help-17

[vast shale]

think of it this way

|x-1| = 2 implies that x - 1 can equal 2 or -2

because both |-2| = 2 and |2| = 2

[quartz mauve]

OH

ok ty

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[exotic dew]

What would the answer to this be

[exotic dew]

the values for a b and c

[outer warren]

have you made an attempt?

[exotic dew]

yes

im so lost though

[outer warren]

show your attempt

[exotic dew]

this all I could get to

[outer warren]

were you using a calculator?

the angles given are on the unit circle, use that or
stuff like reference angles to determine the exact values there

[exotic dew]

yeah

[exotic dew]

ill try

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[vast shale]

Hi I’m really struggling on how to do these problems if anyone could help I’d appreciate it 🥲

[scenic ravine]

what have you tried?

[vast shale]

I’m not sure abt the concepts I was gone from school for a bit and Everytime I try searching up videos I just get more confused

[scenic ravine]

Hint:- in general when you're solving linear equations, try to eliminate a variable by via addition or subtraction

[vast shale]

-10?

[scenic ravine]

for instance for the pair of equations on the left, what happens if you add them?

[paper depot]

btw may i point out the first equation in the problem on the left is written as -2y - 10x = x -- @vast shale are you 110% sure that's how it's meant to be?

with the x on the right?

[vast shale]

I just noticed that

tysm 😭

[paper depot]

it isnt gonna be fatal if it really is like that

but maybe you made a typo?

[paper depot]

ah so in fact the extra x was elsewhere.

ok so. alright.

does the word "substitution" ring any bells to you?

[vast shale]

Kinda 🥲

[paper depot]

ok, so like

with the first equation written as it is, i.e. x = -2y - 10

you may notice that x is expressed entirely in terms of y here

this means you can replace the x in the other equation with (-2y-10)

then it becomes an equation in y only

do you understand what i said here & do you see how to proceed?

[vast shale]

-2y - 10 - x = 0
+ 10 +10
-2y-x=10????????

so after u plug it in would it be -2y -x = 10, -y - 10x = -14????

[paper depot]

that is not what i asked you to do there.

let me repeat what i wanted you to do:

in the second equation -10x - y = -14, replace the x with (-2y-10).

can you do this?

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

-10(-2y-10) - y = 14

[paper depot]

missing minus sign at the very start.

[vocal sleet]

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[stiff oriole]

can someone double check my work please?

[scenic ravine]

,rotate

[boreal condor]

Are we in radians?

It gives me 0.6, but I'd give more workings

[stiff oriole]

no i’m in decimal

so it gave you the same thing huh

[boreal condor]

Decimal?

Degrees?

[stiff oriole]

oh shit yeah

i read decimals on the paper lol

[boreal condor]

Ah

[stiff oriole]

but yeah

[boreal condor]

How come we're getting the same

[scenic ravine]

looks right

thought you may want to explain why the answer is positive and not negative

[boreal condor]

Where did we get sin=xyz

[stiff oriole]

okay i have a question though for future problems. How would i know if it’s positive or nagative?

[boreal condor]

Dont you need to do inverse cos and so on

[stiff oriole]

since it’s in Q4 i take half right

get Q2

sin in Q2 is positive

is that how that works?

[boreal condor]

I've tackled the question differently so I can't help you much there

[stiff oriole]

because i was rewatching a lecture and i kinda got that

oh okay

how’d you do it?

[boreal condor]

Theta(or any placeholder variable) = arccos(0.28)

360-theta

Sin(new value/2)

Does it make sense? It's late for me

As long as you got the answer initially you're fine

[stiff oriole]

hmmm na im not getting it but yeah

thank you both @boreal condor @scenic ravine

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[vast shale]

@mortal anvil

[vocal sleet]

@vast shale Has your question been resolved?

[brisk moss]

@vast shale do you have a question?

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[coarse sequoia]

I don't get slopes at all. The points are 4,0 and 4,5

If we follow y2 - y1/x2 - x1, we get 5-0 = 5 and 4-4 = 0. What am I doing wrong?

[karmic imp]

The points are 4,0 and 4,5
Where are you getting (4, 0) from?

Can you mark where you think that is in the screenshot?

[rigid bolt]

picture it like this, the number of units upwards it goes to a certain point is the numerator of the slope, while the number of units to the right is the denominator

[rigid bolt]

here you can see it moves 5 up and 4 right

[coarse sequoia]

That's definitely a 4,0, all on x with no elevation.

[rigid bolt]

so the slope must be 5/4

[karmic imp]

I don't get slopes at all. The points are 4,0 and 4,5

If we follow y2 - y1/x2 - x1, we get 5-0 = 5 and 4-4 = 0. What am I doing wrong?
Doing this, you're picking two points that are on the line

(4, 0) is not on the line

Slope formula can be used if you pick two points on the line

[coarse sequoia]

So I have to invert my formula?

[karmic imp]

No

[coarse sequoia]

The only one that makes sense is toni's explanation

[karmic imp]

You were trying to use the slope formula, correct?

To apply it properly, you must pick two points that are on the line

Can you identify two points that are on the line?

