help-13

[livid tundra]

can someone explain to me how we got rid of the sqrt and ln

[slate lintel]

this is the answer to a different question

[livid tundra]

sorry i don't understand can you repeat

[slate lintel]

the first thing you posted was question 4

the lined paper with blue ink is question 6

[livid tundra]

bruhh i was like

this shit not solvable

thanks i guess

[cedar kiln]

@livid tundra Has your question been resolved?

[cedar kiln]

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[daring dune]

anyone know how to do this?

[daring dune]

i'll take the answer if u have it

[vagrant elbow]

we'll take the working if you have it

[cedar kiln]

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[cedar kiln]

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[boreal linden]

@crimson sedge

[crimson sedge]

yesss

[boreal linden]

this was my query

[crimson sedge]

okay wait lemme see

done

lemme explain

so , assume a and b be the zeroes

[boreal linden]

okay

[crimson sedge]

sum of zeroes is

7/6

[boreal linden]

ye

[crimson sedge]

this gives , a = 5b/2

from the ratio

done ?

[boreal linden]

didn't get it

[crimson sedge]

ratio of zeroes , a/b=5/2

???? , this implies , a=5b/2

[boreal linden]

yes

[crimson sedge]

put , a = 5b/2 in a+b=7/6

[boreal linden]

b=5/2a

[crimson sedge]

no ,

[boreal linden]

to find b?

[crimson sedge]

yes yes

[boreal linden]

7-15b/6

[crimson sedge]

wait what did you do ?

nnope

[boreal linden]

b=7/6-5b/2

b=7-15b/6

[crimson sedge]

your b will be 1/3

[boreal linden]

how?

[crimson sedge]

don't do it like that , it's confusing

your b = 5a/2 right ?

[boreal linden]

yes

[crimson sedge]

sorry , a=5b/2

(5b/2)+ b = 7/6

[boreal linden]

oh yeah

[crimson sedge]

7b/2=7/6

gives , b= 1/3

got it ?

[boreal linden]

no

wait

7b/2=7/6

transpose 2

it will be multiplied with 7/6 resulting in 14/6

then transpose 7

[crimson sedge]

don't transpose 2 firstly

you see 7 both sides in numerator

cancel them right there

then 2 is transposed , giving 2/6 = 1/3

got?

[boreal linden]

we can't cancel the ones in numerator

like if it's 2/7=7/1

[crimson sedge]

who said that , it goes like this

[boreal linden]

no we can't...

[crimson sedge]

but , bnoth are in numerator

see carefully

[boreal linden]

7b/2=7/6

both 7's are in numerator sides

[crimson sedge]

yes , when you tanspose 7 to the Right hand side , iot goes in denominator

won't it cancel out the 7 in numerator ?

[boreal linden]

yes

[crimson sedge]

that's what i'm saying ,

now , 2 is transposed

[boreal linden]

yes

[crimson sedge]

got 1/3

[boreal linden]

YES

you're right i'm sorry

[crimson sedge]

now put this b = 1/3 in a+b = 7/6

[crimson sedge]

no problem

[boreal linden]

a=5/6

b=1/3

right?

[crimson sedge]

yes

[boreal linden]

next?

[crimson sedge]

multiply a and b

[boreal linden]

5/18

[crimson sedge]

yes , this 5/18 will be equal to 2k/6

(product of roots ) remember

[boreal linden]

Yes

[crimson sedge]

solve it , your k comes out to be 5/6 , which also equals to a

[boreal linden]

Yes

K=5/6

[crimson sedge]

yes

[boreal linden]

got it

thankyou very much

[crimson sedge]

welcome

close this channel

if done

[boreal linden]

okay

once again

tysm :)

.close

[cedar kiln]

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[crimson sedge]

welcome

[cedar kiln]

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[rich yew]

how do u go from 2nd step to 3rd step (line 2-3)

[vagrant elbow]

There are only 2 lines

[frosty ocean]

I suppose the first line was cut out

[rich yew]

wait nvm i forgot to screenshot line 1

yep

this is line 2-3

my bad

[livid hound]

is this chunk the original question

[rich yew]

yes

[livid hound]

then its just direct substitution

when x=1, you get

[rich yew]

i understand that part

[stoic dust]

I think this is laid out kind of weird

[rich yew]

but i don't get 1-x^2+2x^2.... part

[livid hound]

that's just restating the original problem

[stoic dust]

that's the original expression

[rich yew]

ohhhhh

i thought i did something wrong

[rich yew]

yeah it's formatted pretty weird

.close

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[austere plume]

How do I answer c?

