help-42

[ancient helm]

yes 0's inverse is 0 and 1's inverse is 1

[silent sinew]

nice

final one they haven’t asked us to prove but how do you prove it’s associative

[ancient helm]

as for the fact that it's associative

probably just make a table

[silent sinew]

do I need to do a third column?

[ancient helm]

or show all 8 multiplications by hand

or anything of that nature

[silent sinew]

nice

okay will write this proof up

[ancient helm]

yeah, if you make a table make a column for (a + b) + c, as well as a + (b + c)

[silent sinew]

yeh

but ⨁ but ikwym

this is actually so easy it’s just difficult to figure out what your trying to do

[ancient helm]

?

[silent sinew]

yeh wrote +

nvm

[ancient helm]

oh

[silent sinew]

ty for ur help x

.close

[calm coral]

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[ancient helm]

too lazy to write $\bigoplus$ 4 times

[potent lotus]

chebyshev's infinite pee norm

[silent sinew]

lol

[calm coral]

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[devout nest]

can someone help me with figuring this out because I'm not really sure what (a) is saying

like what does it mean, "no difference between the mean scores"?

this is A levels btw

[calm coral]

@devout nest Has your question been resolved?

[calm coral]

@devout nest Has your question been resolved?

[calm coral]

@devout nest Has your question been resolved?

[timid moon]

I think it means H0: Mean Y7 = Mean Y8

then H1 is that they are not equal, therefore its a two tailed test

it’s a bit weird because the table says they aren’t equal but you are testing it to a certain significance level so you are trying to prove whether they differ a significant amount or not

[calm coral]

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[timid moon]

.reopen

nvm

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[jolly crest]

Why does adding (-n+1 c 0) with the extended summation make it zero? (referencing to the large highlighted part)

[jolly crest]

nvm got it now

.close

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[urban musk]

Is there someone who can help me with homework

[urban musk]

I just really forgot how to do this stuff

[pliant horizon]

which one in particular

[urban musk]

The fraction stuff

[pliant horizon]

so yk the general form of the equation of a line

[urban musk]

Nah

[pliant horizon]

yk how you did q1

[calm coral]

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[calm coral]

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as this action is irreversible, and this channel may abruptly lock.

[sinful pasture]

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[calm coral]

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[remote mural]

i derived the formula for complex exponentiation x^(a + bi) as being e^(a * log(|x|) + b * arg(x) * i), is this correct?

[remote mural]

,rotate

[north grail]

what did u do to ln(x) in the first line

[remote mural]

i expanded it via the expansion in the second line

[remote mural]

also this

[north grail]

is x in C too?

[remote mural]

yeah

[north grail]

why use log then

[remote mural]

wait, what are r and k, theta and phi?

[north grail]

r is |z_1|

k is |z_2|

phi is the argument of z1

theta is the argument of z2

[remote mural]

so z^w is just |z| * e^(i arg(z) * w)?

[north grail]

its by law of exponents

[remote mural]

yeah

why didn't i think of that

[north grail]

(2^2)^3 = 2^2*3

[remote mural]

it's so much simpler

[north grail]

yeah

[remote mural]

why are you expanding z_2 there?

[north grail]

u dont have to

i just wanted to write it in exponential form

[remote mural]

ok

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[ionic wave]

Can someone help me with this trig problem

[calm coral]

@ionic wave Has your question been resolved?

[calm coral]

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[honest kite]

in an equation like this, what can I do to solve for x?

[pure kayak]

make use of

2^(ab)=[2^a]^b
[a^b/c^b]=[a/c]^b

[honest kite]

but what happens with the three that is in the power of the 2?

thats the part that confuses me

[pure kayak]

refer to my first identity

[honest kite]

i got it

so it turns out to be 3^x/8^x

and then log and i get x

[pure kayak]

sounds good

[calm coral]

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[blazing coyote]

$\left(\int \frac{\left(x^{2024}-1\right)}{x^{506}-1}dx\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

could I have a hint

using (a-b)(a+b) doesn't work

maybe I sub $x=e^{it}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[swift laurel]

doesn't it?

[blazing coyote]

I don't think so

it's 506

not 512

[swift laurel]

2024 is not a power of 2

[blazing coyote]

oh

2024

right

thanks

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[coral dew]

How should I solve this?

[coral dew]

By factoring

[minor knot]

you have a quadratic in the form

,,ax^2+bx+c

[potent lotus]

Renz

[minor knot]

do you know the method to factor quadratics?

[coral dew]

Kind of

[minor knot]

what do you remember

[coral dew]

We did it a long time ago

There a b and c

We played this game like what multiplied equals this but adds to this

[minor knot]

yea thats right

you need to find factors of c

that add up to b

what two numbers, when you multiply becomes c

but if you add them, you get b

[coral dew]

2 and -1 ?

[minor knot]

yea thats exactly

so your factored form becomes

,,(x+2)(x-1)=0

[potent lotus]

Renz

[minor knot]

you just add it to x

thats how you factor quadratics

[coral dew]

But which one is Y and which is X

[minor knot]

Y?

theres no y there

only x

since its an equation

[coral dew]

Ohh ok I thought the answer was asking for points like on a graph

[minor knot]

in the variable x

[coral dew]

But it’s asking for 2 X numbers?