[rigid bolt]

such as 0,0 and 4,5

oop sorry

[coarse sequoia]

So, two points on the line are.....0,0 and 4,5 possibly

I just didn't understand this at all

[karmic imp]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:linear-equations-graphs/x2f8bb11595b61c86:slope/v/slope-of-a-line

[karmic imp]

That's the method you were trying here

[coarse sequoia]

Now I don't get it

[karmic imp]

https://www.youtube.com/watch?v=UdrUmuLHWvo&ab_channel=Mario'sMathTutoring

[coarse sequoia]

We rose five points to meet the line because it starts on 0, -4, then we run to -4,1

Wait I'm an idiot

[karmic imp]

Slope is in the form of rise over run, between two points that are on the line, you count up then over, then that's the slope, it's a number or fraction

You put a comma

[coarse sequoia]

Ahh I get why

Cuz it was having us do a lot of these problems as commas earlier

Thank you for your help!

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[vast shale]

$$ \int (x-2) \sqrt{25x^2-64x-64} dx$$

[twin meteor]

Brandon H

[vast shale]

$$ \int \sqrt{25x^4-64x^3-64x^2} dx -2 \int \sqrt{25x^2-64x-64} dx$$

[twin meteor]

Brandon H

[vast shale]

Hard

$$\int \sqrt{25x^2-64x-64} dx$$
$$= \frac{2}{3}\abs{(25x^2-64x-64)}\sqrt{25x^2-64x-64}(50x-64) + C$$

Is this correct?

[dull maple]

huh?How ?

[dull maple]

I don't think so.

[vast shale]

oh

yeah

is this one correct?

[dull maple]

Once again, how are you doing that?

That does not seem right.

[twin meteor]

Brandon H

[vast shale]

Reverse power rule

simplify the power

chain rule

[dull maple]

You can't just integrate it like $\sqrt t$

[twin meteor]

Enemagneto

[dull maple]

You need to do a proper substitution first.

If you wanna do that.

[vast shale]

$$\int \sqrt{25x^2-64x-64} dx$$
$$\frac{2}{3} \sqrt{(25x^2-64x-64)^3}$$
$$\frac{2}{3} \sqrt{(25x^2-64x-64)^2(25x^2-64x-64)}$$
$$\frac{2}{3} \abs{(25x^2-64x-64)}\sqrt{(25x^2-64x-64)}$$
$$\frac{2}{3} \abs{(25x^2-64x-64)}\sqrt{(25x^2-64x-64)}(50x-64) +C$$

[twin meteor]

Brandon H

[vast shale]

why?

[dull maple]

That's wrong. 🥲

[dull maple]

Because $\int \sqrt t \cdot dt = \frac {t^{\frac {3}{2}}}{\frac{3}{2}} + c$

[twin meteor]

Enemagneto

[dull maple]

It's only possible when term inside root and variable of integration are same.

your answer will be correct if your integration was like:

$$\int \sqrt{25x^2-64x-64} \cdot \dd{(25x^2-64x-64)}$$

[twin meteor]

Enemagneto

[dull maple]

@vast shale

[vast shale]

I do not understand

maybe

I forgot too much

ooooh

I'm an idiot lol

I understand what i did wrong

I used the chain rule to integrate???? makes no sense

[dull maple]

Okay. Try to remember your power rule. It goes like this:

$$\int x^{n} \cdot \dd{x} = \frac {x^{n+1}}{n+1} + C $$

[twin meteor]

Enemagneto

[vast shale]

I haven't done integration in a long time

[dull maple]

Yes.

Did you confuse it with differentiation ? 😶

[vast shale]

maybe

[dull maple]

Sort of mixed up. Lol. Used half rules of that and power rule of integration.

Anyway

[dull maple]

here, it's important that x in x^n and x in dx be same.

If it's anything different, that rule can't be directly applied.

[vast shale]

I see

[dull maple]

For example,

$$\int (2x)^{n} \cdot \dd{x} $$

[twin meteor]

Enemagneto

[dull maple]

Here, you can't use power rule directly.

[vast shale]

I needed to do some u-sub

[dull maple]

Because base of integrand is 2x while variable of integration is x.

[dull maple]

Yes

[vast shale]

$$u=2x$$
$$du=2dx$$
$$\frac{du}{2}=dx$$

[twin meteor]

Brandon H

[vast shale]

I got it now

thank you

[dull maple]

Yes

[vocal sleet]

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[brisk flax]

4 √7 ÷ 1/2

[brisk flax]

I’m not sure how to do this

[river kettle]

Flip the fraction

[brisk flax]

What

[river kettle]

Whenever you are dividing by a fraction like that you can multiply by the reciprocal of that fraction

What happens when you flip 1/2

@brisk flax

[twin meteor]

MellowDramaLlama

[river kettle]

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@brisk flax Has your question been resolved?