[austere plume]

It says that prove it has no real roots

But -6 is a real root

[iron saffron]

Where is logarithm defined in R?

[austere plume]

Wdym in R?

[vague rapids]

Real

[iron saffron]

R = Real

[austere plume]

Wdym by defined as real?

[vague rapids]

Real numbers

which include rational and irrationals

[iron saffron]

ii

[tropic oxide]

what happens if you try to plug x=-6 into the equation 2 log_2(x+2) + log_2(x) - log_2(x-6) = 3?

[austere plume]

It doesn’t work since log_2 can’t have a negative number

I can’t remember what the rule was

[iron saffron]

so, if it's not defined

what do you conclude from that?

[austere plume]

That it has no real roots?

[tropic oxide]

what rule

[austere plume]

That for log n if n is less than 0 it will be a imaginary number

Something like that

I’m assuming that’s why it has no real roots then?

[cedar kiln]

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[boreal linden]

please help me with the i) one

[cedar kiln]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[boreal linden]

1

[vague rapids]

can you find the roots of the equation x^2-x-30=0

[boreal linden]

sure

1,-30

[vague rapids]

are you sure about that

[boreal linden]

yes ofc

[vague rapids]

Do you know what a root is?

[boreal linden]

sum of roots=alpha+beta

product of roots=alpha*beta

[vague rapids]

That's another thing

[boreal linden]

then?

[vague rapids]

roots are those values of x for which the equation becomes zero

[boreal linden]

now can you please tell me what are the roots for that quadratic polynomial?

[vague rapids]

or wait

you don't need to find the roots

you can use roots coefficient relation

so you have α+β=1 and αβ=-30

let the new polynomial be x^2+px+q

whose roots are 2-α and 2-β

now find the same relation

[boreal linden]

where do we have it

Alpha+beta=1?

oh yeah

sorry

[vague rapids]

from the given equation

[boreal linden]

-30

[vague rapids]

did you get the relation?

[boreal linden]

no

[vague rapids]

you should get -p=4-α-β and q=(2-α)(2-β)

-p=4-(α+β)

[boreal linden]

how

[vague rapids]

roots coefficient relation

we assumed the polynomial to be x^2+px+q

[boreal linden]

yes

[vague rapids]

roots are 2-α and 2-β

[boreal linden]

yes

[vague rapids]

so their sum is -p and product is q

[boreal linden]

yes

[vague rapids]

-p=4-(α+β)

continue to find the value of p

[boreal linden]

How did you get THIS?

[vague rapids]

as stated through roots coefficient relation

-(p/1)=2-α+2-β

-p=4-β-α

[boreal linden]

?

[vague rapids]

what part do you not understand

[boreal linden]

how did you get -p's value to be 4-beta-alpha

and

why is-p/1 said to be 2-alpha+2-beta?

[vague rapids]

do you know what is roots coefficient relation

[boreal linden]

NO

[vague rapids]

like sum of roots = -b/a

[boreal linden]

yes

[vague rapids]

and product = c/a

[boreal linden]

yes

[vague rapids]

same thing is applied

[boreal linden]

-b/a=-p

[vague rapids]

yes

[boreal linden]

why is it 4-b-a

?

[vague rapids]

because the roots need to be 2-α and 2-β

[boreal linden]

so?

[vague rapids]

their sum is just 4-α-β

[boreal linden]

why are u adding them

[vague rapids]

sum of roots = -b/a

SUM

[boreal linden]

yes

but they're not -b/a

they're a+b and ab

respectively-

[vague rapids]

you're confusing with a+b and α+β

If you know how to factor then we can use another process

[boreal linden]

yes

[boreal linden]

please lets do this

[vague rapids]

can you factor x^2-x-30

[boreal linden]

wait

the previous method

i got it

4-a-b

now?

[hard ember]

Or you can use transformation of roots

@boreal linden

[cedar kiln]

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[remote jackal]

is there a way to solve this without requiring brute-forcing the answer? $\log_{2} (x) = \log_{3} (x+1)$ (the solution is obviously ||2|| but i want to see if there is a way to actually solve it. ive tried changing base and making an exponential using base 6 and use the logs as powers)

[remote jackal]

whoops

$\log_{2} (x) = \log_{3} (x+1)$

[wraith dagger]

therealjasonic

[carmine bronze]

Is there a fake jasonic rolling around pretending to be you? 🤔

[remote jackal]

the username "jasonic" was taken, this was the second username i had in mind lol

[carmine bronze]

Anywhos, apply a change of base formula to both log functions.

