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[shadow tiger]

its implicit differentiation which turned into this

now all im stuck on is just isolating dy/dx from here

in which the final answer looks like this

but im really struggling on how dy/dx got isolated

[pale glen]

yea its right

[pale glen]

im getting that

lets start from here

which is from the 2nd line here

[shadow tiger]

sure

[pale glen]

,,let u = e^yxyln3\ \frac{dy}{dx}(u)-y=x\frac{dy}{dx}

[ocean seal]

Renz

[pale glen]

now

we want dy/dx on one side

[shadow tiger]

mhm

[pale glen]

,,\frac{dy}{dx}(u)-x\frac{dy}{dx}=y

[ocean seal]

Renz

[pale glen]

got that?

[shadow tiger]

yea

[pale glen]

,,\frac{dy}{dx}\cdot (u-x)=y

we factor out a dy/dx

[ocean seal]

Renz

[pale glen]

,,\frac{dy}{dx}=\frac{y}{u-x}

[ocean seal]

Renz

[pale glen]

and finally subbing u back shall give you the correct answer

[shadow tiger]

ohh i see

ahh tyvm

[pale glen]

no problem

Ill just suggest that if you have some tons of variables on one side with x and y

such as e^y xy ln3

just let u = e^y xy ln3

it will help you move things around much easier

[shadow tiger]

yeah i could see it def helps to visualize the movement a lot better

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[tepid coyote]

Hello there is this question that I know the answer of, but I can't figure out a trick to solve without trying possibilities randomly :
What is the smallest number that we can write as two different sums of three perfect squares of distinct non-zero integers?

[daring rover]

I dont think theres a trick or formula to it

[tepid coyote]

so it's just me and my intuition

[edgy merlin]

you can systematically try the possibilities until you get a result

for example:

$1^2 + 2^2 + 3^2 = 1 + 4+ 9 = 14$

[ocean seal]

Gapi

[edgy merlin]

so your answer needs to be bigger than 14

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[silent sundial]

Supposed to find the partial fraction decomposition and I’ve got to the part where is 5x+7 = A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1) but no sure where to go from there with this question

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[edgy merlin]

you need to multiply the brackets together to get a polynomial

then you look at the coeficients on front of $x^2, x, 1$ and you get a system of equations

[ocean seal]

Gapi

[edgy merlin]

you solve the system to get A,B and C

[silent sundial]

but its a linear numerator

like

no x^2

so

should the thing

for x^2 be = 0

[edgy merlin]

you should set it up like this:

A(x+1)

B(x-1)

C(x+2)

[silent sundial]

yeah

thats what i have

i think

[edgy merlin]

just solve the system that I talked about above

You should get A+B+C = 0

so you wont have any x^2 in the end

[silent sundial]

okay wait

[edgy merlin]

you might get some x^2 on the top of some fractions, but they should cancel each other out

[silent sundial]

so

like this

5x+7 = (A+B+C)x^2 +(3A+B−C)x+(2A−2B−C)

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[sleek palm]

the questions is asking me 2 things.
a) write and simplify an expression for the volume of a rectangular prism
b) if the height is 6 units what is the volume of the prism

the length is 2h
width is 9-h
height is h

[sleek palm]

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[mint cargo]

hellooo, bases from this statement:

Based from the computations, it implies that the 75% of 50 - which is the highest score that a student can get in the Periodical Test is equal to 38. Therefore, out of 43 Grade 9 - Gemini students, 38 (88%) of students who took the Periodical Test in Mathematics for the Third Quarter got a score that is greater than or equal to 38 in the examination with a determining factor of 75%. With that in mind, thus, contrarily, 5 (12%) students unfortunately failed to pass the Periodical Test in Mathematics.

rewrite it that corresponds to these:

• the highest score that a student can get in the Periodical Test is equal to 46
• **40 **Grade 9 - Gemini students

[mint cargo]

how did the 38 is (88%) T-T

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[rustic coral]

,calc (38/43)(100)

[ocean seal]

Result:

88.372093023256

[rustic coral]

@mint cargo ^

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[frozen lodge]

https://youtu.be/p_di4Zn4wz4?si=q2IpC3mhgfFkRoUV how do solve the similarity thing at 7:28?

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[frozen lodge]

Its just a simple geometry question and it hasn’t been solved in 2 hrs wtf

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[frozen lodge]

.close

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[slate barn]

americans are all eeping and europeans are all on vacation

but the solution is, if you've not figure it out yet: the pink arrow forms a right angle with the blue arrow, and the blue arrow forms a right angle with the white line L; hence pink and white are parallel

but because pink and white are parallel, they form the same angle with any vertical lines; hence the angle between the pink and the yellow arrow is the same as theta

wahoo

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[stark hawk]

is anyone good at drawing bearing diagrams? i need a hand

[alpine sable]

, rccw

[stark hawk]

<@&286206848099549185> drawing bearing diagram help pls with vectors

<@&286206848099549185>

<@&286206848099549185> please ! i really need a hand on how to draw this bearing thing

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[stark hawk]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185> please i am really lost ! drawing bearing diagram and finding bearing vectors

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[strange fractal]

Bcoz they are vectors does it matter where the 020 bearing is

And to turn the 250km/h to a distance to draw a diagram can i just multiply it by t?