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[lament arch]

I have three fractions

[lament arch]

x - 2

x

x^2-2x + x

Are the denominators

How do I figure out the LCD

[outer warren]

factoring would be helpful

[lament arch]

Not possible

Or what do you mean

[outer warren]

did you make any attempt to factor the
x^2 - 2x

[lament arch]

No

But

[outer warren]

try factoring that

[lament arch]

Is multiplying x^2 -2 + x

Would it still work

x(x - 2)

[outer warren]

where's x^2 + x-2 coming from

[lament arch]

Its a common denominator

Eh

Wait fixed

Now?

[outer warren]

you edited to

x^2-2x + x
?

can you double check you have the correct expression
the presence of that in this type of problem is questionable

[lament arch]

Yes thats wrong a mistake

its x^2 -2x

[outer warren]

so what you initially posted?

[lament arch]

Ye

[outer warren]

so factorise that

[vocal sleet]

@lament arch Has your question been resolved?

[vocal sleet]

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[marsh coyote]

Hello, its possible to solve a simultaneous equation of the form:

$axy+bx+cy=d$ and $a'xy+b'x+c'y=d'$ right?

[twin meteor]

Iusgnol

[marsh coyote]

i multiply the left one by a' and right one by a

then i subtract the two equations to get a 3rd one without an xy

then i substitute 3rd equation back either of the original two

[hoary cypress]

yes think it m8 be possible

i shalll try

[mild flower]

to be clear, abcd and a'b'c'd' are constants but x and y are variables?

[marsh coyote]

oh yes

[hoary cypress]

yes

[mild flower]

yeah it'll be possible, although you're not guaranteed to get a solution in all cases

and you might get more than one, up to 4 probably

[hoary cypress]

it shud be possible to find x and y in terms of a b c d in this case

[mild flower]

starting by clearing the xy term is a good idea

[marsh coyote]

Its because after subbing 3rd equation into the original, I get a polynomial that might or might not have real roots right?

[mild flower]

yeah you'll get a quadratic i think? so you'll have up to two solutions

[marsh coyote]

oh yes meant quadratic*

[mild flower]

for one of the variables, each of which might give you two solutions for the other

you'll always get 4 solutions if you include complex numbers i think probably maybe

[marsh coyote]

just to confirm, wouldn't subbing this solution back into the original just give me 1 solution each

[mild flower]

oh right you won't get a quadratic there

maybe it's only two solutions

[marsh coyote]

ah okay makes sense i understand what you mean

thanks for the clarifcation thats all

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[vast shale]

.close

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[mild herald]

help

i tried bringing sinx inside root and writing cos2x in terms of x but like

i dont know what to do after

[sudden compass]

Can u show what u have gotten?

[mild herald]

i have zero idea what to do next

[scenic ravine]

perhaps re-write cos2x in terms of sin

[sudden compass]

Use a different identity for cos 2x

[outer warren]

<@&268886789983436800>

[vocal sleet]

@mild herald Has your question been resolved?

[mild herald]

wait

[mild herald]

1-2sin^2x

then divided by sin^2x

like that?

ill get root cosec^2x-2

not sure what to do there

[scenic ravine]

re-write that as $\sqrt{cot^2{x}-1}$

[twin meteor]

physicsrocks

[mild herald]

right

i dont see how that helps 😭

[scenic ravine]

now what is the differential of cot(x)?

[scenic ravine]

try a u -sub

[vocal sleet]

@mild herald Has your question been resolved?

[mild herald]

-cosec^2x

[mild herald]

ill try

one sec

[mild herald]

any way i can rid of the square root?

[scenic ravine]

$cotx=sec(\theta)$

[twin meteor]

physicsrocks

[mild herald]

that gives tantheta dx

but how would i get dtheta

[vocal sleet]

@mild herald Has your question been resolved?

[vocal sleet]

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[tidal trail]

I uhhh
Did
0 = t^2 - 4t + 4

(t - 2)^2 = 0
t = 2

[gaunt isle]

Make sense I guess

[scenic ravine]

looks right

[tidal trail]

This doesn't look right

[scenic ravine]

why are you finding the limit?

trying to differentiate?

[tidal trail]

Ye

[scenic ravine]

don't you know the rules for differentiation?

[tidal trail]

13 has 2 parts
i) Find time
ii) Find speed

[tidal trail]

Hahaha... which one did I fk up again?

[scenic ravine]

why the limit, just differentiate

so v(t) =25t-1000

[tidal trail]

Cuz we hvn't learnt it yet

"not in syllabus" 🥲

[scenic ravine]

two mistakes

you haven't cancled the h

[tidal trail]

Even if, it's 25h = 0

[scenic ravine]

how

you don't equate the limit to 0

[tidal trail]

Approaching 0

[scenic ravine]

that doesn't mean you equate it to 0

[tidal trail]

Ehh

No shot it's h = .01 ?

[scenic ravine]

no,no. When you differentiate you don't equate it to 0

heck you never do tha

[tidal trail]

But I've always been doing this tho

[scenic ravine]

how did you manage to get the correct answers then

[tidal trail]

No idea

[scenic ravine]

instead , find the derivative of each part seperately as follows

can I work out an example for you?

just to explain the process

[tidal trail]

Sure

One of the h->0 ans

[scenic ravine]

ok,i'll differentiate 100t(as its the easiest)

this is the definition of the differential at any arbitrary x :-$\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}$.

