help-28

[craggy tapir]

kinda similar yeah, but i still dont quite know the general solution

[granite torrent]

similar? is it not the exact same

[low echo]

🙂

[craggy tapir]

is it?

i remember accompanying it with a graph

this one has no illustration given

right

[low echo]

Hm ! Then you can figure out another point from the cubic

[craggy tapir]

wait actually, only A and C is not enough

inputting both A and C only gives 2a + b = 0

[low echo]

Same eqn ? Both ?

[craggy tapir]

yeah, since its symmetrical across y-axis

[low echo]

Oh yeah

Ok

Np

You diffrenetiate it another timw

[craggy tapir]

we still havent used the given area of the region

[low echo]

And see if both a and c give minima or maxima

No need now

[craggy tapir]

ok so 12ax^2 + 2b

[low echo]

You get same value for both right again ?

[craggy tapir]

wdym

i differentiated it the second time

[low echo]

replacing x with A and C

I mean putting A , C value

It mean both are maxima or both are minima

[craggy tapir]

12a + 2b

[low echo]

Whatever it is

[craggy tapir]

but not sure if it = 0

[low echo]

inflection

[craggy tapir]

whats inflection

[low echo]

Leave it

[craggy tapir]

im not sure im following you

[low echo]

now construct g(x) first

A, C must satisfy gx

[craggy tapir]

right

[low echo]

Let g(x) = ax² + bx + c

[craggy tapir]

cant use a b c, already taken

why not d e f

[low echo]

Use other def

Yeah

[craggy tapir]

ok so dx^2 + ex + f

so

[low echo]

Yea

Satify with A and C

[craggy tapir]

$d + e + f = -\frac35$

[craggy tapir]

$d - e + f = -\frac35$

[craggy tapir]

we conclude e = 0

[low echo]

Yeah

[craggy tapir]

so its really dx^2 + f

[low echo]

Oh yes

Now see

Differentiating this quadratic f(n)

We get only maxima or minima of

This eqn

Which is the required B point

[craggy tapir]

we know that d < 0

according to my diagram

[low echo]

So x = 0

Your diagram can be reverse also

[craggy tapir]

reverse?

its already given a > 0

[low echo]

oh sry i didnt see

B(0, k) consider

[craggy tapir]

yes B(0, f)

[low echo]

Yeah

Now 4ax³ + 2bx satisfies this

[craggy tapir]

but sub B in it gives 0

wait i notice

[low echo]

F = c

[craggy tapir]

f as a variable is unnecessary, f literally = c

yeah

so

$g(x)=dx^2 + c$

[low echo]

So we can solve

The equations

Dx² + c = ax⁴ + bx²

+c

So d = ax² + b

[craggy tapir]

wait hang on, can we recap everything we've got until now

[low echo]

Yeah

[craggy tapir]

so we have:

• 2a + b = 0
• g(x) = b/2 x^2 + c
• a > 0 (therefore b < 0)
• d = a + b

again we still have not used the area = 2/5 given

[low echo]

d = ax² + b is satisfied by A, C

D = a+b

[craggy tapir]

hm, i notice something

i was able to find the general form of g(x)

https://www.desmos.com/calculator/gc9csjzjob

[low echo]

Hm

We have figured out B (0, c)

Now area will be used

To find c

[craggy tapir]

actually this makes sense

[craggy tapir]

we got 2a + b = 0, d = a + b

so therefore d = b/2

[low echo]

yes

[craggy tapir]

so now that we have expressed g(x) entirely in terms of f(x) coefficients

what now

[low echo]

Integrate along y axis

to find the area

bwtween curves

[craggy tapir]

you mean wrt x or y

[low echo]

wrt y will be hard

[craggy tapir]

yeah, so x axis

brb

[low echo]

Hm but how

[craggy tapir]

ok so i got

$\frac16b + \frac15a = -\frac25$

[craggy tapir]

i think with this and 2a + b = 0, we can solve for a and b now

so a = 3, b = -6

g(x) = -3x^2 + c

[low echo]

?

[craggy tapir]

(1, -3/5) satisfies g(x), sub this in and i find c

c = 12/5

[low echo]

Yeah

So its done

[craggy tapir]

seems to check out

this yields me 1 as the final answer

thanks!

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[kindred sundial]

what is the concept behind this question

does the original graphs negative and positive slopes affect its derivative?

[fast peak]

the slope is the derivative

[kindred sundial]

so then

at the max and min

because the slope is zero at those points

the derivative graph will have that line on x=0?

[fast peak]

you might mean the correct thing but what you said is wrong

[kindred sundial]

and then because its a negative slope at x=0

on the derivative graph

the y value

is at -1?

or like

around there

jus tsomewhere negative

[fast peak]

yes

[kindred sundial]

what did i get wrong

[fast peak]

did you mean y=0? aka the derivative has a root at that point?

