2x so this is simply

.5x so our solution here would be answer

D so I think they're just trying to get

across um can you basically set up these

kind of simple U equations from words

SAT Math Question #5

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! A sample SAT math question taken from a practice exam of the Princeton Review.
For more free math videos visit http://JustMathTutoring.com


all right in this video we're going to

do one more example of integrating uh a

function using substitution so um in

this case we've got the function 12x + 8

* 3x^2 + 4x + 1 all raised to the 9th so

again the thing that makes me think you

substitution um is I see something in

parentheses that's to the second power

and then you know if I take the

derivative of that I'm going to get

something to the first Power and

hopefully it's set up so that everything

will work out correctly so I'm going to

pick U to be the

3x^2 + 4x + 1 so the stuff that's being

raised to the

power our DU well uh the derivative of

3x^2 we would do 3 * 2 which would give

us 6X to the 1 and then our positive 4X

the derivative of that will just turn

into

pos4 um and then we still have our DX

left over well I don't want just 6X Plus

4 I want 12x + 8 but notice if you

multiply the right side by

two well we would have to multiply the

left side by two as well well then if we

distribute we do get 12x + 8 DX so

everything is set up nicely to work out

so it says I'm going to the stuff in

parentheses I'm going to replace that

with

u and then we still have to replace the

12x + 8 DX

so the 12x + 8 DX it says we can replace

that with 2 * du and now we're in

business okay we can just

integrate so the anti-derivative of U to

the 9 will be well U to the 10th over 10

plus C well 2 over 10 that's going to

give us 1 U is our

3x^2 + 4x + 1 so now I'm just replacing

our U with what we originally picked it

as and plus C and now again we found our

anti-derivative so again it's it's

certainly common when you calculate your

du for it to not work out exactly um you

know so just kind of look especially

maybe you can just multiply or divide to

get you know the the the numbers that

you

need

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Calculus : U-substitution Involving Polynomial Functions - Ex 3