Same with the 35

Sry my test is in 30 minutes

nooo im driving into school

cya guys

[jagged cobalt]

good luck

you can multiply with t, yes

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[old sluice]

Yo I need help with an equation rq

[alpine sable]

solved

[fleet crown]

Yo I can help you with that equation rq

[alpine sable]

vacuously

[old sluice]

I need help with this

[fleet crown]

casts greater mind read

[old sluice]

Does my drawing match the equation so far?

[alpine sable]

looks good

whats the eq of the area of a rectangle

[old sluice]

L x w

Right?

[alpine sable]

L*w=…?

[old sluice]

I don’t understand what your asking me

[alpine sable]

whats on the other side

L * W = Area

[old sluice]

Ohhh yes ofc😭

I’ll try to solve it

Tell me if I did something wrong

Pls

On the right track?

[alpine sable]

… + 0 = 45

… - 45 = 0

[clever wind]

kind of, you do x(2x+1)=45

foiled out is 2x^2+x-45=0

[alpine sable]

$2x^2-x-45=0$

[ocean seal]

4C

[clever wind]

because l=2w+1

wait

[alpine sable]

?

[clever wind]

I did it wrong lol

I read it wrong

yeah, 2x^2-x-45

[old sluice]

I’m cooked where did I go wrong

I did the square root of 368 and got something weird

[alpine sable]

why -46

we moved +45 to the other side

[old sluice]

Because it was -1-45

[alpine sable]

wheres the -1

[old sluice]

No we can’t add it since it’s already a positive

2x-1

That makes the 1 a -

[alpine sable]

one step at a time; we have (2x-1)*(x)

what is that

[old sluice]

2x2 -1x

[alpine sable]

right

[smoky mango]

Help me plsss

[alpine sable]

$2x^2-1x=45$

[ocean seal]

4C

[alpine sable]

start a new one

[old sluice]

Yea so from here we subtract the 45 by the -1 right?

Since the 45 is a +

[alpine sable]

what does a+ mean

[old sluice]

A positive

[alpine sable]

we can only subtract and add the same types of values

-1x is just -x

[old sluice]

So I add the 45?

[alpine sable]

here

[old sluice]

Yes please show me how to do this part I’m so confused rn

[alpine sable]

if x - 2 = 4 how do we get x

[old sluice]

By adding 2 to 4?

[alpine sable]

good

if x - 2 = -x how do we get x

[old sluice]

Omg so I’m supposed to add the 1 smh

[alpine sable]

what

[old sluice]

Add the 2 to the -x

[alpine sable]

no

[old sluice]

??

[alpine sable]

we can only add and subtract x’s with other x’s

2x + 3x = (2+3)x = 5x

2x + 1 cannot be changed

[old sluice]

This is a whole different equation but as you can see I subtracted 55 from 1 and got the right answer I don’t get it?

[alpine sable]

yes this one is good

[old sluice]

So what wrong with my other one it’s the same exact thing?

The 1 is just a negative now

[alpine sable]

but 2x^2 -x =45 is not equal to 2x^2 -x -46 =0, it is equal to 2x^2-x - 45=0

you somehow got an extra -1

[old sluice]

-1 -45= -46 though

[alpine sable]

theres no -1

[alpine sable]

when u did this, you didn’t add the +1x to -55

[old sluice]

Yea right I subtracted the 55

[alpine sable]

right

[old sluice]

Because it was a positive

[alpine sable]

so u subtract the 45 same way

theres no -1

[old sluice]

And so is the 45 but there’s a -1

Please type this in a calculator

[alpine sable]

$2x^2 - 1x = 45$

[old sluice]

Yup that means the 1 in that equation is a negative is it not?

[ocean seal]

4C

[alpine sable]

$2x^2 -1x {\textcolor{red- (45) }}= 45 {\textcolor{red -45) }}$

$2x^2 -1x - (45) = \not {45} -\not{(45)}$

[old sluice]

What am I looking at

[ocean seal]

4C

[alpine sable]

-45 on each side

are u following?

[old sluice]

I do not understand man

[ocean seal]

4C
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[old sluice]

All I want to know is why wasn’t I supposed to subtract the 45

[alpine sable]

u are

[old sluice]

Or actually I was but your telling me there is no -1 when there is I don’t understand??