$\log_{a}{x} = \frac{\log{x}}{\log{a}}$

[wraith dagger]

kookiemon

[remote jackal]

yeah but even applying it to both sides still makes it awkward to work around (you cant factor out x, or at least thats what i think)

$\frac{\ln(x)}{\ln(2)}=\frac{\ln(x+1)}{\ln(3)}$

[wraith dagger]

therealjasonic

[carmine bronze]

I would perhaps make a common denominator.

[remote jackal]

i guess im trying to find if theres an answer to a more general problem or something?

[carmine bronze]

What do you mean?

A general method for finding a solution?

[remote jackal]

yeah in like this type of equation:

$\log_{a}(x) = \log_{b}(x+c)$

[wraith dagger]

therealjasonic

[carmine bronze]

Ahh, a formulaic answer?

[remote jackal]

yeah

[carmine bronze]

I would surprised if there wasn't.

Hmm, the obvious answer to your original question is n = 2. That can be done by observation.

And that screams that there should be a formulaic solution.

[remote jackal]

there probably isnt a solution considering wolfram gives up when a=3,b=5,c=1

[carmine bronze]

That would be a non-integer solution which there are iterative methods for solving.

I don't think WA "knows" to use an iterative method in all cases.

[cedar kiln]

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[austere plume]

Can someone help me work through bronze b

[austere plume]

For set A

[cedar kiln]

@austere plume Has your question been resolved?

[open nacelle]

I believe you use the sign of the second derivative to determine its concavity?

I think thats what its asking for

Ive never heard it referred to as just "concave" tho i thought it was usually concave up or concave down

[cedar kiln]

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[timber turtle]

Can someone help me with question 5?

[violet flume]

my thought is brute but

distribute and recollect

youll get 4 equations by equating coefficients

[timber turtle]

can you give me an example of one of the equations

coz I don't know what distribute and recollect is

[violet flume]

say like

$5x+2=a(x-1)+b$

distribute

[wraith dagger]

janniku

[violet flume]

so

$5x+2=ax-a+b$

[wraith dagger]

janniku

[violet flume]

collect

[timber turtle]

so I have to expand everything

[violet flume]

$5x+2=(a)x+(b-a)$

[wraith dagger]

janniku

[violet flume]

so a has to be 5

and b-a has to be 2

[violet flume]

this is my way yea

there may be a more clever way

but this will work

[timber turtle]

hmm alr

[violet flume]

you know a and d by inspection

maybe you can just find the x¹ term

if youre careful

[timber turtle]

yeah i found a and d

im trying to do what my notes say where you sub in some things and get 2 equations based on b and c

[violet flume]

sub in?

[timber turtle]

one of them you sub in a zero of the function

ok nvm

I sub in x = 2 and x = 0

idk why but

I remember I need to make the (x-1) equal to 1

im tripping

. close

.close

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[crimson sedge]

How do I solve the volume and area of a cylinder?

[surreal cave]

you plug in what you know and out comes the answer :)

[crimson sedge]

Alright, could you give me a quick rundown of the entire equation please? Just a quick one

[solemn pulsar]

area of a cylinder, split the shape into 3 differnt shapes

you got 2 circles and a rectangle

the rectangle is curved to form the circumference of the circles

so the rectangles area is pi x d x h

now you have 2 circles and the area is pi x r^2

now add this together

and you have the area of the cylinder

volume on the other hand is simply the area of the circle multipled by the height of the cylinder due to its 3-dimensional nature

[crimson sedge]

Ohhhhhhh

Thank you, algebra and letters have me extremely confused

.close

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[solar vector]

https://clipart-library.com/img/1469801.gif

When you first learn sine and cosine you define them as ratios of side lengths of right triangles.

Sine of the angle A is the ratio of the side length opposite of the angle A, to the side length of the hypotenuse. So, you'd say sin(A)=a/c here.

Cosine of the angle A is the ratio of the side length adjacent to the angle A, to the side length of the hypotenuse. So, you'd say cos(A)=b/c.

People usually remember this with the mnemonic soh-cah-toa.

You have to change around sines and junk to extend this definition to the unit circle you'd see in a trig class.

And there are more advanced definitions for these things you might see later on down the road.

But this is probably a good starting point.

Oh yeah, tangent of A would be the ratio of the side length opposite the angle A to the side length adjacent to the angle A too. So, you'd say tan(A)=a/b in addition to your sine and cosine values.

[solar vector]

That completes the mnemonic here

You compute sine by taking the opposite over the hypotenuse, cosine from the adjacent over the hypotenuse and the tangent from the opposite over the adjacent.

Mmm

Pick one of the two angles in your triangle. Call it A.

So, sin(A)=opposite/hypotenuse, cos(A)=adjacent/hypotenuse, tan(A)=opposite/adjacent.

