help-0

[violet nova]

^This is what I have down for #2.

^This is what I have down for #6.

If someone could help, I would greatly appreciate it.

<@&286206848099549185>

[ancient lantern]

Yes?

[violet nova]

I need someone to check my answers

[ancient lantern]

Ok.

[violet nova]

I put the questions and my responses above

If you could help, I would appreciate it

[ancient lantern]

The notes look good, but I would provide reasoning on why the point would be so close to the asymptote, ok?

[violet nova]

What would that reasoning be?

[ancient lantern]

Prove that the point is getting very close to the asymptote, but does not lie on the asymptote. Try using a limit notation.

[violet nova]

...

Could you possibly show me what that would look like?

Sorry, I'm a bit rusty.

[ancient lantern]

Where does the asymptote and the point lie?

[violet nova]

The asymptote lies on 150, 450

[ancient lantern]

Ok, and the point?

[violet nova]

Let me check

[ancient lantern]

Where x is the point, and A is the asymptote.

Goes back to basic Calc 1.

[violet nova]

(150.00166, 449.99)

I think...

Tbh

[ancient lantern]

I have the limit up there for you.

Does that help?

[violet nova]

So this?

[ancient lantern]

Plus or minus infinity at the end, but yes.

[violet nova]

Whenever I put that into my calculator, it says "Illegal Input"

Since there's also a y in f(x)

[ancient lantern]

This symbol: ±

[violet nova]

Still shows an error

[ancient lantern]

No, it has to approach the the value of the asymptote, which is 150, not 3x.

My fault, it should approach 150, not infinity.

[violet nova]

It still doesn't work

Since the y is unaccounted for.

Ohhhhh

[ancient lantern]

On the asymptote, y = ±∞

[ancient lantern]

YES

THAT'S IT

[violet nova]

This too?

[ancient lantern]

Yes.

[violet nova]

Wow

That took longer

Than expected

[ancient lantern]

Did I exceed your expectations?

[violet nova]

You actually helped me learn.

[ancient lantern]

I'm in 10th grade, lol.

🙂

[violet nova]

Ok, so now that I included those two limits, is the question 100% correct?

[violet nova]

Knowledge is ageless

[ancient lantern]

Yes.

[violet nova]

Were you responding "yes" to my question about correctness or...

[ancient lantern]

Both.

[violet nova]

Nice

Well, what about #6?

[ancient lantern]

Use the same method.

You will be shocked.

[violet nova]

You want me to use limit notation again?

[ancient lantern]

Indeed.

[violet nova]

For part b?

[ancient lantern]

Yes.

[violet nova]

Interesting

I got 4.00036...

[ancient lantern]

You should.

[violet nova]

I mean, is limit notation necessarily required though? I got the same fraction by just plugging in 20.

[ancient lantern]

So you know that the point is 4 units above the asymptote, right?

[violet nova]

Wdym?

[ancient lantern]

Well, if the vertical distance of the asymptote is 0.00036, 4 plus that is 4.00036, which can be rounded to 4.

Correct?

[violet nova]

Well, isn't the asymptote at y=4?

And the point is (20, 4.00037)

So can't I just do 4.00037 - 4 to determine the vertical distance?

So how is the point 4 units above the asymptote?

That would make it (20, 8)

[ancient lantern]

You can, and I just did.

[violet nova]

No?

[ancient lantern]

Yes, it would.

[violet nova]

But then the vertical distance would be 4, not 0.00037.

I'm confused

[ancient lantern]

Meet the requirements of this limit, and there you go.

Find a way to make the limit equal 4.

[violet nova]

x approaches 70?

Nah

Nvm

x approaches infinity

[ancient lantern]

Yup.

[violet nova]

So why would I need to write that again?

What does this prove?

Isn't my original answer acceptable?

[ancient lantern]

This shows how it could be possible for the point to equal the asymptote.

Yes, it is acceptable, but this is for undrerstanding.

[violet nova]

Ahhhh

So I don't actually have to write it down.

Or can I?

[ancient lantern]

Not if you don't want to.

[violet nova]

Could I just write both of them down and subtract the limit evaluations?

[ancient lantern]

Sure.

[violet nova]

Ok, so the question is absolutely correct now?

[ancient lantern]

Exactly.

[violet nova]

Ok, thanks

[ancient lantern]

Yw.

Just say ".close"

[violet nova]

Yup

.close

[lone heart]

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[past bloom]

if i move this cos2x to the left of the equality sign, does this eqn become homogenous differential eqn?

[lone heart]

@past bloom Has your question been resolved?

[subtle mango]

no, if so, every equation is homogeneous

you can think of the functions only depend on x is the coefficient of y (y^0, y, y', y", etc.)

Then homogeneous means that a1y+a2y'+a3y"+...=0, where a1, a2, a3... is the coefficient function

[slender marten]

A nice introduction to help you. 🙂

[lone heart]

@past bloom Has your question been resolved?