[alpine sable]

but you are subtracting it incorrectly

there is no -1

[old sluice]

Alr can you show me how to do it correctly pls

[alpine sable]

2x^2 - x + 0 = 45

now subtract 45 from each side

[old sluice]

It would be a positive 45

Idk

[alpine sable]

exactly the way u did it with 55

what did u get

[old sluice]

I got -55

[alpine sable]

good

now subtract 45

what do u get

@old sluice does this step make sense

[old sluice]

Man oh man

All you had to tell me was I don’t actually subtract the 45 from -1💀💀

I’m pretty sure this is right

[alpine sable]

i said that like 20 times man💀

[old sluice]

It’s like a has been lifted

I’m a little slow

You should’ve just said it straight up😭

But thanks man🙏

[alpine sable]

np

[old sluice]

Also do you know what I should label the 5 and 9

Like should I say units or something

[alpine sable]

hold on whats the 9 from

[old sluice]

2(5)-1

Since I found the x

Which equals 5

[alpine sable]

your good i thought u were doing something else

usually the units in the question

so here W=5 cm

L=9 cm

[old sluice]

Thanks🙏

Have a great night

[alpine sable]

.close

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[vapid shuttle]

I was hoping someone could look over my proof for this problem

Please ping me if you do !

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@vapid shuttle Has your question been resolved?

[vapid shuttle]

<@&286206848099549185>

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.close

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[formal ginkgo]

$\lim_{x \to \infty} \frac{x + \sin(x)}{x}$

[ocean seal]

RulzerFly

[formal ginkgo]

how to solve it

[thorny copper]

squeeze theorem

[formal ginkgo]

can u check my steps ?

$-1\le\sin\left(x\right)\le1$

[ocean seal]

RulzerFly

[formal ginkgo]

first i did this

$0\le\frac{1}{x}\le1$

[ocean seal]

RulzerFly

[formal ginkgo]

$0\le\frac{\sin\left(x\right)}{x}\le1$

[ocean seal]

RulzerFly

[formal ginkgo]

$1\le\frac{\sin\left(x\right)}{x}+1\le2$

[ocean seal]

RulzerFly

[formal ginkgo]

and here comes the problem

[thorny copper]

looks right but prob too complex

id split the limit up using limit sum rule

[thorny copper]

then clearly using this, just say $\lim_{x \to \infty} \frac{-1}{x} \leq \lim_{x \to \infty} \frac{\sin(x)}{x} \leq \lim_{x \to \infty} \frac{1}{x}$

[ocean seal]

AlphaNull

[formal ginkgo]

yeah yeah

thanks for help

.close

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[sinful cosmos]

This isn't a usual question but I'm going to ask it anyway. I need help with studying
I've tried:
Reading all my notes and summarizing them
Working out past examination papers which helped a little but I couldn't learn from my mistakes because you have to pay for the marking scheme
Watching maths youtube videos
And I still feel unprepared

[sinful cosmos]

If anybody could offer some suggestions or tips I'd really appreciate it

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[clear relic]

to not feel unprepared, practice as many questions as possible

try finding free sources

and your mistakes, google those questions to see the working

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[cinder compass]

do u know the rational root test

[torpid roost]

probobly not idk

[cinder compass]

skill issue

u kinda need that for this question

[torpid roost]

i might know it

[cinder compass]

if r/s is a root then r divides -3 and s divides 2

[torpid roost]

i thought it was factors of -3 divided by factors of 2 and then you try all the possibilities of that as x to see if they equal 0

[cinder compass]

yes

ur right

[torpid roost]

so i just gotta try things like 1 and -1?

thats the only whole numbers

i think

[cinder compass]

yh but i don't think u need to try them

just list the possibilites

that's how i understand the question

+-1, +-3 etc

[torpid roost]

do i need to after i find a zero tho i need to do long devision

[cinder compass]

um idts

[torpid roost]

but -3 isnt an answer of r/s

[cinder compass]

okay?

still a possibility

a priori

[gray isle]

question is worded poorly
should be saying something like candidates for rational roots

any rational roots will be in this list

doesn't necessarily mean that all values will be roots

[torpid roost]

isnt a root tho where the equation goes through the x axis

[cinder compass]

bro yes it is

no one's saying it isn't

u just gotta list the fraction u get from the rational root test

[torpid roost]

1,-1,-3,3,-3/2,3/2

[cinder compass]

yes

good job

[torpid roost]

what do i do now?

[gray isle]

use the specified symbol

[cinder compass]

thank me

[gray isle]

$\pm 1, \pm 3, \pm \frac32$

[ocean seal]

ℝαμΩℕωⅤ

[gray isle]

and that's it

[cinder compass]

i think u gotta type it like +- but yeah

[torpid roost]

?