[twin meteor]

physicsrocks

[tidal trail]

Ye

[scenic ravine]

so the differential from first principals would be $\lim\limits_{h\to 0} \frac{100(x+h)-100x}{h}$. , which gives a differnetial of 100, not 0

see?

[twin meteor]

physicsrocks

[tidal trail]

Eh

No I'm lost

[scenic ravine]

should I do the same for $25t^2$, would that help?

[twin meteor]

physicsrocks

[tidal trail]

50t

Ye shld help?

[scenic ravine]

yeah, but you got 0

somehow

I'm trying to understand why, so that I can help

[tidal trail]

Like

Ik how to do the instant one

Idk where I fk up the first rule

The long one

[scenic ravine]

wdym long one?

[tidal trail]

$\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}$

[scenic ravine]

yeah, what's wrong here

[tidal trail]

How is it not h=0 at the last step

[tidal trail]

Liek this

[scenic ravine]

assume h were not tending to zero at all, but instead that this were an algebric expression, what would your answer be?

[tidal trail]

25h

[scenic ravine]

no you have to square it

[tidal trail]

? :D

[twin meteor]

Kai

[scenic ravine]

yes

so for 25t^2

$f(x+h)= 25(x+h)^2$

[twin meteor]

physicsrocks

[tidal trail]

Y do I need to square it again?

Wait is this even correct?

[scenic ravine]

sort of, what happened to the 100h and 100?

[tidal trail]

25(4h) = 100 h
25 (4) = 100
-100 (2) = -200

Cancelled out everything

[scenic ravine]

that's wrong

they don't just cancel out

the derivative would be calculated like so:- $\lim\limits_{h\to 0} \frac{25(x+h)^2-25x^2}{h}= 25 \frac{2xh+h^2}{h}$.

[tidal trail]

Is the x=2?

[scenic ravine]

no, I'm finding the derivative a an arbitrary point

[twin meteor]

physicsrocks

[scenic ravine]

now ,when you simplify this , you get 50x+25h. And it's here, you substitute h=0

[tidal trail]

I m very sry but I still don't get this part

[scenic ravine]

I'm applying the definition of the derivative

[tidal trail]

Uhh then y can't simplify f(2+h) ?

[scenic ravine]

because the derivative is defined as the slope, you first find their difference

techicnally the slope at any point I should say

[tidal trail]

$\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}$

[twin meteor]

Kai

[tidal trail]

Cuz I let x = 2

[scenic ravine]

ok, sure

[tidal trail]

Then it becomes

$\lim\limits_{h\to 0} \frac{f(2+h)-f(2)}{h}$

[twin meteor]

Kai

[scenic ravine]

yes

[tidal trail]

f(2) = 0

f(2+h) = 25h^2

$\lim\limits_{h\to 0} \frac{25h^2-0}{h}$

[twin meteor]

Kai

[tidal trail]

$\lim\limits_{h\to 0}25h$

[twin meteor]

Kai

[scenic ravine]

you're only differentiating teh first part of the function here

the entire function is 0 at 2 yes

25x^2 is 100 at 2

[tidal trail]

Wait m I not doing everything?

[scenic ravine]

I'm just explaining teh derivative of the 25x^2

part

[tidal trail]

Is it this one?

[scenic ravine]

sort of , yes. But again, when you're finding the derivative, you can't directly susbtitute the point

[tidal trail]

I can't directly sub?

[scenic ravine]

no

as for why , give me a few minutes, I'm trying to find a resource that can explain it better than I can

[tidal trail]

Aight

Dem this is new... ok

[scenic ravine]

Ok, I can't find any explaination online, sorry?

[tidal trail]

Hahaha

[scenic ravine]

<@&286206848099549185> , can someone help here? I'm unable to give an explaination

[tidal trail]

I m hell to teach

[scenic ravine]

I've pinged people who have given themselves this role, someone will hopefully respond

[naive sable]

Polynomials are differetiable functions thus lim_x-->a P(x) = P(a)

Ahh wait we are trying to find the derivative of h(t) = 25t^2 - 100t + 100 using the definition of derivatives ?

[scenic ravine]

yeah

at 2

I'm trying to explain why we can't directly substitute while finding the derivative, but I'm unable to come up with anything

[naive sable]

lim_h --> 0 ( f(2+h) - f(2) / h )

[dull maple]

What's the issue? speed at t=2 is indeed 0. Isn't it?

[meager hearth]

What's the ques

[naive sable]

We can substitute t= 2 , v(2) = lim_h-->0 ( f(2+h) - f(2) / h )

[tidal trail]

Yes

[dull maple]

How are you getting this?
25t^2 - 100t + 100 differentiates to 50t - 100. Right?

[tidal trail]

Ques in pins
A little further down is the working I currently hv

[scenic ravine]

yeah, calculation eror

[dull maple]

This is indeed correct. What's the issue ?

[tidal trail]

Correct?

[dull maple]

It's not ?

[meager hearth]

Ig u should have used 2-h

But still 2+h proves right

[tidal trail]

I mean that's what I got following the formula

[dull maple]

It seems correct to me. Speed is indeed 0 at t=2.