[past bloom]

thanks

[past bloom]

.close

[lone heart]

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[gray isle]

properties of the discriminant

i.e. no x-intercepts

show what exactly

the work is right there

we can use that

the graph of y=x^2+kx+k+3
is an upwards facing parabola
to determine when that's always positive we can consider when that has no real solutions / no x-intercepts

[lone heart]

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[bleak island]

can anyone tell me how they got this answer, my brain is not working rn

[bleak island]

like what did they do to get 3-x

OHHHHHH they added 1 + 2 and then x - 2x = -x

im so sorry ppl

thanks

.close

[lone heart]

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[opal merlin]

I need help solving b-e

I’m confused

[wind cloak]

Do you know the condition for parallel lines

[opal merlin]

nope

[swift shore]

@opal merlin what does it mean for two lines to be parallel?

[opal merlin]

I know it’s the same slope

[swift shore]

Oh ok

[opal merlin]

but what would I do for b?

[swift shore]

Ok so you have y = 2x + b right

[opal merlin]

yup

[swift shore]

And it passes through (-8,6)

[opal merlin]

right

[swift shore]

So we should be able to plug in -8 for x and 6 for y, and the equation would be true

6 = 2(-8) + b

6 = -16 + b

22 = b

Hence b = 22

[opal merlin]

right

[swift shore]

So final answer y = 2x + 22

[opal merlin]

appreciate it

[swift shore]

🙂

[opal merlin]

but what’s perpendicular mean

I’m assuming uh

[swift shore]

Perpendicular is a bit tricky

It means they intersect at 90°

But what does this mean about their slopes?

[opal merlin]

oh

isn’t it uh

M=y2-y1/x2-x1

to find slope

or idk

[swift shore]

that gives the slope if you have two points

But that doesn’t help here

[swift shore]

Try to explore this a bit

[opal merlin]

right angle?

y=-1/3x+2

but how would I show my work

I found when you multiply the slope of two perpendicular lines you get -1

how does that check out?

[swift shore]

Fantastic

Yes

[opal merlin]

can’t I plug it into an equation to find b

similarity to b

[swift shore]

If the slope of the first line is m, the slope of the perpendicular line is -1/m

So yep it’s -1/3

[opal merlin]

for b

same thing as before?

equation

[swift shore]

Wait you mean the b in y = (-1/3)x + b right

[opal merlin]

yeah

[swift shore]

Yeah same thing, you can just plug in the given point

[opal merlin]

might’ve been confused on how I worded it

alright

just checking

-1=(-1/3)(4)+b

I want to make sure I have the right equation

[swift shore]

Yep nice

[opal merlin]

that’s actually right?

idk I got -1.33333

then you’d add it right

would be 1/3

[swift shore]

Yes

So b = 1/3

[opal merlin]

For problem d would it just be y=-5
and for problem e would it just be -2

can anyone confirm this?

[fathom kite]

can you post original photo again

[opal merlin]

.

[fathom kite]

no

lets start with problem d

[opal merlin]

alright

[fathom kite]

a vertical line

is x = some number

this is a vertical line

[opal merlin]

right

y axis

[fathom kite]

can you see a value that never changes in this function?

[opal merlin]

yeah that’s what I figured

[fathom kite]

the x value never changes

[opal merlin]

right

but the y does?

[fathom kite]

yup

goes from -infinity to infinity

[opal merlin]

For d

[fathom kite]

yup

[opal merlin]

what is something I’d write

[fathom kite]

since they gave you a point

they told u its vertical

once you know its vertical, you just write x = ?

[opal merlin]

x=-5

right

[fathom kite]

? would be the x value of the point

no

y is -5

[opal merlin]

9

my bad

[fathom kite]

9

yeah

percect

lol

perfect

x = 9

[opal merlin]

for e

[fathom kite]

and e is the exact same thing, except that you don't do the method we did for x, you do it for y

[opal merlin]

can’t it go from negative to positive tho

[fathom kite]

y= ? is a horizontal line

the y never changes

don't touch the x in this one

once you know the y never changes, and they give you a point with coordinates, you can tell what ? should be

you need to have the intuition that a vertical line is x = ? and a horizontal line is y = ?

[opal merlin]

x doesn’t change

[fathom kite]

it does

[opal merlin]

yeah idk why I’m making this super complex

[fathom kite]

let me see smth

this is the functio

this is the function y = 5

you can see two points

look at the coordinates

what is changing and what is constant?

[opal merlin]

oh

the 5 stays the same

and the x changes

[fathom kite]

exactly

x changes

y doesn't

[opal merlin]

that’s for question e

since it’s a horizontal line

[fathom kite]

yes

[opal merlin]

what would I write though since it can change

[fathom kite]

well u dont touch x

u put y =

and then since you know y won't change

you take the coordinate of y they gave you

[opal merlin]

7

[fathom kite]

yup

y = 7

[opal merlin]

so that would be all for that

[fathom kite]

yeah

[opal merlin]

mind helping with something similar

[fathom kite]

not rn i gotta sleep

[opal merlin]

oh gotcha

Thanks though for the help

.close

[lone heart]

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[dawn ridge]

#help-0

[old horizon]

,rotate

[dawn ridge]

HELLO, this is a sequence problem and how to deal with such..

