help-27

[bright ridge]

you can see it by plotting

[restive river]

What math class is this for? Algebra or calculus, because there's two different ways to do this

[bright ridge]

yep you can do it without differentiation as well

as this is just a quadratic

[cyan sentinel]

its just general math, the topic is factoring simple and nonsimple trinomals

[restive river]

Yeah so differentiation is not the answer haha

[bright ridge]

do you know how to graph quadratics

[cyan sentinel]

yes

[bright ridge]

the vertex is at x = -b/2a

do you know this too?

[cyan sentinel]

no

[woven radish]

QueenOfDireWolves

[bright ridge]

made a tex mistake oops but nvm

[cyan sentinel]

oh

.close

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[frozen elm]

The only one Id disargee with is e

Both seconds and temperature are continuous values right?

It doesnt specify any kind of rounding

Ye

Or 45.94385389246312489632489632489638924638924689324896324896... fahrenheit

Theres no inherent limit on a measure of the energy an object stores in heat

Ye

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[frozen elm]

Np

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[wanton harness]

Is there a simpler way of simplifying? I find it hard for me to simplify fractions normally with numbers like 235 or 659, basically 2 or 3 digit numbers.

[devout snow]

@wanton harness Has your question been resolved?

[wanton harness]

<@&286206848099549185>

[wanton harness]

OMG NOOOOOOOOOOOOOOOOOOOOOOOO

yayyyyyyyyyyyyyyyy

I didnt need help from this server

[brave osprey]

Hey

You need help in simplifying equations?

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[wanton harness]

.reopen

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✅

[wanton harness]

Not rlly thanks though.

.close

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[worn herald]

I drew a diagram of the a to b road

But got confused after that

Could I please have a hint?

Not the answer tho

Thanks

[steady mango]

can someone pls say 4+1

[worn herald]

4+1?

[steady mango]

Yes

[worn herald]

5

[steady mango]

Thanks

[worn herald]

Uh

What

[steady mango]

the answer is 1 WRITE AT UR OWN RISK

[worn herald]

I'm so confused

[steady mango]

Ok

bye

[worn herald]

Bye?

[jade atlas]

giv me few min I'm abt to get it

[worn herald]

Thanks

Take your time

[jade atlas]

is it 12 ??

[worn herald]

Idk

[jade atlas]

no answers ?

[worn herald]

Yeah

[jade atlas]

oh k

[worn herald]

Can you tell me the method pls

[jade atlas]

I'll explain my method

[worn herald]

Thanks

[jade atlas]

1. A to B = 30 M and lets take speed is V
2. B to A = 36 M and speed is V+6
   ig uk how to find time
   find time from A to B + time from B to A = 4.5 hrs

solve for V

understood @worn herald ??

[worn herald]

Gimme a sec

[jade atlas]

k

[worn herald]

Don't get the ending

[jade atlas]

use time = distance by speed

from A to B and from B to A
then add them which is given to be 4.5 hrs

[worn herald]

So what is the actual calculation then?

Is it just like

66/v smth

[jade atlas]

no no
u need to solve
30/v + 36/(v+6) = 4.5

[worn herald]

Oh

I think I get it

[jade atlas]

30/V = A to B time
36/(v+6) = B to A time

[worn herald]

Yeah lemme try

Thanks sm for the help

You're rlly good lol

[jade atlas]

k tell me if u didn't get I'll send

[jade atlas]

thnx lol

btw I'm new here

[worn herald]

So should I leave the channel open then or close it?

[worn herald]

Oh cool

[jade atlas]

try it out
if u didn't get or any doubt u can ask here or even dm I'm open for anything

[worn herald]

Ok thanks

[jade atlas]

np

[worn herald]

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[neon hemlock]

<@&286206848099549185>

I need help not sure whats wrong here

[strange swan]

can you explain the question?

[neon hemlock]

Need help w interval notation

Im not sure why its marked wrong

Because the values are definitely right

[strange swan]

what't marked wrong?

are you able to represent the function as a power series?

https://tutorial.math.lamar.edu/classes/calcii/PowerSeries.aspx

this might help

[neon hemlock]

Wait a minute

Ok ig it wasnt the interval

It was the power series which was wrong

Thanks @strange swan

Honestly I am lost on how to find the 10th derivative

[strange swan]

the hint is given

sorry

[neon hemlock]

I used the maclaurin series

thats what im getting

[strange swan]

https://socratic.org/questions/how-do-you-use-a-maclaurin-series-to-find-the-derivative-of-a-function

use this link for help

find the n'th term, using the Maclaurin expansion of arctan(x)

then use the formula directly

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[willow mortar]

can someone help me w this

[willow mortar]

<@&286206848099549185>

[devout snow]

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@willow mortar Has your question been resolved?

[willow mortar]

<@&286206848099549185>

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[jade atlas]

is it -4 ?