okay viewers we want to simplify this

you're going to make expression the very

first thing I want to do is to write

this decimal in a index form as log 10

to the power negative 3 base 10 because

10 to the power negative 3 is same as

this expression we have here so of

course technology will affect the power

negative three so I have negative three

you know 10 this 10 and log 10 to this

10 is just one so I have minus 3 as my

final answer thank you

Maths: Logarithmic Expression #mathematics #students #shorts

#mathematics #shorts #online #onlineclasses #logarithm #education


okay so one more example of graphing a

complex number and finding its absolute

value or its modulus so here we have 1

minus I over 2 plus I've to write this

in the more familiar a plus bi form what

we do is we multiply by the conjugate of

the denominator which means we just flip

the sign in the middle so I'm going to

multiply the denominator by 2 minus I

but that also means we have to multiply

the numerator by 2 minus I so let's see

if we distribute things out we'll get 1

times 2 which will be 2 so I'm working

at the top we'll get 1 times negative I

which will be well negative I will get a

negative 2i then we have negative I and

negative I which is just going to give

us a positive I squared okay in the

denominator we'll have 2 times 2 which

is 4 we'll get 2 times negative I which

will be well negative 2i on the inside

we'll get a positive 2 I

then we have positive I times negative I

which will give us negative I squared so

let's see in the top we have two we've

got negative I and a negative 2i

that will give us negative 3i we have

plus I squared

but that's plus negative 1 which we can

write as minus 1 notice in the

denominator the negative 2i and the

positive 2 I cancel we would have 4

minus I squared which is negative 1 so

in the numerator we've got 2 minus 1

which is 1 minus 3i all over 4 plus 1

which is 5 and we can now rewrite this

as 1/5 minus 3/5 times I so now by doing

all this I've got it written as a more

recognizable complex number well all you

would do so maybe this is 1 and we'll

stick down here negative 1

it says we're graphing well how about

negative one how about we put positive

one on the right side

so here's positive one I was already

thinking of the negative so here's a

positive one and negative one

well it says for the real part you're

only going over to one fifth and then we

have to go down so let's see one two

three four five so here would be

negative three-fifths we can put a dot

right there so the real part is positive

one-fifth the imaginary part is negative

three-fifths again we can just make a

vector pointing at that little dot and

that would now be the graph of our

complex number to find the modulus again

to find the modulus we just take each

component the real part and the

imaginary part and square it so plus

negative 3/5 squared

and now just a little bit of a

arithmetic so we'll get 1 over 25 plus 9

over 25 when we square everything

that'll give us 10 over 25 we can

rewrite this as the square root of 10

over the square root of 25 which would

just simply I think reduce down to the

square root of 10 over 5 and we can't

really simplify the square root of 10

any further so I think I would leave it

right there so we would say the modulus

of this complex number is root 10 over 5

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Complex Numbers: Graphing and Finding the Modulus, Ex 2. In this video, I show how to graph another complex number and how to find its modulus.


all right in this video I just want to

do another another problem where we're

going to solve an equation containing a

radical so here we have the problem < TK

of 4 - W - 6 = -2 again the first thing

you want to do is just get the radical

by itself so you don't well I mean

that's really almost always the best

thing to do so I'm going to add six to

both sides on the left I'll have the

sare < TK of 4 - W on the right side -2

+ 6 is going to give me

pos4 and now again to get rid of the

square root I simply

Square both sides if I square the square

root of 4 minus W I'm just left with 4

minus

W and on the right side 4 squar is going

to give me 16 well I can subtract four

from each

side and I'll get that- W = 12 and then

simply If I multiply both sides by -1

we'll get W = -12 and we can always

again you should always go back and

check your answers to see if they work

so notice if we plug it into the

original

problem we would have the square < TK of

4 - -12 which would make 4 + 12 - 6 and

again we're we're asking ourselves does

that equal -2 well notice 4 + 12 is 16

the 16 is 4 4 - 6 does in fact equal -2

so this solution W = -12 is correct and

we found the the answer

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving an Equation Involving a Single Radical (Square Root) - Example 2


So, if 16 + 4x is 10 more than 14, we want to know what is the value of 8x?

So, we've got answer choices of 2, 6, 16 or 80, so I guess what I would do on this one is just turn this into

an equation, and then solve for x, and then you could multiply by 8 and get the solution.

So that's going to be the approach I'm going to use here.

Okay, so 16 + 4x is 10 more than 14. We have to be careful and make sure

we set up the correct equation there, so 16 + 4x.

Okay, I know it's not equal to 14, right? This quantity is 10 more,

so to make this side equal 14, since right now 16 + 4x is 10 more, I would have to subtract 10 from that quantity

to make it equal to 14.

Okay, so be careful. I think this would be the one place to make a mistake, maybe you could, you know,

make this a plus sign and then, you know, maybe something weird would happen.

But, since again, 6 + 4x, since that's 10 larger, I would have to subtract 10 away from it to make it equal 14.

And again, now we just have a nice simple equation. So, 16 minus 10, that's going to leave us with 6.

Plus 4x equals 14. We could subtract 6 from both sides.

So we'll have 4x equals 14 minus 6, which is going to be 8. And I said I was going to solve for x,

but I'm not going to do that, actually, because I want the value- I mean, you could solve for x,

you could divide both sides by 4 and then get the value of x. I want the value of 8x.

Well I've already got 4x on the left side, so what I'm going to do is I'm just going to multiply

both sides here by 2. I'm going to multiply the right side also by 2. And I'm multiplying the left side by 2

because hey, I recognize that that's going to give me- 2 times 4x is going to give me 8x.

And 8 multiplied by 2 is going to give me 16. So the answer choice of C, of 16 is going to be the correct answer.