, rotate

[dawn ridge]

Little changed

<@&286206848099549185>

[slender marten]

Just calculate the limit as n goes to infinity.

(-1)^(n + 1) goes between -1 and 1 infinitely but the other part of the product clearly goes to zero.

[lone heart]

@dawn ridge Has your question been resolved?

[dawn ridge]

let me tr

What I get is this

@slender marten

[slender marten]

Your work is very messy.

I think you've followed the idea I was suggesting though. :)

[dawn ridge]

oh.. yea i tried on rough

What i have did, is it a actual solution?

[small helm]

are you familar with the alternating series test?

[dawn ridge]

yea.. it goes changing sign

[small helm]

oh wait I just realized your question is about a sequence

regardless, for these types of sequences/series, as long as the limit as n goes to infinity is 0, then it converges

so I think yours is a correct solution

[dawn ridge]

aaah..okey

naru👍

@slender marten hey man do you also agree this a right solution?

[slender marten]

The paper like this. No way. Look at the mess.

[small helm]

I'm pretty sure you made a mistake in your calculations

I can't quite tell

but you got the correct answer

[dawn ridge]

@slender marten ..

[slender marten]

I think that would work.

[dawn ridge]

well. very much thank you both.

.close

[lone heart]

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[lament basalt]

Prove that: (Trigonometric identities)

[lament basalt]

Is what I did correct? Thanks

[slender gull]

The third last step has an error, though you corrected it shortly afterwards.

What you did is correct.

[lament basalt]

What's the error there?

[slender gull]

It should be 1-sinx

You wrote (1+sinx)

[lament basalt]

Ohh okay I just imagined what would happen if I separated it

Would be 1 + sin x if I multiplied the negative sign to it

[slender gull]

It would not.

It would be sinx-1

-(1-sinx) = sinx - 1

Is what I just said.

[lament basalt]

Rly how?

[slender gull]

Wdym how, that's how it is. Distribute the multiplication.

$-(1-\sin(x)) = -1(1 + (-\sin(x)))$\
Distribute the multiplication,\
$-(1) \cdot 1 + (-1)\cdot(-\sin(x))$

[ocean seal]

What the hell am I doing here?

[slender gull]

That surely gives you sin(x) - 1

[lone heart]

@lament basalt Has your question been resolved?

[lone heart]

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[compact lintel]

i need help

[compact lintel]

ive spent a long time trying but idk how to do it

[vernal holly]

Do you know what an inverse equation is?

[compact lintel]

yes

i just dont know how to turn it into an inverse

i tried a couple things

but it hasnt worked

[compact lintel]

apart from swapping x and y

what steps do i do?

[vernal holly]

Well once you swap x and y, you solve for y

[compact lintel]

so i got the sqaure root

what do i do now

[vernal holly]

Is y isolated?

That's the goal

[compact lintel]

um i have

x = (y^2 -9y +15)/3

then i solved for y in that quadratic equation

i got

(9+-√21)/2

thats the roots

but im not sure what to do know

[vernal holly]

You should have x somewhere

[compact lintel]

wdym

i equated (y^2 -9y +15)/3 to x

then i solved for y

thats what i do right

[ocean seal]

AlgebraManiacABC

[compact lintel]

oh okay

what do i do after

[vernal holly]

Well, you keep going until y is alone on the left

And there are no "y's" on the right

[compact lintel]

so then i minus 15

divide by 9

then i have y^2 -y

y^2 - y = (3x-15)/9

well yea that

[vernal holly]

(add the $ to both ends of the eq)

[compact lintel]

oh okay

$y^2 - y = (3x-15)/9$

[ocean seal]

lozzby

[vernal holly]

You've made one mistake

Go one step back

[compact lintel]

$y^2 - 9y = 3x-15$

[ocean seal]

lozzby

[compact lintel]

this?

[vernal holly]

Right

Now divide by 9 again

[compact lintel]

sure

$y^2 - y = (3x)/9 - 15$

[ocean seal]

lozzby

[compact lintel]

is that what i do?

[vernal holly]

You were closer in your first one

[compact lintel]

oh

oh

[vernal holly]

You just keep forgetting to divide y^2

[compact lintel]

how do i do that?

[vernal holly]

Well it would be $(y^2)/9$

[ocean seal]

AlgebraManiacABC

[vernal holly]

But now I'm afraid about having the y^2 with the y. You may have to shove everything on the left and use quadratic formula

[compact lintel]

uh

so thats what i tried before

[vernal holly]

But you somehow lost x

[compact lintel]

$(y^2 -9y +15)/3 = x$

[ocean seal]

lozzby

[compact lintel]

i did that

and used the quad formula to solve for y

[vernal holly]

You can't use quadratic formula like that

[compact lintel]

oh

[vernal holly]

the right side has to equal 0

[compact lintel]

ohhh ok

$(y^2)/3 -3y +5 = 3$

[ocean seal]

lozzby

[compact lintel]

so i find from that?

