help-39

[sage patrol]

nvm

.close

[pearl pond]

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[old sable]

Solve for x.

I have done other questions with this type of layout but I’m not sure where to start here.

[tall mural]

is that 13.5?

[old sable]

yes

[tall mural]

use the concept of similar triangles

ratio

[old sable]

X + 8 should just be 8 sorry

[deep grove]

https://byjus.com/maths/similar-triangles/#:~:text=If all the three sides,%2FXZ then ΔABC ~ΔXYZ.

[merry stirrup]

It's a right angled triangle?@old sable

Or not?

[old sable]

no

[merry stirrup]

Ok

[deep grove]

x/(x+8) = 13.5/18

[tall mural]

hm?

[deep grove]

bruh

[tall mural]

x/13.5 = (x+8)/18

[deep grove]

still this

wrong

[merry stirrup]

Oh oh oh

You have to talk about the whole triangle forget it I am wrong

[old sable]

Where do I go from here?

[tall mural]

cross multiplication

then elementary algebra

[pearl pond]

@old sable Has your question been resolved?

[pearl pond]

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[quartz path]

hey im curious how you would simplify this type of equation

[buoyant panther]

it's not an equation

[quartz path]

i feel like its only possible if you somehow squared the bottom

what is an equation and why isnt this an equation

[buoyant panther]

in equation sign "=" must occur

[quartz path]

ah, so what would this be called?

[buoyant panther]

an expression/a function

[quartz path]

ok, so how do you simplify this type of function

[buoyant panther]

if you're talking about rationalizing denominator (let's say it simplifies) then multiply by conjugate of the denominator

[quartz path]

ive heard of that, whats a conjugate denominator

[buoyant panther]

the conjugate of a two-term expression is just the same expression with subtraction switched to addition or vice versa

[quartz path]

so if i had x+y, i would get -x-y right

[quartz path]

for this expression i should just switch all the signs for the numerator and denominator and this would become the conjugate

[buoyant panther]

nope

then it would be x - y

[quartz path]

but realistically we have +1x + 1y

why is the y special

[quartz path]

if that was the case, switching the orders around would change what answer we got, if x+y became a conjugate aka x-y we could also claim y-x giving a new answer

[buoyant panther]

we don't switch order

[quartz path]

but it doesnt matter what order its in initially, we could have x+y initially or y+x

the conjugate would be two different things even though x+y=y+x

[buoyant panther]

no matter if we multiply both numerator and denominator by conjugate, since we multiply by it to get (a-b)(a+b) = a^2 - b^2

[quartz path]

oh

what about for x + y + z is the conjugate x - y - z?

[buoyant panther]

it depends on how we treat terms

[quartz path]

wdym

what does treat the terms mean

[buoyant panther]

but do not overthink, just multiply by sqrt(x+5) + 2 and you'll see

[buoyant panther]

it can be (x+y) + z or x + (y+z)

in fact it depends on situation

[quartz path]

what situation would change it

[buoyant panther]

sometimes it would be better to choose 1st option, but 2nd option would be useful in other case

[quartz path]

better or correct?

[buoyant panther]

better I mean easier to simplify/solve something

[quartz path]

but regardless you could chose either x-y+z or x+y-z or even -x+y+z and you could get a correct answer

[quartz path]

for this problem would the conjugate be this?

image above is conjugate, image below is og

[buoyant panther]

we consider conjugate of the denominator

not whole expression

[quartz path]

wait also i screwed that up, allow me to try again

[buoyant panther]

look

$$\frac{x+1}{\sqrt{x+5}-2}=\frac{x+1}{\sqrt{x+5}-2} \cdot \frac{\sqrt{x+5}+2}{\sqrt{x+5}+2}$$

[quartz path]

this is conjugate

[jolly parrot]

Modus

[quartz path]

oh and this keeps it equal

[buoyant panther]

yes

this is the point of doing it

[quartz path]

wait so is my screen shot called the conjugate if so what is yours called

when they are stacked

[quartz path]

is it called a fractional conjugate?

[buoyant panther]

I'd say just conugate

[quartz path]

so both are the conjugate

[buoyant panther]

simple thing:

[jolly parrot]

Modus

[quartz path]

is there no term for the conjugate in a fraction

[buoyant panther]

in fact you're right it's conjg/conjg (because we don't want to change an expression)

[quartz path]

also if the first screenshot is the conjugate, is the second screenshot also true?