SAT Math - Question 4, Section 2, 2016 Edition and Beyond

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Section 2 - A Calculator IS allowed!


all right in this video and I do another

example of solving a word problem

involving inequality so in this problem

there's suppose you have two options

you're going to rent a car and either

option a charges you have twenty five

bucks a day and five cents a mile or

option B charges you zero dollars a day

but they charge you a lot more per mile

they charge you 70 cents per mile and we

want to know how many miles must one

drive and one day for option a to be

cheaper because certainly right

originally freely gonna drive a few

miles better to pay the seventy cents a

mile and not have to pay this twenty

five dollars a day so we're trying to

figure out here when is the cost of

option a we want that to be less than

the cost of option B okay well let's

think about option A suppose so we'll

let X here denote the number of miles

driven in one day so if you drive X

miles how much would it cost to use

option A well you've got to pay that

initial twenty five dollars plus you pay

five cents per mile so we'll just take

point O five times X and that will give

us the you know the the total number of

the cost incurred by driving X miles

okay option B though okay we don't have

any upfront cost like we did here but it

is gonna cost seventy cents per mile

okay and now we've got a nice little

just simply a nice little

linear inequality to solve so alright

the first thing I'm going to do is I'm

simply going to I think I'm gonna put

the X's on the left side well let's put

them on the right so minus 0.05 X minus

0.05 X so we have 25 less than okay so

0.7 zero minus 0.05 is 0.65 X and now

all we're simply gonna do is we're going

to divide both sides by 0.65 and that

it's gonna give us our answer so if I

take 25 and divide it by 0.65 I'm

getting that to be thirty eight point

four six miles

so maybe assuming you have to drive I

don't know well let's just leave it at

that

so suppose they charge you based on you

know the fraction of the mile that you

drive it says basically if you drive a

little bit over 38 miles almost thirty

eight and a half miles if you drive more

than that in one day it's going to be

cheaper to use option a

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Word Problems Involving Inequalities - Example 3


all right in this video I want to do a

couple examples of solving absolute

value equations and to solve the

absolute value equation if we have the

absolute value of some quantity a so

this could be a lot of stuff in between

the absolute value bars if that if we

want that to equal B for this equation

to have solutions B is going to have to

be greater than or equal to 0 otherwise

there won't be solutions so that's one

thing to kind of keep in mind to solve

this equation we basically have to solve

two different equations we just remove

the bars and say you know solve a equals

B this equation and the other equation

we solve is a equals negative B so let's

solve the absolute value of 3x plus 2

equals 8 this will have a solution

because we've got a number that's

greater than or equal to 0 on the right

it basically says just remove the bars

solve that equation also we have to just

remove the bars and solve the

corresponding equation where we make the

value negative ok well this is pretty

easy so now we just have two simple

linear equations I can subtract 2 from

both sides that will give me 3x equals

positive 6 if I divide both sides by 3

I'll get x equals 2 is a solution and

for the other equation I'll simply

subtract 2 from both sides and get 3x

equals negative 10 and then again I'll

just divide both sides by 3 and get my

second solution of x equals negative 10

thirds and again I definitely encourage

you go back and check both of these

numbers do the arithmetic on the inside

take the absolute value at the end and

you'll see that you get something that

equals 8

so another one here real quick suppose

we have the absolute value of 7 minus 2x

equals 10 ok so again we're just going

to do the exact same thing here I just

remove the bars 7 minus 2x I set that

equal to 10 and then again I do the same

thing I just removed the bars but I set

it equal to negative 10 ok so to solve

the first equation I'm going to subtract

7 from both sides that will leave me

with negative 2x equals positive 3

simply divide both sides by negative 2

and that will give me one of my

solutions of negative 3 halves and for

the other one I can subtract 7 from both

sides that will give me negative 2x

equals negative 17 if I divide both

sides by negative 2

I'll get my second solution of x equals

so two negatives make a positive

17 halves

Solving Absolute Value Equations (2 Quick Examples)

(https://www.youtube.com/watch?v=wE3Us-d5KJo)
(https://youtu.be/qRFX_HK-6Ik)
(https://youtu.be/OzmqAZZd88g)
(https://youtu.be/C7de7f2LER0)
(https://youtu.be/HlQZ6XkRIb0)
(https://youtu.be/rC4bRSQetvQ)

In this video, we explore how to solve absolute value equations focusing on two examples:

Solving an absolute value equation: ∣ 3x + 2 ∣ = 8
Solving an absolute value equation: ∣ 7 − 2x ∣ = 10

We go over the step-by-step process for breaking down these equations, including isolating the absolute value expression, setting up two separate cases (positive and negative), and solving for the variable. By the end of the video, you'll understand how to approach both simple and complex absolute value equations and inequalities.