[minor knot]

yea

now, they are equal to 0

meaning, its either the first factor or second factor must equal 0

[coral dew]

And I don’t need to go further like x +2 = 0 and then -2 -2 so x = -2 ?

[minor knot]

yea thats right

,,(x+2)(x-1)=0

[potent lotus]

Renz

[minor knot]

you have this right

you can say that:

,,x+2=0\x-1=0

and solve for each x

[potent lotus]

Renz

[coral dew]

To get the answer ?

Or do I already have the answer when I have (x+2)(x-1)=0

[minor knot]

yea you gotta solve each to get the answer

[coral dew]

Ok I got -2 and 1

[minor knot]

yea thats right

[coral dew]

Ok awesome! Also I’m having trouble with another question

[minor knot]

let me see

[coral dew]

This

I tried and it’s not -5 , 0

[calm coral]

@coral dew Has your question been resolved?

[calm coral]

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[snow garden]

hello

[snow garden]

quick question

is this correct? or do i need to do something else afterwards

<@&286206848099549185>

[finite iris]

Seems fine

[snow garden]

Thank you!

Do you mind checking this as well?

[calm coral]

@snow garden Has your question been resolved?

[calm coral]

@snow garden Has your question been resolved?

[calm coral]

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[jade wing]

howdy! i need help trying to isolate a variable where x1 and y1 are constants and xc and yc are variables. i need to get to either an xc = or a yc= but i don't know how

[jade wing]

this is my other work, trying to expand everything and replace variables with other variables to make it easier to read but I can't think of a next step\

the 3rd line was trying it with the xc =1, yc=1 but that didn't get me anywhere

[untold compass]

move this over then square root

[jade wing]

woah i can't believe i didn't think of that 😭 thank you so much

have a great day!

.close

[calm coral]

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[jade wing]

.reopen

[calm coral]

✅

[jade wing]

okay follow up question, this one is a bit harder. how do i now get the yc on its own

i tried expanding the left side but that just ended up with square roots which is exactly what i was trying to avoid

i realize also that you can sqrt both sides and move the sqrt over but then im stuck with this which I am not sure how to get out of

the y2 and y1 seem to be getting in the way of everything

and there's this as a next step too

[untold compass]

I would give up

[jade wing]

hm

good point

alright, well, thanks for your efforts i guess

.close

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[novel cipher]

for this question where i have to factor the quadratic expression how do i find a number which is a product of ac but sum of b when the numbers aer so big. i know i factor out -2 but thats still 49*32 which is 1568 and i have to find a number which is the sum of 84 and product of that. how do i do that

[thorny tulip]

if you focus on only 98 and 168

[drifting seal]

so find the prime factors

and rearrange them

a few times

to see how you can make 168

[drifting seal]

of 98 and 64

i’d pull out the 2 first for sure though

-2

so you’d get -2(49x^2+79x+32)

the prime factors of 49 and 32 are 7,7,2,2,2,2,2

how can you rearrange the order of multiplication to get 79

like maybe multiply a couple 2 with a 7 and the rest of the twos with the other 7

and see if they add to 79

and just find the configuration

oh wait

this isn’t factorable with integers

[calm coral]

@novel cipher Has your question been resolved?

[calm coral]

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[fast stump]

please explain how they did line 3 they turned all of the xs and zs into t

[muted lotus]

They're using the parameterization

[calm coral]

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[fervent fox]

quick qn just got into differentation and im not sure how to differentiate 14/5x

[clever ruin]

Is that $\frac{14}{5x}$ or $\frac{14}{15}x$?

[potent lotus]

Max

[fervent fox]

1st one

tho 2nd one shld just be 14/15 right? if im not wrong

[clever ruin]

Okay what rules have you met in differentiation so far?

[clever ruin]

Yea good

[fervent fox]

power rule

am i supposed to use quotient rule?

[clever ruin]

Quotient would work but it's very inefficient

[fervent fox]

er cld be nice if u teach me abt chain rule too havent rlly went through it in clas

[fervent fox]

oh

then how would i do it?

[clever ruin]

$\frac{14}{5x}=\frac{14}{5}\frac{1}{x}=\frac{14}{5}x^n$$

[potent lotus]

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

[clever ruin]

What would n be?

[fervent fox]

0?

after differnatiating

[clever ruin]

I'd suggest studying indices first

[fervent fox]

oh

accl er

-1

[clever ruin]

How can you write $\frac{1}{x}$ as a power

Geah

[potent lotus]

Max

[clever ruin]

Yeah

So you are differentiating $\frac{14}{5}x^{-1}$

[potent lotus]

Max

[clever ruin]

You can use the power rule now

[fervent fox]

so it would be -14x^-2/5?

[clever ruin]

Watch some videos on YouTube on it.

Essentially you differentiate the outside normally. Put the inside back in. Then multiply by the derivative of the inside. But first what a video on YouTube to understand it better.