[left axle]

ln(ab) = ln a + ln b

so ln(4e) = ln(4) + 1

then ln(4)+1-3-ln(4)-1 = 1-3-1 = -3

[jade atlas]

2ln2=ln4

[left axle]

2ln2 = ln 2^2 = ln4

ln(e^-3) = -3ln(e), and ln(e) = 1, so ln(e^-3) = -3

this is not a rule, this is wrong in almost all cases.

it should be -2 my bad

well, -2ln(2e) = -2(ln(2) + ln(e)) = -2 - ln(4)

so you do get -4 in teh end

🙂

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[gritty creek]

the hypotenuse of right triangle is 10m and one of cathets Is 8m question is an area of a triangle

[devout snow]

@gritty creek Has your question been resolved?

[gritty creek]

<@&286206848099549185>

[scarlet osprey]

did you get how the 3rd side is 6m?

[gritty creek]

oo yes yes i was just thinking that i missed something

cos 6<10+8 and etc right?

[scarlet osprey]

do you know pythagoras theorem

[gritty creek]

may be

[scarlet osprey]

lol ill tell u
pythagoras therem states that
(hypotenuse)^2 = (base)^2 + (altitude)^2

do u remember now

[gritty creek]

mmm yes

[scarlet osprey]

so we get third side by
base^2 = hypotenuse^2- altitude^2
therefore 3rd side^2 = 10^2-8^2= 36

as 36=6^2 we get 3rd side as 6m

[gritty creek]

get it now

so for now i should use a+b+c/2

?

[scarlet osprey]

to get the area?

[gritty creek]

yyes ..

[scarlet osprey]

a+b+c/2 wont get you the area

btw do u know the final answer?

do you know the formula to find the area of a triangle?

[gritty creek]

bh/2?

[scarlet osprey]

yess

ok so u have b and u have h

now whats stopping you

[gritty creek]

so 6x8/2?

[scarlet osprey]

yess correct

[devout snow]

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[gritty creek]

yeeee

thank you a lot

.close

bruh

[scarlet osprey]

np 🙂

[devout snow]

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[scarlet osprey]

.close

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[restive river]

<@&286206848099549185>

[restive river]

i need some advice for this

what is this topic called

haven't learnt it and am expected to learn it tomorrow

[timber pebble]

this looks like a lot of different topics

you mean you have a class on it tomorrow?

[restive river]

i'm doing a test to get into Cambridge university have to hand the work into tomorrow

[timber pebble]

idk what to say what are you asking exactly

this looks like a study guide

what are you confused about

[inner heron]

Are you doing an interview Assessment by any chance? @restive river

[restive river]

i am asking what the topics i need

to learn how to complete this assignment

[restive river]

pretty much to get into Cambridge you need to take a test to get in and i have to do topics like these which i don't get taught in school

[inner heron]

integration and series/sequences will be taught before you take the STEP and the STEP is taken in June so you will know the whole A-level syllabus if you plan on taking it. If you arent doing A-levels, you will be taught those topics nonetheless beforehand

these foundation papers are meant to be hard so dont stress if you cant figure them out

[devout snow]

@restive river Has your question been resolved?

[restive river]

could you please give me tips on how i can learn this since i haven't been taught this

[devout snow]

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[restive river]

Is d/dx(sec^2(x)) the same as d/dx(sec(x)) squared?

[neat marsh]

is that square around the sec or the d/dx ?

[exotic stump]

$\sec^2(x)=(\sec(x))^2$

[woven radish]

a disappointing son

[restive river]

it’s around the d/dx(sec(x))

Oh thanks

[neat marsh]

then no

$\frac{d}{dx} \sec^2 x = 2 \tan x \sec^2 x \neq (\tan x \sec x)^2 = \left(\frac{d}{dx} \sec x \right)^2$

[woven radish]

ofthegoats

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[latent ruin]

Let $a$ be a positive, constant, real number and $R$ the internal region to the circumference with equation $(x-9)^2+(y-6)^2=a^2$, and knowing that: $\int \int_R$ \left[ \frac{(x-9)^2}{81} + \frac{(y-6)^2}{81} \right]dA=\pi \cdot \alpha \cdot a^{\beta}, find $\frac{1}{\alpha}+\beta$.

[latent ruin]

Let $a$ be a positive, constant, real number and $R$ the internal region to the circumference with equation $(x-9)^2+(y-6)^2=a^2$, and knowing that: $\int \int_R$ \left[ \frac{(x-9)^2}{81} + \frac{(y-6)^2}{81} \right]dA=\pi \cdot \alpha \cdot a^{\beta}$, find $\frac{1}{\alpha}+\beta$.