[kindred sundial]

oh ya

y=0

okok

tysm

.clsoe

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[cunning hatch]

If theta is the angle between vectors a and b, find the exact value of cos theta given that: ii) a = [2 2], b = [3 -1] and iii) a = [-6 4], b = [-8 -2]

[somber niche]

How would you approach this?

[cunning hatch]

I’ve got no idea mate

Don’t think we did this in class

If you can start the process

[somber niche]

Are you familiar with vector products?

[cunning hatch]

Maybe I can continue tho

we started dot products in class

but he gave us this exercise which had this question in it

[somber niche]

How do you calculate the dot product of 2 vectors?

[cunning hatch]

isnt it abcostheta where theta is the inside angle

[somber niche]

Yes

Now, is there another way to calculate that?

[cunning hatch]

not that i know of at this point

would you like to enlighten me?

[somber niche]

There are 2 equations for it

[cunning hatch]

we didnt do the first one

let me try it with this

thanks mate

so for the top one

a.b = 2x3 + 2x-1 = 4?

[somber niche]

ye

[cunning hatch]

so therefore 4 = √8 x 2 x cos theta

so cos theta = 1/√2

so theta is 45?

bc the mod of a is √(2^2 + 2^2) which is √8

its not x 2 im a bimbus

its √10

so its 4/√80 = cos theta

which is 1/√5

is that right mate?

[somber niche]

bimbus lol

Let me recheck

[cunning hatch]

what i did there was maximum bimbus level mate

i read it as root 3 instead of 3

[somber niche]

It's correct now

[cunning hatch]

beauty

is iii's answer

[somber niche]

Uh, wait a sec actually

[cunning hatch]

cos theta = 10/√221

[somber niche]

How did you get this?

[cunning hatch]

this line here

replace the x 2 with x √10

[somber niche]

I didn't look at your process I just checked the answer

[cunning hatch]

thats it

[somber niche]

ye but how did you simplify

4/root 80

to be 1/root 5

[cunning hatch]

root 80 is root 16 x root 5

root 16 is 4

cancel the 4's

[somber niche]

ye it's good

nvm

https://onlinemschool.com/math/assistance/vector/angl/

You can check your answers with this

[cunning hatch]

thanks mate

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[mossy palm]

very simple question but I just don't understand where the minus that I underlined comes from, is this an error in the book?

[somber niche]

Talk about doublespeak

Did you try to calculate the derivative yourself?

[mossy palm]

what is that

[somber niche]

Some people don't ask questions at all

you asked it twice

[mossy palm]

ooo yes

my bad

[pearl trout]

1 - exp(- R1 * t / L)
^

it's this minus moved to the front of everything

[spiral vigil]

i enjoyed watching that caret move in real time

[mossy palm]

ooo yes i get it

ty

[pearl trout]

no problem ^^

[mossy palm]

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[torn jolt]

It says my abs max is wrong

[dull seal]

12x^2-12x-72

Continue from here

Factor out 12

[torn jolt]

I did no?

[dull seal]

I see u factored out 6 unless im reading wrong

[torn jolt]

Oh okay so your saying instead of using 6

I couldve used 12

[dull seal]

Oj u did in 2 steps

But i think i did wrong

[torn jolt]

Brains not braining

[dull seal]

It is 12(x-3)(x+2)

You have 12(x^2-x-6)

[dull seal]

Then this

[torn jolt]

Sir

I am not following

😭

[dull seal]

Is this what u wrote? I cant read very well your writting

[dull seal]

Yes that part is hard for me to read but i assume u made a mistake there

That is a +

[torn jolt]

Okay wait

Lemme see this

Okay

[dull seal]

Just as an advice

[torn jolt]

I see what you mean

[dull seal]

Always try to factor out the biggest possible

So u have 12(x^2-x-6) instead of 6(2x^2-2x-12)

You will have less mistakes

[torn jolt]

True

I was using a video and it seemed much easier

But okay so

If its x-3 and x+3

Then is it going to work still?

[dull seal]

Where it is x+3?

[torn jolt]

Wait

Lmao sorry

Okay

Ao instead

Of x-3 it would be x+3

Rught

<@&286206848099549185>

[hollow granite]

I didn't get what your question is

[torn jolt]

You see the max?

On the table i made

With max and min

The min is right

But the max is wrong

[hollow granite]

Oh you have to find max and min value of the function?

[torn jolt]

Yes

And i had a little error https://cdn.discordapp.com/attachments/903486480985489478/1237386664351891476/1237384458391257098remix-1715086368461.png?ex=663b75a2&is=663a2422&hm=7bc5dbbeb93ddfaafe107d353ae6a57275dc7ff6fea978f3cf3e751852e45334&

Here

So if i change that to +

Cuz its supposed to be +

Does that change the rest of my steps or?