[meager hearth]

It is correct it proves the same meaning at t =2 it touches the planet

[naive sable]

Both 2+h and 2-h works here because the limit exists at that point

[tidal trail]

What abt the lim= 0 part?

[meager hearth]

Yes

[dull maple]

-f(2) in numerator*

[scenic ravine]

ok, yeah, my bad. I'm terribly sorry, I was so focused on the equation for speed I didn't realise you wanted its value at 2 @tidal trail . I'm very sorry for wasting your time

[dull maple]

Limit is equal to zero. What's the issue with that ?

[tidal trail]

No worries 🤣

[naive sable]

Yeah typo

[tidal trail]

With what speed does it land if lim = 0?

[dull maple]

zero.

[tidal trail]

Ah.

[dull maple]

It's almost like it stops.

Like putting brakes.

[tidal trail]

It's still so weird

[dull maple]

It must be some weird force field. Lol

[tidal trail]

So I'm not tripping right?

[meager hearth]

Gravitation man

[meager hearth]

There is no applied speed instead of that constant one other wise only force is acting here only

[dull maple]

Indeed. Falling object having terminal velocity zero. So, this clearly is not only gravitation. Some extra forces as well.

[tidal trail]

Lmao this is what will happen if u give a sci student a fked up maf ques

I do be questioning my prior knowledge

[meager hearth]

Just imagine bruh , when rockets wants to land there it should apply negative gravitation force to make it stable

At landing

Where velocity comes to be zero

[dull maple]

Yeah. So extra forces still. Lol. My point was that it's not free fall.

[meager hearth]

Yeah

[meager hearth]

Btw it is freefall ig ,

Velocity at end , should be 100m/s

[tidal trail]

And this is...?

[meager hearth]

Free fall

Velocity with which rocket strike

Planet

[tidal trail]

Aight

Thanks a bunch guys

Rly appreciate it :)

[meager hearth]

Welcome

[tidal trail]

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[astral pebble]

Idk where to start

I am stuck on both

<@&286206848099549185>

[rotund junco]

For the second you just have to find the area under the function on the interval -3,6

[astral pebble]

I forget how

lol

[rotund junco]

Ok do it together ok?

The area under the curve is the area between the function and the x-axis

Lets look at each piece separetly

[astral pebble]

-3 to -1 area is -1.5 right?

[rotund junco]

Yes

[astral pebble]

area -1 to 2 is 1.5

so now we are at 0

is the area 2pi3^2

[rotund junco]

No

[astral pebble]

?

0 to 2 is 1

[rotund junco]

Oh wait mb

[astral pebble]

-1 to 0 is 0.5

[rotund junco]

Yes it is 1.5

What did you get for the semi circel?

[astral pebble]

oh shoot

its only half a circle

(pi*r^2)/2 right?

r=3

[rotund junco]

r=2

[astral pebble]

huh?

oh

shoot

yeah

[rotund junco]

So the final area..

[astral pebble]

so its justt 2pi?

[rotund junco]

Yep

[astral pebble]

ok

for this one you subtract the area on the bottom from the area on top right?

so for the first its f(x)-g(x)

and the second is g(x)-f(x)

[rotund junco]

Yes

[astral pebble]

ok

how do I do this one?

[rotund junco]

No clue at first

But im thinking

[astral pebble]

here is what chat gpt said to try

but I did that and got 2 that could be overestimates

a and d

wait

I am dumb

its b

The function -x^2 is a downward-opening parabola, which means it is decreasing over the interval [2, 5]. Hence, the left-hand sum approximation will overestimate the integral.

[rotund junco]

Oh yeah its b youre right

[scenic ravine]

I wouldn't trust chat GPT. It isn't built for math

[astral pebble]

it gives good guidance sometimes

@rotund junco can you help me with this one?

the base functions are y=x on -2<=x<=1 and y=-x+2 on 1<=x<=2

<@&286206848099549185>

[heady laurel]

Find it's function

Get integral

Plug it in

And Ur good to go

[astral pebble]

I found 2 functions for it

but dont I need to take the integrals bounded by its domain

I get -1 when i solve both and add them together

the integral for the first is -1.5

and the second is 0.5

[heady laurel]

It's an absolute value function flipped and moved to the right

What other functions u found ?

[astral pebble]

huh?

I am so confused

[astral pebble]

none?

[heady laurel]

Find the integral for that

Change the +c to something that gives u 8 at -1

Then get the answer by pluging In 2

[astral pebble]

from 0 to 2=1

[heady laurel]

?

[astral pebble]

wait

I am dumb

this right?

for our integral

[heady laurel]

Yes

Now plug in -1 and add

Wtv to get 8

After that u have the full F(X) equation

[astral pebble]

-5/2

[heady laurel]

And u just plug in

[astral pebble]

huh

but if I plug in 2

wait

[heady laurel]

No wait

Not

It's 0.4

0.5*

-1.5+1

=-.5

[astral pebble]

but if I plug in 2 everything in the absolute value area cancels

[heady laurel]

Man nvm iam stupid

I forgot was -1

Not 1

It's -2.⁵

Now to get 8 u just add 10.5

So Ur new equation is this +10.5

[heady laurel]

.