[vernal holly]

1. Where did x go? 2. Right hand side must equal 0

[compact lintel]

oh i meant

$(y^2)/3 -3y +5 = x$

[ocean seal]

lozzby

[compact lintel]

i meant this

[vernal holly]

Okay, now move the x to the left

[compact lintel]

oh okay

$(y^2)/3 -3y -x +5 = 0$

[ocean seal]

lozzby

[compact lintel]

this?

[vernal holly]

Looks better!

Now you can use quad. formula

Before you do, what will be your "c?"

[compact lintel]

-x + 5?

[vernal holly]

Yeah, or, easier, (5-x)

[compact lintel]

oh ok

so far i have

$y = (3+-√(9 - (4/3)x - (20/3))$

[ocean seal]

lozzby

[compact lintel]

/2/3

forgot that part

[vernal holly]

I would use $\sqrt{num}$

[compact lintel]

oh okay

[ocean seal]

AlgebraManiacABC

[compact lintel]

$y = (3+-/sqrt{(9 - (4/3)x - (20/3)}$

[ocean seal]

lozzby

[compact lintel]

uh

wrogn slash

[vernal holly]

backslash

[compact lintel]

$y = (3+-\sqrt{(9 - (4/3)x - (20/3)}$

[ocean seal]

lozzby

[compact lintel]

well it looks like that

just divided by 2/3

[vernal holly]

Give me the quadratic formula real quick

[compact lintel]

3

[vernal holly]

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[ocean seal]

AlgebraManiacABC

[vernal holly]

Therrrrre we are xD

[compact lintel]

lol

[vernal holly]

Okay lemme plug in stuff

[compact lintel]

oh okay

$y = (3+-\sqrt{(9 - (4/3)x - (20/3)})/2/3$

[ocean seal]

lozzby

[compact lintel]

this is what i have to far

[ocean seal]

AlgebraManiacABC

[compact lintel]

wouldnt it be x-5

[vernal holly]

But you had -x+5 right?

[compact lintel]

oh nevermind

yea mb

[ocean seal]

AlgebraManiacABC

[compact lintel]

wait wouldnt the -3 be 3

[vernal holly]

$x = \frac{3 \pm \sqrt{9 - (\frac{20}{3}) + \frac{4}{3}x}}{\frac{2}{3}}$

[compact lintel]

cuz in the formula its -b

[vernal holly]

Dang

[compact lintel]

so -(-3) = 3

[vernal holly]

Yeah

[ocean seal]

AlgebraManiacABC

[compact lintel]

ok then what i have is

4/3x - 7/3

[vernal holly]

Would be plus

9=27/3; 27-20 = 7

[compact lintel]

oh right

[vernal holly]

You could also continue with:

$y = \frac{3}{\frac{2}{3}} \pm \frac{\sqrt{\frac{7}{3} + \frac{4}{3}x}}{\frac{2}{3}}$

[ocean seal]

AlgebraManiacABC

[vernal holly]

which equals

$y = \frac{9}{2} \pm \frac{3\sqrt{\frac{7+4x}{3}}}{2}$

[ocean seal]

AlgebraManiacABC

[vernal holly]

Either way that isolates y and gets rid of most of the ugliness

It's still ugly though

[compact lintel]

lol yea

but it looks better now

[vernal holly]

Just a bit xD

[compact lintel]

lmao

wait so

do i plug this this in?

[vernal holly]

This would be your inverse equation

Just replace y with f(x) and boom

[compact lintel]

oh

ohhhhhhhhhhhhhhhhhhhhh

oh so this is the answer

i was kinda lost lmao

[vernal holly]

I think so, unless we've made a mistake

[compact lintel]

thanks for the turtorial

[vernal holly]

Np xD

[compact lintel]

tyty

even if it is wrong

the guide was helpful

ty

.close

[lone heart]

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[alpine sable]

Would someone please explain me this problem?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[slender marten]

It's just polar coordinates using u and v instead of the regular r and theta.

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[agile wedge]

this is a really simple

[agile wedge]

question, just some basic probability / economics math

let us say, i have a game of chance, on winning, i gain my bet back, + the value of the bet, on losing, i lose my bet, and on tying, i keep my original sum

if i double my bet, what now happens to this EV calculation?

[lone heart]

@agile wedge Has your question been resolved?

[lone heart]

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[glacial dagger]

The first photo is the question and second is the answer. How can there be a value such as (5,-2) when x must be less than 3?

[old horizon]

I cant see the question

I cant see the answer

zoom it a bit

[lone heart]

@glacial dagger Has your question been resolved?

[lilac nest]

The point (5, -2) is not a solution of the inequality (so it's not in the blue part of the graph)

[lilac nest]

The part colored in blue here is the solution

[glacial dagger]

Ok but when you do something like y=<x+1, you count the line and the shaded area. Why do you not count the line in this case?

[lilac nest]

The problem is that the point (5, -2) is well outside the blue area

It's not on one of the lines that limits that plane

If the point was on the purple line then it would have been a solution

But it's not

[glacial dagger]

Reply to this pls

[lilac nest]

.

[lilac nest]

.