[hollow granite]

I am getting maxm at -2

Minm at 3

@torn jolt

[torn jolt]

Might be cooked

[hollow granite]

Did you not understand anything?

[torn jolt]

No its not that

I factored with

6

Instead of 12

[hollow granite]

Well that won't cause any problem as well

[torn jolt]

Play wait

Okay*

Your right

I understand now

I shouldve just factored with 12

😓

Lovely handwriting btw

🙂‍↕️

[hollow granite]

Well the graph I have drawn in incorrect.

*is

[torn jolt]

Yea i dont mind that

Just

The numbers

But thank you again

[hollow granite]

Np

[torn jolt]

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[lean hornet]

What are the steps necessary to solving for g(x) if f(x) is $- \sqrt{x (2 - x)} + 1$ on the domain [0,1]

[glossy valve]

AboutCatHelper

[somber niche]

This is such an interesting problem

Is g(x) supposed to be a semicircle?

[lean hornet]

Yeah

It can be denoted by $\sqrt{r (2r - x)}$

[glossy valve]

AboutCatHelper

[lean hornet]

I personally am stumped and I'm in calc

[lean hornet]

I feel like I'm in the process of a brainfart but I wanted get an experts input

[somber niche]

Doesn't seem right

[lean hornet]

wait

no sorry its x(2r-x)

mixed up r and x on that initial variable

[somber niche]

Like this maybe?

[lean hornet]

Yeah, you can simpify it further

And also, I understand r ends up being 0.25, but I'm not sure how to prove that with f(x) initially

[somber niche]

oh I think I figured it out, it's kinda lame though

Wait nvm

Hmm... let me give this a crack

f(x) and g(x) have the same tangent at the point they touch

so maybe the slope functions will tell us something

also we know they touch there

so we have f(x) = g(x)

and f'(x) = g'(x) if I'm not mistaken

2 equations with 2 unknowns, should be solvable

[lean hornet]

Okay I'll try that

[somber niche]

Try to solve for x and r, you should get the radius and x coordinate of the touching point

I think

The solution of the system does indeed give the right answer

the system is in green

y is actually r in the system, had to write it that way for desmos

idk how to solve this system analitically though

Maybe try to rewrite in terms of like polar coordinate or something... idk

[lean hornet]

Alright thanks

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[bright glade]

Guys why did we only do limits approaching from the positive side

[solemn gate]

i dont think you need to from both sides

to fin vertical asymptote

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[stone trench]

isn't this wrong

[stone trench]

look at my work

$let\ \arcsin\left(x\right)\ =\ t
\cos\left(2t\right)\ =\ \cos\left(t\right)^{2}-\sin\left(t\right)^{2}
\sqrt{1-\sin\left(t\right)^{2}}-\sin\left(t\right)^{2}
\sqrt{1-x^{2}}-x^{2}$

first let arcsin(x) = t

[low echo]

Wrong

Formula is 1-2sin²x

[stone trench]

$\cos\left(2t\right)\ =\ \cos\left(t\right)^{2}-\sin\left(t\right)^{2}=\ \sqrt{1-\sin\left(t\right)^{2}}-\sin\left(t\right)^{2}\ =\ \sqrt{1-x^{2}}-x^{2}$

[glossy valve]

RulzerFly

[low echo]

Why put root

[stone trench]

yeah but it's 2arcsin(x)

[stone trench]

ahh yeah yeah

[low echo]

1 - 2sin²arc sinx

[stone trench]

the result is right

[low echo]

== 1 - 2x²

[stone trench]

yeah

[low echo]

Yeah result is corrsect step wrong

[stone trench]

and the root ?

[low echo]

No root

cos 2x = 1- 2 sin²x

Cos²x = 1- sin² x

So why you get root ?

[stone trench]

yeah yeah

[low echo]

You wrote sin x² in equations messed up there

[stone trench]

no i meant by sin(t)^2 all of the sin

[low echo]

Oh okk then solved ?

[stone trench]

yeah all clear thanks for help !!

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[little fractal]

Hello, can you help me with this exercise pls?
Find the general equation of the plane that passes through the points (1, -1, 1) and (1, 3, -2) and is parallel to the abscissa axis.

[little fractal]

I don't know what to do with the fact that it is parallel to the abscissa axis.

