help-39

[eternal tulip]

so you can just treat (u^2+u)(u^2+1) as one function for example

[foggy dome]

btw how would you do it

[eternal tulip]

well, if f(x)=(u^2+u)(u^2+1) and g(x)=(u^2-u+1)

how would you do product rule with these?

[foggy dome]

let me see

give me 30 seconds

[eternal tulip]

well, do you know the formula for product rule at least?

[foggy dome]

i can use it

$(u^2 + u)(u^2 + 1) dx(u^2 -u + 1) + (u^2 - u + 1) dx(u^2 + u)(u^2 + 1)$

[jolly parrot]

odokawa

[foggy dome]

musy

okay never mind

i wont do that shit

[eternal tulip]

?

[foggy dome]

!close

.close

[pearl pond]

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[night viper]

Been trying to solve this all morning, I'm not sure if there's a trig identity or rule I'm missing that would help me solve for x. Everything becomes a mess if I try to go forward from here.

[pearl pond]

@night viper Has your question been resolved?

[pearl pond]

@night viper Has your question been resolved?

[pearl pond]

@night viper Has your question been resolved?

[inland lantern]

none of the solutions look particularly nice, are you sure this is the right problem?

,w 2sin(2x) = 4cos(x)+3

[jolly parrot]

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Requested by tuzis

[inland lantern]

you arent going to be getting any of these from trig simplification

[night viper]

Yes, it was given to me in a test. I've managed to get a solution but I'm not sure if it's right.

Uh hold on let me rewrite it since my working is messy

There @inland lantern

[inland lantern]

i dont think that has the same solutions (per wolfram)

in fact, that doesnt have any (real) solutions

(your cos^4 expression)

[night viper]

Well I'm stumped then, if there's no way to get this through trig simplification then I'll just go ask my lecturer for help tomorrow, as we're not doing advanced stuff so any complex solution would be off the table as far as I'm aware

.close

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[pearl pond]

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[quick quiver]

Hi, I'm doing surface integrals. What are they doing here?? 'do an integral. but actually don't do an integral jk' Well this is a calculus class and I am here to do integrals

[quick quiver]

My guesses for the integral they are setting up vs. the actual answer

This is the example problem

as in this is all the example problem, none of my work

Trying to learn to study for the final exam, where I am required to do integrals instead of plugging it into memorized area equations derived from integrals 🤔

[sharp vigil]

when you are integrating surface area, that is the same as integrating the function 1 as a surface integral

similarly, when you find 2D area, you are integrating the function 1 as a double integral

[quick quiver]

well, yes, but I am not sure what to do with that information here as those aren't getting me the right answer either

and the problem appears to be using the z^2 = 16r^2 derived earlier (x^2+y^2=r^2)

but in a different way

[sharp vigil]

so we have 2 different equations to worry about here:

• equation of the surface (z² = 16r²)
• function to be integrated (1)

[sharp vigil]

this integral formula is a general formula for surface integrals when the surface is given by z = g(x,y) like we have here

[quick quiver]

yeah, but I am not sure what f(xyz) is here and it doesn't mention it further into the problem (it sets it to 1) so I don't know how to solve this one

I can solve it when given an f(xyz) function, but I am confused as to what to do without one

[sharp vigil]

f(x,y,z) can be any function, but remember to find areas/volumes/etc we integrate the function f(x,y,z) = 1

[quick quiver]

ok well it's pranking me then because in my next question f(x,y,z) is given as 5xy and used

but still... how are they setting up the last part? even given that information

I am not sure where the pi comes from, as there doesn't seem to be a factor of 2

the '9' comes from the bounds conversion, the sqrt(17) is the other part of the given integral equation

the bounds of the cone SHOULD be the new bounds of r + a circle doing one full rotation, and sin and cos in there somewhere to do that rotation

i think

well its surface so no

man can I just do a rotating line integral

[sharp vigil]

so $\sqrt{z_x^2+z_y^2+1} = \sqrt{17}$

[jolly parrot]

cloud

[quick quiver]

yes here it does

[sharp vigil]

so we're left with $\sqrt{17} \iint_R dA = \sqrt{17} \iint_R r ,dr,d\theta$

[jolly parrot]

cloud

[quick quiver]

god freakin damn it is that what we're doing

sorry for my language

ok yes I am going on with this train of thought

o_o

oh my bounds are backwards

yeah nevermind I understand now thanks we are doing the r conversion so we do the r conversion integral bookend thing replacement

thank you 😭

what is the dA or dr called

like the end of the integral

that sounds like a useful word

[sharp vigil]

a differential

[quick quiver]

thank you

i understand now

.close

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[gusty prism]

hi

[gusty prism]

what are these numbers he put on the tpp right

and how was 9 put where opp is

[regal herald]

pythagorean triples i believe

[gusty prism]

o

[regal herald]

theyre just 3 numbers that fulfill c^2=a^2+b^2

[gusty prism]

Ohhh

[pearl pond]

@gusty prism Has your question been resolved?