[woven radish]

bortoletto
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[restive river]

Let $a$ be a positive, constant, real number and $R$ the internal region to the circumference with equation $(x-9)^2+(y-6)^2=a^2$, and knowing that: $\int \int_R$ \left[ \frac{(x-9)^2}{81} + \frac{(y-6)^2}{81} \right]dA=\pi \cdot \alpha \cdot a^{\beta}, find $\frac{1}{\alpha}+\beta$

[woven radish]

Matejp1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[latent ruin]

i have no idea on how to start this

[main gull]

$\int \int_R$ \left[ \frac{(x-9)^2}{81} in that line, you either forgot another $ or added one by accident

That's why it didn't compile correctly

[latent ruin]

Let $a$ be a positive, constant, real number and $R$ the internal region to the circumference with equation $(x-9)^2+(y-6)^2=a^2$, and knowing that: $\int \int_R \left[ \frac{(x-9)^2}{81} + \frac{(y-6)^2}{81} \right]dA=\pi \cdot \alpha \cdot a^{\beta}$, find $\frac{1}{\alpha}+\beta$

[woven radish]

bortoletto

[latent ruin]

there ya go

this is the integral i came up with

after changing to polar form

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[ruby talon]

Would a vertical line (whos equation looks something like x=3) be considered linear?

[stone locust]

why not?

[ruby talon]

Well the slope is infinity (which is just a concept) so the only other way you would write it in "slope-intercept" form is 0y = -x + b which you can simplify to x = b
And even here b is not even being represented as the y intercept

[exotic stump]

x=3 is not a function

[restive river]

Linear in what sence

[ruby talon]

exactly

[restive river]

Yeah not lineae function

[cursive smelt]

Would vertical line test and horizontal test be useful here?

[ruby talon]

im just using this for a program im making dealing with linear lines

[ruby talon]

I mean vertical line test def fails but I mean if you can still consider it a linear equation then sure

[cursive smelt]

[ruby talon]

but I think I got my answer (no)

.close

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[ruby talon]

.reopen

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✅

[ruby talon]

what you wanna say lol

ok then

.close

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[restive river]

Task: Get all x where x > 0 and where the functions f(x) = e^x and g(x) = ln(x) have the same slope. though the slope can differ +-0.001

my current approach:

f'(x) = e^x
g'(x) = 1/x
e^x = 1/x

Here (with my approach) the slopes have to be equal but it can also differ (+- 0.001) (i curerntly do not consider that) how could I solve this?

[restive river]

<@&286206848099549185>

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[languid ocean]

this works?
you might have to bring 1/x to the LHS as well

[restive river]

i dont consider the 0.001 difference

[languid ocean]

i mean

e^x - 1/x = 0

[restive river]

yes but 0.001 is possible to it doesnt have to be 0

+-0.001 and in between

I do not search values where f'(x) = g'(x) only I search values where f'(x) = g'(x) or where the max difference is 0.001

@languid ocean

[mortal haven]

so -0.001 <= e^x - 1/x <= 0.001 ???

[languid ocean]

then just find the values where the difference is -0.001 and 0.001 (IVT)

[restive river]

right

[restive river]

everything in between -0.001 and 0.001 is fine too

[languid ocean]

i mean
use intermediate value theorem
IVT

[restive river]

how is that helpful here?

if I want to solve this

-0.001 <= e^x - 1/x <= 0.001

[languid ocean]

just one quick google search would do

in this case, specifically

[restive river]

yeah still asking how does that help in any way?

[languid ocean]

if we have an interval [a,b] where f(a)=-0.001 and f(b)=0.001 that doesn't have any extrema in between
then the interval would only contain values between -0.001 and 0.001

[devout snow]

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[weak grove]

help

[exotic stump]

don't give people answers

[devout snow]

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[fluid stag]

A computer screen measures 40 cm by 65cm. An image on the computer is projected onto a whiteboard with a screen area of 7750 cm squared. Determine the length and width of the whiteboard.

[exotic stump]

have you drawn a picture?

[fluid stag]

I have not

[exotic stump]

i'd start there

[fluid stag]

Is there some sort of a formula that allows me to figure out the length and width from the surface area

[exotic stump]

have you drawn a picture?

[pseudo basin]

@fluid stag no

area alone is not enough info to find the length and width

think about it: if there was such a formula, what would you get from it for an area of 10? (1,10) or (2,5)? or maybe (0.2, 50)?

[fluid stag]

Ah

So more of a trial and error

[pseudo basin]

no.

consider this: the aspect ratios of the screen and the projected image ought to be the same.

or to put this another way, the projected image results from scaling the screen image by some factor

perhaps you could find that scale factor

[fluid stag]

Ok that makes a lot more sense I should’ve realized that info wouldn’t be there for no reason

So the scale factor is about 2.98

So how do I use the scale factor in finding out my length and width

<@&286206848099549185>

[pseudo basin]

'about' 2.98?

well, ok, that's the area scaling factor

which is equal to the square of the linear scaling factor

to recover the linear scaling factor you will need to take the square root.

[fluid stag]

I got 69.2cm and 112.45cm

But if I used the 2.980769231

I would get a more exact number

Thanks for the help @pseudo basin I really appreciate it!

[pseudo basin]

there is no such thing as "more exact"

[fluid stag]

I just used sig figs

[pseudo basin]

exactness is a binary property: either something is exact or it isn't

you are thinking of "more precise"

[fluid stag]

Fair enough

Anyways have a good rest of your day and thanks again

.close

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[fair crescent]

whats the "central angle" of a hyperbola?