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[broken tangle]

can anyone help me understand the concept of a singular solution with the following problem? $$\frac{dy}{dx} = \frac{y+1}{x^2y^2}$$ i've got that the general solution is $\frac{y^{2}}{2}-y+\ln\left|y+1\right|+\frac{1}{x}+C=0$ and possibility of the singular solution at $x=0,-1$

[glossy valve]

horizon2.0

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[broken tangle]

<@&286206848099549185>

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[unborn vessel]

i don’t know how to go about this question at ALL (this is the polynomials unit of pre-calc 12)

<@&286206848099549185>

[ember shadow]

let x be the age of the triplets

then figure out the equation, solve for x

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@unborn vessel Has your question been resolved?

[unborn vessel]

i don’t understand how to form the equation.

[ember shadow]

if x is the triplets' age, what is Janice's age in terms of x

[unborn vessel]

i’m not sure

[unborn vessel]

i’m fully lost

how do you make the equation

<@&286206848099549185>

[polar valve]

"they were born 3 years after their sister" - if someone is - for example - 5 years old and was born 3 years after her sister, how old is the sister?

[unborn vessel]

8

[unborn vessel]

8

[unborn vessel]

Okay I get it so it’s x+3 but what about the 60687.

<@&286206848099549185>

[unborn vessel]

can someone please help

<@&286206848099549185>

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[slow merlin]

Hey 👋, I'm struggling with question. Was wondering if anyone could provide help.

[quartz gale]

It is actually

[bright cargo]

nk +k(k-1) ish right?

[quartz gale]

Not really

[quartz gale]

Try to see it as the sum of a(k)b(k) where k goes from 0 to n. a and b are function, like a(k) could equal 5k for example

[bright cargo]

k(n+1-k)

[quartz gale]

You should let svabby look for the solution instead of giving her some solutions

[bright cargo]

ôhhh shit sorry:/ mbmb

[slow merlin]

sorry, I'm a little lost what do you mean by that?

[quartz gale]

Do you know how to use sigma notation ?

$\sum$

[glossy valve]

alex

[slow merlin]

Yes, but I'm also not the best at it

[quartz gale]

Okay so try to write the sum using sigma notation : the sum of k for k in [1,n]

So that I know at where we start

[slow merlin]

Mean this or am I misunderstanding?

[quartz gale]

Exactly

Just forgot the k inside the sum

It would be 1 + 2 + … + n

[quartz gale]

When I say inside I mean at the right of the sum

[slow merlin]

My bad, trying to write math from text message is a bit hard for me but I'll try

[slow merlin]

Ahhh kk

[quartz gale]

Continue to write on paper it’s easier

Okay so do you know what is a function ?

[slow merlin]

I know clac functions

[quartz gale]

Okay

[slow merlin]

So I'm assuming the same thing

[quartz gale]

Indeed but it might be better to talk of about sequences

Forget about function

It’s my bad

[slow merlin]

All good

[quartz gale]

So do you know sequence ?

I guess yes

Since your using sigma notation

Here we are trying to find a sequence $(u_k)$ such that we have your thing equal to $\sum_{k=1}^nu_k$

[glossy valve]

alex

[slow merlin]

yup yes sir

[quartz gale]

Could you try to find the sequence ?

Look for pattern

The goal is not to find it on first try, guess and check as long as it’s not correct

If you don’t find I’ll give you a hint

[slow merlin]

I can find the sequence for basic ones. but the pattern with this one seems to go in reverse after the +...+

[quartz gale]

There is some inverse indeed

[slow merlin]

I haven't delt with inverse sigma so I'm not sure if there is any difference

[quartz gale]

See the terms as a multiplication involving two number

And look what happens for each number

[quartz gale]

Idk if it’s clear

[slow merlin]

ummm let me check the question real quick

within the brackets it seems to increase with a negative number by -1, and outside the brackets it seems to increase by +1. Not sure if that is what you meant

[quartz gale]

Indeed

That’s good

Now try to find the sequence u_k = ?

[slow merlin]

ill try

[quartz gale]

The easiest is to guess something that feels right and then try for k=1,2,3 and k=n

Any guess yet ?

[slow merlin]

one sec it's uploading

I know it ain't right, but it's a guess

Hahaha

I just know

[quartz gale]

You know what ?

[slow merlin]

That it's wrong

[quartz gale]

Yeah it ain’t right but it’s a guess so that’s always better than nothing

Okay so $u_k=a_k*b_k$

[glossy valve]

alex

[quartz gale]

So $a_k$ is the bracket before the +…+ and $b_k$ is the bracket after +…+

[glossy valve]

alex

[slow merlin]

Okay that makes more sense

[quartz gale]

So for example $a_0 = ?$

[glossy valve]

alex

[slow merlin]

could I multiply out the brackets then replace N with K

ill send a picture of what I mean

[quartz gale]

Just so I am sure we’re clear :

$\sum_{k=0}^nu_k = u_0 + u_1 + … + u_{n-1} + u_n$

[glossy valve]

alex

[slow merlin]