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[amber niche]

Once I set up the integral I am fine, but I have trouble setting it up.

This is just asking for the integral from 0 to 1 of f(x) right?

As in should be setup like this?

sweet, thanks

.close

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[subtle ginkgo]

try to investigate by the definition existence of f' (0)

[marsh acorn]

I feel like this is one of those questions where you just think about it and reason it out

like yes there won't be a value at x=0 for the first part of the piecewise because it will be undefined

but because it defines f(0) to be 0, you can say that it will have a tangent line equal to the x-axis

Well because the function will be approaching the origin from both sides

left limit and right limit

so at x=0, the value of the function is 0, and the limit of the other portion of the piece wise is just approaching that from both sides horizontally

f(0) = 0 is a point though

and both sides of the other portion of the piecewise are just asymptotically approaching this point

no

wouldn't be tangent to the point directly at x=0

y

y=0

the piecewise is basically just completing the point of the function that cannot exist

[subtle ginkgo]

you have to use this definition

$f'\left( x_{0} \right)\overset{def.}{=}\lim_{\Delta x \to 0} \frac{f\left( x_{0}+\Delta x \right)-f\left( x_{0} \right)}{\Delta x}$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

if this derivativce exists, then tangent line exists

35 or 36 ?

exists

36does not exist

one minute i write it fo royu

$f'\left( x_{0} \right)\overset{def.}{=}\lim_{\Delta x \to 0} \frac{f\left( x_{0}+\Delta x \right)-f\left( x_{0} \right)}{\Delta x}=\\\lim_{\Delta x \to 0} \frac{f\left( 0+\Delta x \right)-f\left( 0 \right)}{\Delta x}=\lim_{\Delta x \to 0} \frac{\Delta ^{2}x\cdot sin\frac{1}{\Delta x}}{\Delta x}=\\\lim_{\Delta x \to 0}\left( \Delta x\cdot sin\frac{1}{\Delta x} \right)=0\\\text{because}\\0\le \left| \Delta x\cdot sin\frac{1}{\Delta x} \right|\le \left| \Delta x \right|$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

you can use " h " , instead of my notation, but that doesnot change anything

and in 36 you get:

$\lim_{\Delta x \to 0} sin\frac{1}{\Delta x}\text{ }\text{ }\text{ doesn't exist, due to Heine's definition of the limit}$

[jolly parrot]

Joanna Angel

[subtle ginkgo]

you may indicate two diffeent sequences, convergent to zero, but if we plug them into sin(1/x) , they convergent to differnent limits

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

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[jaunty barn]

hi i need help with an assignment

[jaunty barn]

im pretty sure none of these fit the criteria of neither symmetric nor antisymmetric

the empty set is symmetric, the 2nd to 4th option all contain (a, b) and (b, a) so they are symmetric

wait the 2nd and fourth option needs to have (c,a) for it to be symmetric

and (a,b) (b,a) makes it not antisymmetric

i think

<@&286206848099549185>

[pearl pond]

@jaunty barn Has your question been resolved?

[pearl pond]

@jaunty barn Has your question been resolved?

[pearl pond]

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[foggy dome]

i was solving a derivative
$\frac{(2x +1)(2-x)}{2}$ how from this $//$
you get
$\frac{1}{2} d/dx([(2x+1) (2-x)]$

[jolly parrot]

odokawa

[midnight haven]

Pull out the 1/2 it’s a constant

And derive the rest

That have x’s

[foggy dome]

j :/ sorry

but this is something important

it would be better if you go to bother other people

[midnight haven]

?

[foggy dome]

ill open another thread but please dont respond this is something important

.close

[pearl pond]

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[copper kestrel]

he wasn’t trolling..

[midnight haven]

What did I do

Lmfao

I legit gave the answer

☠️☠️

[copper kestrel]

yeah, constant multiple rule allows you to factor out the constant

[midnight haven]

Fuck me ig

[foggy dome]

i think i already close this channel please

[pearl pond]

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[rocky axle]

Where did 525 come from?

[copper kestrel]

f(x) = kx
21 = k(0.04)
k = 525

[rocky axle]

nvm got it

[copper kestrel]

dang

[rocky axle]

damn lol

thanks though

[copper kestrel]

lol np

[rocky axle]

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[thorny forge]

Two circles C(O1,R1) and C (O2,R2) intersect at two points I and J.
Show that the line (O1,O2) is perpendicular to the line (IJ)

In the figure below, the lines (AB) and (CD) are perpendicular in I.
Let J be the middle of [B, D]. The line (IJ) intersects [A, C] at H.
This involves showing that (IH) is the height of triangle IAC.
1° Justify the equalities of the marked angles
Indication Also think about the inscribed angle theorem

2° We are only interested in the AIC triangle. Justify the ehalities of the marked angles: Deduce the desired result

Hello, I need help with these math exercise. Can someone help me.