[devout snow]

@fair crescent Has your question been resolved?

[devout snow]

@fair crescent Has your question been resolved?

[fair crescent]

so do i just subtract it then what?

<@&286206848099549185>

[strange swan]

@fair crescent send qn

[fair crescent]

number 3 @strange swan

the one on the pics before this one were just example questions

[strange swan]

so you must find the equation for the parabola

?

[fair crescent]

yes then i solve for all the parts of the parabola

[fair crescent]

<@&286206848099549185>

[devout snow]

@fair crescent Has your question been resolved?

[fair crescent]

<@&286206848099549185>

[devout snow]

@fair crescent Has your question been resolved?

Channel closed

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[quiet sedge]

In the rectangle below the width is x cm and the length is 5 cm greater than double the width.

how to find the perimeter of rectangle

[winter patrol]

what have you tried?

[quiet sedge]

um i dont know how to do the question

In the rectangle below the width is x cm and the length is 5 cm greater than double the width.

a. For this rectangle write an simplified expression, in x, for
i. the length in cm of the rectangle.
ii. the perimeter, P cm.
b. If the perimeter is 46 cm, find the value of x.

this the entire question

[winter patrol]

are you able to do part a)

I)

[quiet sedge]

im kinda stuck on it rn

😦

[winter patrol]

what about

the length is 5cm greater than double the width
is confusing you

[quiet sedge]

ye

like i dont know how to frind the w

width

[winter patrol]

you are given that the width is represented by the variable x

[quiet sedge]

so do u minus and devide?

yes

width = xcm

[winter patrol]

for the purposes of part i) the width in cm is x

[quiet sedge]

how much cm?

[winter patrol]

x

[quiet sedge]

ight ty and what abt part 2

[winter patrol]

did you complete part i)

[quiet sedge]

ye i put width - x cm

[winter patrol]

whut?

[quiet sedge]

did u said that?

just now

[winter patrol]

no

[quiet sedge]

oh

[winter patrol]

that's not what I said at all

[quiet sedge]

oh lol

[winter patrol]

the width is x cm
that was given in the question

[quiet sedge]

oh yes

[winter patrol]

part i)
is asking you to express the length in terms of x

and you are told that the length is 5cm greater than double the width

[quiet sedge]

ye

so like whats the length

[winter patrol]

what about

the length is 5cm greater than double the width
is confusing you

[quiet sedge]

i dobt understand if u have ti devide and minus

[winter patrol]

what if I asked you for the number that is
4 greater than the triple of 3
would you be able to do that

[fallen swallow]

that's trivial

[quiet sedge]

i dont understand lmao

[quiet sedge]

💀 how

[winter patrol]

do you know what it means to double something

[quiet sedge]

x2

[winter patrol]

apply that to the information about length

and what would you have

[fallen swallow]

Just construct 2 equations and there you go, you have a system of linear equations

[fallen swallow]

solve them and then find your answer

[quiet sedge]

ok ty you 2

[fallen swallow]

opps, sorry you deal with him @winter patrol

[quiet sedge]

khan and ram

and i have 1 mnore

more

A light-year is defined to be exactly 9 460 730 472 580 800 metres. Express the value in standard
form giving the significand to 2 significant figures.

[fallen swallow]

state it

[winter patrol]

have you successfully finished the previous question?

[quiet sedge]

i am going to work it out

i want to finish this sf question

[fallen swallow]

width = x
length = 2x + 5

[quiet sedge]

ok i have that

[fallen swallow]

just solve it and find 2(x+l)

[quiet sedge]

3x+5

[winter patrol]

are you able to do part ii)

now

[quiet sedge]

yes and ty

[winter patrol]

what's your answer to part ii)

[quiet sedge]

this

[winter patrol]

is incorrect

[quiet sedge]

= 46 cm

[winter patrol]

3x+5 is NOT the correct expression for the perimeter

recall the formula for perimeter of a rectangle and/or how perimeter is calculated

[quiet sedge]

l+w

[fallen swallow]

perimetre = 2(l+w)

[quiet sedge]

times 2

ye

so u r taking 46-5 and devide by 4

?

[winter patrol]

whut

[fallen swallow]

bruh

[quiet sedge]

💀 ye

[winter patrol]

don't skip ahead before successfully completing previous parts

applying the correct formula for perimeter, what is the perimeter of the rectangle in terms of x?

[fallen swallow]

BTW @winter patrol How did you got that honourable role?

[quiet sedge]

perimetre = length add width multiply 2

[winter patrol]

missing parentheses

[quiet sedge]

?

[winter patrol]

P = **(l + w)***2

what you wrote implied
P = l + 2w
which isn't correct

[quiet sedge]

well i meant

[quiet sedge]

.

[winter patrol]

ok. apply that to what you have

[fallen swallow]

lol

He got muted

.close

[devout snow]

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[restive river]

.open

@winter patrol hw again ;-;

[restive river]

<@&286206848099549185>

ehm anyonen please explain

[harsh harness]

For 1st exercise just multiply.