[gray isle]

yeh

[torpid roost]

ok

.close

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[full garnet]

H

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[orchid portal]

I need help understanding completing the square\

[crimson field]

Send an example problem

[orchid portal]

y^2 + y - 7 = 0

[crimson field]

What binomial squares to something that starts with y^2+y

So like

$(ay+b)^2=y^2+y+…$

[ocean seal]

coak

[orchid portal]

I mean the only number is the invisible 1 attached to the y's

[crimson field]

Yes so what should a and b be

[orchid portal]

I think it should be 1? Since square root of 1 is 1

[crimson field]

Try it and find out

[orchid portal]

nope that comes to y^2 +2y+1

oop moved it over to the other side so its y^2+y = 7

[ocean seal]

faiyrose

[crimson field]

Just ignore it

[orchid portal]

well a would be 1 in this case

[crimson field]

Now try to get 2ab to match up

[orchid portal]

???

[crimson field]

Oh we’re missing terms here this is confusing

Try to get 2aby to match up

[orchid portal]

match up to what?

[crimson field]

Do you remember the goal

[orchid portal]

solve the square but 1 doesnt square!

[crimson field]

Okay you got that a is 1

Now what should be be

For it to line up

[ocean seal]

faiyrose

[orchid portal]

im lost now x.x

[ocean seal]

faiyrose

faiyrose

faiyrose

[orchid portal]

.close

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[shy valve]

A car on the cliff is pulling a boat in the water. If the car travels at 5m/s, will the boat travel the same speed, faster or slower than the car?
Initial rope length from cliff to boat = 105m
Cliff height = 20m
Justify your answer throughout with mathematical reasoning

I’m doing trigonometry right now

[shy valve]

I’m not sure how to measure or get the speed of the boat

[fossil laurel]

try making a sketch of it

[shy valve]

Like a diagram?

[fossil laurel]

yeah

[shy valve]

Ok

[fossil laurel]

fancy

looks good

now see where you can find a triangle lmao

[shy valve]

The rope to the boat to the bottom of the cliff and back to the top of the cliff

[lone heart]

@shy valve Has your question been resolved?

[shy valve]

No

<@&286206848099549185>

[blazing sky]

Problem?

[shy valve]

Hi this is the question I’m stuck on

[blazing sky]

I know what to do. I'm solving it. I'll explain when I'm done.

[shy valve]

Thanks so much

[blazing sky]

@shy valve

If you need further explanations, you can ask me.

[shy valve]

Ok I’m just looking right now

What’s Vx and v?

[blazing sky]

V is the velocity of the car, which is also the velocity of the rope and Vx is the velocity of the boat in the x axis.

[shy valve]

Oh ok and also do I need to know the angle of theta

[blazing sky]

In this case, no. We were already going to use cosθ in solving so we didn't need to know θ. All we had to do was substitute cosθ.

[shy valve]

Ohh ye cause it’s 20/105

[blazing sky]

Correct.

[shy valve]

So the boat is slower than the car

[blazing sky]

Yup.

[shy valve]

Ok thank you

[blazing sky]

It's OK.

[shy valve]

How do I close this now

[blazing sky]

.close

[shy valve]

.close

[lone heart]

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[alpine sable]

hi so idk how to go about this i know that the

U - L <= max f * Δx_max

that was from this problem

.close

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[fleet elbow]

when is this fuction increasing and decreasinng

[steel olive]

check its derivative

[fleet elbow]

i got the derviate

[steel olive]

show

[fleet elbow]

I got that as the derivative

[steel olive]

wait nvm

[gray isle]

derivative is incorrect

[steel olive]

this function can be simplified first

[fleet elbow]

how

[livid sage]

factor the numerator and denominator

[steel olive]

yea

[gray isle]

where's the (x-2) on the numerator coming from

[fleet elbow]

k got it

[lone heart]

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[alpine sable]

Explain me how to derive stem

I am not sure which identity is used

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[sand sapphire]

how do i use symetries to find all the solution of x for this doman

[sand sapphire]

<@&286206848099549185>

[exotic belfry]

"find all the solution of x" - there is no x in your picture.

[sand sapphire]

theta

i meant

sorry

[swift crystal]

you can move the sin to the right

[sand sapphire]

what do you mean

[swift crystal]

2theta = arcsin( -sqrt(3) / 2 )

[sand sapphire]

i got -30

but there are more answers

[swift crystal]

need to follow the trigo quadrant circle to find the correct angle

[sand sapphire]

aight

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[hushed warren]

help

@here

@every

@everyone

@mighty cosmos

<@&286206848099549185>

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[hushed warren]

<@&286206848099549185>

[exotic canopy]

you are given that AB is 12, therefore AL is half of that, i.e. AL=6 cm

you have AL and CL, hence you can find AC using pythagoras

same logic for CB

[blazing sky]

Solve the missing sides of all the right angled triangles you see.

[hushed warren]

what

i dont

understand

[exotic canopy]

do you get this?

[hushed warren]

yes

[blazing sky]

I'll solve it and send it to you.

[hushed warren]

ok

@blazing sky whre>

[blazing sky]

Here.

[blazing sky]

L is not in the middle of AB.

[hushed warren]

wheres the answer

....

and the way

[blazing sky]

I'm not done.