[pearl pond]

@thorny forge Has your question been resolved?

[pearl pond]

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[vital moth]

Okay

[pearl pond]

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[toxic fractal]

not enough data

Plus the angles given have a notation error, since if they dont specify angular unit they would be radians, not degrees as it seems to use.

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

So, as they are both congruent, the set of angles are the same

If you assume that the angle in S and G are the same, so you'll get the equation 4x+16=7x-2

@midnight haven

If you take the value of x you get as a result and put it back onto the STU trangle, you'll see that they, in fact, sum 180

So, you also can do thee same with the 7x-2 and get the value of this angle

The opposite of the angle in G must be equal to 32, as the same happens with the angle in the S value, as they are congruent

So we just need to equivale 32=3(y-4) and get the answer

By the way, with the HI distance and the angle around G, we can also equivale 57=3w and get this answer as well

Sorry for the not-formal mathematical terms

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

@midnight haven , did you get it?

NO

What did you not understand?

@midnight haven ?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

@midnight haven , what did you not understand?

Please, I need to know it for help you

STOP YELLING

Sorry

I didn't mean to offend you

Sorry!

you did..

I just wanted to help

Sorry

You said you didn't understand, so I asked why

Sorry, i shouldn't do it

Anyway, good luck!

[marsh acorn]

Lol chill girl he was just trying to help you

[midnight haven]

.close

[pearl pond]

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[ivory shadow]

how to 1/3 square = 2/3

[old marsh]

?

What u tryna ask

[ivory shadow]

how to 1/3 square = 2/3

not 1/9?

[old marsh]

$(\frac 13)^2$

[jolly parrot]

Stephen

[old marsh]

That?

[ivory shadow]

yes

[tacit mulch]

that is equal to 1/9

[ivory shadow]

.close

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[late jolt]

how do i factor this

[late jolt]

yes the x and y term

they are different

8

yes

the 8xy makes sense

where did the xy go from the 3

24xy

yes

yes

is this the same process

[pearl pond]

@late jolt Has your question been resolved?

[pearl pond]

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[pallid fossil]

Can anyone check my work

On the first problem

[pallid fossil]

Or ig answer lol

And also check my work on the second which is uhh

Ye

[regal herald]

check your range

for example:

[pallid fossil]

is the domain correct tho?

and also ye im listening

[regal herald]

,w 7-4^(-10)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by aldrnari_

[regal herald]

domain is fine

thats above 6

[pallid fossil]

oh

i didnt think about that

uhh

@regal herald so would that be 7?

[regal herald]

it is bounded above by 0 yeah, but is it closed or open

7 sorry

[pallid fossil]

wut

[regal herald]

) or ]

[pallid fossil]

oh

open?

[regal herald]

) yeah

[pallid fossil]

kk and can you check my other thing

[regal herald]

sure one moment

seems alright

[pallid fossil]

ok

uhhh

i have like 2 other things if ur interested in helping

[pallid fossil]

i can open a new channel if not

[regal herald]

feel free to blast them here

[pallid fossil]

kk

It's the 21 and 22

I'm like confused on what I needa do there

and that should be all lol

it should be kinda simple ig

[regal herald]

try writing 4^(x+3) as a power of 2

[pallid fossil]

ohhh lemme show u my work

[regal herald]

seems alright so far

what do you think you could do next to make the fact that 2^x=9 useful

[pallid fossil]

uhhh

i was planning on doign it manually

like uhh

its like x+6 so u just

uhh

put 9^6

or something

but you needa do 2x first

so idrk

[regal herald]

not too sure i follow

[pallid fossil]

uhhh for example

[regal herald]

thats not quite true

[pallid fossil]

idrk what to do then lol

[regal herald]

well, can you split $2^{2x+6}$ into the product of two things

[jolly parrot]

AℤØ

[pallid fossil]

oh ok

so uh

ok holdon

[regal herald]

no need to rush, take your time

[pallid fossil]

likee

2 to the power of x plus 2 to the power of x plus six?

[regal herald]

not quite how exponents work

[pallid fossil]

oh ok

[regal herald]

$a^b+a^c\neq a^{b+c}$

[jolly parrot]

AℤØ

[pallid fossil]

oh

[regal herald]

however

$a^b \cdot a^c = a^{b+c}$

[jolly parrot]

AℤØ

[pallid fossil]

ohh

so the equation would just be multiplication?

[regal herald]

wdym by that?