[lilac nest]

It's not on the plane or on the purple line

[lilac nest]

Which are all your solutions

(5, -2) is here

It's on the red line, but it's neither a solution to the inequality y>3x/2-3 nor to the inequality x≤3

The red lines do not indicate solutions, they're there as a limit between the solutions and non-solutions of each inequality

[lilac nest]

If you don't understand what I explained just tell me, I can try to explain that in another way

[glacial dagger]

I understand. But why did the answer not just delete the other parts?

[lilac nest]

Wdym?

Which other parts?

You have these three inequalities

The solution is the intersection of all of them

Which is that part in purple

[glacial dagger]

The parts of the line that are incorrect

So why did the not delete everything except the purple?

Or are you just supposed to know that the parts that are not surrounding the shaded area are wrong?

@lilac nest

[lilac nest]

The purple part is the intersection of the 3 pictures I sent you before

Which are the graphs of each inequality you have

@glacial dagger

[glacial dagger]

Ohh I get it now

Can you help me with sum?

How can I find the b

For y=2x+b

[lilac nest]

Depends what the problem asks. What is the whole question?

[glacial dagger]

Find the b

Is finding b possible with jsut that?

[lilac nest]

Do you have a graph?

[glacial dagger]

But u cant tell what b is from here

OHHH

[lilac nest]

This is another problem though

[glacial dagger]

I CAN JSUT SOLVE ALGEBRAICLY RIGHT?

JUST PLUG IN 2 REAL POINTS

And solve for b?

[lilac nest]

Yes, but that graph you sent here is not going to be the graph of y=2x+b for any b

Is that the correct graph

[glacial dagger]

Ik

No

[lilac nest]

Oh ok

[lilac nest]

In general b would be the y value when x is equal to 0 (they intersection of the graph with the y axis)

[lone heart]

@glacial dagger Has your question been resolved?

[lone heart]

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[alpine sable]

Prove that the following are irrational.
(i) √5
(ii) 3+2√5

[mortal trellis]

do you know the proof that sqrt(2) is irrational?

[alpine sable]

I know the proof but I think the process I learnt is a bit wrong so I would like to learn it again.

proof by contradiction:
assume √2=p/q, where p and q are natural numbers and minimal (!), now it follows:
2=p²/q²
2q²=p²
therefore, p is a multiple of two and even, so p=2r for a natural r:
2q²=(2r)²
q²=2r²
can you finish the proof from here? and why is p even?

[lone heart]

@alpine sable Has your question been resolved?

[analog basalt]

This is easy using the Rational Root Theorem, but I suppose you haven't learned that yet?

I assume I found an elementary proof but I'm rusty on these things, I'll leave it up to @alpine sable to judge me:

||Assume, by way of contradiction, that $\sqrt{5} \in \mbb Q$, hence write $\sqrt{5} = \frac{a}{b}$ with $a, b \in \mbb N \setminus { 0 }$ and $\gcd(a,b) = 1$. Rewrite this as $5b^2 = a^2$, which means that $5 \vert a^2$, but because $5$ is prime this means that $5 \vert a$. Hence write $a = 5c$, then we obtain $5b^2 = 25c^2$, or otherwise, $5c^2 = b^2$ and using the same argument we obtain that $5 \vert b$, which would imply that $\gcd(a, b) \geq 5$, yielding a contradiction.||

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[ocean seal]

Yuese

[regal sapphire]

What?

[regal sapphire]

Can someone tell me what to do

. . .

[mellow tusk]

this is a definate integration problem do uk that topic?

[regal sapphire]

Integration?

[mellow tusk]

yea

[regal sapphire]

No idk the topic

Welp

This is hopless

.close

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[alpine sable]

how does this turn into squrt 49^x

[mellow tusk]

can it?

[alpine sable]

isnt it supposed to be sqrt 49^x / 49^-x ?

[vale wigeon]

7^((x+2)-(2-x)) = ?

[alpine sable]

yeah that makes sense

aight thanks

.close

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[flat saddle]

How do I solve n

[flat saddle]

(without bruteforcing)

I mean how do I find out the largest n for which n satisfies this equation

[mortal trellis]

the largest?

the LHS grows quicker than RHS so eventually this will always be true

much more interesting would be the smallest

[analog widget]

You can solve by log

Pass the two as a base of the log and solve it,

Then after you finish apply the inequation property to determine the solution set

[mellow tusk]

would the smallest work? cause n=any -ve integer satisfies the inequality

[analog widget]

Let me solve

done

Ill just give a real check and send a pic

wait, the problem is that you sill remain with a n

[flat saddle]

idk I just brute forced it

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[ornate condor]

doesnt it also neccessarily diverge whenever |z-z0|=|z1-z0| ?

[raw latch]

no the boundary is always a special case

[ornate condor]

why so?

<@&286206848099549185>

[lone heart]

@ornate condor Has your question been resolved?

[sage summit]

x+x²/2+x³/3+... diverges for x=1 but converges for any other x with |x|=1

[ornate condor]

..i just realised idk how to show or tell if something is convergent in complex, how do i do that?