I would probably explain it along these lines.

Then work out examples.

[solar vector]

Idk if this makes more sense to you @crimson sedge?

[iron saffron]

3 year old usually don't have to know about this. You must start understanding some concepts. Maybe you should check first the definitions of what you're being told.

[cedar kiln]

@crimson sedge Has your question been resolved?

[daring portal]

@crimson sedge SOHCAHTOA

S=Sin

O=Opposite

H=Hypotenuse

C=Cosine

A=Adjacent

H=Hypotenuse

T=Tangent

O=Opposite

A=Adjacent

Basically just look at this and do the first thing that shows up after S, C, or T divided by the next letter

So Sin=O/H

C=A/H

T=O/A

[cedar kiln]

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[vague rapids]

Is that the complete question?

[livid hound]

seems to be missing a letter

specifically here:

well i suppose you can actually assume that its from B based on everything else in the question

[slate lintel]

seems to be having a stroke

[cedar kiln]

@restive carbon Has your question been resolved?

[restive carbon]

Not yet

[slate lintel]

sigh
<@&268886789983436800> spam

[cedar kiln]

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[onyx knoll]

Part i

[south tundra]

Are you familiar with polar form of a complex number?

[slate lintel]

you don't need that, just take $a+bi$ and square it then figure out what $a$ and $b$ need to be so that the square is equal to $5 + 12i$

[wraith dagger]

Hayley

[south tundra]

That works as well, yeah

[cedar kiln]

@onyx knoll Has your question been resolved?

[cedar kiln]

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[austere plume]

Quick question. If I have a point x=4

[austere plume]

How would I show its a stationary point?

[livid hound]

x=4 isn't a point

[austere plume]

4,0

[livid hound]

show that the first derivative there is 0

[austere plume]

Ok thank you

.close

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[livid hound]

reverse of determine the location of a stationary point

[cedar kiln]

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[obtuse frigate]

hey

[obtuse frigate]

the question wants me to express p in terms of q

but how?

[dire bronze]

note that y = q when x =0

[obtuse frigate]

the question is meant to be y=|pcos^2(x) -1| btw the k is a typo

[obtuse frigate]

yes

i sub x=0 and y=1

q*

but i got p=q+1 or p=q-1

but answer says p=2q

[dire bronze]

when x = pi, y = q

[obtuse frigate]

why use that instead of x=0 tho

it also gives same answer as when x=0 no?

[dire bronze]

it keeps touching q

[obtuse frigate]

yeah

which value of x do we choose?

i tried all but i cant manage to get p=2q as an answer

<@&286206848099549185>

[dire bronze]

when x = pi/2 y = q

what is cos (pi/2)

[obtuse frigate]

0

doesnt that remove p

[floral obsidian]

chai

[dire bronze]

ye

[obtuse frigate]

but we need to express p in terms of q tho

[dire bronze]

this implies that q=|-1|

[cedar kiln]

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[crimson sedge]

@dense hornet

[crimson sedge]

What if I do this?

[dense hornet]

again, the simple question is, how do you know that the projection of T2 onto the vertical axis is t?

this is not given, t is the magnitude of T1, and that's it, it has nothing to do with T2

[crimson sedge]

t is the tension in object A

So it’s the vertical axis

[dense hornet]

that's not how it works...

note here the projection of N2 onto the vertical axis is nsqrt(3)/2

why is the vertical projection of T2 not equal to this?

[crimson sedge]

Oh

Okay

Thank you again

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[crimson sedge]

$x=\sin(2\arctan(\alpha))$ \ $y=\sin(\frac{1}{2}\arctan(\frac{4}{3}))$
\ If S is a set where $\alpha \in \mathbb{R}$ such that $y^2=1-x$ then $\sum_{\alpha \in S}^{}16\alpha^3$ is equal to?

[wraith dagger]

mydickdoesbackflips

[crimson sedge]

the answer i got is -126, but the book says 130

cant find out where i'm wrong

<@&286206848099549185>

[cedar kiln]

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@crimson sedge Has your question been resolved?

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[cedar kiln]

@delicate patio Has your question been resolved?

[delicate patio]

how can i approach this question?

ik that 28n^2 + 1 must be a square number

but i can't think how to proceed now

[astral bay]

well sqrt(28n^2 + 1) must be odd, and equal to either 1 or 6 mod 7, but none of that feels helpful

(just by analysis of "x^2 = 28n^2 + 1" modulo those numbers)

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[steady imp]

I think I got it?

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[austere plume]

For b

[austere plume]

If they wanted to find the minimum

Would they find the second derivative

Then solve when it’s greater than 0

Why did they only use the first?