So plug in that

[astral pebble]

I am so lost man

[heady laurel]

Ok are we good til this ?

[astral pebble]

idek anymore

[heady laurel]

U know how we got the equation right ?

[astral pebble]

I am still confused what the final equation is

[heady laurel]

No mean the integral one

Not the final

[heady laurel]

This

[astral pebble]

I am so confused

I just gotta move on from this

its multiple choice

[heady laurel]

U want the answer?

Iam afraid I can't give u that kind of information

[astral pebble]

I am so confused man

so is it +10.5x

or what

because when I plug in 2, everything insides the absolute value area cancels

and you get -2

wait

is it just 8.5

10.5-2

[heady laurel]

Yessss

Finally

So it's 8.5

[astral pebble]

bet ty

there is one more I am stuck on

@heady laurel

I know the answer to b

but idk a, c, d

since F(7)=0 doesnt that mean it was a critical point on the og graph

nvm

I got it

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[vast shale]

How did the 2cos^2C become -CosC

[errant briar]

-2cosc has been factored out

[vast shale]

Thanks

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[trim walrus]

While proving using induction, can't you take n=0 or any other number...?

[trim walrus]

For base case

[hard rose]

Use n=1

[trim walrus]

Should it always be n=1?

[hard rose]

Yeah it's easiest to be consistent

[river minnow]

The base case doesn't really matter

[hard rose]

But n can be anything based on the given constraints

[river minnow]

If you pick n = 0 as the base case, you will be proving the statement for all integers >= 0

If you pick it to be -3, then the statement will be proven for integers >= -3

[trim walrus]

That's what I want

[hard atlas]

the base case is the smallest number for which you have to prove the claim

[trim walrus]

I get it...

[hard atlas]

if you have to prove the claim for all numbers n>=0, then base case is n=0. if you have to prove it for all n>=17, then base case is n=17

[trim walrus]

Alr alr I get it

Thanks mates

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[still anvil]

4
| g(x)dx
-1

is the answer 1.5?

having trouble bc x=4 isnt shown on g(x)

so i assumed the area of
4
| g(x)dx is -.5 and added that to
3

the area from 1->3 which is 2

[halcyon compass]

i think you can assume and extrapolate value of 4 since it's a straight line

[still anvil]

yeah

[halcyon compass]

although funny how they just cut it there 💀

[still anvil]

fr lmao

[halcyon compass]

so you need to find the integral integrating or by geometric definition?

cause you can just find the are of the rectangles and triangles and you are done

[still anvil]

yeah that’s what i did

[halcyon compass]

if your calulations are correct than you are good

keep in mind for y<0 it's a negative area

[still anvil]

yeah

[halcyon compass]

aight than you gucci

[still anvil]

alr i think i got it it was just the cutoff throwing me off lol

yu

ty

[halcyon compass]

xdd

[still anvil]

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[trim walrus]

I've a recurrence question...

[trim walrus]

I want to prove it using induction...

[sudden compass]

i would subsitute n=2^k * y and see where that gets (y,k are integers)

if by induction

[trim walrus]

First I'll like to solve it...

[sudden compass]

#❓how-to-get-help

[trim walrus]

I've tried using induction but I'm unsure

I think this is correct...

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[obtuse radish]

how do i make sense of this?

I have an operator but nothing to operate in on...

my end goal is integrating v1 i think

[vocal sleet]

@obtuse radish Has your question been resolved?

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@obtuse radish Has your question been resolved?

[cyan talon]

not sure how you got that equation

$$m\dv{v}{t} = 500 \cdot 1300$$

[twin meteor]

aPlatypus

[cyan talon]

that's what you should have @obtuse radish

it's a standard physics thing, you can search "rocket motion differential equation" you'll find a ton of resources on your favorite search engine

ping me if you reply

[obtuse radish]

umm

[cyan talon]

ah

re

[obtuse radish]

v is refering to/

?

[cyan talon]

the speed of the rocket, m its mass (fuel included)

[obtuse radish]

so thats supposed to replace my md/dt equation?

[cyan talon]

yes

[obtuse radish]

could u point out why mine is wrong?

it seems to me like thats what the question is literally saying just in maths language

[vocal sleet]

@obtuse radish Has your question been resolved?

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@obtuse radish Has your question been resolved?

[wraith venture]

Differentiating the conservation of momentum is a good idea (it's basically the 2nd law of mechanics), but you need to remember that the acceleration is changing (at least until you proved it isn't, which I hope you won't).
So you should have a term in dm/dt, and a term in dv/dt.

Then you get dv/dt and you can integrate that to find v(t)

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[lost yarrow]

.reopen

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[jade horizon]

Hello, could you help me to find convex and concave? Thank you

[civic otter]

You have to study A''(x) ≥ 0, do you agree?

[jade horizon]

Yes I put what’s inside equals to zero right

[civic otter]

You should study also the change of sign if you want to see concavity

[jade horizon]

I don’t know how

[civic otter]

You don't know how to solve that ≥ 0?😬

[jade horizon]

I tried but failed.