Regular algebra to reduce the terms, then swap in terms of k

[quartz gale]

You can’t replace the n with k

Because n it fixed

[slow merlin]

Ah k

[quartz gale]

This form is easier

[slow merlin]

👍

[quartz gale]

I’ll give you $a_k$

[glossy valve]

alex

[quartz gale]

It’s k

So you have to find $b_k$ such that you have your sum equal to :
$\sum_{k=1}^nk*b_k$

[glossy valve]

alex

[slow merlin]

could you explain how you came to the conclusion that it's just K?

for Ak

[quartz gale]

I used what you said

It’s increasing by +1 each time

So at first it’s 1 then 2 then 3 then .. then n

k does exactly that

When it goes from 1 to n

You see ?

[slow merlin]

Yeah I think I get what you mean

[quartz gale]

Okay so maybe you can find b_k

Using the fact that it increase of -1

find something that mimic “n then n-1 then … then 2 then 1” using k and n

You can do it here without sending me a picture

[slow merlin]

k, let me check the question

k * (n - 1 - k) ?? since: (n -1 - 1 ) will be (n -2), but then (n - 1 - 2) will be (n - 3) which is incorrect

[quartz gale]

Looks good

But not exactly

it’s almost that

[quartz gale]

What could you add into the bracket so that it match the result ?

[slow merlin]

can I start the index at k = 2

[quartz gale]

let’s keep it at 1 you’re almost there

[slow merlin]

k

[quartz gale]

Not really because you try it with k=1 and k=2 so removing the k you could not do that anymore

Add a number instead

[slow merlin]

k one sec

[quartz gale]

And you seemed to be happy with n-2 but remember that it is n at the beginning not n-2

[slow merlin]

any number I add doesn't seem to hit the correct result

[quartz gale]

What do you need to add to n-2 to get n ?

[slow merlin]

• 2

[quartz gale]

yes

Try that

k*(n-k+1)

[slow merlin]

wait I tried that but I thought it didn't work

[quartz gale]

k goes from 1 to n

[slow merlin]

OOHHH

[quartz gale]

Eureka

[slow merlin]

I can't believe that took so long, I'm good at calc but not linear algebra hahaha

[quartz gale]

No no it’s just practice

And it’s not easy to be fast when there is an app between smn and a helper

[slow merlin]

yeah the text sometimes aint the best

[quartz gale]

But you got there in the end

[slow merlin]

Do you know if there are some YouTube channels or practice questions that help with these kinds of questions?

[quartz gale]

I would just search on the internet I guess

[slow merlin]

fair, I'll try my hand at a few more. Thanks for your help!

[quartz gale]

This for example

http://maths.mq.edu.au/numeracy/web_mums/module4/Worksheet410/module4.pdf

[quartz gale]

You’re welcome !

[slow merlin]

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[violet minnow]

considering 1/4 is .25 this confused me

[craggy verge]

it's a mixed number because they're terrible people

5(1/4) is 5.25 -> 1.0525 multiplier

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@violet minnow Has your question been resolved?

[violet minnow]

I see...

[violet minnow]

Ok

Thanks again

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[obtuse geyser]

How do I this

[obtuse geyser]

Write the equation of a transformed exponential function whose asymptote is y = 4, and whose y-intercept is 6.

[torn jolt]

e^x

Had an asymptote at y=0

Has*

So how can you transform that asymptote up 4 units

[obtuse geyser]

Y+4

[obtuse geyser]

is that what u mean?

[torn jolt]

Yea just add 4 to the function

That’s good

Now that gives u the correct

Asymptote

But u still need to tweak something so that

It’s y intercept is 6

[obtuse geyser]

yeah

but idk how to

is 6 supposed to be the like the base

[torn jolt]

Well our new function is

e^x + 4

When we plug in x=0

We get 1 + 4

What can we do to the 1

To make it 2+4 which = 6

[obtuse geyser]

so it’s 2^x+4?

but isn’t gonna make y =5?

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[craggy tapir]

Given a complex number $z=a+bi$ $(a,b\in\mathbb{R})$ and its conjugate $\bar{z}$, whose points on the complex planes are A and D respectively. $(2+5i)z$ and its conjugate have geometrical representations on the complex plane as B and C respectively. Given ABCD is a rectangle and $|z+3-i|$ is at its minimum value, calculate ab

[craggy tapir]

so i wrote down the coordinates for all the points

$A(a, b)\B(2a-5b,5a+2b)\C(2a-5b,-5a-2b)\D(a,-b)$

[craggy tapir]

i can also find the conditions for them to create a rectangle (vector AB = vector DC)

but not sure what to do with |z+3-i| is at its minimum

[craggy tapir]

for this, 5a + b must = 0

[low echo]

ABCD rectangle

[craggy tapir]

uh huh

[low echo]

What condition u get by AB = CD

[low echo]

Let z = -3 + i

Then min value gets 0 , does this satify rectangle

[craggy tapir]

no

where did you even get that value from

[low echo]

Just equated to zero to minimize it

Ok 5a + b = 0

Then z = a -5ai

|a+3 -5ai-i|

√(a+3)² +( 5a +1)²

Ok now consider the square of this another function

And find extrema

[craggy tapir]

ok got it, thanks

brb

[low echo]

Yeah

[craggy tapir]

got my answer, -80/169

looks good

thanks!