[livid hound]

first derivative indicates where there are local extrema

[austere plume]

By that do you mean stationary point?

[livid hound]

yes,

[austere plume]

But it ask for the minimum point

Couldn’t that point also be a maximum

[livid hound]

you could do more work if you want to confirm whether it's a max or min

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[simple harbor]

hi how do i solve for the domain of this

[vapid lotus]

are you able to graph it?

[simple harbor]

i could but id like to show work

[granite eagle]

root(y) + y^(-1) cant be negative right?

[simple harbor]

no

[granite eagle]

can it be 0?

[simple harbor]

im pretty sure Dy{y/y>0}

i dont think so

cos 1/y

[granite eagle]

and upper bound on the value of the exponent?

[simple harbor]

whats an upper bound

[granite eagle]

whats the highest i can be

it

[simple harbor]

upper bounmd of y?

[granite eagle]

of root(y) + y^(-1)

[simple harbor]

im not sure

[granite eagle]

does it not go all the way up?

[simple harbor]

the graph goes up

it doesnt stop

[granite eagle]

yep

so u think u get an idea abt the domain now?

[simple harbor]

y>0?

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[lucid skiff]

So I know that it could be 4,3,2,1 for quantity A

But I don’t know how to do quantity B

[tropic oxide]

is that h or n right at the beginning of the problem statement?

[lucid skiff]

n

[tropic oxide]

right... your handwriting threw me off a bit

also, consider: how can you tell whether a natural number is divisible by 9?

[lucid skiff]

If it’s divisible by 3

?

[tropic oxide]

no, divisibility by 3 and divisibility by 9 aren't the same.

do you know how to test a natural number for divisibility by 9?

if you don't then say you don't

[lucid skiff]

No

I don’t

[cedar kiln]

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[civic sentinel]

<@&286206848099549185>

[civic sentinel]

hello?

https://tenor.com/view/mr-bean-waiting-still-waiting-gif-13052487

I may need someone to help with domain and range

and how Linear and function work

Just explain me how it work

[crimson sedge]

Wdym

[split harness]

Linear what

[civic sentinel]

linear on graph

[split harness]

As in a line?

Or do you mean a linear function

[civic sentinel]

wait 1 min

sorry

like how do you put Y=2x-1 on a graph

[crimson sedge]

first find the y intercept

[split harness]

You take two points and draw a line through them

[civic sentinel]

oh

what do you call the line crosses twice on y-int

[split harness]

Twice?

[civic sentinel]

yea or more

[split harness]

There is one line that does that

Y=0

But it crosses it an infinite number of times

Are you sure of what you wrote?

[civic sentinel]

honestly i have no idea

lets do different question

(2x^3y^6)^-2
--------- Can u help explain to solve this without using calculator?
(3x^5 y)

[split harness]

I think you're missing part of it

[civic sentinel]

?

[split harness]

Maybe you missed an = sign?

[civic sentinel]

I dont think it has a sign

[split harness]

Oh

[civic sentinel]

it just solving for exponent

[split harness]

Maybe you meant simplify?

[civic sentinel]

no idea i dont speak math

[split harness]

Weird

I have no idea what the question is

[civic sentinel]

oh damn

its alright

maybe another time anyway gtg

how do you delete this chat?

[split harness]

.close

[civic sentinel]

.close

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[civic sentinel]

.close

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[tacit juniper]

Hey! Not sure where to go with 53:

[tacit juniper]

Namely, I'm not really sure what to do with the provided hint, and what rs means.

Obviously r is the common ratio, but what's the s?

[dawn junco]

typo, they meant S

[tacit juniper]

So its S - rS?

[dawn junco]

yes

[tacit juniper]

Wow lol

Thank you haha

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[tacit juniper]

.reopen

[cedar kiln]

✅

[tacit juniper]

Sorry, got stuck again lol

This is what I have so far, but I'm not sure how to get to 1 - r^{n+1} from what I have in the numerator

[crimson sedge]

Hi,
i have a question, how can i prove that:
If p is a prime number such that p=4q+3, then p|x²+y² ==> p|x and p|y.

[tacit juniper]

You'll probably wanna take that to a free channel

2, 5, 16, and 24 are available

[crimson sedge]

Okay ty

[tacit juniper]

I gotchu bro

[dawn junco]

sorry gtg to sleep very soon

hope someone else can continue

[tacit juniper]

No worries, thanks for your help ^^

Wait, I just realized that the fourth equation I wrote probably isnt correct

Since S is equal to the series, then i can't say S(1 - r) is equal to the same series, huh?