[sage summit]

well here the proof for x= -1 is easy and that's enough to give you a counter example

[ornate condor]

oh, yea i guess

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[lament basalt]

Prove that:
(trigonometric identities)

[wide niche]

Try rewriting cos^2x

In terms of Sinx

And factorise the sinx s on the numerator and denominator

[wide niche]

When you cancel the cos2x over cos2x remember it becomes 1

This is the answer, we want to deal with sinx since that is what the answer wants

[wide niche]

Also 1 over cos^2x is sec^2x not cot

[lone heart]

@lament basalt Has your question been resolved?

[lament basalt]

Sry for late reply but may I ask what to do next?

I absolutely don't know how to answer this other than the factoring method used in trinomials

[wide niche]

So rewrite the denominator in terms of sinx

Using the cos squared x + sin squared x = 1 rule

Since the answer we want to prove has only sinx terms, we want to get rid of all the cosx

[wide niche]

Then you factorise the fraction and cancel

[lament basalt]

Sry wdym on this?

Also

[lament basalt]

This, right

[wide niche]

This rule

[wide niche]

Ok so only change the denominator at first, leave the numerate alone

*numerator

[wide niche]

You should not have a secx at the top

[lament basalt]

Oh only changing the denominator to 1- sin^2 x?

[wide niche]

Yes

[wide niche]

Then you can factorise the top and bottom, and cancel. you no longer need to use any trig identities

[lament basalt]

Sec^2 x = 1 + tan^2 x

[wide niche]

Yea, but you don’t need to use that for this question

[lament basalt]

I got this instead

I just simply made a 1+ sin x, separating it from the rest

[wide niche]

Treat the top and bottom part of the fraction as quadratics

Then factorise as if it was x^2 +4x + 3

[lament basalt]

No I mean there's still some additional sin's, coefficients, and sec^2 x

Even if I factored out 1 + sin x/ 1 + sin x

[wide niche]

Check this, you see how I factorised it like a quadratic

[lament basalt]

I did different in which there's a sec^2 x

Is that the only way?

[wide niche]

Possibly, but I haven’t seen any other way. You don’t really want to make a secx when it asks for the answer to be in sinx only

[lament basalt]

Oh okay thanks

.close

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[inner kayak]

Hi in class we didn't study linear graphs and what's a linear equation but we did by accident get a question with a linear equation so I wanted to check if I solved it right and if my graph is right too

[inner kayak]

The equation is (x-2) (x+3)=0 btw

[ornate condor]

i dont see what is the question...

[ionic trail]

Your calculations look good. You calculated the roots of (x-2)(x+3). But thats actually not a linear equation.

[inner kayak]

Then what type is it ?

[ionic trail]

If you multiply that out, you see that it's a quadratic.

[ornate condor]

oh is the question points of intersection of (x-2) (x+3) = 2x + 1 = y

[inner kayak]

So I should use the roots formula ?

And this also means my graph is completely wrong ?

<@&286206848099549185>

[small fog]

x^2+x-6

is what the function should look like

i think

[inner kayak]

Like because x×x and 2×(-3)

[small fog]

so now you just solve with quadratic formula

[small fog]

yeah

[wanton junco]

what exactly does the question want you to do tho ?

[inner kayak]

Solve the equation and draw a graph

[small fog]

he wrote (x-2)(x+3)=0

so i think that's a quadratic

[wanton junco]

a graph of what ?

[small fog]

so your graph is wrong

[wanton junco]

Its just an equation though with no context

its not a function

[small fog]

x^2+x-6 aka (x-2)(x+3)

[small fog]

i mean it looks ike a quadratic

[wanton junco]

can you show the question ?

[inner kayak]

OK thanks

[wanton junco]

this is the solution though, not the question

[small fog]

the question was solve for x and draw a graph

[wanton junco]

the equation you solved can be used to pinpoint the roots of the quadratic (x-2)(x+3) but it all depends on the question

[small fog]

i think

[small fog]

.

[inner kayak]

The question is on top (x-2)(x+3)=0

[wanton junco]

do you have the picture of the segment itself maybe ?

[inner kayak]

And I should solve the equation and draw graph

[small fog]

he literally typed the question

[wanton junco]

I might as well solve for x=2 and x=-3 and draw 2 vertical lines instead as well

[inner kayak]

It's not in English

[wanton junco]

post it either way

[small fog]

????????

bro

it's obvious what he is asked to do

draw a graph of x^2+x-6

and solve the quadratic

[wanton junco]

I'm sorry but it's not that obvious to me

[small fog]

it is when he typed out "I need to draw a graph and solve for the roots" 100 times

[wanton junco]

and besides, how can he be asked to graph a quadratic when he hasn't even studied linear functions ?