[hushed warren]

oh ok

[alpine sable]

cold L

🥶

[exotic canopy]

oh yeah

cuz AM and KB are not the same

the diagram fooled me

[alpine sable]

how is this jev guy in an undergrad math course but cant solve this

[exotic canopy]

lol

[hushed warren]

ee

[alpine sable]

this is wrong

[exotic canopy]

yeah i realised

[alpine sable]

oh L alrdy said that

[exotic canopy]

woops

[alpine sable]

sry

[exotic canopy]

you can solve for MB also using pythagoras in triangle AMB, as well as the length AK in triangle ABK

[alpine sable]

also you don't need to pythagoras all the lengths

you can use similarity

[exotic canopy]

that's also true

[alpine sable]

i think it's less calculation but both work

close this thread when

hasn't the question basically been resolved

[hushed warren]

i dont get any answer yet...

[hushed warren]

how is this wrong

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[lone heart]

@hushed warren Has your question been resolved?

[boreal nymph]

stap taggin

[lone heart]

@hushed warren Has your question been resolved?

[lone heart]

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[lost flax]

why is sqrt(logx) greater?

[mortal trellis]

zoom out

[lost flax]

OH SHITTt

ok thanks

.close

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[hot rose]

I have a pretty easy exercise

[hot rose]

I have to verify that 0 < Un < 2

But how tf i can verify that

!status 2

[lone heart]

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

[topaz lake]

maybe show 0< f(x) < 2

derivating f

[hot rose]

Btw i verify it for U0 and suppose that Un verify this condition so i need to verify it for Un+1

No way

[hot rose]

How?

[topaz lake]

derivate f

to know its max and min

[hot rose]

Um remind me how

Its - infini and 1.

[topaz lake]

determine $\frac{d}{dx}f(x)$

[ocean seal]

vipe

[hot rose]

Its - infini and 1

[topaz lake]

so you know that f(x) <= 1

so f(un) <= 1

[hot rose]

Oh that right

So i only need to verify it for 0

Yeah yeah tysm

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[sinful cosmos]

How do you solve a quadratic using a graph

[vapid drift]

do you have an example?

[runic glacier]

You sketch the graph and see where it intersects the x axis

[vapid drift]

if you have a full graph of a parabola the x intersects are the solutions to the quadratic

[sinful cosmos]

I dont have an example but i always get stuck on questions that are first like

You draw a graph using an equation

and then you are given another equation and it asks you to solve that equation using your graph

[vapid drift]

hmmm

[sinful cosmos]

So there are 2 lines

Ill try find an example

[vapid drift]

yeah if you have 2 functions for example f(x) and g(x) you can solve an equation f(x)=g(x) by checking for what x the graphs of both functions intersect

finding the example and trying to solve it will be the best way to help you

[sinful cosmos]

Something like 10b

[vapid drift]

ok

lets try to solve it

start with 10a and graph the y=2x-3

[vapid drift]

here f(x)=5/x and g(x)=2x-3

lets find first the solutions to f(x)=g(x) the way i said

after this we will show that 2x^2-3x-5=0 has the same solutions as our f(x)=g(x)

[sinful cosmos]

Ohhh

Is x=2?

[vapid drift]

i dont think so

5/2 =/= 2(2)-3

so its not a solution

have you drawn the graph of 2x-3?

[sinful cosmos]

I found that example online so i dont have the paper but i sketched it

[vapid drift]

show the sketch then

it needs to be pretty accurate for the method to work

[sinful cosmos]

Oh i just did a rough sketch

[vapid drift]

the 5/x is already drawn so its easier

and linear functions are easy to draw accurately

[sinful cosmos]

💀

[vapid drift]

oh gosh

[sinful cosmos]

i dont havbe time the exam is in 2h ours

lmao

[vapid drift]

just use desmos for now to graph it

or any other graphic calculator

and get solutions

to not waste time

[vapid drift]

and we will go for this step

[sinful cosmos]

Okay i got it

it looks like this

[vapid drift]

ok

what are the solutions?

lets just list them here

[sinful cosmos]

Where they intersect?

[vapid drift]

yes

for what x

[sinful cosmos]

2.5 and -1

[vapid drift]

alright

[vapid drift]

howd you approach this then?

[sinful cosmos]

I put these values into x

and try to solve it

[vapid drift]

no

start with 5/x=2x-3

and rewrite it to get 2x^2-3x-5=0

[sinful cosmos]

What do I do next

[vapid drift]

actually, you can do that way too

[vapid drift]

but stick to this first

[vapid drift]

have you rewrite it?

[sinful cosmos]

Yea i did

[vapid drift]

show how

[vapid drift]

there is a rule, if two equations are equivalent (aka you can rewrite one equation into the other like you did now) they have a same solutions

so you just showed they have the same solutions

but thats the more general case

[sinful cosmos]

so that answers 10b?