[clever ruin]

Yep

[fervent fox]

ohh

[clever ruin]

$-14x^{-2}/5_

$-14x^{-2}/5$

[potent lotus]

Max

[fervent fox]

ohhh

hmm i see alright thanks a lot!

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[minor knot]

what have you tried

[molten latch]

c = 4n

-8c = 10n

[minor knot]

thats kinda not it

lets start from nina

We want to

let n = nina

if nina is n, and charles is four times as old, what would be charles age in terms of n

[molten latch]

4c = n?

[minor knot]

we can express charles age as 4n

because n is just nina's age

[molten latch]

oh

i thought u said its wrong

[minor knot]

you dont need to have a seperate variable for charles

Alright

so we have

Nina: n
Charles: 4n

Nina: n$\$ Charles: 4n

[potent lotus]

Renz

[minor knot]

now, the problem states that charles was 10 times as old as nina 8 years ago

meaning, we have to create expressions for their age, 8 years ago

if nina is "n" years old today

what will be the expression for that age 8 years ago?

[molten latch]

n-8

[minor knot]

correct

how about charles

charles is 4n

so 8 years ago would be?

[molten latch]

4n (n-8)?

[minor knot]

nope

we simply just subtract 8 again

since its 8 years ago

so what would it be?

[molten latch]

4n - 8

[minor knot]

ye thats right

Nina: $n\$ Charles: $4n\$8 Years ago:$\$Nina: $n-8 \$Charles: $4n-8$

[potent lotus]

Renz

[minor knot]

now, what does the problem state about their ages

8 years ago

[molten latch]

x10

[minor knot]

yea

charles is 10 times as old

meaning for their ages to be equal

you have to multiply nina's age by 10

since we are talking about 8 years ago

we are going to use the expression for their ages 8 years ago

so we have

,,4n-8=10(n-8)

[potent lotus]

Renz

[molten latch]

8, 32

I just got confused i thought it was a simultaneous eq in the beginning

Thx for the help though

[calm coral]

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[tranquil salmon]

is there a relation between a spacial dimension of a special orthogonal group and the dimension of the group
the dimension of SO(2) is 1
(only one parameter)

[tranquil salmon]

now i want to demonstrate that the dimension of SO(n) is
n(n-1)/2

how to start?

[remote mural]

Lie groups are probably a bit too advanced for the help channels sorry

try ur luck with #advanced-algebra perhaps?

[tranquil salmon]

Alright thanks

.close

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[thick mica]

If you take the absolute value of a series and it converges it is absolutely convergent

[glacial inlet]

root test does so as well

but the alternating series test only proves that the alternating version of the series converges, not the absolute value of it

[thick mica]

Ye

[glacial inlet]

although it does seem that we are taking the absolute value when we take the limit, we’re essentially doing the divergence test on the absolute value of the series, which is inconclusive if the limit equals 0

that condition is essentially just the divergence test on the absolute value of the series

but if the limit evaluates to 0, that doesn’t necessarily prove that the absolute value version of the series converges, the test is inconclusive

yes

the original series converges because it satisfies the alternating series test

but the absolute value of the series is inconclusive

yw!

[calm coral]

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[weak rapids]

guys i spent 2 ufking hours doing this

and this is all what i could do

and at last my cubic is wrong somehow

[remote mural]

The best answear ever you can get for this problem: https://youtube.com/shorts/JOVfP0AKz2k?si=PJUM8KjLLu83CzK0

[weak rapids]

bruh

[worthy heath]

[cursive linden]

im not reading all that

[weak rapids]

bruh

pls

[cursive linden]

just try to make two elements of a particular row/column zero

[weak rapids]

dont do this to me

[cursive linden]

and you can calculate the determinant easily

[cursive linden]

(revenge for what you did to my preamble)

[weak rapids]

bro i didnt

you messed it up

lmaoo

[cursive linden]

now do the problem 💀

[weak rapids]

this is one of the step

do you think i should apply operations here

and make them 0

second row of the first det and third row of the second det

[cursive linden]

yeah

[weak rapids]

okay

[cursive linden]

idk what you did there and if that is correct

but yeah

[weak rapids]

it should be correct

ok lemme do that rq

HO GYA

@cursive linden thanks

thankss

love you

love you

so much

.close

[calm coral]

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[cursive linden]

https://cdn.discordapp.com/emojis/779081117944119336.webp?size=64&quality=lossless&name=Flushed_Moyai

[weak rapids]

lmao

[calm coral]

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[split wharf]

0 * (-οο), indeterminate what do i do

[tired lichen]

pray

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

A better way is to multiply ln(x-1) in

[split wharf]

it just creates more indeterminate forms lol

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

yea I noticed too

[split wharf]

its not lnx - ln1 or sth

ln(x-1) cannot be simplified

[eternal shard]

I just realized something else

ln(x)ln(x-1)

[split wharf]

i cant make it work

what if i know the first ( ) is >= 0.

does it change anything

[eternal shard]

Try this maybe

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

Then you have $\lim_{x \to 1} \left ( \ln \left( \frac{x}{e} \right) + \frac{1}{x} \right ) \cdot \ln(x-1)$

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

Then it's what?