[small fog]

i have no idea

[wanton junco]

@inner kayak if you have the question itself from the book or source post it here even if its not in english, that way we can get a better understanding of what you need to do

You can't graph an equation that you're given

you can graph a function, but I don't see a function in your solution

so its a bit weird to me

[inner kayak]

It's supposed to be for 8th grade our teacher accidentally let this question slip into our homework
When I googled the equation a bit I falsely understood that it's a linear equation and tried to draw a graph for it

[small fog]

8TH GRADE

[inner kayak]

I now understand that I was wrong

[small fog]

holy shit in what education system do you not learn linear equations until grade 8

[ornate condor]

stop right there doc u might get banned for underage

[small fog]

?

grade 8 isn't underage

[ornate condor]

disord tos is like depending on region

[small fog]

it's like 14

[ornate condor]

well

its supposed to be, implying

[inner kayak]

I'm over 13

[ornate condor]

some countries its 16+

haha

but its ok say no more

[small fog]

anyway

try studying basic algebra and functions on khan academy

[inner kayak]

K ty

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[lone heart]

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[surreal meteor]

Bit confused on how I would solve this by hand?

[surreal meteor]

I have to calculate the integral

[wanton junco]

Substitution

[surreal meteor]

Hmm, okay!

[wanton junco]

Its very handy to identify functions and their derivatives, this allows you to see immediate subs and even solve the integral immediatley

[surreal meteor]

I'll watch a video rq on substition a bit tim since I did it last time

[surreal meteor]

The video shows this formula?

Is that correct?

[wanton junco]

thats the general theorem of substitution yea

but try reviewing notes and example of substitution

[wanton junco]

this should be pretty easy if you understand how to do subs and how they work

[surreal meteor]

yep but i have held sabbatical year

So haven't touched math for a whole year

hence why im a bit rusty

but actually watching this video 1 minute already made me remember more

[wanton junco]

yea ofc
which is why im recommending to review the method

If you after reviewing you still struggle you can come back or even ping me

[surreal meteor]

Is it okay i ping you and tell you what I'm going to try and do?

And then you can just say no or yes

[wanton junco]

sure

[surreal meteor]

Okay thanks! then in 5 minutes I'll write something haha!

[ionic trail]

Also consider integration by parts!
It's simpler for this problem imo.

[wanton junco]

hmm .. ?

is it ?

[ionic trail]

Altough I'm bad at integration by substitution😅

[wanton junco]

if it had been xlnx i would agree with you

[ionic trail]

Well I'm not sure actually. 😅

[wanton junco]

well you can try it

[wanton junco]

. have this in mind

[mellow tusk]

lnx=t, dx/x=dt, becomes integral 0 to 1 {2t}=1^2 - 0^2=1

[lone heart]

@surreal meteor Has your question been resolved?

[surreal meteor]

Something like this?

Do i then put -2/x*dt at the end?

[wanton junco]

with your sub you can replace 2dx/x - > dt

and then you're left with lnx when according to your sub you can replace with 1/2*t

after that you need you have to replace the integrals limits according to your sub

[surreal meteor]

you mean like because its a constant i can just take it out the integral

something like this

[wanton junco]

hmm ?

im not sure I follow ?

[surreal meteor]

Sorry

I thought u meant like this

[wanton junco]

I mean you're right, but I wasn't talking about that

[surreal meteor]

oh welp

xD

[wanton junco]

I just said how to rewrite your integral according to your sub

[surreal meteor]

but that changes my sub now

it's just 1/x

no9w

t=1/x

Now right?

t=g(x)=ln(x)

[wanton junco]

It doesn't chage your sub
you can maintain your current sub you just need to rewirte the integral accordingly

[wanton junco]

This sub will also work (and its actually a better one)

[surreal meteor]

Yes so now

[wanton junco]

you can always do whatever sub you want
you just need to make sure you rewrite your integral accordingly, and the integral limits

[surreal meteor]

dt=1/x*dx

[wanton junco]

the new limits as written here

[wanton junco]

correct

[surreal meteor]

But don't I need to get it on the other side

[wanton junco]

But notice that the expression 1/x*dx already exists so you can replace it with dt

[surreal meteor]

Yes that's the outer function?

Right

Or is that a big

Thought it was like since this is the same as the red

[wanton junco]

it is

but it doesn't matter the constant doesn't affect anything

[surreal meteor]

Then I'm a bit confused on what you mean 1/x*dx already exists so i can replace it with dt?

Don't I have to isolate dx

So I can replace dx with dt

and substitute t

[wanton junco]

after you did your sub you've correctly expressed the differential dt=1/x *dx

[alpine sable]

hey im kinda confused and not really sure about my solution for:

[wanton junco]

So essentially you can write your integrand as 2lnx/x dx -> 2*lnx * (1/x *dx) => 2lnx * dt

[wanton junco]

#❓how-to-get-help

try asking your qeustion in an available channel @alpine sable

[surreal meteor]

Ohh!

[alpine sable]

which?>

[surreal meteor]

So we are trying to isolate dt

I see

[wanton junco]

read #❓how-to-get-help

[alpine sable]

yeah im reading

[wanton junco]

we essentially want to scale our dx differential to match our sub's differential with in this case is t (the differential would be dt)

[wanton junco]

hmm.. why did you write t/x ?