[pallid fossil]

2 to the power of x times 2 to the power of x plus six

[regal herald]

well, yes but thats still not as helpful yet

$2^{2x}\cdot 2^6$

[jolly parrot]

AℤØ

[pallid fossil]

ohhh

ic

what would 2x be tho

because like uhm

its 2 times x

so depending on. what x is

idrk

like uhh

it would be different

[regal herald]

$a^{bc}=\left(a^b\right)^c$

[jolly parrot]

AℤØ

[regal herald]

so you have $\left(2^x\right)^2 \cdot 2^6$

[jolly parrot]

AℤØ

[pallid fossil]

ohh

is the answer 72? @regal herald

[regal herald]

72 seems a bit low

[pallid fossil]

uhh

wait im confused

[regal herald]

its 2^6, not 2^3

[pallid fossil]

dont you square root it tho?

[regal herald]

ah i see whats happened here

you cant square root everything, otherwise youll have the square root of the answer you actually want

[pallid fossil]

ohh

wait is it 9 times 9 times 8?

[regal herald]

9^2 * 2^6

[pallid fossil]

uhhh

yeah

so its 648??

[pallid fossil]

81 times 8

so

ye

i think

[regal herald]

2^6 isnt 8

8 would be 2^3

[pallid fossil]

ohh

ic

81 times 64?

[regal herald]

indeed indeed

[pallid fossil]

5184?

[pallid fossil]

i think like thats it

i did it on paper tho

so idk

[regal herald]

seems right

[pallid fossil]

ok and what about the second problem

D

D

:D

dammit my colen isnt working

lol

[regal herald]

any thoughts

[pallid fossil]

i can show my work lo

lol

[regal herald]

sure

[pallid fossil]

Does this make sense?

[regal herald]

its slightly hard to read

[pallid fossil]

ok lemme resent it

resend

[regal herald]

im not sure what youve done

[pallid fossil]

oh

[regal herald]

the 3rd line especially

[pallid fossil]

lol idk whwat i did either

[regal herald]

also that x^2 is in the exponent of the 3

[pallid fossil]

uhhh

lets jsut start from the beginning

ill erase it all

[regal herald]

so you started off
$$\frac{3^{x^2}}{3^{2x}}=3^3$$

[jolly parrot]

AℤØ

[regal herald]

you then did
$$3^{x^2}=3^3 \cdot 3^{2x}$$ which is fine

[jolly parrot]

AℤØ

[regal herald]

after that you went a bit off

[pallid fossil]

oh ic

so uhh

[pallid fossil]

What did I do wrong lol

[regal herald]

you said that 3^3 * 3^2x= 3^5x

which isnt at all true

[pallid fossil]

oh

uhh

idk rlly what to do from here

[regal herald]

remember this?

[pallid fossil]

yes

[regal herald]

use it here

[pallid fossil]

oh okk

[regal herald]

although thats true

i dont see the usefulness in doing that

[pallid fossil]

oh

[regal herald]

i was thinking more on the lines of combining

rather than separating

[pallid fossil]

oh

3 to the power of 2x plus 3?

[regal herald]

yeah

and if $$3^{x^2}=3^{3+2x}$$ then what can you infer?

[jolly parrot]

AℤØ

[pallid fossil]

uhh

i can do that

is the answer

√-3

[regal herald]

cant be im afraid

[pallid fossil]

oh

[regal herald]

how did you come to that?

[pallid fossil]

uhhhhh....

stupid stuff

[regal herald]

hmmmm

[regal herald]

the intention was right, the process was not though

[pallid fossil]

oh

[regal herald]

im not too sure how you did that

[pallid fossil]

idk either lol

[regal herald]

so, x^2=2x+3
so x^2-2x-3=0

do you know how to solve quadratics?

[pallid fossil]

No lol

Sorry

[regal herald]

hm, that may make things difficult

[pallid fossil]

Yeah

[regal herald]

do you know anything of quadratics?

[pallid fossil]

Not rlly sret

Srry

[regal herald]

seems odd
the stuff youre doing would generally be done after learning stuff like that

[pallid fossil]

It's an extracurricular

😅

[regal herald]

ah i see

i can get some videos if youd like

[pallid fossil]

Oh uhh

[regal herald]

https://youtu.be/qeByhTF8WEw

https://youtu.be/LKm3JVH5P7o

[pallid fossil]

wut

😭

[regal herald]

theres not much you can do to finish the question if you cant do one of these two

[pallid fossil]

im in pain

[regal herald]

maths requires a foundation, it builds on its self

youre missing a few things

theyre not as bad as they may look though

give them a go

[pallid fossil]

uhhhh

[pallid fossil]

im so confused

[regal herald]

how so

[pallid fossil]

its uhh

idrk

how to explain

[regal herald]

give it your best shot

[pallid fossil]

Was it uhh

[pallid fossil]

Negative three and 1

[regal herald]

very close

[pallid fossil]

uhh can you check?

[regal herald]

did you factorise it or use quadratic formula?