[cedar kiln]

@tacit juniper Has your question been resolved?

[tacit juniper]

<@&286206848099549185>

Apologies if I used the mention incorrectly here

[crimson sedge]

hey you wanna check again what S - rS equal to

$S=a+ar+ar^2+\dots+ar^n$

[wraith dagger]

mist9912

[crimson sedge]

$rS=ar+ar^2+\dots+ar^n+ar^{n+1}$

[wraith dagger]

mist9912

[tacit juniper]

Okay, I see now. That's where the collapsing sum comes in, right?

[crimson sedge]

uhhuh

[tacit juniper]

Got it. Always a silly little oversight lol

Thanks a bunch ^^

[crimson sedge]

np

[tacit juniper]

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[ripe cradle]

there is no place to put the x value as -4

wait nvm i put the values wrong

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[thin roost]

Hello

[thin roost]

I got a different answer

[surreal cave]

Can you show your work? :)

[thin roost]

p+q=-p

pq=q


From here i got p=1, q=-2

Our quad becomes


x²+x-2=0

Min = -b/2a = -1/2

[buoyant latch]

Shouldn’t it be

Oh you fixed it

[hard ember]

Least value of y is -D/4a

-1/2 is min for x

[thin roost]

Ye I found the least of x

So question is asking for min value of y

How are you suppose to know that?

[hard ember]

Min value of x^2+x-2

Read the question again

[thin roost]

Oh ye x²+x-2=y

Yep

Alr

I need help with 2 more question

9 and 10

What is the condition for having a common root??

[hard ember]

I don't exactly remember that

[thin roost]

Well I search on Google and It has some lengthy relation

Any easier way to find it out the value of 'a'

[hard ember]

Its something like

What is the answer?

[thin roost]

A)

[hard ember]

I can't think of any now

For 10th no. Just put x=1

[thin roost]

Alr let me

They all cancel out

[hard ember]

So 1 is the root

[thin roost]

Yep

[hard ember]

Now apply prod of roots

[thin roost]

Alr

I get the other root as c(a-b)

[hard ember]

What is the answer?

[thin roost]

H.P

C)

?

[hard ember]

Both roots will be 1

[thin roost]

Ye

C=1/a-b

[hard ember]

Now apply prod of roots

[thin roost]

Alr

c(a-b)=1

[hard ember]

No

c(a-b)/a(b-c)=1

[thin roost]

Oh ye

I'm dumb

So

[hard ember]

Now simplify it

[thin roost]

ab,bc,ac are in AP

[hard ember]

Or 2bc= ab+ac

[thin roost]

Ye

[hard ember]

Now divided both sides by abc

[thin roost]

2/a=1/c+1/1/b

Yep

[hard ember]

It is hp

[thin roost]

Yes

Thanks mate

I appreciate your time!

[hard ember]

Welcome

[thin roost]

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[scenic delta]

I thought it was d, but it says that's wrong

[scenic delta]

I tried taking the laplace on both sides, and since F(s) was given I just plugged in the conditions and got something similar to d

[foggy merlin]

oh they gave u F(s) already

you should have :

s² Y(s) - sy(0) - y'(0) + Y(s) = F(s)
Y(s) = F(s)/s²

$Y(s) = \frac{1-e^{-\frac{\pi}{2}s}}{s^3}$

[wraith dagger]

herels

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[thin roost]

Hi

[thin roost]

Any help with this?

[buoyant latch]

What do you know about the roots of a quadratic

[thin roost]

Well A+B = -b/a

AB = c/a

A,B are roots

So I get

[hoary chasm]

Let's roots be p and 2p

[thin roost]

3A = -b/a, 2A²=c/a

[muted bear]

Let them finish

[hoary chasm]

Ok sorry sir

[thin roost]

That is all I gotta know

[muted bear]

Isolate A on the first equation

[thin roost]

A=-b/4a

*3

[muted bear]

Then substitute

[thin roost]

2b²/9a²=c/a

2b²=9ca

Alr

.cloae

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[muted bear]

:)

[hoary chasm]

Bro are these dpps from ioqm pw course?

My lil bro in 8th...will he be able to follow the course if he starts today?

[thin roost]

Ahh

Ye

It's module

Dpp are hard

[hoary chasm]

Is that also free?

[thin roost]

Ye

[hoary chasm]

Cool

[thin roost]

Here's a glance

[hoary chasm]

Noice it's just like the ones I used to do

[thin roost]

Alr

[hoary chasm]

Except that I did a cuet course lmao

Alr

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[steady elk]

part (iii) says neither reflexivenor symmetrric but TRANSITIVE

[steady elk]

but how transitive?

this is my thoughts

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[fossil pollen]

.