[civic otter]

So show me what you tried

Maybe there was just a wrong sign

[jade horizon]

There cannot be two plus like this

[civic otter]

Well, do you know how to draw that parabola y = -x² - 2x + 4 ?

[river minnow]

Why are you plotting -1 and 5 though, they aren't the roots

[jade horizon]

I should be

[jade horizon]

How do I get the roots from that

[civic otter]

So draw it and see when it is ≥ 0 and when negative

[sage wind]

or you can get the roots then do table of sign

get the roots by quadratic formula

[jade horizon]

This maybe is correct

[jade horizon]

Is it by doing what I just did

[sage wind]

Do you know the quadratic formula ?

[jade horizon]

This one I should use?

[sage wind]

Yes

Ok from here do you know what is the table of sign of second degree polynomial ?

[jade horizon]

Could this be correct?

[jade horizon]

Mm I don’t.

Oh I have mistake at second root, it should be -2

So it should be 3,2 as well

[sage wind]

Keep the roots in terms of square roots

$-\sqrt(5)-1 and \sqrt(5)-1$

[twin meteor]

calculus is fun

[sage wind]

These are the roots

[jade horizon]

Oh okay

[sage wind]

Now the sign of the second degree polynomial is the same as the sign of the leading coefficient for x<x_1 and x>x_2 and opposite to the sign of the leading coefficient for x_1<x<x_2 where x_1<x_2 and x_1,x_2 are the roots of the polynomial

Use this to get the sign of this polynomial over $\mathbb R$

[twin meteor]

calculus is fun

[jade horizon]

Thank you.., I will study it

[sage wind]

Np

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[vast shale]

How do i solve this problem without guessing

[vast shale]

I know that

$$h'(x) = \frac{1}{f'(h^{-1}(x))}$$

which is

[patent nymph]

Notice that h(1) = 0

[twin meteor]

Brandon H

[vast shale]

What if i don't notice that

[patent nymph]

You could graph it

[vast shale]

no

[patent nymph]

at least part of it

substitute a few values in and see what you get for h(x)

[vast shale]

there has to be a better way

graphing takes too much time

gotta go fast

[astral shadow]

This is the way to go.

[patent nymph]

if you try substituting values like -1,0,1 in, you’ll be able to find that h(1) = 0

[vast shale]

what if it's not that easy

[patent nymph]

Then apply this

[vast shale]

what if h(5123.312) = 0

[astral shadow]

They don't expect you to solve it then.

Or you'll get a calc.

[patent nymph]

usually is, because finding the inverse of the value should be easy for you to be able to use that inverse derivative formula

[vast shale]

😞 okay

I have exam tomorrow

I no take chances

TY

for

help

[patent nymph]

If you encounter any problems of this type, try substituting -1,0,1

[vast shale]

Will do

[patent nymph]

and other small integer values

then apply the inverse derivative thing

[vast shale]

mhmmm

ty

I go back to solving problems

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[vast shale]

.reopen

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✅

[vast shale]

Am i just supposed to use L'hopital rule for this?

I can't use squeeze theorem right?

is there supposed to be another way to solve this?

[patent nymph]

Use the limit definition of e

[vast shale]

i forgor

[patent nymph]

do you remember now?

[vast shale]

Yeah idk

I just use L'hopital for these problems but idk if that'll keep working

[astral shadow]

It'll keep working for 0/0

[patent nymph]

You can put this into the problem

[vast shale]

$$\lim_{x \to 0} [\lim_{n \to \infty}(1+\frac{1}{n})^{xn} -1] x^{-1}$$

[patent nymph]

Still need to account for e^x

[twin meteor]

Brandon H

[patent nymph]

You can make the substitution t = e^x - 1 here

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@vast shale Has your question been resolved?

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[elfin moon]

This should be -1/3

[paper depot]

#1 or #2?

i guess you mean #1

why should the answer be negative 1/3?

[elfin moon]

A^3+b^3+c^3 =-3abc ? When a+b+c =0

[paper depot]

,w factorize a^3 + b^3 + c^3 + 3abc

[twin meteor]

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[paper depot]

,w factorize a^3 + b^3 + c^3 - 3abc

[twin meteor]

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[paper depot]

so it looks you're wrong about this

in fact a + b + c = 0 implies a^3 + b^3 + c^3 = **+**3abc. not -3abc as you wrote.

just to check with a simple example, taking a = 1, b = 2 and c = -3 we have:
a^3 + b^3 + c^3 = 1 + 8 - 27 = -18
3abc = 3 * 1 * 2 * (-3) = -18

[elfin moon]

ohh i was doing wrong at (a+b+c)^3 =a^3+b^3+c^3+3(a+b)(b+c)(c+a)

3(a+b)(b+c)(c+a)=-(a3+b3+c3)

Ohh this one is not used here

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[vast shale]

I need help for this

[scenic ravine]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[vast shale]

Just give me a working

and a answer

[hasty pulsar]

no

[vast shale]

Fine

[karmic imp]

We're not here to do your work for you

[vast shale]

I do not know how to do it

[vast shale]

I mean how to do the working

I will do it

[scenic ravine]

What's the formul a for volume?( This figure is a cuboid)

[vast shale]

Well

width

length

and height

Multiply all those

[hasty pulsar]

whats the formula for area of one side?