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[low echo]

Haha

[low echo]

Welcome /!

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[restive quiver]

can someone explain that how am i getting two different answers here

[umbral dome]

there are two different numbers satisfying x^2 = 1, which are 1 and -1. we could say that either of them could be values of 1^(1/2)

of course, we usually pick 1 to be the "principal root", abut different algebraic manipulations (especially with complex numbers) can leave you with different roots. so here you found both of them

the "principal root" is -i, in this case, but both -i and i are roots

[restive quiver]

are both of these answers correct?

what if i do the +ve i one

and leave the other one

how do i find out that which one is the pricipal root

@terse terrace

<@&286206848099549185> rs

[torn jolt]

uh.....

root of -1 aint 1 homie

oh thts i mb

[restive quiver]

im getting different answers dude

idk y

[final folio]

The left one, you would take the principal sqrt first then raise to the power of 15

unless you are solving an equation where i and -i satisfy the answer

[restive quiver]

is it a rule that i hv to take the sqrt first

[restive quiver]

no its just an expression which i hv to simplify

[final folio]

then it is the left one. You have that $n^{\frac{a}{b}} = (\sqrt[b]{n})^{a}$

[glossy valve]

TomB

[final folio]

I'd say 'convention' rather than 'rule'

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[craggy tapir]

$\int\frac{\dd x}{(x+1)\sqrt{x}+x\sqrt{x+1}}$

[craggy tapir]

taking sqrt(x(x+1)) as factor in denominator

but not sure what to sub

[wide wind]

x+1

[craggy tapir]

wouldnt change anything tbh

except introducing minus

[wide wind]

$u\sqrt{u-1}+(u-1)\sqrt{u}$

[glossy valve]

Good

[wide wind]

I guess this would work.

[craggy tapir]

not sure im following, i cant see how it would change anything

[wide wind]

Could be, I don’t got a lot of time, so I did it at a glance.

Gtg.

[granite torrent]

you seem unfamiliar with integration

[granite torrent]

that's a good thing to notice

next i would recommend writing everything out

as a hint, there's a trick to this integral where once you see it it becomes extremely easy

[craggy tapir]

$\int\frac{\dd x}{\sqrt{x(x+1)}(\sqrt{x+1}+\sqrt{x})}$

[craggy tapir]

maybe usub u = sqrt(x(x+1))?

but i believe that would be messy

[granite torrent]

beyond "messy", that wouldn't even work

[craggy tapir]

hm, you got any idea?

[granite torrent]

i've essentially just said i know how to do this

[craggy tapir]

hint?

[granite torrent]

if you're asking, then consider ||how you normally rewrite expressions with roots in them||

[craggy tapir]

rationalize the denominator

oh wait so

multiply both numerator and denominator by sqrt(x(x+1))(sqrt(x+1)-sqrt(x))?

[granite torrent]

have u tried

[dull seal]

Check that

Not exactly that

[craggy tapir]

ok wait

[dull seal]

Ur denominator is (x+1)sqrtx + xsqrt(x+1)

[dull seal]

Assuming this is the original

[craggy tapir]

yes thats the original

[dull seal]

So rationalize with (x+1)sqrtx - xsqrt(x+1)

[craggy tapir]

yeah thats the same with my suggestion, and this is what i got

right so the denominator is rationalized, whats next?

[dull seal]

Is the blue thing?

[craggy tapir]

yes

[dull seal]

That looks wrong

[craggy tapir]

oh whar

[dull seal]

Numerator must be this

And denominator

X(x+1)^2 - x^2(x+1)

[narrow helm]

its the same thing, they just wrote the simplified version of it

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@craggy tapir Has your question been resolved?

[dull seal]

,w is ((x+1)sqrtx - xsqrt(x+1))/(x(x+1)^2 - x^2(x+1))=(sqrt(x(x+1))(sqrt(x+1)-sqrt(x)))/(x(x+1))?