[split wharf]

lol

then nothing

[eternal shard]

Oh it's still 0 times inf

Do we know that the limit converges?

[split wharf]

since there is ln(x-1) ig its x -> 1+

[eternal shard]

You should use this

I tried it

$\lim_{x \to 1} \frac{\ln(x-1)}{\frac{1}{\ln(x)+\frac{1}{x}-1}}$

[potent lotus]

𝔸dωn𝓲²s

[eternal shard]

L'Hopital twice

[potent lotus]

𝔸dωn𝓲²s

[split wharf]

i did 3 times hopital found it to be 1/2, when acording to graph it's 0. Awesome

[eternal shard]

hmm

[eternal shard]

Maybe this works out above

If you apply L'Hopital a second time

[calm coral]

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[calm coral]

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[fickle fiber]

not sure what to try here

[calm coral]

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[jolly crest]

I don't even know where to start with this

[calm coral]

@jolly crest Has your question been resolved?

[jolly crest]

<@&286206848099549185>

[warm horizon]

yo

[jolly crest]

yeah could you help me out with this question

I've tried to the best of my abilities but can't figure out just how I can make 2n-1 from the prefixes

[warm horizon]

You are gonna have to give me a solid 20 minutes to make some sense of this 😂

[jolly crest]

alright lmao

[warm horizon]

so

correct me if im wrong

but

this is asking for sig-

sig->usual

[jolly crest]

i didn't get you

[warm horizon]

sigma notation

to usual notation

[jolly crest]

ah yeah

yeah that's what it means

[warm horizon]

fuckkkkk man

that is such an odd question

[jolly crest]

for real though

gotten me so hung over this

[warm horizon]

i am beyond lost

[calm coral]

@jolly crest Has your question been resolved?

[calm coral]

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[calm coral]

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[fathom thistle]

I HAVE DOUBT IN THIS PLEASE HELP ME

[fathom thistle]

<@&286206848099549185> CAN YOU HELP ME IN SOLVING THIS

[formal pulsar]

damn, no need to yell

nah, but what you having trouble with more specifically?

[formal pulsar]

you trying ta prove this or smth?

[fathom thistle]

C OPTION

I HAVE DOUBT IN C OPTION

[formal pulsar]

can you turn the caps lock off pls??

also, why do you have doubt, is there a counter-example you could find?

[fathom thistle]

ok sorry

i got 'a' value =1

[formal pulsar]

so, a = 1 and b=0?

[fathom thistle]

no i dont get any b value as if i solve lhd and rhd they will never be equal for any value of b

rhd = 2b(2-2)

and lhd =1

so for any value of b this can never be equal

[formal pulsar]

how'd you get lhd = 1?

[fathom thistle]

in this case we got option a and b correct

[formal pulsar]

yeah sorry i dont really get what you mean. but i think its false too tbh

cos if you tke the counter example a = 0.7 and b= 0.3

[fathom thistle]

(a-a-1)-(a-h-a-1)/h=1

lhd

[formal pulsar]

are you saying that |x-a| - 1 = 1 at x=2 for all a? cos thats really only the case when a = 0 tho

[fathom thistle]

i have done like this

if at 2 it is not differentiable so the function may be continuous or may not be

and if function to be continous and non differentiable a=1

for function tobe non continuous and non diffrentable a is not =1

[calm coral]

@fathom thistle Has your question been resolved?

[fathom thistle]

No

According to it 2 option are correct option a and d but it is a single choice question

[calm coral]

@fathom thistle Has your question been resolved?

[vocal sigil]

@fathom thistle

Actually only a is correct

There are two non differentiable points

[calm coral]

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[rain copper]

@fluid edge Idk where that channel dissappeared

[fluid edge]

Bruh

Anyway

The equation I think u differentiate is z=x+iy

[rain copper]

But why not z = u+iv

I am confused 🙃

[fluid edge]

Because f is differentiateble by z=x+iy and not z = u +iv

[rain copper]

Z = x+iy
Then differentiate z wrt r ?

@fluid edge

[fluid edge]

Bruh

I confused my self 🙃

[rain copper]

🥲

[fluid edge]

@true glade can u help us out

[rain copper]

This thing is bothering me for a while

[fluid edge]

Partial differentiation is always annoying

[rain copper]

Yes 😦

Should I tag helpers ?

[strange lichen]

can you show your original problem?

[rain copper]

Yes

The second Equation

@strange lichen

[strange lichen]

Sorry I'm not familiar with analytic functions. Just wait for someone to help, or tag helpers

[rain copper]

Ok

<@&286206848099549185>

[calm coral]

@rain copper Has your question been resolved?

[rain copper]

Welp it's Emergency 🙃

[calm coral]

@rain copper Has your question been resolved?

[calm coral]

@rain copper Has your question been resolved?