[pallid fossil]

yeah

[regal herald]

careful
its x^2-2x-3=0
so a=1, b=-2, c=-3

youve written the formula with b as 2

[pallid fossil]

Yeah what

[regal herald]

$\frac{-(-2) \pm \sqrt{ (-2)^2-4(1)(-3)}}{2}$

oops

[jolly parrot]

AℤØ

[pallid fossil]

Yeah no ic what I did

[regal herald]

it just caused a sign issue

should be x=-1 and x=3

[pallid fossil]

Yeah

I realized

[regal herald]

snazzy

[pallid fossil]

Loll

Tysm

.close

[pearl pond]

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[trail vine]

can anyone assist me as to how to approach this question

[trail vine]

.close

[pearl pond]

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[naive flicker]

more funny limits

[naive flicker]

let me write equation

$$
\lim_{x \to 2}{\frac{\sqrt{x^3+1}-3}{x^2-7x+10}}
$$

[jolly parrot]

kisek

[naive flicker]

0/0

so i need

1. get rid of irrationality
2. find critical multipiler

and when i try to do it

im kind of stucc

$$
\lim_{x \to 2}{\frac{(\sqrt{x^3+1}-3)(\sqrt{x^3+1}+3)}{(x^2-7x+10)(\sqrt{x^3+1}+3)}}
$$

[jolly parrot]

kisek

[naive flicker]

so far so good

then i use rules of squares~ (a-b)(a+b) = a^2 - b^2

and get:

(ill not write limit this time)

$$
\frac{(x^3+1)-9}{(x^2-7x+10)(\sqrt{x^3+1}+3)}
$$

[jolly parrot]

kisek

[naive flicker]

is it ok that root chills on bottom?

[rigid barn]

yeah

but have you learned about L'Hopital's rule?

[naive flicker]

intuitively, we dont have this yet, but i feel like dividing everything by highest power does something with it

if i get something like infinity/infinity, then i need to just find how different derivatives are, right?

like, if one rises real fast, like x^2, and other is just x, then on infinity x^2 / x is going to be just x, and one infinitely large number is infinitely larger than other

[rigid barn]

right

L'Hopital's rule is take derivative of top and bottom, which is similar to what you're saying

[naive flicker]

i find derivative approach cooler, because it has some sence behind it, it isnt just blunt algorithm telling you to divide everything on highest power

[rigid barn]

well dividing everything by highest power can be useful for evaluating convergence but in this case it seems like that would get really messy

and actually where you left off if you divide everything by x^3 you get it converging to 1/0

I think

[naive flicker]

to do this i need to get rid of root, i think

[rigid barn]

but actually if you take the inital thing and divide everything by x^2 it looks like (without me writing it all out) that it will converge to 0

[naive flicker]

oh wait

i can use critical binom

lim x-->2

so it is (x-2)

[rigid barn]

I just looked at the graph and that was completely wrong lol

[naive flicker]

and if i extract (x-2) from upper and bottom part, i can use fraction properties to eliminate them and then just substitute x with 2

[rigid barn]

how would you extract (x-2) from upper without just getting left with 1/(x-2)?

because subbing in 2 for x then you get 1/0 right?

[naive flicker]

x^3 + 1 - 9 = x^3 - 8
nice
then ill just divide one on another:

x^3 - 8 / x-2 = (x^2-2x+4)

[rigid barn]

oh I see

ok it looks like you've got this pretty well under control, I have to go to bed

[pearl pond]

@naive flicker Has your question been resolved?

[cosmic charm]

@naive flicker still need help?

[naive flicker]

close it

[pearl pond]

@naive flicker Has your question been resolved?

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[pearl pond]

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[proven abyss]

Got a different answer from a friend who is also good at math, dk who's right, maybe I'm wrong?

[solemn fog]

try to remove variable from base of logarithm using properties

[proven abyss]

already done everything my answer was 8/u + 3/w

I just need confrimation

[pearl pond]

@proven abyss Has your question been resolved?

[pearl pond]

@proven abyss Has your question been resolved?

[pearl pond]

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[next dragon]

Hey can y help me

[pearl pond]

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@next dragon Has your question been resolved?

[pearl pond]

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[zenith ocean]

is it 64

i stiil found 64 with considering duplicatein colars

[zenith ocean]

ow

lemme try again

we can only see 3 sides right

damn

thank you

yep

sounds correct

sory for not helping you out

[pearl pond]

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[brazen mauve]

need help rq

[brazen mauve]

how can I solve it I have no idea

[echo oxide]

x-1+1 =x

try using that and solve for b

[brazen mauve]

for right side? right?

wait solving for b

ill try

..

i didnt change anything

its still x

:/

[echo oxide]

show work

[brazen mauve]

you said x

[brazen mauve]

how can i use it for the left side function

it will remain the same..

[sturdy imp]

well you know a is 4
because the coefficient of x² in
a(x-1)^2
is the same as in ax²

then just plugin x-1, find the new function

in terms of b

[brazen mauve]

so i replace the coefficients then plug (x-1)

all of the coefficient

right?