[restive river]

for example 3(4c-5) + 6(c-3)

how do i do this?

[harsh harness]

3(4c-5) + 6(c-3)
= 12c - 15 + 6c - 18.

= 6c -33.

And for 2.

Just substitute.

So m = 4 and n = -2.

[restive river]

i see in here that 3 * 4c, makes 12c so 3 * 5 makes 15 so 12c-15

[harsh harness]

2m + 3n = 2(4) + 3(-2).

[restive river]

and then another one in the same way?

[harsh harness]

= 8 + -6.

= 2.

[harsh harness]

Yes exactly.

Now where is my cookie.

[restive river]

so the answer is 2

[restive river]

after u help me to finish my hw

[harsh harness]

Yes.

Now you see how to do it?

[restive river]

lemme do all of them and then amma send them to you so that you could check!

[harsh harness]

Am I even supposed to do that.

[lunar hazel]

You can if you want since you're not giving answers, but it's not like you have to.

[harsh harness]

Yeah correct.

[restive river]

4a + 8b - 3b + 6a

in the second one

2a - 11B

[restive river]

sooo this is the answer?

to b)

[harsh harness]

Wait let me check.

[harsh harness]

+6a not -6a.

Because - * - = +.

[restive river]

alright

so it makes

10a + 11b

@harsh harness

[harsh harness]

Yeah.

No what.

Dude it is addition I am not supposed to help you in that now.

[restive river]

no like im asking if its correct

@harsh harness

[harsh harness]

So sir what is 8b - 3b?

Read again.

[restive river]

5b

[restive river]

but its 8b + 3b not -

[harsh harness]

No?

[restive river]

oh 8 is -

i though it was +

so the answer is 10a + 5b?

or it makes 10a - 5b

@harsh harness can you please explain that why it is - or + in there

[harsh harness]

Grade 5 Multiplication.

[restive river]

bruh that was along time ago ;-;

its already been like 3 years

@harsh harness

[harsh harness]

Well.

See.

• • • = -.

• • • = +.
• • • = -.

• • • = +.

[restive river]

ehm

[harsh harness]

That is it.

[restive river]

i asked for another thing

[restive river]

like look does it make 10a + 5b or it makes 10a - 5b, if it makes 10a + 5b then can you explain why?

@harsh harness

please lets do this quick i got other homeworks too

[copper mango]

You can't do addition yourself...?

[restive river]

no i mean like

why it makes + not -?

[copper mango]

That was explained

-3*-2=6

[restive river]

look 4a + 6a makes 10a, 8b-3b makes 5b but why 8b-3b not 8b+3b?

[copper mango]

Cause you have-3*1

[harsh harness]

The reason I left the chat.

[copper mango]

(-3*1)b=-3b.

[restive river]

yes

[copper mango]

So what's the problem

[restive river]

look

[restive river]

like why cant i do 8b+3b?

[copper mango]

You just said yes to my explanation of why it's-3b

refer to what I just said.

You don't have +3b

You have-3b

[restive river]

ohhhhh

lemme try to do the thing again

[copper mango]

But yeah... you should understand basic arithmetic before doing any algebra tbh

[restive river]

4a + 8b - 3b - 6a

[copper mango]

No

-3*-2 is 6

Not -6

[restive river]

ohhhhhhhh so its with ()

[copper mango]

??

[timber tapir]

what do you think -2(-3) is?

[restive river]

so its (4a + 8b) (-3b + 6a)

cause - * - makes +

sry i was afk

so the answer will be

10a - 5b?

[lunar hazel]

If this is what you think #2 is, you're incorrect.

4(a+2b) - 3(b-2a)

Is this what you're trying to expand and simplify?

[restive river]

ye

[copper mango]

• in the middle, but yes

[restive river]

where did you take that + from?

[copper mango]

From your error.

In forgetting it

[restive river]

ohhhhh

• • • = +

thats what i've been asking for the whole time Lol

[lunar hazel]

4a + 8b is correct, but then you've got parentheses indicating multiplication. Which isn't how the problem is set up. For 1b, anyway.

[restive river]

(4a + 8b) + (-3b + 6a)

so the answer will be

10a + 5b

????

yay

my brain burned but it was worth it

[copper mango]

You shouldn't be asking that... that's stuff you learn in like grade 2/3

[restive river]

i know but my memory is bad

so lemme calculate c

[restive river]

c) (12s+18v) - (5s-5v)

@copper mango

and it makes

[copper mango]

That's wrong.

[restive river]

bruhhhhhhhhh

@copper mango why???

[copper mango]

Cause it's not right

[restive river]

lemme calculate again!

[copper mango]

K.

I also don't plan on teaching day 1 algebra btw.

[restive river]

6 * 2s = 12s, 6 * 3v = 18v so (12s+18v) and then 5 * s = 5s, 5 * v = 5v, soo (12s+18v) - (5s-5v)

cause + * - makes -

@copper mango ????

[copper mango]

The minus in the middle should be a plus.

[restive river]

but why

[copper mango]

Read the initial question

Now stop pinging. I have class.