[calm coral]

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[unique lynx]

this is rank 1 not rank 2 right?

a,b,c are real

[unique shale]

what is question

[unique lynx]

linear algebra

[swift laurel]

you're right that the matrix is rank one (provided one condition: what is that?)

[calm coral]

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[novel herald]

i actually feel so dumb i dont undesratnd this i think im just drawing the diagram wrong this is in my vector chapter 😢 how do i solve this

[novel herald]

<@&286206848099549185>

huh

did i do something

[glossy niche]

What the hell t rex?

<@&268886789983436800>

[novel herald]

bruh he deleted it

[novel herald]

<@&286206848099549185> anyone able to please help 🙏

<@&286206848099549185>

[calm coral]

@novel herald Has your question been resolved?

[novel herald]

<@&286206848099549185>

[tiny monolith]

what is your diagram?

[tiny monolith]

your diagram looks fine

use sine law

[novel herald]

ok this is what I’ve done but tre answer is 1h and 34 min also for a how would I find the direction 😢

[tiny monolith]

oh oops

the diagram isnt fine haha

they dont lie on the same line, the instead both lie on different lines

get two lines pointing north amd the angle is the reference angle to those lines

[calm coral]

@novel herald Has your question been resolved?

[novel herald]

Is this what u mean ?

[tiny monolith]

yess

slay queen

u got this

[novel herald]

girl im actually dying whys this so difficult 😭

its still not the right answer tho :(

[unique shale]

ı thınk they have a formula for this

ı will write when i found it

[novel herald]

okk thank you 🙏

[unique shale]

Crosswind Component: 55.15 m/s

Headwind Component: -9.72 m/s

What department are you studying?

[novel herald]

this is just year 11 maths 😭

[unique shale]

17 ?

[novel herald]

yes

[unique shale]

me too

but this

is hard

i think

[novel herald]

bruh

it actually is

idk

i feel like im overcomplicating it but then again how could i i feel like ive done some simple stuff

[unique shale]

true

where r u from

[novel herald]

aus

hmm

i feel like its the diagram but im not sure

ahh

i think i know

wait

[unique shale]

if u find please let m know @novel herald

[novel herald]

the question says the wind is FROM 100 degree so it should be facing the other way.. it’s pretty dumb 💀

bruh ur username lol

[unique shale]

☠️

?

[calm coral]

@novel herald Has your question been resolved?

[tiny monolith]

the formula not the answer!!

[tiny monolith]

hahaha

[novel herald]

Can’t believe it took us so long

.close

[calm coral]

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[rare bane]

Guys.. what do you think.. the answer will be.. 😁

[lofty cedar]

Compare row 2 and 3

@rare bane

[calm coral]

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[calm coral]

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[latent ocean]

how do i solve this

[grand sinew]

do you know what the symbol is?

[calm coral]

@latent ocean Has your question been resolved?

[latent ocean]

Sum symbol

[grand sinew]

sorry I didn't see you replied

[grand sinew]

ok so you can first seperate the 2j + 5

because you are just adding them up multiple times

[calm coral]

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[grand sinew]

@latent ocean do you still need help?

[calm coral]

@grand sinew Has your question been resolved?

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[cosmic coral]

So i'm learning how to reduce radical expressions where exponents are larger and indivisible by the index. and example i'm working on is the square root of A^11 (11th power). I understand that i need to split the radican up so that the exponent can divide by the index-- a^10 and a^1 so that simplified id get a^5 * square root a. My question is can I split up the original expression multiple ways and it still be concidered simplified? such like a^11 split to a^8 and a^3?

[warm warren]

Usually when we want to simplify an expression it’s because we want to do something else afterwards

So it depends on what your goal is

The point of learning to simplify is to learn to see how these different looking expressions are really the same thing

And if we have some tools to solve a particular expression, perhaps we could change this new problem into something that looks familiar

So perhaps making a¹¹ into a⁸a³ is useful, but it depends on what you want to do next

Example: $\frac{a^{11}}{a^3} = \frac{a^3a^8}{a^3} = a^8$

[potent lotus]

Frosst

[cosmic coral]

I see! so it depends on the context... I'm currently learning basic simplification and i'm learning i suppose the properties of simplification

[warm warren]

Yes, you want to be good and converting between different forms of the same thing, that’s the goal

As you go further into maths, simplifying is only a means to an end

[cosmic coral]

so I guess then I can simplify a radical multiple ways?

I'll post a picture of my work

I think I figured out my own question. I think that square root a^8 and square root a^3 doesn't work as a split is because although 8 is divisible by 2, 3 is not, so in regards to the first rule this split does not work due to 3 being unable to be divided by 2 and still being larger than 2

[calm coral]

@cosmic coral Has your question been resolved?

[rose olive]

What else do you need help with?

[cosmic coral]

I was wondering if my assumption was correct? if it made sense logically

[rose olive]

As already stated earlier, simplification is done if we want to do something with it later. However, if we are ONLY finding the simplified final answer, it has to have the lowest amount possible

[cosmic coral]

okay, thank you. that clarified/verifies my question. yes, I was only looking for/trying to understand finding the simplified final answer only.

[rose olive]

You're welcome, anything else?