[vapid drift]

yeah

oh wait

[vapid drift]

you forgot about x next to -3

[sinful cosmos]

Oh oops

[vapid drift]

we can also show that\ this way

lets put those values into 2x^2-3x-5 and if you get 0 it means they are solutions to this

thats the other way

[sinful cosmos]

Can I put any one of them

[vapid drift]

start with x=2.5

if 2(2.5)^2-3(2.5)-5=0 x=2.5 is a solution

after that check for x=-1

[sinful cosmos]

I got it 0=0 les goo

[vapid drift]

and if 2(-1)^2-3(-1)-5=0 x=-1 is a solution

[vapid drift]

so its good

[sinful cosmos]

I got it with -1 too

[vapid drift]

great

so were finished

[sinful cosmos]

Great cause I gtg

Thanks a lot

[vapid drift]

glad i helped

[sinful cosmos]

Have a good one

[vapid drift]

thx u2

[sinful cosmos]

.close

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[carmine moat]

For each value of b with 0 ≤ b ≤ 100, points A(0 | 0) and B b( | 0) are given, as well as point C. Point C has the x-coordinate b and lies on the graph f(x)
Determine the value of b for which the area of triangle ABC is maximal, and provide the corresponding area.

[carmine moat]

Can anyone help me?

[mossy bronze]

whats up

[carmine moat]

This is the function.

[mossy bronze]

sorry i would be no help sorry

im rubbish at maths

[vocal dirge]

john think of a function to determine the area of said triangle

[carmine moat]

Yeah, the function is:

[vocal dirge]

great

thats true because we have a right angled triangle to work with

[carmine moat]

But how can I determine the extreme points of it?

[vocal dirge]

take the derivative

well

lets do it right

we're gonna name our function A(x)=x^3*e^(-0.2)x

lets not use b for the function because we want to solve for the optimal b anyway. lets just sollve it over x and call the specific value we find b

so now we just have to take the derivative, and see where it has roots so we can determine the maximums and minimums

[carmine moat]

is that the correct derivative of the function?

[lone heart]

@carmine moat Has your question been resolved?

[carmine moat]

What can i do?

.close

[lone heart]

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[uneven depot]

Hey guys, just wanted to ask if this is correct or not:
b|x+c| can be done but multiplying b|x-c| can't

B, c are constants

[gray isle]

wdym by "multiplying"

[uneven depot]

Sorry, edited

[gray isle]

wdym by can / can not be "done"

[vocal dirge]

im more confused now

[uneven depot]

Can I do it while solving an absolute inequality or equation

[gray isle]

do what

exactly

show the specific manipulation you're trying to do

[vocal dirge]

b|x-c| and b|x+c| are practically the same thing if u just write b|x-(-c)| instead of b|x+c| and just say (-c) = d

[uneven depot]

Let's say there is a question similar to this but, there is a minus sign before 1/2

[uneven depot]

But can I do it with b|x-c| always?

[gray isle]

do what

exactly

what's "it"

[uneven depot]

Write |2x-1|

[uneven depot]

In this question, if it had a minus sign before 1/2

[gray isle]

sign inside there is irrelevant

2 = |2|
and
|a| |b| = |ab|

[uneven depot]

Oh, so I can't do it with -2 then, got it

[gray isle]

you can leave the - on the outside

[uneven depot]

Thanks!

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[alpine sable]

how do i find an expression for this

if there was no r multiplying in the expression, id be able to solve it

[lone heart]

@alpine sable Has your question been resolved?

[near temple]

What is C_r in this context?

[alpine sable]

thats the combinations symbol

2n _ C_ r

sorry idk how to type that

[near temple]

Ah right, apologies, though it was $r^{2n}\cdot C_r$

[ocean seal]

JJP

[alpine sable]

ohh

[near temple]

Gimme one sec, you can probably use the binomial expansion for this or something like that

What class was this asked in?

[alpine sable]

this is the original question

i know i dont need this

i dont need this value ik

but i dont remember seeing something of this sort before

so i was curious as to what is going on

[near temple]

You can take the derivative of the binomial expansion twice, set x = 1 in both equations then manipulate them to get that

I'll write it down

[alpine sable]

and what i have to do in case i have to find value of that

[alpine sable]

wym

sorry i dot follow

[near temple]

Just writing it down

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@alpine sable Has your question been resolved?

[lone heart]

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[fair gazelle]

I am currently working on a Quadratic Word Problem. The world problem with be provided below with the answers.

My question is where did I go wrong?

Word Problem: A rectangular floating dock measures 4 m by 5 m. A new dock is to be made by increasing each side length by the same amount. The area of the new dock is to be 50 square metres. Find the dimensions of the new floating dock, to the nearest 100th of a metre.