[restive river]

ok sorry

[lunar hazel]

How did you get -5s? Where did your - come from?

It's ...+5(s-v)

So, +5*+s = +5s

[restive river]

ohhhhh so i + it

so it makes

(12s+18v) + (5s-5v)

[lunar hazel]

yup

[restive river]

answer is 17s - 13v

[lunar hazel]

No.

18v - 5v = ?

[restive river]

13v

@lunar hazel it makes 13v and i wrote 13v in there too

[lunar hazel]

Just answer the question. You'll see.

It makes 13v, right?

[restive river]

yes

[lunar hazel]

So....why do you have -13v?

[restive river]

oh it must be +

i accsidently wrote -

[lunar hazel]

Yup.

[restive river]

Thanks for the help everyone

my school is like we learn 1 thing one day and then we get homework about thing that we havent learned

[restive river]

@lunar hazel can i have the answeres for d, e and f please only for them i got like 5 homeworks

[lunar hazel]

No. I don't give answers.

[restive river]

oof, i know i must do them myself i just have so many homeworks and its night time + i promised to everyone that i will always do my homeworks

[lunar hazel]

That doesn't change my answer.

[restive river]

alright

but thanks for explanation

[lunar hazel]

Good luck

[restive river]

thanks

[devout snow]

@restive river Has your question been resolved?

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[devout snow]

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[velvet notch]

What I did is that I know ak = 1/8k+9

right?

and we know k = n+1

a(n+1)=1/8(n+1)+9

Since it it asked for less than 10^-3

We set up 1/8(n+1)+9 < 10^-3

Then we have 8(n+1)+9>10^3

8n+17>1 * 10^3

(1000-17)/8 whatever that is the answser

correct?

[long fog]

It's not just for one term in sequence it's for sum

[velvet notch]

Ye that's what I did

n+1 is sum of squence

[long fog]

Is (n+1) in numerator?

We finding $\displaystyle\sum_{k=0}^{n} \frac{(-1)^k}{8k+9} +\displaystyle\sum_{k=n+1}^{\infty} \frac{(-1)^k}{8k+9} = \displaystyle\sum_{k=0}^{n} \frac{(-1)^k}{8k+9} + R_{n}$ and That $R_n < 10^{-3}$

[velvet notch]

Yup that's what I have done

[woven radish]

it's Sam

[velvet notch]

Oh WAit

there is (-1)^k

I didn't see that k

lmao

[long fog]

No but I still don't understand what you did

[velvet notch]

1^k/8k <1*10^-3

We know a=1

[woven radish]

it's Sam

[velvet notch]

Isn't that what I got previously?

[long fog]

Nope I'll tell you what to do here

[velvet notch]

You set arrow 1/8(n+1)+9 < 1 * 10^-3

[woven radish]

it's Sam

[velvet notch]

Yup you replace an+1 with the top

substitution

cancel out fraction and flip the arrow 8(n+1)+9 > 1*10^-3

We know that 8(n+1)+9 must be bigger 1000

at what n

so 123?

[long fog]

Are you doing this by partial sums

[velvet notch]

Ye

similar to sn

And comparison

Yup

[woven radish]

it's Sam

it's Sam

[velvet notch]

So I got 123 since at n it's going to be bigger than 10^3

123(8)+17 is bigger than 1000

So that should be it

right?

[long fog]

,w (991/8)-1

[woven radish]

Results provided by WolframAlpha

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

Requested by it's Sam#5028

[long fog]

123 will be right

[velvet notch]

bruh

[long fog]

How come

,w 8(124)+9

[woven radish]

Results provided by WolframAlpha

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

Requested by it's Sam#5028

[velvet notch]

Ohhhhhhhhhhh

I get it

[long fog]

So n=123

[velvet notch]

because n

it's not N+1

cuz there no 1

yuppp

at least i get to retry

[long fog]

What?

[velvet notch]

Because k=0

so there n+1

It's n+0

[long fog]

We want $a_{n+1} < 10^{-3}$

[woven radish]

it's Sam

[velvet notch]

it's just n

[long fog]

Ok n terms

[velvet notch]

ye

at least i get like 6 attempts at this

but different quesiton

[long fog]

Ok good luck on next

[velvet notch]

also

for this one

would it be ak= 1/k^2/3

right?

[long fog]

Yes

[velvet notch]

1/(k+1)^2/3 < an

an+1 < an

to see if it's converge or divergence

[long fog]

Is this same question or is this about limit

[velvet notch]

Limit

[woven radish]

it's Sam

[velvet notch]

yup

that's an

[long fog]

Just find the limit

[velvet notch]

Ohhh

that was simple

[long fog]

Yes

[velvet notch]

i thought you supposed to compare

[long fog]

No the question was entirely different

[velvet notch]

anyways ty

.close

[devout snow]

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[clear granite]

Could someone check my work for number 5? I feel like I'm doing something wrong.
I currently have this:
$\\begin{aligned}
&\int -xe^{-x^2} dx\
&=-1 \cdot e^{-x^2} - x \cdot -2e^{-x^2}\
&=-e^{x^2} + 2xe^{-x^2}\
\end{aligned}$

[woven radish]

@clear granite

[storm gyro]

Looks like you've tried to do ibp

if you have youve done it poorly

there's also a much better way of doing it, that being u-sub

[clear granite]

Ibp?