[cosmic coral]

that's it for now, thanks!

.close

[calm coral]

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[calm coral]

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[remote mural]

a 120 foot tall tower is to be erected on the side of the hill with a 6 degree angle from the horizontal. find the length of each of two support wires attached to the top of the tower, if the first wire is anchored 135 feet uphill and the second wire is anchored 135 feet down the hill, from the base of the tower

[remote mural]

can somebody just draw a picture for this

[stiff spear]

We took this in our current class

it can be solved using trigonometry

[calm coral]

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[calm coral]

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[worn hedge]

Can anyone explain how they’ve done this

[calm coral]

@worn hedge Has your question been resolved?

[worn hedge]

<@&286206848099549185>

[rare bane]

What is your question exactly?

[worn hedge]

Why is there x^h-1

I don’t get how they’ve differentiated

[calm coral]

@worn hedge Has your question been resolved?

[upper sparrow]

Looks really similar to something we did before

If I recall right then u(kx, ky) can have that k pulled out so something like noticing that u(x, y) = u(x * 1, x * y/x) - do you have the context from last time?

Oh here it is! #help-23 message
@worn hedge

[calm coral]

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[calm coral]

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[buoyant mirage]

dy/dx = 0 ??

then what to do

[timber forge]

ah mb ignore me, mis understood what you meant

[buoyant mirage]

@ me

[calm coral]

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[tidal hawk]

I have this homework problem to determine whether the series converges or diverges (and I know the right answer is that it converges)
It should be simple enough to solve using the limit comparison test but the problem I have is the fact that the series starts at k = 0 and so I don't think I should be able to use 1/k^(5/2) for comparison as it isn't defined at k = 0

[pale blade]

Why not prove it for k=1 then?

Start at k=1, once you’ve established that the identical series starting at k=1 converges. Showing that the k=0 term gives a constant is sufficient to show that the series converges

[calm coral]

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[calm coral]

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[quick tiger]

how didi they go from the first line to the second

[pure zephyr]

tan(x+π/2) = -cot(x)

[quick tiger]

ohh so they did this

[pure zephyr]

ye

[quick tiger]

$\tan\left(\frac{\pi}{2}\cdot\left(x+1-1\right)\right)\ = \tan\left(\frac{\pi}{2}\cdot\left(x-1\right)\ +\frac{\pi}{2}\right)\ = -\cot\left(\frac{\pi}{2}\cdot\left(x-1\right)\right)$

[potent lotus]

RizzlerFly

[pure zephyr]

yep

[quick tiger]

thanks for help !!

.close

[calm coral]

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[pure zephyr]

n

p

[calm coral]

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[fierce cradle]

could someone help me with these pro blems?

[calm coral]

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[calm coral]

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[smoky badger]

So I'm trying to get the x intercepts for this quadratic formula but so I put it into a quadratic equation. Seeing as I cant square root -236... what do I do now?

[flint portal]

Make use of ‘i’

i = sqrt(-1)

[smoky badger]

I have no idea what this is or what this means

[clear delta]

if you don't know about imaginary numbers then the answer is that there are no x-intercepts

you can visually verify this with a graph

[flint portal]

^

[smoky badger]

Got it, should have done that honestly

Im trying to use this to solve an inequality

should be good for now then, thank you for the help!

.close

[calm coral]

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[floral gale]

how would I solve this???

I understand that x can't be negative

or a square root can't be negative rather

so would the answer to this be undefined or something???

[vivid vine]

but sqrt(4-4)=sqrt(0)=0 is well defined

[viral carbon]

yo how do i make one of these channels

[vivid vine]

so dont know where the confusion comes from

[meager dome]

but approaching it from the positive side of 4 its undefined

[flint portal]

#❓how-to-get-help

[meager dome]

there’s no available channels

[vivid vine]

!help

[calm coral]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[floral gale]

wait I thought a square root could not be 0

[meager dome]

it can

[vivid vine]

since when?

[flint portal]

0^2 = 0

[floral gale]

oh shoot

I never even tried it in a calculator

[vivid vine]

lmao

[floral gale]

but when putting it in from the right side

like

we choose values greater than 4

here

so if I chose a value greater than 4 it would be undefined

[meager dome]

using the epsilon delta definition of a limit it would indeed be 0 but i am also confused by this

because you would get the square root of a negative infinitesimal

0 not 4 mb

[vivid vine]

yeah its strange, my guts tell me you can just put in 4 and you get solution

[bleak citrus]

no

[flint portal]

The limit DNE

Since we’re taking the limit from the right

4-( a number > 4) = a negative number

Sqrt(neg number) does not exist

[fleet verge]

Bigger not smaller

[flint portal]

Thxs

[floral gale]

I'm a little ocnfused on how to do this though

I haven't ever had a question with inifinity

[vivid vine]

infinity means, just when you take bigger and bigger x

they ask you to what the f(x) tends to

[flint portal]

Formal way: multiply numerator & denom. by 1/x^3 (or whatever the highest power is)

Informal: only look at the highest powers in num. & denom. . If highest power in num. = highest power in denom. then the limit will be the quotient of their coefficients

[floral gale]

so if it gets bigger and bigger can I simply just write infinity?