What You Will Learn:

How to solve absolute value equations by setting up two cases
How to solve absolute value inequalities
Techniques for checking solutions using substitution
Make sure to leave a comment if you find the video helpful, and don’t forget to like, share, and subscribe! Also, let your friends, teachers, or parents know about this channel if you think they would benefit from these lessons.

Support me on Patreon if you’d like to see more content: https://www.patreon.com/patrickjmt?ty=c

#AbsoluteValue #Equations #Inequalities #MathHelp #SolvingEquations #Algebra #PatrickJMT


all right in this example we're going to

find the equation of a hyperbola that

has vertices at 0 comma positive ne5 and

the fosi are located at 0 comma positive

-20 so again uh I'm just going to make a

little rough

sketch uh of my my hyperbola so the

vertices are at 0 comma 15 and 0 comma

15 and again the

fosi are at 0 uh 0 comma 20 and 0 comma

-20 so okay so that tells me that

my hyperbola will be opening

upwards and

downwards So based on that I've got a

little bit of information already I know

that I'm going to use the equation y^ 2

over a 2 - x^2 over b^2 = 1 one so

that's kind of the the first thing I

think

about and again to get the value uh you

know the the

value for a that's just the distance

from wherever our hyperbola is centered

which is at the origin okay so uh it's

centered at the origin we look at the

distance to one of our vertices and

that's going to be our a value so in

this case well a simply equals the value

15 okay so the only thing that we're

really missing is we need to know the

value for B once we know the value for B

hey we're done we don't quite know that

though um we do have this distance from

again where it's centered to one of the

fosi and since that goes from0 up to 20

uh the distance to this the fosi is 20

and we label that uh as C and the

relationship the equation that we'll use

the relationship is the fact that this

distance this c^2 = a^2 +

b^2 okay so now we can use this equation

to solve for b squared and hey we'll be

finished okay so C is 20 we're squaring

that a is 15 we're squaring that and

then we'll figure out the value of

b^2 so 20 s is 400 let's see 15 squar is

going to be

225 + B ^2 well we can subt subtract 225

from both sides 400 minus uh 225 will

leave us with

175 = B ^2 and now I'm finished we can

uh just fill in our equation

so we have y^2 over a 2 and again that's

going to be 15

s 15 2 is

225 - x^2 over B squ which we just said

is

175 equal pos1 and that will be the

equation of our hyperbola

Conic Sections, Hyperbola : Find Equation Given Foci and Vertices

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, Hyperbola: Find, Equation Given Foci and Vertices.  In this example, we are given the vertices and the foci of an ellipse.  We then find the equation of the ellipse.


all right here we're going to find the

parametric equations of a line that

passes through a point -1 comma 2 comma

3 and it's parallel to the vector with

components 4 five six so this is just

kind of one of those knowing a formula

type of thing so the equation of a line

we can write that as r equal

rb0 + T * uh we'll use a little V here

so R sub z um you can think about this I

guess is being um it's really going to

be um the position Vector whose tip lies

on the plane okay so uh so if you think

about the uh position Vector pointing at

this point 1 comma 2 comma 3 that's what

we're going to plug in for our little r

sub Z we're going to use -1 comma 2

comma 3 and then we take T * uh

V and again this is just the direction

Vector of the line L the direction

Vector so we know it's parallel to this

Vector so we can use

that and now we've basically got the

equation of our line so we can

uh simplify this a little bit we can

write it as

4T 5T and

6t when I distribute the T and now we

can just add respective components so we

would have negative 1 + 4T comma 2 + 5 T

comma 3 + 6 T and now uh to get the

parametric equations uh we just look at

each individual component so the

parametric equation for x or sometimes

we'll write X of T is going to be -1 +

4T uh y of

T would be 2 + 5 T and then our Z of T

is going to be 3 + 6t and these are now

going to be the uh parametric equations

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Parametric Equations of Line Passing Through a Point.  In this video, I show how to find the parametric equations of a line that passes through a given point with a given direction vector.