[storm gyro]

integration by parts

[clear granite]

Oh. D:

I thought I was using the product rule or something.

[storm gyro]

the product rule is a rule for derivatives so its not gonna be much help here

[clear granite]

Oh.

Wait, was there no product rule for integrals?

[storm gyro]

No. Integration by parts is kinda analagous, but its use case is much more specific than the product rule

[clear granite]

Oh, huh. D: I think I mixed up derivative stuff with integral stuff.

[storm gyro]

sounds like it lol

[clear granite]

I just realized I used the derivatives' power rule was well.

[clear granite]

With u = -x^2?

[storm gyro]

yes

[clear granite]

What would I do with the -x, though?

[storm gyro]

it would get dealt with by du

[clear granite]

-x = sqrt(u)?

Wait, no.

I think I should go re-learn that.

Thank you. :'D

.close

[devout snow]

Channel closed

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[clear granite]

.reopen

[devout snow]

✅

[clear granite]

I'm not sure I fully understand how to do u-substitution stuff. D:

Could someone help go over the steps with me? Preferably with this problem (question 5).

[graceful cosmos]

u-sub is the opposite of chain rule

Chain rule can differentiate a composition by multiplying a derivative

u-sub can, when a derivative is being multiplied, undo a differentiation on a composition

[clear granite]

Oh, right.

[graceful cosmos]

Mind you, u = x^2 works fine here

[clear granite]

I don't see something and its derivative though.

[graceful cosmos]

The negative won't matter, try to see why

[clear granite]

Wouldn't the derivative of x^2 be 2x?

,,-\frac{u'}{2} \cdot e^{-u}

Like this?

Oops, forgot the square root.

Wait no.

That.

[woven radish]

@clear granite

[graceful cosmos]

So note this gives you two algebraic equations:
u = x^2
du = 2x dx
(Where du and dx should be though of as variables being multiplied here)

And we're converting the integral of:
-xe^(-x^2) dx

We would like to get rid of all x, AND convert dx to du using our two equations

[clear granite]

Hmm

[graceful cosmos]

Remember there is a dx in the integral

[clear granite]

,,\frac{du}{dx}-\frac{u'}{2} \cdot e^{-u}

[woven radish]

@clear granite

[graceful cosmos]

Well, you can add that, sure. I did say what I meant though, that a dx exists in the integral and you should include it in the original problem

I see an x dx.
What a coincidence, one of our equations is du = 2x dx

[clear granite]

Splitting the integral?

[graceful cosmos]

That is, du/2 = x dx

[clear granite]

,,-\frac{du}{2} \cdot e^{-u} , dx

[woven radish]

@clear granite

[graceful cosmos]

Perfect except for that dx haha

[clear granite]

I thought I only replaced x. D:

Does the dx also go away?

[graceful cosmos]

du/2 = x dx

[graceful cosmos]

I know it is unintuitive to think of dx and du as "just normal variables" but that's how they act in u-sub.

[clear granite]

Oh.

,,-\frac{du}{2} \cdot e^{-u}

[woven radish]

@clear granite

[clear granite]

,,-\frac{1}{2} \cdot e^{-u}, du

[woven radish]

@clear granite

[clear granite]

Are these two the same thing?

[graceful cosmos]

Yes, and both are equal to your original integral (Except for boundary stuff but we're ignoring that atm)

[clear granite]

^

Oh, okay.

,,\frac{-u^2 \cdot -\frac{1}{2} \cdot e^{-u}}{2} + C

[woven radish]

@clear granite

[clear granite]

Or am I mixing this up with derivatives again?

D:

[clear granite]

,,-\frac{1}{2}\int e^{-u} , du

[woven radish]

@clear granite

[clear granite]

Oh. It looks a lot nicer now.

Does the negative in front of the u affect anything?

As of integrating it, I mean.

Another u-sub?

v = -u?

v' = -1.

,,\frac{1}{2} \int e^{v} , dv

[woven radish]

@clear granite

[clear granite]

Integral of e^v dv is still e^v + C, right?

,,\frac{1}{2} \cdot (e^v + C)

[woven radish]

@clear granite

[clear granite]

,,\frac{1}{2} \cdot (e^{-u} + C)

[woven radish]

@clear granite

[clear granite]

,,\frac{1}{2} \cdot (e^{-x^2} + C)

[woven radish]

@clear granite

[clear granite]

,,\frac{e^{-x^2}}{2} + C

[woven radish]

@clear granite

[clear granite]

Or should I put $C_1$?

[woven radish]

@clear granite

[clear granite]

Could I not put just C?

,,\frac{e^{-x^2}}{2} + C_1

[woven radish]

@clear granite

[clear granite]

Ah, okay.