[granite axle]

When it comes to taking limits to infinity or to a discontinunity there are many tricks that you kinda of just have to "learn". In a case with polynomials in a fraction the most common trick is to divide everything by the largest power of x. as said by kysquared

[floral gale]

so like can I sub in 1, 10, and 100 and show that it's getting bigger hence it's approaching infinity or negative inifinity

[flint portal]

Not necesarily

[floral gale]

wait how would mutiplying by highest power help solve this equation

[granite axle]

you are multiplying by the recriprol of it

so technically dividing

give it a whirl

and see

divide all the numbers by x^3

and show us what u get

[floral gale]

wait so I'd get (4+1/x^3) / (5-1/x^3)?

I'm confused on that part

[granite axle]

yeah

and now what happens

if you plug in larger and larger numbers for x?

[floral gale]

I'll try it rn

[granite axle]

well actually techniqually you'll be seeing the same pattern as before if u just done it without dividing

but the trick is that 1/A as A -> infinity

is a lot easier to deal with

[floral gale]

oh it gets smaller and smaller

[granite axle]

than everything else

[floral gale]

so it approaches 0??

or negative infinity

[granite axle]

is that with the whole fraction?

or just 1/x^3

[floral gale]

I just did the equation I mentioned earlier

(4+1/x^3) / (5-1/x^3)

only subbed in

one and 2

[granite axle]

okay yeah lets take a step back, and see what happens to each of the 1/x^3 as we put in larger x's

that was mb i mentioned the wrong thing

[floral gale]

it gets smaller and smaller

[granite axle]

yeah, and eventually it basically is zero

anything like 1/x or 1/2x or 1/x^23 will apporach zero as x gets bigger and bigger

if you think of the 1/x graph for example

as you go all the way along the x axis

is looks as if it will aproach zero

so now knowing that as 1/x^3 -> 0 in this case

can you deduce what your answer should be

[floral gale]

yes it will be 0

as it approaches closer and closer to 0

right?

[granite axle]

hang on

so 1/x^3 goes to zero

but what about the toher numbers

[floral gale]

ther is still a 4 and 5

oh wait

are they to small

to make a difference

so it would approach 1

[granite axle]

nope

let me try and write something out to make it clearly what we are doing

[floral gale]

really i thought it would begin to approach 1 or something

ok

[granite axle]

$lim\ \frac{4+\frac{1}{x^{3}}}{5-\frac{1}{x^{3}}}$ so this is where we are at

[potent lotus]

SollyPolly

[granite axle]

i cant be bothered to write x-> infity under neath lim

but just know its there

[floral gale]

ok

[granite axle]

now what we are basically asking is

$\frac{4+lim\ \frac{1}{x^{3}}}{5-lim\ \frac{1}{x^{3}}}$ this right?

[potent lotus]

SollyPolly

[granite axle]

and as you have already said

the $lim \frac{1}{x^3}$ = 0

[potent lotus]

SollyPolly

[granite axle]

so why dont we sub zero in

now all of a sudden

we have no more x's

[floral gale]

so when we have 4/5

[granite axle]

and are just left with a lonely 4 and a 5

yup

and that is what it approaches

[floral gale]

oh tysm

that makes sense now

[granite axle]

no problem

[floral gale]

how would I actually answer this would I substiute multiple values to show that it is true?

[granite axle]

i wouldnt, just this is how you show it is true

like this method itself. Breaking it down into easier to answer limits

[floral gale]

ok tysm

appreciate everything you've done to help

.close

[calm coral]

Channel closed

Closed by @floral gale

Use .reopen if this was a mistake.

[granite axle]

np

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
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[nimble parrot]

2b please

[calm coral]

Please don't occupy multiple help channels.

[nimble parrot]

Isn’t it 7? Why do we have to do so much work

[inner zealot]

What is your reasoning for that?

[nimble parrot]

Bc u can have any y intercept if u add or subtract on from the equation

[inner zealot]

Yes, but the question is asking about the minimal value that you could have, not just any value.

[carmine sorrel]

it's saying the minimum of the whole graph is 7, not just the intercept.

[nimble parrot]

Oh

😭

But it’s a sin equation

Wait nvm

Wait do I plug in x= 0 and y = 7

[inner zealot]

Just x = 0.

[nimble parrot]

Oh yeah

Ok then pi/2 is 0

0,1

[inner zealot]

sin(-pi/2) is -1

[nimble parrot]

Oh yeah cus sin y value

And it negative

So 5(-1) + 1.5 + q

Now I’m lost

[inner zealot]

That's because you went about solving the problem the wrong way.

[nimble parrot]

😭

Whats the right way

[inner zealot]

So let me just state that the minimum value is 7. Your methodology to try and prove that is incorrect. The problem states that you would need to orally, and in writing, be capable of justifying your answer.

[nimble parrot]

Yeah

[inner zealot]

The first step you need to do is determine a minimum value of q that ensures the sine wave is always greater than or equal to 7.