[sturdy imp]

compare coefficients to find a

then plugin and compare f(x-1) in terms of b to f(x-1) to what they give u

[brazen mauve]

im sorry but i have 3 mins can u give me the answer
and ill try to solve it after

nvm i have 1 more 24h

i can think of it

alr solving time

[brazen mauve]

yep
got it

i did it like that

[brazen mauve]

alr

i did something diff i think it work
i got b = 6
is it right?

i used (x+1) in the equation f(x-1)
because f((x+1)-1) equals to f(x) :D

is that right?

[pearl pond]

@brazen mauve Has your question been resolved?

[brazen mauve]

not sure yet

need an expert confirm :)

[pearl pond]

@brazen mauve Has your question been resolved?

[pearl pond]

@brazen mauve Has your question been resolved?

[brazen mauve]

no i got it

b=(-6)

some signs mistakes

[misty galleon]

f(x-1) = a(x-1)^2 + b(x-1) + c = 4x^2 - 2x + 4

try this

[brazen mauve]

ooh
i finished already
forgot to close the chat, it was as i said, plugging (x+1) in the right side function
we get -6

thank u for trying anyways

.close

[pearl pond]

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[neat cave]

what am i supposed to do in the inductive step

[pearl pond]

@neat cave Has your question been resolved?

[pearl pond]

@neat cave Has your question been resolved?

[pearl pond]

@neat cave Has your question been resolved?

[potent hedge]

Compute $1 - \frac{F_{n+2}}{2^{n}} + \frac{F_{n}}{2^{n}}$

[jolly parrot]

black_couscous#1618

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[glacial badger]

<@&286206848099549185>

man any help

[strong atlas]

!15m

[pearl pond]

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.

[glacial badger]

i did 15 Minutes

but my channel has gone idk

[pearl pond]

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[tall galleon]

Help with 2

[pearl pond]

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[scenic flame]

whats wrong with my solution?

[scenic flame]

this is original question **

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[minor fox]

Can anyone help me?

[pearl pond]

Please don't occupy multiple help channels.

@minor fox Has your question been resolved?

[lucid idol]

can u specificy in which questions u need help with?

[lucid idol]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[pearl pond]

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[severe steppe]

i im not sure if my answers are correct for the factoring perfect square trinominals worksheet

[stoic path]

Which one?

[severe steppe]

13

my senses tell me its 5 squared and 8 squared

15 is (4x - 5) squared right

or no

[stoic path]

Are you allowed to use the quadratic formula?

[severe steppe]

yes i think

i heard that word in class

[stoic path]

Or or...
You could factor 5 out
5(25x² - 40x +16)

And then factorise whatever's left

[severe steppe]

yes

wait itsnt it

5(5x-4)^2

[stoic path]

Is it?

Oh yeah

[severe steppe]

OH YEAHH

so basically

you have to find 2 numbers that go into eachother>

?

[stoic path]

But in general, you could either split the middle term or use quadratic formula

[stoic path]

Wdym

[severe steppe]

like 5 and 4

5 goes into 125

[stoic path]

For perfect squares that works, yes

[severe steppe]

4 goes into 200 ? i think

yes

[severe steppe]

mkay mlkay

ok thank you i need to do it know

[stoic path]

I just found the highest common factor of 125,200 and 80

And then factored 5 out

[severe steppe]

yes

then i do that for all

ok ok

thank you sir or maam or pal

have nice day

[stoic path]

You too

[pearl pond]

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[urban raft]

oxbridge interview question pretty lost

[urban raft]

usually there is some key insight in questions like this that helps you lol

[eternal tulip]

plugging in -x or -y tends to give you a good idea of symmetries, and -x -y

[urban raft]

yea fair i guess its symmetric i.e if (x,y) work, so do (y,x)

but im a little lost on how to plot this on paper

desmos gives a weird graph

ps: anyone have a good lecture on sketching implicit? functions like this

[lament dagger]

You can insert polar coordinates

[urban raft]

brilliant lets see

[lament dagger]

and then you get a function $r\left(\phi\right)$

[jolly parrot]

Gapi

[urban raft]

is this pre uni math? personally never seen this but ill give it a shot thanks bro

[eternal tulip]

yeah, sometimes its taught in precalc

$x=r\cos(\phi)$, $y=r\sin(\phi)$

[jolly parrot]

Moosey

[eternal tulip]

sometimes instead of phi is theta

most of the time i've seen theta

[urban raft]

yea phi and theta is just difference in notation tho right

[eternal tulip]

ye

[lament dagger]

Yeah

I've always seen phi

[urban raft]

never seen it weird i guess different curriculum and its not that important lol

[lament dagger]

Its just one letter, it doesnt matter that much

[urban raft]

yea no im saying ive never learnt to plot graphs using polar coordinates

[lament dagger]