all right in this video I'm going to do

another example of graphing a rational

function so here we're going to graph F

ofx equal x^2 - 9/ x - 3 um I'm going to

just rewrite this as y = x^2 - 9 over x

3 uh so at this point um notice a couple

things um notice this actually factors

the numerator factors is x - 3 over x +

3 and then in the denominator we've got

our xus 3 so notice we could actually

cancel out the x - 3 factors and we

would be left with Y = X + 3 which is

nice Rec uh nice and recognizable

hopefully um so yal MX plus b this is

just an equation of a line but we have

to be careful before we graph the line Y

= X + 3 and that's what it's going to

look like almost you know identically

there's going to be one small

change notice the domain of the original

function that we start with remember if

you have a fraction um the only thing

that's kind of bad for a fraction is we

don't want to divide by zero so the

domain of this function would be

um we could say it's all reals all real

numbers

except X = pos3 we'll have to leave that

value out of our domain

okay so even though kind of

algebraically we simplify it down to

this Y = X + 3 which is the equation of

a line you still have to take the

original you know the original function

into consideration you have to take its

domain into

account so okay so same thing down here

we've got the graph of y = x + 3 um and

again the

domain is all X or all reals

except x =

3 so let's see the Y intercept of this

is positive three 1 2 3 the way I kind

of think about it is if I plugged three

in we would get six

out but since it's undefined there's

going to be a little circle

there and notice if you plug three into

the

denominator we get zero but we also get

zero in the numerator if we plug three

in

since we get 0 over 0 what that means is

you just have a hole in the graph at

that point so basically our our our

rational function looks exactly like the

line Y = X +

3 um except for this big hole in the

graph otherwise it would be

identical okay and that's all there is

to it in this

case

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Graphing a Rational Function - Example 2


all right in this video I want to talk

about finding the components of a vector

and what we'll do in this case is we'll

be given the magnitude and the direction

angle and uh from there we'll have to

figure out the components so the way

typically I like to do these is just use

a little bit of you know trigonometry

here um

so the idea is okay we've got a a vector

uh it makes an angle of

60° uh from the origin so here's our

angle of

60° and our Vector has a magnitude of 8

so again magnitude uh in in these cases

just correspond to the length of the

vector so really what I imagine is uh

just making like a little right triangle

and what I'm trying to find is I'm

trying to find the value for x and I'm

trying to find the value for y um the

value for x will be the first uh the

first component and the value for y will

be our second component so recall for a

triangle so let's see well we've got 60°

we've got8 we've got y we've got X we're

trying to solve for x and y so again

recall for a triangle I think we can use

S and cosine so let's see notice cosine

of

60° remember cosine is the adjacent over

the hypotenuse so that'll give us X over

8 if we multiply both sides by8 we'll

get 8

cosine of 60° is X well let's see cosine

of 60° um I believe cine of 60° is going

to give us

12 so 8 * a half is going to be

four so that's going to be the first uh

our first component

um so the first component is simply

going to have value four and we can do

the same thing to figure out the Y

component um the second component we can

just do well well s of

60° s is the opposite over the

hypotenuse so y/ 8 again we can multiply

both sides by 8 so we'll get 8 * s of 60

y s of 60° that's just < tk3 over

2 well 8 / 2 is 4 we'll get 4 < tk3 as

our value for y so that that'll simply

be our second comp component for our

Vector here so that's it this original

Vector would have components 4 comma 4 <

TK 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Components of a Vector, Ex 1.  In this video, we are given the magnitude and direction angle for the vector and want to express the vector in component form.


here we're going to determine the

midpoint between the points 4 comma 7

and -6 comma 6 so recall the midpoint if

we have two points X1 comma y1 and X2

comma Y2 to get the midpoint we simply

add the x coordinates together and

divide by two and we do the same thing

for the y-coordinates we add those

together and divide by two you're

basically averaging the X coordinates

and you're averaging the y-coordinates

it doesn't matter what point you call X1

y1 and X2 Y2 so let's assume that you

know X1 y1 is 4 comma 7 and X2 comma Y 2

is -6 comma 6 so if we add the x

coordinates we'll have 4 +

-6 / 2 and then to get the midpoint the

the y-coordinate of the

midpoint We'll add 7 + 6 and then divide

that by

pos2 so let's see 4 + -6 that's going to

be -2 -2 / 2 will be -1 7 + 6 is 13

divided by two that doesn't really

reduce so I'm going to leave it as 13

over2 and that would be our

midpoint

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Midpoint of Two Points - Another Example.
In this video I find the midpoint of two points