[nimble parrot]

Aye 10.5

Right

That’s when I plugged in y = 7 and x = 0

Or > 7

[inner zealot]

Good, so now you have a fuction that you can work with. From there, now it's just a matter of plugging in x to fine y = r.

Which is 7.

[nimble parrot]

Wait u plug in x to the original equation?

[inner zealot]

No, the original equation ended with + q. The constraint, g(x) >= 7, allows you to find a minimal value for q.

[nimble parrot]

So I plug in 10.5

To q

[inner zealot]

Yes

[nimble parrot]

But the minimum was 7 and the y intercept was 7 too because pi/2 has a 0 x value

But if the transformation was like 2pi/2 the minimum and y intercept would be different

[inner zealot]

Yes. The fact that the trough of the sine wave was at x=0 is "coincidental". If the horizontal shift were different, you would likely end up with a different y = r value.

[nimble parrot]

Oh

I would do the same process if the horizontal shift was different?

[inner zealot]

Yes.

[nimble parrot]

Ohh

Ok thank u so much

[inner zealot]

yw

[nimble parrot]

.close

[calm coral]

Channel closed

Closed by @nimble parrot

Use .reopen if this was a mistake.

[nimble parrot]

.reopen

[calm coral]

✅

[nimble parrot]

😭 I tried it with pi/6 and it didn’t work

[remote mural]

whats up?

what you got?

[nimble parrot]

5sin(x-pi/6) + 1.5 + q
5sin(0-pi/6) + 1.5 + q
5(-1/2) + 1.5 + q > 7
q > 8

[remote mural]

if its TRIG IM OUTTA HERE

[nimble parrot]

It’s trig 😭

[remote mural]

AW HELL NAW YOU ON YOUR OWN BUDDY😭

[nimble parrot]

Boyy

Help 🙏

[remote mural]

wait

type the question

[nimble parrot]

Can u see the stuff before

Cus I reopened the ticket

[remote mural]

here???

[nimble parrot]

Yes

[remote mural]

I CANT ZOOOM

[nimble parrot]

Boy

Lemme repost it

Consider f(x) = 5sinx + 1.5 and g(x) = 5sin(x - *) + 1.5 + q, where x € * and g
The graph of g is obtained by two transformation of the graph of f.
(a) Describe these two transformations.
The y-intercept of the graph of g is at (0. +).
(b) Given that g (x) = 7, find the smallest value of ri

Oh hn that didn’t work

[remote mural]

let me try wait

okay here

for a, it is horizantal and vertical shift

[nimble parrot]

Yuh

2 im cooked cus bro said its a “coincidence its 7”

[remote mural]

The term x - pi/2 indicates a phase shift to the right by pi/2 units, hence, horizantal shift

[nimble parrot]

Yuh yuh

[remote mural]

WAIT YOU SOLVED IT?

[nimble parrot]

Do 2b

[remote mural]

OOOH

OKIE

[nimble parrot]

No not b 😭

I solved it but then I tried a different variation and got it wrong yfm

[remote mural]

it IS 7 tho

[nimble parrot]

Yes

But like the y incercwot and minimum are the same

I think I did it wrong

I switched pi/2 with pi/6

5sin(x-pi/6) + 1.5 + q
5sin(0-pi/6) + 1.5 + q
5(-1/2) + 1.5 + q > 7
q > 8

[remote mural]

to find the smallest value of r given that g(x) > 7 we need to consider the y-intercept of g(x), which occurs at ( x = 0 ).
g(0) = 5sin(0 - pi/2 + 1.5 + q = 5(-1) + 1.5 + q = -3.5 + q

[nimble parrot]

Wait

[remote mural]

since g(x) > 7 we set the y-intercept greater than 7 and solve for q
-3.5 + q > 7
q > 10.5

[nimble parrot]

Wait

[remote mural]

the smallest value of q that satisfies this inequality is 10.5

[nimble parrot]

Do it wit pi/6

Cus I got q = 8 and then I put it in Desmos and the equation went lower than 7 😭

[remote mural]

so the smallest value of r which corresponds to the y intercept of g(x) is r = -3.5 + 10.5 = 7
r is 7.

BOY STOP GOING CRAZY

[nimble parrot]

Hoy

[remote mural]

THE ANSWER IS 7

[nimble parrot]

Boy

Nah but i did it wrong but i somehow got the right answer

I dont know what I did wrong

So I used a different horizontal shift

[remote mural]

i solved it for you just copy it down😭

[nimble parrot]

Bru I swear if the test has a different horizontal shift it’s ur fault

Aye but solve it wit pi/6 instead of pi/2 tho 🙏

[remote mural]

😭😭😭😭😭

[remote mural]

BUT WHY

[nimble parrot]

Cus bruh I got it wrong that way

I only got 7 right cus it’s the same as y intercept so I got lucky 😭

I’ll solve it then

5sin(x-pi/6) + 1.5 + q
5sin(0-pi/6) + 1.5 + q
5(-1/2) + 1.5 + q > 7
q > 8

BUT LOOK