[glossy valve]

Results provided by WolframAlpha

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

Requested by _arkimond_

[dull seal]

Im eitherway not following or not reading correctly (both options are normal because i am still in the bed xD)

[glossy valve]

Dyssrupt

Dyssrupt

[narrow helm]

lol

[dull seal]

Yep, i need stay in bed with eyes closed xd

But

No but, my brain still thinking bad things xd

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[prisma thistle]

Hi, can someone please explain to me why the answer is E. I worked it out and thought it was D.

[torn jolt]

Probably because
f(x) =x^4
If x=x/5
It will be
x^4/625

[prisma thistle]

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[real musk]

why is it 4p1 for part a and b?

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[rocky saddle]

There are 2 ways

1 is by using lines and the other is by using areas

so which one did you do?

oh alr

uuh let's do the line one then

so to do the line method, we need to use 2 points to find its gradients first

so let's use (1,3) and (-1, -1)

oh

what grade are you?

[spiral vigil]

gradient = slope

once you've plotted them it'll be pretty clear whether they make a line

[ashen crescent]

Collinear just means they lie on the same straight line

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[spiral vigil]

yeah

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[willow sedge]

methman

[torn jolt]

Need to show if this series is bounded or not

[tired sedge]

I feel that it is bounded

It smells like that

[devout valley]

(Any idea how you’d check for it being bounded?)

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@torn jolt Has your question been resolved?

[torn jolt]

no

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@torn jolt Has your question been resolved?

[static forge]

@torn jolt you can use the fact that the general term for that series is inferior to 2^(-k) for all 1 <= k <= n

[torn jolt]

Ok

What next

[static forge]

you compare the sums

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[queen abyss]

list 2 integers that are congruent to 4mod12 I don't even know where to start

[meager minnow]

i could be wrong but thats just 4 + 12k no?

so 16, 28...

also hi @queen abyss

[queen abyss]

wait but a congurent to 4mod12
so a-4=k12

ohh yeh so a=4+12k

so we replace k by any Natural number

and boom there we have it

[queen abyss]

best helper ever😍

[meager minnow]

YEAH!

[meager minnow]

thanks broski

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[torn jolt]

my question is on part (iii)

[torn jolt]

this is what i've tried. but im not sure how to justify that v must be always in the span of u1 and u2

[fast peak]

you can argue by dimension

[torn jolt]

how

[fast peak]

what dimension can the orthogonal complement of Im(P) have

show that span(u1,u2) is a subspace of the complement

[torn jolt]

no idea tbh

the solution says that i only need to show that Im(P) = span(u3). why does that work?

[fast peak]

well just general knowledge about orthogonal bases

but based on the arguments I hinted at above

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[unkempt rampart]

how do you factor by grouping

[unkempt rampart]

and by substitution

with examples

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[slender pond]

Can someone help me with 2 please

[tight glen]

!status

[full forum]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[slender pond]

This is all I did

I think I need to use sugar

But I’m confused on how to start

Or if my diagram even correct

[tight glen]

second ball was not thrown upward

It was released from rest if I’m not wrong

Yeah it was released from rest, that’s equivalent of “dropping” something

[tight glen]

I also recommend you designate signs for directions of your vectors

[slender pond]

Ok

I try redrawing them

Would this be more accurate?

[tight glen]

For part one, v is not equal to 0

[slender pond]

Yea mb

Forgot to remove that

Would this be correct,

[tight glen]

One more thing

What is the direction of your acceleration?

for first one

[slender pond]

Oh yea

They should both be negative

[tight glen]

And time for second part

[slender pond]

They then both hit the floor at T seconds?

[tight glen]

We care about for how long did the motion happen

[slender pond]

Yes i think for T

[tight glen]

For example

Say it’s 12:30 pm

And I am at rest

[slender pond]

Yea

[tight glen]

I start motion at 12:40 pm

And then stop at 1:00 pm

For how long was I in motion?

[slender pond]

20 minutes?

Or no?

[tight glen]

@slender pond

[slender pond]

Yes

[tight glen]

Oh hell sorry I think my internet was lagging, I didn’t receive your messages

[tight glen]

Yes

You are correct

[slender pond]

So I should subtract 3 from T for the second ball?

[tight glen]

It didn’t matter that I was not in motion for the first 10 minutes

[tight glen]

Yes

[slender pond]

Ah ok

[tight glen]

Once you do that can you form the equations?

Try

[slender pond]

Yea

Would this be correct

From using s = ut + 0.5 at^2

[tight glen]

Where is s?

[slender pond]

A for acceleration

So -9.8

[tight glen]

You did 1/2 (a) on RHS

[slender pond]

Mb ignore the first equation

[tight glen]

Also keep one variable, t or T

[slender pond]

Its second one I’m talking about

[tight glen]

^

[slender pond]

Yea mb for that

Im assuming after find T

I can just sub it into one the equations to find H?