Answer Key: 6.59m by 7.59m

My work:

[steel olive]

the part (4+2x)(5+2x) = 50 is correct

if you answer isnt matching then maybe you are doing calculation mistake somewhere

(also final answer,i.e. dimensions will be 4+2x and 5+2x)

[fair gazelle]

I used a calculator for the multiplication and it all checks out

18 x 18 = 324
4 x 4 = 16
16 x 30 = 480

[lone heart]

@fair gazelle Has your question been resolved?

[lone heart]

@fair gazelle Has your question been resolved?

[radiant crag]

Why did you write (4 + 2x) (5 + 2x)?

[fair gazelle]

4 and 5 are the width and length of the rectangle, since they are expanding it you write + 2x. Then it is equal to 50 area as that’s the new rectangle size

[radiant crag]

But you’re increasing each side’s length by the same amount

not twice the same amount

[lone heart]

@fair gazelle Has your question been resolved?

[fair gazelle]

???

[radiant crag]

Instead of writing 2x I think you should’ve written just x

[shell prawn]

When did the start using alphabets

[radiant crag]

“A new dock is to be made by increasing each side length by the same amount”
If we define that same amount with x, then the side lengths would be (5 + x) and (4 + x)

I’m not sure why 2x is there

[fair gazelle]

?

There are two widths and two lengths?

So it would make sense to use 2x as you have two widths and lengths increasing by the same length

[radiant crag]

no, because S = b * h

[fair gazelle]

Never mind

I did it the right way

If you use a calculator and plug the numbers into the original equation you get the answer key answers

[radiant crag]

Oh?

Interesting, I had done it my way and got the same answer

[radiant crag]

Maybe I’m dumb but I’m getting x1 = -5,79 and x2 = 1,29 with the original equation and quadratic formula

Yeah I’m dumb, interesting though, thank you

I’m just used to setting everything with x, but 2x in this case works, so will anything else that multiplies x

trying it with 4x in this case, the results from the quadratic formula are going to be 2 times smaller than using 2x, but you’re still gonna multiply it by 4 in the original equation instead of 2, so you’ll still get the same answer

basically you just defined the increment of each side length by 2x instead of just x

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[alpine sable]

Find CH

[gray isle]

apply geometric mean

[alpine sable]

What is that?

[gray isle]

look up geometric mean triangle

[alpine sable]

Thanks!

[daring rover]

area of triangle = base * height / 2

[gray isle]

that works too

[alpine sable]

Looking to find Height over the hypotenuse

[gray isle]

simpler here

[alpine sable]

Are these true:

<@&286206848099549185>

[lone heart]

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[past solstice]

I know the coordinates of A and B also the length of AC how do I find the lengths of CN and CM?

[gray isle]

construct horizontal and vertical lines from B

then you can use similar triangles

[past solstice]

can you help me construct a formula for CM and CN

[median shuttle]

you have the length for AC so you can find the y-coordinate of M

you need to smh find coordinate of C

since the y-coordinate of C and M seem to be equal

[past solstice]

how?

[median shuttle]

are there some hidden thnigs like

AB is 2x AC

or same

idk

[gray isle]

is A at the origin?

[median shuttle]

ig

[past solstice]

no

[past solstice]

no

[gray isle]

(doesn't really matter, calculations are simpler if it is though)

[median shuttle]

oof

[gray isle]

ok, distance formula for AB

[median shuttle]

waiiit

we have angles

we can use sohcahtoa right?

smh

idk how

and pythagoras later

nvm we don't know angle

AN=MC if that helps

[gray isle]

then
AC/AB = CN/(difference in y coord of B and A)

same idea for CM

[median shuttle]

don't we need to know if it's bisector or not then

since diagram not accurate i thinl

we also don't know ab

[gray isle]

where did i say anything about bisection

[median shuttle]

im asking

[past solstice]

so CN=(AC*(B.y-A.y))/AB

[gray isle]

i didn't say anything about bisection because its not relevant/doesn't matter

[median shuttle]

yes

[gray isle]

yeh

[median shuttle]

we don't know ab and cn though

we have 2 unknowns

we can find ab though

[past solstice]

we know AB form coordinates

[gray isle]

coordinates of A,B and length of AC are said to be known

all required info can be determined

[median shuttle]

yeah im blind 💀

CM can be found with pythagoras

atleast I know that

[past solstice]

@gray isle dosen't seem to work

[median shuttle]

btw

i have a really cool tip

since diagram is to scalre

scale

just measure it

🙂

if in a ratio compare ratios

dumb tip but best irl usually

[gray isle]

don't know how to use geogebra notation

to grab either coordinate

[past solstice]

desmos

A.y

[median shuttle]

desmos.com

[past solstice]

A is the point

[median shuttle]

why not find angle and measure it?

or measure sides

since you are in desmos after all

[past solstice]

I use it for the visualisation the actual use of this is in programming

[median shuttle]

oh

idk

[gray isle]

can you link what you did

[past solstice]

https://www.desmos.com/geometry/uouyjjiyfi?lang=ru

[gray isle]

did you intend to have a right angle at A?