okay in this example we're going to find

the area and perimeter of a

parallelogram and to find the um the

area all you have to remember is that

the area formula for a parallelogram we

just take the length of the

base and multiply that by the height so

the length of the base multiplied by the

height and fortunately we are already

given both of those things we're given

that it has a base of length eight and

the height is the um this uh this dash

line here of

3.6 so we have to multiply those two

values together so it says the area is

going to equal

8 multiplied by

3.6 and I'm just going to do this by

hand so let's see I've got 3.6 I'm going

to multiply that by 8 so let's see 8 * 6

that's 48 8 so we'll drop down the 8 and

carry the 4 8 * 3 is 24 24 + 4 is going

to give us

28 and if we count we've got uh one

decimal place so I'll put the decimal

place right there and it says that we're

going to get an area of

units all right so not too terrible

again just just formulas to

know the other part is is the perimeter

and again the the perimeter is just the

distance around and since we have a

parallelogram the left side has a length

of 4.2 that means the right side is also

going to have a length of

4.2 the base we said has a length of 8

that means the top part is also going to

have a length of eight so we just have

to add up those values 4.2 + 8 + 4.2 + 8

and that's going to give us the

perimeter so we said 4.2 + 8 +

4.2 +

8 well let's see

4.2 and 4.2 if we add those together

that's going to give us

8.4 if we add the 8 and

8 that's going to give us

16 and let's see so 8 + 16 that would be

24 and then we still have to remember

the point4 so 8.4 + 16 that's going to

give us a value of

24.4 and that's going to be the

perimeter of our parallelogram

GED 'Most Missed' Math Practice Problems , Code Q.4.c,  #2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The GED Testing Service keeps a list of the most missed math topics and can be seen here: http://www.gedtestingservice.com/uploads/files/126de0283c94ff6323b3b5b3f310da5d.pdf
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 do