[vast shale]

One side of cubiod?

[hasty pulsar]

yes

[vast shale]

Weight and length multiply

[hasty pulsar]

so volume would be area * height, u agree?

with area=width* length

[vast shale]

Yeah

So basically

Volume is Height x the area?

[hasty pulsar]

base area yes

[vast shale]

Huh?

So it is 9x+10 X 81x^2+90x+100

[civic otter]

Use the parentheses

Otherwise it's wrong

[vast shale]

WHAT THE HELL IS PARENTHESES

[hasty pulsar]

did you mean (9x+10) * (8x²+90x+100) or literally what you have sent

[vast shale]

Which working?

[ebon rapids]

Never use x when multiplying, especially involving algebra

[civic otter]

Brackets

[hasty pulsar]

theres a big difference between the two

[karmic imp]

Brackets is another name for them

[vast shale]

Oh so

[vast shale]

this working?

[hasty pulsar]

which one do you think

[vast shale]

The one with parentheses

[civic otter]

Yep, but do you know why?

[vast shale]

Yes

But anyways

That’s a long ass working…

[hasty pulsar]

luckily you just hgave to give an expression, not solve it or whatever

though im not sure if you have to expend it or let it so

[vast shale]

is this correct

[hasty pulsar]

,w (9x+10)*(81x²+90x+100)

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by fractalogist

[vast shale]

…

[hasty pulsar]

yes

[vast shale]

Well okay

My answer is there

which I think is correct

[karmic imp]

You could have just included the word expand

,w expand (9x+10)*(81x²+90x+100)

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by dldh06

[hasty pulsar]

i thought expand would be one of the first things to show up

[vast shale]

Okay I am done

[hasty pulsar]

yes

[karmic imp]

Roots are normally the first thing that shows up with wolfram as it sees an expression with a variable so it assumes that you are solving for that unknown

[vast shale]

Btw

The questions wasn’t for me

my best friend asked me to help her

Because she was lazy

so yeah

Thanks

❤️

✌️

.close

[vocal sleet]

Channel closed

Closed by @verbal cove

Use .reopen if this was a mistake.

[karmic imp]

Then next time you should get her to do her own work

That's how people learn

By doing it on their own

[vocal sleet]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[tame smelt]

im not which one of my answer is correct, the top or bottom

[vocal sleet]

@tame smelt Has your question been resolved?

[tame smelt]

.close

[vocal sleet]

Channel closed

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Use .reopen if this was a mistake.

[vocal sleet]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hoary swift]

Where do I go from here?

Is there a way to simplify this?

[wraith venture]

I'm afraid that's not the way

[scenic ravine]

I've a suggestion, let 5^x=u

[hoary swift]

Ok

[wraith venture]

however, I'd like to hint that 5^2x = (5^x)²

[hoary swift]

I’ve got a quadratic

[wraith venture]

good

[scenic ravine]

now solve that, and before you select the answer, remember the substitution you made

out of curiosity, what is NSAA(Your document has NSAA 2019 in its name)

[hoary swift]

Cambridge natural science entrance exam

How did you know that was how you were meant to do it and nothing to do with log rule?

That’s really impressive

[scenic ravine]

I studied for JEE lol, this was something I had to do fairly often

[hoary swift]

I see

[scenic ravine]

oh, I really envy you. All the best!

[hoary swift]

What’s JEE

[scenic ravine]

An engineering entrance exam in India, considered to be one of the hardest in the world.

[hoary swift]

Oh wow

Did you pass?

[scenic ravine]

We just get a rank , most colleges in the country accept it. I technically can get a college , in some engineering fields with my rank, but not in the course I'm interested in(Maths).

[hoary swift]

Oh wow you’re very smart then?

What’re you studying now?

[wraith venture]

experience

these are also classic expressions to bring yourself back to

[hoary swift]

If I’m trying to sit questions like these do you know any exams of similar difficulty?

[wraith venture]

no I'm french

[scenic ravine]

I've seen oxford's entrance test, maybe that?

the MAT

PAT

and CAt

[hoary swift]

I see I’ll try that too thanks

[wraith venture]

the only non-olympic exam I know/took was the concours général, and it's quite different

and it's french

[hoary swift]

I see

[wraith venture]

and its curriculum is not as advanced, i think

[scenic ravine]

I's suggest JEE advanced too(this is crazy hard), but only if you really know what you're doing

[hoary swift]

Ok I’ll try it

What age do you do the JEE?

Just before university?

[scenic ravine]

after 12th, and if you want to try again, the year after 12th.(there are typically 2 attempts a year for the easier version of the exam (mains) which is used for admission for most colleges , which serves as a qualifier for the harder exam (jee adv), which is used for teh elite unis.

[hoary swift]

Ah I see

Anyway thanks a lot my friend

I will see you around

.close

[vocal sleet]

Channel closed

Closed by @hoary swift

Use .reopen if this was a mistake.