[past solstice]

yes

[gray isle]

are those lines coming out of A supposed to be perfectly horizontal/vertical?

[past solstice]

yes

[gray isle]

your lines aren't horizontal/vertical

[past solstice]

oh ok thanks

.close

[lone heart]

Channel closed

Closed by @past solstice

Use .reopen if this was a mistake.

[lone heart]

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[turbid narwhal]

Hello

[faint lynx]

hai

[turbid narwhal]

So I was trying to model the following spiral and couldn't figure out which tecnique I needed to use. I tried using polar functions and other spirl formulas however none of them seem to catch a good fit with my shape. Can you hrlp me to figure out how to turn the following shape into a function. Are there any formulas I can use or any techniques

or what else can I do if there are no r(t) functions fitting this shape

[lone heart]

@turbid narwhal Has your question been resolved?

[turbid narwhal]

<@&286206848099549185>

[uncut furnace]

Looks fairly polar to me. Sin or cos graph moving in the z plane

[turbid narwhal]

I tried most sin or cos graphs but they didn't work the further I could go was this

This shape is the best I could done so far however is there a formula?

[lone heart]

@turbid narwhal Has your question been resolved?

[turbid narwhal]

<@&286206848099549185>

[lone heart]

@turbid narwhal Has your question been resolved?

[old sluice]

Can someone tell me if I got this correct?

<@&286206848099549185>

[lone heart]

@turbid narwhal Has your question been resolved?

[lavish cave]

Your answer is correct but your last step seems like you got stuck

$x = \frac{-2 \pm \sqrt{4} \sqrt{3}}{4}$

So you can divide both top and bottom by 2

That's literally it

[ocean seal]

south

[lavish cave]

Yes you were so close cause you had 2 sqrt(3) already

[turbid narwhal]

bruh

no one to answer my question?

[patent abyss]

like this perhaps?

[turbid narwhal]

Nope I also tried that curve. I want to know if there is a methodology I can follow to reach to the shape of this curve

Or should I just divide it into 2

[patent abyss]

like a regression?

there are tools for that

[lone heart]

@turbid narwhal Has your question been resolved?

[marsh star]

Sabc=80. Sbcd=?

<@&286206848099549185>

[toxic kiln]

what do you have to do here

whats the question

@marsh star

[marsh star]

Sbcd

[lone heart]

Channel closed

Closed due to timeout

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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.
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[white marsh]

gradient of the line = slope of the segment between P(x,y) and (-2,1)

[faint monolith]

yeah i understoodf that part

[white marsh]

y = \sqrt{2x-4} is the second equation

[faint monolith]

but i dont understand the part under it

how is 1/√2x-4 = to that

[white marsh]

because that is the equation of the half parabola

for a point to lie on it (like P) that must be hold

[faint monolith]

what is √(2x-4) -1 / x+2

how

where did they get that frmo

[white marsh]

it is $\frac{y-1}{x+2}$ and y = $\sqrt{2x-4}$

[ocean seal]

penguin

[faint monolith]

oh

[white marsh]

$\frac{y-1}{x+2}$ is the gradient of the line between (-2,1) and P(x,y) \newline
since P(x,y) also lies on the half-parabola, y = $\sqrt{2x-4}$ must hold

[ocean seal]

penguin

[faint monolith]

ok i understand it now

thanks

waitwait

one more question

[white marsh]

yes

[faint monolith]

why did they reject z=4

x=4

why cant it be (4,2)

[white marsh]

at x = 4, x-6 = -2

so we cant consider it because square root of a positive integer is never negative

[faint monolith]

oh

how am i suppose to notice that though

like i dont just randomly look back on my working

[white marsh]

oh theres a known thing, whenever you square or increase powers to solve a problem - youre bound to get more solutions that there actually are

same if you multiply something to solve the equation

[faint monolith]

ok thanks

[lone heart]

@faint monolith Has your question been resolved?

[faint monolith]

.close

[lone heart]

Channel closed

Closed by @faint monolith

Use .reopen if this was a mistake.

[lone heart]

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.

[strange fractal]

how do i do dis

[strange fractal]

if they are independent then P(AnB) = P(A) x P(B)?

[prime badge]

yes

[strange fractal]

how come if P(A|B) = P(A)?

[prime badge]

you solved it wrong because P(A & B) is not given

[exotic canopy]

recall that P(AuB) = P(A) + P(B) + P(AnB)

[strange fractal]

]

[prime badge]

yes

[strange fractal]

how come P(A|B) = P(A)

[exotic canopy]

P(A|B) = P(AnB)/P(B) = (P(A)*P(B))/P(B) = P(A)

[strange fractal]

uhhhh

can u pls use venn diagram ?

[exotic canopy]

one sec I'll type it out in latex

[strange fractal]

i thought P(A|B) was probability of A happening in B