another example of solving a word

problem involving some inequalities so

in this example suppose we've got our

buddy here Jason and suppose he makes a

living uh by making and selling fishing

poles suppose he he it costs him $12,000

a year um plus a cost of $4 per each

poll that he makes so maybe he pays you

know $1,000 rent a month in his little

Factory um and then he has costs

occurred to make each one of these

fishing po

we want to know how many fishing poles

he must make assuming he can sell them

all um if he wants to have a profit of

$48,000 in one year and if we know that

he sells the fishing poles for $28

each okay so to me to kind of really get

me started uh you know intuitively what

I'm thinking is kind of you know the

money that he brings

in minus his

costs right so whatever selling minus

his costs that's going to be his profit

um that's how much money he gets to keep

and we want that to be um so suppose we

want him to make at least

$488,000 here so we said uh to make a

profit of

$448,000 um let's just make let's let's

modify this a little bit and say uh to

make a at least a

profit cuz most people probably uh you

know once they hit some number as long

as they're over that they're happy so

okay so well how much money does he make

so let's let X represent the number of

fishing poles that he sells um if he

sells X of these fishing poles well he

gets $28 per each one okay so that's how

much money he would make if he sells

them but then we have to subtract away

his costs well his costs are

12,000 plus he got to Shell out $4 per

each one that he

makes greater than or equal to

48,000 and hey now we've got our

inequality set up that we need to solve

okay so this is 28x if we distribute the

negative we'll get -1 12,000 we'll get

-4x that has to be greater than or equal

to

48,000 and again now I've just got a

nice little in uh linear inequality so

28x - 4X is going to be

24x I can add the 12,000 to both sides

so that'll give me 24x is greater than

or equal to 48,000 + 12,000 is going to

be 60,000 and now the last thing we have

to do is just simply divide both sides

by

24 so it says the number of fishing

polls that he has to make if he wants to

make

$48,000 per year or more I'm just taking

60,000 dividing by 24 and I'm getting

the number

2500 so it says he's got to make and

sell 2500 fishing poles a year if he

wants to make um at least

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


okay so just one more example here of uh

finding a derivative using implicit

differentiation so here we're going to

find the derivative uh we're going to

find

dydt of Y Cub + sin of y * t = t

5 so again the idea in this video is

we're taking the derivative with respect

to T so again we're going to kind of

treat the T kind of like the x that we

you know the way we did it in the first

video so think about T and X as being

your uh you know the independent

variable in this

case so all we have to do in this case

again so when we take the derivative of

y 3 well we'll get 3 y^2 but then we

have to tack on our dy over

DT plus okay we have to use the uh the

chain Rule and we take the derivative of

s of YT so the derivative of s is just

cosine

we'll leave the inside alone but then

when we take the derivative of the

inside we'll have to use the product

rule so let's see when we take the

derivative of one y we'll just get one

but we have to tack out tack on the dydt

term um and then we'll leave the T alone

and then plus okay so now uh we'll leave

the y alone when we take the derivative

of 1T we'll just get

one and then on the right right side

when we take the derivative of T 5th

we'll get 5 T 4th so now what we need to

do is just solve for our dydt term so um

all I'm going to do is just get rid of

the brackets

here and I'm going to have to distribute

this cosine of YT to you know both both

terms so let's see we'll get plus so I'm

going to write this as T time cine of

YT * dy over DT and then when we

distribute again we'll get plus I guess

we'll just get y * cosine of

YT on the right we still have our 5T to

the 4th just hanging out and now I'm

going to move everything that doesn't

have a dydt attached to it I'm going to

move that stuff over to the other side

so let's see here there's our 5T

4th um I could subtract the Y

cosine of YT from both sides so we've

basically moved that over to the other

side notice on the left all we would be

left with would be the 3 y^2

dydt plus T * cosine of YT *

dydt and what I'm going to do now is

just factor that dy over DT

out well let's see if we pull the dyt DT

out we would need a 3 y^2 and then if we

pull it out from the um second term we

would need the positive T * cine of

YT and now all we have to do is just

divide both sides by the 3 y^2 + T

cosine YT and we'll get our solution so

it says Dy DT is going to be 5 T

4 - y * cosine of

YT all over 3 y^ 2 + T * cine of

YT and we've now got our

derivative

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Implicit Differentiation - Basic Example 3 / 3 .  Just a fairly straight forward example of finding a derivative using implicit differentiation.


All right, in this video we're going to do some examples of going from,

basically, our polar form back to our complex form. So a couple examples here...

And I think these are certainly a little bit easier

to do than putting them into polar form because really all you have to do is just kind of do the arithmetic. So

part a) here. We've got 3 times cosine of 135 degrees.

Well, 135 degrees is going to be in quadrant two.

Cosine of 135 degrees is negative square root of 2 over 2.

Sine of 135 degrees is positive square root of 2 over 2.

So now all we have to do is just distribute out the 3. So we'll get negative

3 root 2 over 2

plus 3 root 2 over 2 ... I'm going to stick my i out to the right side... And now we've taken our

number that was in our polar number and put it in complex form. So

that's all there is to do. You really just have to evaluate whatever

cosine of the angle is, whatever the sine of the angle is, and just multiply it out.

So we can do our other example here, I think, real quick as well.

So, let's see, 270 degrees. If you think about 270 degrees, well,

270 degrees is just going to put us at the negative y-axis.

And, well, cosine of 270 degrees ... the x-coordinate on the unit circle there would be zero.

Sine of 270 degrees... that would be the value of the y-coordinate on the unit circle, which would just be negative one.

Well, if we distribute 4 times zero, we'll get zero. A positive 4 times a negative one will be negative 4i.

So,

typically, we'll just write it as negative 4i. We could write it as zero minus 4i

if you also want to see the real part, but that's all there is to it. Again,

just simply evaluate cosine of the angle, sine of the angle.

Distributed out, in this case, there wasn't a lot to cancel or simplify,

But if there was that would be all you'd have to do, and you would get your solution.

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Complex Numbers: Convert From Polar to Complex Form, Ex 1. In this video, we take a number in polar form and write it in complex form.