Thank you. :D

So the dx and du things are somewhat like they're multiplied, right?

So then you can manipulate them however you would manipulate terms that you are multiplying?

Ohhh, okay. That makes a lot more sense now.

Thank you. :D

.close

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[devout snow]

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[full gulch]

What is it called when the order of operation doesnt matter?

[full gulch]

Nvm I was able to find it

.close

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[restive river]

how would i know if the area and perimeter are preserved?

[restive river]

is it something to do with determinant?

its a linear transformation btw

I think they want you to determine the transf. Matrix T such that X' = T*X

yeah i got the matrix T

Ok then

so how can i show that its preserved?

after the transformation, that is

There are two ways: compute the area after the transformation, or use the general formula to compute the area of the figure

A' = f(X') = f(T*X) something like this

Then prove f(T*X) = A

The matrix T should disappear somehow

While you do the computations

ahhh ok ill give that a go

thanks so much!

Ur welcome

.close

[devout snow]

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[lime zodiac]

why is there a minus sign infront of this integral? i thought integral of cosx=sinx right?

[torpid cairn]

(1+cos(x))'=-sin(x)

[lime zodiac]

oh right thank you!

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[safe pendant]

How do I approach this question?

Oh i have a mean and sd

So I think Id get a number now but Ive been stuck on it since last night

Yep

Mm ok

[woven radish]

KurtDee

KurtDee

[safe pendant]

Hmmm

[woven radish]

KurtDee
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[safe pendant]

I would get p(bar xnx)/p(x)

P(x) would just be probability of getting 170?

I dont get the bar x part

[woven radish]

KurtDee

[safe pendant]

Yep

[woven radish]

KurtDee

[safe pendant]

I think so yes

Thank you

[devout snow]

@safe pendant Has your question been resolved?

[devout snow]

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[restive river]

can someone help me squeeze thereom this? idk i was staring at the unit circle for a while but it didn't help lol

[narrow plank]

https://www.sfu.ca/math-coursenotes/Math 157 Course Notes/sec_TrigLimits.html

This may help

I've never gone through this process myself, unfortunately

@restive river

[restive river]

thats acc super helpful

thanks so much!

[narrow plank]

👍

Np - close if applicable

[restive river]

.close

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[latent veldt]

would the answer be 7/4 or 7/3 ?

[restive river]

Show work

But I’m going to guess 7/3 if it is the case that one of those two is correct

[latent veldt]

thanks

.close

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[restive river]

Ok then…

[latent veldt]

i’m not sure how to work it out

but, i want to make and educated guess

I was stuck between those two guesses

[restive river]

Use substitution

Also reopen this if you don’t want it to get locked

Let u = x^3, change the bounds, substitute where appropriate, etc

[devout snow]

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[full gulch]

How can I prove that real numbers between 0-1 have a larger cardinality than the Natural numbers?

[devout snow]

@full gulch Has your question been resolved?

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[idle knot]

how do i find the inverse of (x+3)/(2x-1)

[jade atlas]

equate this to y and find x
and replace x by y

[idle knot]

I got it

Had to divide

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@idle knot Has your question been resolved?

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[limpid lantern]

Hello

[limpid lantern]

does −2×57 come out to a rational number?

i got 10/7

i think thats rational

idk tho

<@&286206848099549185> sorry to ping

[languid ocean]

you mean negative 2 times 57?

[copper mango]

what's -2*5?

[limpid lantern]

-2 times 5/7

[copper mango]

Yeah, and I asked you what -2*5 is

cause it isnt 10 as you suggest

[limpid lantern]

Im not sure, I put that in a calculator and got 10/7

[languid ocean]

no...

[copper mango]

you don't know grade 2/3 math?

[languid ocean]

do youknow how to multiply fractions?

[limpid lantern]

Guys, Its been a few years, I dont remember much

[languid ocean]

wow, grade two-thirds, certainly my favorite year in elementary school
(jk)

[languid ocean]

anyways

to multiply a number by a fraction

you multiply the numerator of the fraction by that number

[limpid lantern]

yes

[languid ocean]

so

-2 times 5/7

= (-2 times 5)/7

and what's -2 times 5?

[limpid lantern]

-10

[copper mango]

yes, so what is the actual answer... and thanks for finally answered what I asked.

[limpid lantern]

-10

oh shoot

I remember now

lol

[copper mango]

the actual answer to -2*(5/7) is what?

[limpid lantern]

-10

[copper mango]

??

are you suddenly adverse to the number 7?

[limpid lantern]

What do i do with the 7

[copper mango]

.

[limpid lantern]

oh

ok

[copper mango]

so, I ask again, what is the actual answer to -2*(5/7)?

[limpid lantern]

-1.4285714285714285714285714285714?

[copper mango]

no

I did not ask for a decimal approximation

[limpid lantern]

Fuck im not smart huh

[copper mango]

.

and given you just said -2*5 is -10... -2*(5/7)=-10/7

[limpid lantern]

OH SO THATS IT?