[surreal meteor]

I substituted ln(x) with t

[glossy current]

but dx≠dt

[surreal meteor]

since we said t=g(x)=ln(x)

[wanton junco]

correct

[glossy current]

You can't replace dx with dt

[wanton junco]

correct*

but you still retained 1/x

[surreal meteor]

Oh so I can't do it unless dt=dx

[wanton junco]

even though you expressed that 1/x *dx = dt

dt != dx in this case but rather 1/x* dx= dt

[glossy current]

I think this is first time you do integration right

[surreal meteor]

No no, but like a year and 3-4 months ago since

Can't even open my notes since my computer is broken

[glossy current]

Hm

[surreal meteor]

since that could've ignited something inside of me

https://tenor.com/view/lighter-ignite-fire-flame-gif-8668350

https://tenor.com/view/elmo-fire-burn-flame-gif-5042503

[glossy current]

The problem you send, only need 2 lines if you have been doing enough integrals

[surreal meteor]

Well, that is good to know haha, don't see how that helps though 😛

[glossy current]

Also you are wrong at this for not just t/x but

Also the bound

you set t=ln(x) right

[surreal meteor]

Yes

[glossy current]

Then the new bound for t

Not x anymore

When x=1, t=ln(1)=0
When x=e, t=ln(e)=1

$\int_0^1 f(t)dt$

[ocean seal]

Darkness

[surreal meteor]

Wow - but why can we just plug it into t=ln(x)

And not the whole equation?

[glossy current]

We just substitute, you haven't done it enough tho

Where are you at

[wanton junco]

Its how the sub is defined

[wanton junco]

its written here

[surreal meteor]

g(a) and g(b)

Oh yes!
I see

[wanton junco]

the new bounds are g(a) and g(b), where g(x) is the subbed function

[surreal meteor]

Yes and t=g(x)

So just insert the bounds into them!

That actually makes sense!

[wanton junco]

yea essentially

That's waht the theorem states

After replacing the bounds you need to rewrite your integrand using your new sub

[glossy current]

Yes, to recap

[wanton junco]

so if t=lnx
dt=1/x *dx
you've received the expression of the differential in relation to t

[glossy current]

,tex

$I=\int_1^e \frac{2\ln(x)}{x}dx\\
t=\ln(x)\Rightarrow dt=\frac{1}{x}dx\Rightarrow dx=xdt\\
x=1\to t=\ln(1)=0\\
x=e\to t=\ln(e)=1$\\

[wanton junco]

so you can directly repalce the epression 1/x * dx with dt directly

[ocean seal]

Darkness

[glossy current]

I think we should go another line where dx=xdt so he can substitute correctly

[surreal meteor]

Okayu

the only line I didn't understand on that tex picture

was dx=xdt

[glossy current]

It's just multiply both side for x

[surreal meteor]

how do we go from 1/xdx to xdt

Oh

[wanton junco]

multiplying both sides by x

[surreal meteor]

And then multiply with t

after

[glossy current]

So now instead of writing dx, you can write xdt to the original integral

and where there is ln(x) you write t

It's how substitution works

[surreal meteor]

I'll definitely read up upon this lmao once im done with the assignments, just wanna get done with the assignments before uni start, but then once im done, i'll read up on the weakness and this is definitely one without a doubt

[glossy current]

Can you do next step

[surreal meteor]

yes

give me a second

[glossy current]

Next step of this

[surreal meteor]

yep yep

but writing on my own

But i didn't substitute x out fully?

[glossy current]

No

They will cancel them de

[surreal meteor]

oh because it goes to the nominator the top 😂

right? Or no

[glossy current]

Because the question is designed so

[surreal meteor]

so is that right?

[glossy current]

Yes

[surreal meteor]

missing the 2*

outside too right

so like this?

[glossy current]

Yes

[surreal meteor]

So it ends up being 2

[glossy current]

Yes

[surreal meteor]

Nice

Okay so thanks man!

I'll definitely read up on it

Luckily they added a bunch of pdfs

Do you have any videos you could recommend to watch?

[glossy current]

Well integration is just practice

[surreal meteor]

No doubt

[glossy current]

You need practice first

Watching videos now ain't help and may confuse you more

Watching videos is when you have enough training

[wanton junco]

khan academy is not shabby

[surreal meteor]

Oooor it might actually make me remember what I was taught lmao

[glossy current]

Because videos usually show you

Other techniques

[surreal meteor]

Yes, that is true though

[glossy current]

But like if you are new at ordinary technique's

[surreal meteor]

But perhaps once my pc is working I can go check my old notes on it

[glossy current]

Getting exposed to new techniques just confuse you mkre

[surreal meteor]

Ye, true

Well, khan academy has math too?

[glossy current]

Yes

So this is 2 line solving

[surreal meteor]

Got it

Anyway

Thanks guys!

Appreciate the help once again! 😄

[ocean seal]

Darkness

[glossy current]

Actually you are wrong at this step
Integrate of t is not t

[surreal meteor]

Oh so you still integrate t 😂

1/2t^2

so its 1

?

[wanton junco]

looks right

For the recrod btw
notice that the integrand was in the form of g'(x)f'(g(x))

which is the direct derivative of the function f(g(x))^2

[surreal meteor]

ye chain rule