There might be a trick to this exercise (so that you dont need polar coordinates), but I dont see it

[urban raft]

im pretty sure there should be because that doesnt seem very interview like but eh

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[glacial badger]

hello

[glacial badger]

is there a way to tranfer a fraction in of of the sides in prove the identify

[pearl pond]

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[pearl pond]

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[glacial badger]

<@&286206848099549185>

[midnight haven]

I didn't understand your question

If you wanna prove it, so I think it's useful to multiply both sides by cos²x

So it'll be equal to $\frac{tan^2(x)+sin^2(x)+1}{cos^2x}{cos^2(x)}=(1+sin^2(x) \cdot cos^2(x)) \cdot cos^2(x)$

So, it'll be equalto

[jolly parrot]

Palahoo

[midnight haven]

$\frac{sin^2(x)}{cos^2(x)}+sin^2(x)+1=cos^2(x)+sin^2(x) \cdot cos^4(x)$

So, we can multiply the top and bottom by cos²(x) without affect the term

So it'll be equal to:

$\frac{sin^2(x)}{cos^2(x)}+\frac{sin^2(x) \cdot cos^2(x)}{cos^2(x)}+\frac{cos^2(x)}{cos^2(x)}=cos^2(x)+sin^2(x) \cdot cos^4(x)$

[jolly parrot]

Palahoo

Palahoo

[midnight haven]

It is equal to:

$1+sin^2(x) \cdot cos^2(x) = cos^2(x)+sin^2(x) \cdot cos^4(x)$

[jolly parrot]

Palahoo

[midnight haven]

Isn't this wrong? $1+sin^2(x) \cdot cos^2(x) = cos^2(x) \cdot (1+sin^2(x) \cdot cos^2(x))$

[jolly parrot]

Palahoo

[midnight haven]

Does it have the cos²(x) on the bottom?

@glacial badger

[pearl pond]

@glacial badger Has your question been resolved?

[pearl pond]

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[crisp mango]

I tried using this theorem for this problem

[jolly parrot]

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

[crisp mango]

i tried doing this

and i got x=45/62

which isnt correct

is this not suppose to work?

[autumn topaz]

hmm. maybe try figuring out how long it'd take for pipes 1 + 2 together. then use that value for "a" and you want to solve for "b" (which will be the time for the third pipe)

[crisp mango]

i got 3.2

still isnt correct

the correct answer was 2.8

oh wiat

yeah

i got it

thanks

[autumn topaz]

2.8?

[crisp mango]

yeah

.close

[pearl pond]

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[carmine root]

how am i supposed to show this:

[carmine root]

when i dont get the same values out

I am supposed to show it is equal but i dont the same out

[humble lintel]

It is probably easier if you look at det(B_21) and det(B_31) instead , what you wrote here doesn't look right

[carmine root]

if i remove the first row and second column

i am left with b,h,c,i?

or am i wrong

[humble lintel]

Yes, but the coefficient is then d not b

[carmine root]

oh

i thought that was random constants

so a is the same a that is the matrix

[humble lintel]

No, you cross out row 1 column 2

the constant is the overlap which is d

[carmine root]

ok nice

so the constans are

a

d

and g?

[humble lintel]

Yes

I was suggesting you go down the column instead so that the constants are still a b c

but that works too

[carmine root]

but now that i have the right constans

the matrices still arent the same

oh wait

this is the same?

only the first one seem to be similar

[humble lintel]

You'll need to do some rearranging

Ignore the first term because they are the same

Now you have -bdi -bfg on top

and -dbi -dch on the bottom

[carmine root]

doesnt it become +bfg

[humble lintel]

-bdi is the same as -dbi so that's fine. But there is no bfg. However on the third term, there is a gbf

[carmine root]

when there is a - in front

[humble lintel]

Yes that is a typo

You should get the idea though

[carmine root]

now we have this

ought tihs to be equal to anything

[humble lintel]

Just take your previous image and circle the matching terms

[carmine root]

you mean the fact we have an i in both equations?

[carmine root]

i think i got it

[humble lintel]

Honestly, if you just draw those equations on 2 lines and just draw arrows connecting them like I did, whoever is marking this will be happier than reading the actual gory details

[carmine root]

if you want it like that

then this line should satisfy you

at any rate, thanks for the help

[pearl pond]

@carmine root Has your question been resolved?

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[shell dove]

just need some help checking, are events V and C independent? I have that they are not.

[shell dove]

not independent because i have found that P(V n C) does not equal P(V) * P(C)

[pearl pond]

@shell dove Has your question been resolved?

[shell dove]

<@&286206848099549185>

[shell dove]

.close

[pearl pond]

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[rapid marten]

Uhh I did

[rapid marten]

7sin(π/4(x))-3

WHAT THE

WHY DID IT SEND LIKE THAT

7sin(π/4(x))-3

I have no clue what the +c means

[pearl pond]

@rapid marten Has your question been resolved?