[exotic jungle]

In the step where both numerator and denominator were multiplied by 2 to remove the 1/2 fraction, I didn't do that step and just integrated it with the denominator as 1/2 + t and got log|1/2 +t|. Would that be correct or not?

[muted bear]

Yes, it falls under the +C in the way they transformed

just looking at $\int \frac{1}{\frac 12 +t}$, you can get either $\ln(\frac 12 +t)+\ln 2 + C=\ln(1+2t)+C$

[wraith dagger]

garlicbredfries

[muted bear]

The ln2 is a constant, so it falls under +C

[exotic jungle]

why is there an extra ln2?

[muted bear]

Using log laws, ln(0.5+t)+ln2=ln(1+2t)

[exotic jungle]

oh right i see

oh so it just gets incorporated into the C

okay thanks alot

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[proud spindle]

how do i prove this

[royal loom]

mess around with it

do you know the double angle identities?

use those

rewrite cot in terms of sin and cosine

and just play around with it

[gritty galleon]

i have seen this question twice on the server, in my time here 😐

[proud spindle]

ye cuz i got costheta-sintheta/costheta+sintheta

but then what

[inland ocean]

Mulitply divide by conjugate

[proud spindle]

idk wut that means sorry 😞

ohh

[inland ocean]

Mulitply costheta - sintheta to numerator and denominator

[proud spindle]

i looked it up

i get it now

[gritty galleon]

identities before conjugates?

[proud spindle]

so i divide both sides of the fraction by sintheta

so the top is costheta/sintheta - sintheta/sintheta

= cottheta - 1

[proud spindle]

i didnt know dividing by sintehta was called the conjugate sorry 😢

thx guys! :)

[gritty galleon]

thats not what a conjugate is

[proud spindle]

The conjugate is where we change the sign in the middle of two terms

but how is dividing by sintheta the conjugate since u dont change the sign

[gritty galleon]

he means to multiply and divide by $cos(\theta)-sin(\theta)$

[proud spindle]

but why not divide both sides of the fraction by sintheta instead

isnt that easier

[wraith dagger]

itzkraken.

[gritty galleon]

how does that help?

[proud spindle]

cuz u get (costheta-sintheta)/sintheta

= costheta/sintheta - sintheta/sintheta

= cottheta - 1

and then the denominator is

costheta+sintheta/sintheta = costheta/sintheta + sintheta/sintheta = cottheta + 1

[gritty galleon]

huh? you have $\frac{cos(\theta)+sin(\theta)}{cos(\theta)-sin(\theta)}$

[wraith dagger]

itzkraken.

[proud spindle]

ye but then u can separate the farctions right

wait wut

nah

i mean divide the numerator by sintheta and then the denominator by sintheta

[gritty galleon]

you'd have $\frac{sin(\theta)cos(\theta)+sin^2(\theta)}{sin(\theta)cos(\theta)-sin^2(\theta)}$

[wraith dagger]

itzkraken.

[proud spindle]

nah u wouldnt tho

costheta-sintheta/sintheta =

this is the numerator fam

[proud spindle]

costheta/sintheta - sintheta/sintheta

= cottheta - 1

[gritty galleon]

Ohh

[proud spindle]

ye

[gritty galleon]

i was thinking of smthing else

[proud spindle]

all good blud

[gritty galleon]

well you could do that

[proud spindle]

isnt that easier

[gritty galleon]

Ig yeah

[proud spindle]

thx g

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[spice oracle]

Can somebody please help me with solving triangles in geomtery? I have an image for my problems that I cant figure out

[cedar kiln]

@spice oracle Has your question been resolved?

[spice oracle]

<@&286206848099549185>

[heavy briar]

Any question in particular?

[spice oracle]

44 rn

i did 45

[heavy briar]

Trig, right?

[spice oracle]

yes

i have these formulas

[spice oracle]

this is the one needed tho im pretty sure

sine i mean

but i cant figure out how to get the sides

[heavy briar]

I would solve it like this:
sin(A) = AB/BC

and then substitue your values in / rearrange

[spice oracle]

i have
angle A is 35
angle B is 55
and angle C is 90

[spice oracle]

but how to i get AB

[heavy briar]

You substitute BC and sin(A) in

because you know those

[spice oracle]

ok

[heavy briar]

You can write it in the form:
AB = BC * sin(a)

[spice oracle]

ohhh

i got 5.73576

[heavy briar]

So yeah, SohCahToa states that the sine of an angle is O / H (opposite length over hypotenuse length)

[heavy briar]

hmm

I don't think that's right

[spice oracle]

oh

then what would it be?

cuz this is what I got

[heavy briar]

oh yeah duh sorry yep that's right

[spice oracle]

ok thank you

also i just dont want to doubt myself for 45 but I got angle A equals cos 45

[heavy briar]

So yeah, just to revise, you plugged the info you were given (the angle which was 35, the length of BC (opposite) which was 10) into the equation sin(a) = AB/BC 👍

thanks

[spice oracle]

that makes sense now

also I was unable to figure out 46 and 47

would u be able to help me on those too?