[tight glen]

Yes

[slender pond]

Got T as 4.5

Seems right enough

Thank you for the help

[tight glen]

Sorry I didn’t solve for the answer

[tight glen]

Do you have any other questions?

If not, please close the channel when you are done

[slender pond]

Ok

.close

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[silver thorn]

Hello I need help with general mathematics studying and would like to get a general review over multiple topics (geometry grade 10)

[silver thorn]

Small example of what I need help in mostly circles and stuff

[tight glen]

Which one?

[silver thorn]

Can we start with #2

?

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[silver thorn]

@tight glen?

[robust slate]

The circumcentre is defined as the intersection of the perpendicular bisectors of the sides of a triangle

Recall you can find the perpendicular bisector of a segment AB as follows:

• Compute the slope of AB. Then, take the negative reciprocal of this - this is the slope of your perpendicular bisector (since perpendicular lines have slopes that multiply to -1)
• Find the midpoint of AB - The perpendicular bisector must pass through this point

This allows you to find the eq. of the perpendicular bisector of a side

do this for two sides and find their intersection

[silver thorn]

Alright

👍

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[cloud loom]

Hi, I have a question regarding to finding the eigenvalues. From lecture my prof mentioned that we shouldn't row reduce to find the eigenvalues because it changes the eigenvalue. Maybe I heard it wrong, but currently I am confuse on why is that the case because I tried finding eigenvalue here and I reached the solution.

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[kindred grove]

I'm late, but what your prof prolly meant is that you shouldn't row reduce the matrix (i.e. the original C, the one without -lambda*I) and expect the result to have the same eigenvalues as C

if you're just using row-reduction to compute det(C-lambda*I) it's completely fine

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[cloud loom]

Gotcha Tysm!

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[undone beacon]

I’m trying to find the maximum area, but am stuck when writing the equation into one variable, as my w would cancel out. I set the perimeter equation equal to 600 since that’s how much fencing we have, then I would figure out what the length and width were to find the area.

[devout valley]

What did ya do there

(Also be careful of the area of the pens )

[undone beacon]

i think i got it, i saw where i messed up and i'm fixing/working it out real quick

Okay uhhhh…. I’m stuck again

@devout valley

[devout valley]

You've found the area as a function of w

[undone beacon]

Yeah but when I tried to find the length it equaled zero

[devout valley]

You can either differentiate that with respect to w to find its maximum, or otherwise find the vertex/complete the square

Don't equate the actual area to zero (because, of course, if you had zero area )

[devout valley]

(Also the area you found is for one of those pens )

[undone beacon]

oH

[undone beacon]

how would i do thisss

(also my mom is here i will be back)

[devout valley]

For the differentiation, find where da/dw = 0 and then find what the area corresponding to what you found was

(Might get some rest )

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[hollow tide]

Hi:) I need help finding the range on this piecewise equation

[hollow tide]

my first try would give me (-5,-5/4)U[-1,1)

but that's not an option

I having a hard time understanding why it would go above 1, but the only option where it doesn't, it doesnt seem right either

I'm starting to consider the exercise is wrong?

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[hybrid zephyr]

first find whether the functions are increasing or decreasing in the given domain of x

and then find maxima and minima

that would be the range

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[twin wolf]

A particle moves along the x-axis with acceleration $\frac{d^{2}x}{dt^{2}}=3t-2$. Initially it is positioned 4 units to the right of the origin with a velocity of 2 units per second. What is the position of the particle after 5 seconds?

[glossy valve]

water beam

[twin wolf]

all i know was to get x = (t^3)/2 -t^2 + ct + k but idk wha tto do after that

[thick hedge]

you have to find both the constants

first find C

then k

[twin wolf]

how do i find c

[thick hedge]

x(o)=4

[twin wolf]

why

[thick hedge]

v(0)=2

[twin wolf]

?

[thick hedge]

that's the given information

[twin wolf]

no

where it say that

[thick hedge]

read the question again

[twin wolf]

i did but i dont get it

[thick hedge]

Initially it is positioned 4 units to the right of the origin

what does this tell you

[twin wolf]

x-4

[thick hedge]

when?

at what t?

[twin wolf]

2

[thick hedge]

at t=0

Initially positioned

[twin wolf]

um

so what i do with that?

x-4?

[thick hedge]

what's the expression for x as a function of timr

[twin wolf]

x = (t^3)/2 -t^2 + ct + k

[thick hedge]

good

sub t=0

[twin wolf]

ok

x=k

[thick hedge]

good

so what is k

[twin wolf]

x-4

[thick hedge]

k=4 you mean?

[twin wolf]

okay

how i find c?

[thick hedge]

Well you know the expression for velocity?

[twin wolf]

ys

[thick hedge]

so can you solve it now?