[rapid marten]

<@&286206848099549185>

[opal pawn]

the c term will say how much your graph is shifted horizontally, similar to how the d term gives how much the graph is shifted vertically

[rapid marten]

Do I need anything there

[rapid marten]

Wat ur thoughts on my thing fr

@opal pawn

Yo wait

Idk

Wouldn't that only work with like cosine idek

[opal pawn]

i think it's a good start, though we should add a c term and maybe change our b term
if we think about an u altered sin(x) graph, the (0,0 point is our midpoint and we increase
given that, is there a horizontal shift in our given graph?

[rapid marten]

Umm

Maybe

Probably

Yes

Possibly

@opal pawn

I have no clue

Yo I don't remember what a sine graph looks like

OK I SEARCHED IT UP

Uhh maybe like +2.5 for C

[opal pawn]

yeah we would need to end up shifting it to the left, though 2.5 is not quite the right value. Could you explain how you got 2.5?

[rapid marten]

uhhhh

I went like backwards from the 0

Oh wait

Maybe like

+1.25

So if u reversed it it would go through the middle point like normal

[opal pawn]

yeah, that sounds right to me

[rapid marten]

FR

OK now the period

Maybe like

2π/5

@opal pawn

Wat are ur thoughts on halothane

[opal pawn]

yeah so if you know that one fourth of a period is 1.25, i think it would be reasonable to say 2pi/5 is b

[rapid marten]

Yo I didn't even know that

OK SO

7sin(2π/5(x)+1.25)-3

[rapid marten]

Wat u think

Plz return

[rapid marten]

@opal pawn

@opal pawn

PLZ CONFIRM

@opal pawn

[rapid marten]

<@&286206848099549185>

Is this right

@opal pawn

[rapid marten]

<@&286206848099549185> is this right

<@&286206848099549185>

[sick creek]

You could calculate a few points and see whether they fall on the plot.

Checking is good, even if you have no reason to doubt yourself.

[rapid marten]

Guys I got it wrong😭

I don't even know wat they did😭

How did they get π/2@sick creek

@g.spark888

[weak stratus]

the +1.25 was inside the parenthesis

[rapid marten]

Oh

[weak stratus]

so they distributed the 2pi/5 in

[rapid marten]

😭

.close

[pearl pond]

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[crisp mango]

Tom left Paris with a speed of 68 mph. Fred also left at the same
time in the opposite direction at a speed of 64 mph. Find how many hours
Fred must travel before they are 218 miles apart.

The distances they travel must add up to 218, so 64x+68x=218, x=1.65

[crisp mango]

i just wanna check if my solution is correct

[pearl pond]

@crisp mango Has your question been resolved?

[dull pond]

Its rather close

But its not exactly right

64(1.65) + 68(1.65) appears to be 217.8

0.2 error

Which is

Less than 1% error

But its there

[crisp mango]

huh

[crisp mango]

why though

that means my solution was wrong lol

what did i do wrong

[dull pond]

64x + 68x = 132x , 132x = 218 ==> x = 218/132

1.65 is a good approximation of 218/132

But its not exactly equal to it

[pearl pond]

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[sullen leaf]

Could someone help me with this specific problem?

[hollow cobalt]

!status

[pearl pond]

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

@sullen leaf Has your question been resolved?

[sullen leaf]

Sorry, for part A I got a revision back that said, for the very first statement, I need to "back up one more step: What must be shown to see that by definition, f is well defined" For the second statement, where it says "which shows that n divides ka-kb", I had to explain why that was the case. i've since edited my work for part A, and I'd like to share it to see if I satisfied expectations.

[hollow cobalt]

What had you shown for part a?

[hollow cobalt]

Perhaps they wanted you to mention that you need to show f([a]n) = f([b]n) first?

And for the second statement I guess I'd put because n divides kmn and ka - kb is equal to the latter

[sullen leaf]

I'm getting this;

Is this better to put before the statement "I need to show that ka_n = kb_n"

[hollow cobalt]

Yeah

[sullen leaf]

I also must ask, do you know what they mean when they say "First, need a 'Let...' statement"

[hollow cobalt]

Probably "Let [a]n, [b]n in Z/nZ such that f([a]n) = f([b]n)"

[pearl pond]

@sullen leaf Has your question been resolved?

[pearl pond]

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[pearl pond]

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[river aurora]

For question 42: Am I solving this integral correctly? I have hit a bit of a dead end and don't know where to go from here (attaching work shortly)

[toxic lichen]

why not recognize that e^(2t) + 2e^t + 1 = (e^t + 1)^2 lol

you've screwed yourself over kinda

[river aurora]

Sorry?

So I shouldn't distribute?

[toxic lichen]

from here

factoring out e^t was kind of pointless

[river aurora]

I did it so that I could substitute U but that didn't really work

Oh wait I see what you mean