[heavy briar]

let's find out

[spice oracle]

ok

[heavy briar]

okay so just like the last one, we have our formula, in this case, it's the Sine rule
all we need to do is plug in the values that we know

[spice oracle]

ok

[heavy briar]

a / sin(A) = b / sin(B) = c / sin(C)

so if we know B = 50, C = 40, and c = 5

[spice oracle]

also isnt angle A automatically 90 then?

[heavy briar]

🤷

yes!

[spice oracle]

so i have 3 angles and one side like in question 44

so i have to find side BC and AC

[heavy briar]

kk

so I guess we'll use trig functions for that?

[spice oracle]

so sin(40) = BC/5?

and sin(40) * 5 = BC?

[heavy briar]

frankly I'm rusty at this so this is helping me as much as you

but I think soooo

it's probably better to draw the triangle to visualize it

[spice oracle]

ok

sry its sideways but i have a triangle here

and BC is this

[heavy briar]

I think it's supposed to be O / H

and I think you put 5 as H

when 5 should be O

[spice oracle]

oh

[heavy briar]

is that makes sense, could be wrong

oh wait

nope

I'm wrong

I thought b was the hypotenuse

[spice oracle]

i went off of this

[heavy briar]

okay gn, too late for this, good luck
ty

[heavy briar]

👍

[spice oracle]

but yea ur right 5 is opposite

but its the same answer anyways right?

sin(40) = 5/BC

[heavy briar]

But A is the 90 degree angle

[spice oracle]

so its still sin40 * 5

[heavy briar]

so a must be the hypotenu~

okay, cool

[spice oracle]

yea

[heavy briar]

okay gn 4 real now

[spice oracle]

ok gn

[heavy briar]

professor dave has good videos btw

[spice oracle]

thank you

[cedar kiln]

@spice oracle Has your question been resolved?

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Use .reopen if this was a mistake.

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Send your question here to claim the channel.

Remember:
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[crimson sedge]

Hi again

[crimson sedge]

$${\alpha}^2 = 5\alpha-3
,,, ,{\beta}^2=5\beta-3$$
Find the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$

[wraith dagger]

.fire007

[crimson sedge]

$x^2-\frac{A^2+B^2}{AB} +1=0$

[wraith dagger]

.fire007

[crimson sedge]

It looks smth this

Rn

Not sure How to get AB

Here's the orginal question

[fading narwhal]

is this for jee prep?

I swear I did smthn like this before

[crimson sedge]

Nah

[fading narwhal]

aight

[crimson sedge]

Olympiad

[lyric cargo]

polynomials...

did this last month

find the actual roots first

oh wait they didn't give u an equation... mb

[crimson sedge]

Ye

Maybe

A²-5A+3=0

This is a quad

Can I evaluate this

Then plug in values and see??

[lyric cargo]

yeah

find Alpha and beta of that

i think it's just trial and error

[south tundra]

It's either option 2 or 3

[crimson sedge]

,w a²-5a+3=0

[wraith dagger]

Results provided by WolframAlpha

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

Requested by .fire007#0

[south tundra]

Note that (A^2 + B^2)/AB is always positive

So option 3 is the only possibility

[crimson sedge]

How can you AB can be positive??

[south tundra]

a^2 = 5a - 3
a^2 is nonnegative, meaning a has to be positive

Same thing for b

[crimson sedge]

Oh

,w x^e+2x^5+9

Alr

[south tundra]

If a and b are positive, then so is ab

[wraith dagger]

Results provided by WolframAlpha

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

Requested by uewhksbkdbqsk#0

[crimson sedge]

Yes

[lyric cargo]

true, squares r always positive

[crimson sedge]

Alr mate

Thanks.

.close

[cedar kiln]

Channel closed

Closed by @tall fiber

Use .reopen if this was a mistake.

[vague rapids]

wait

what about D

[lyric cargo]

oh-

[south tundra]

Hmm

[vague rapids]

reopen

Let me find a way

[south tundra]

.reopen

[cedar kiln]

✅

[livid hound]

consider subtracting the equations

and multiplying the equations