[quartz path]

ok that makes sense

so this simplified version of the original problem is this?

or the 3 would be 1

in the denominator

that doesnt look like simplist form

well you can get it here... but how do you get it to the last step?

oh wait i figured it out, factor out (x+1)

cancel

and you have your answer

thank you!

.close

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[primal bluff]

Why is sin(pi/2) 1, but when I type it in the calculator its 0.027?

[buoyant panther]

maybe you've typed smth wrong idk

or check if you've chosen radians

since

look

[coarse dawn]

Depends on if you're in radian or degree mode

Looks like you might be in degree mode

[pearl pond]

@primal bluff Has your question been resolved?

[pearl pond]

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[subtle arrow]

so i have (4y-3)"2, and the answer is 16y"2-24y+9, but where does the 24y come from

[pearl pond]

@subtle arrow Has your question been resolved?

[elfin laurel]

does " mean the power of?

[subtle arrow]

Yes

[pearl pond]

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[rare totem]

A plane is flying away from a control tower. At a certain point in time the plane is 200 km/ away from it and flies with a speed of 800km/h and with a positive acceleration of 20km/h^2. The plane type does not allow for more then a +- 120km/h^3 change of acceleration.
let f(t) be the distance away from the control tower, measured in km, after the point in time. what can we say about f(t)?

[rare totem]

https://cdn.discordapp.com/attachments/1018703193473024061/1036306004092141608/IMG_1053.jpg

this is a multiple choice question and i think its 10 but i feel like ive come to that conclusion for the wrong reason

[grim patrol]

LOL THIS MCQ

[rare totem]

and someone earlier said i need to use the Taylor-Lagrange's equality but i dont think i should know at this point

[rare totem]

?

[grim patrol]

just looks like deth to me

[rare totem]

it looks a bit wonky sure but its perfectly readable isnt it?

I’ll repost just in case

.close

[pearl pond]

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[spark mesa]

.close

[pearl pond]

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[noble sigil]

hey can someone help w this

idk which test to use

[bold ledge]

first try ast

to check if div or not

then try ratio test?

idk it's been a while since I've done this

[noble sigil]

is this ok for ratio test?

[bold ledge]

not quite

you need to make sure it's (2(n+1)-1)!

and not (2n)!

because it needs to be odd

[noble sigil]

Ohhh ok is this better ?

[bold ledge]

double check your cancellation on the first step

why is there still a (2n-1) in the denom?

[noble sigil]

how does that cancel?

[bold ledge]

there's a (2n-1) in the numerator

it's

1 3 5, ..., 2n-1, 2n+1

divided by

1 3 5, 2n-1

so you're left with only 2n+1

not (2n+1)/(2n-1)

[noble sigil]

huhw here did this come from

isnt it 1 3 5 ... 2n+1

[bold ledge]

it is

but right before the 2n+1 there's a 2n-1

[noble sigil]

ohh hok

do the (-1)^n and (-1)^(n-1) just cancel each other out?

or would it make the denom negative

[bold ledge]

well you're taking abs

so it doesn't matter at all

[pearl pond]

@noble sigil Has your question been resolved?

[pearl pond]

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[desert forge]

I need to prove that for every natural n

[shrewd basin]

what have you tried

[desert forge]

I tried to do it through induction, but got stuck trying to figure out the right side.

[shrewd basin]

show your work

[desert forge]

I don't have anything yet... I'm trying to get an idea to go with

[shrewd basin]

Well if you want to try this by induction youd need base case etc.

[desert forge]

doing it with n=0 is easy

i couldn't get the n + 1

[potent hedge]

Express $4^{n}$ using binomial coefficients

[jolly parrot]

black_couscous

[potent hedge]

That might help

[desert forge]

I looked at that. That's what I got, but I'm not sure how it helps

[potent hedge]

What happens if you divide by $2n+1$?

[jolly parrot]

black_couscous

[potent hedge]

Use $k\binom{n}{k}= n\binom{n-1}{k-1} $

[jolly parrot]

Pure

[desert forge]

I'm sorry, my semester's just started and I'm not getting it

[pearl pond]

@desert forge Has your question been resolved?

[potent hedge]

$\binom{2n+2}{n+1} = \frac{(2n+2)(2n+1)}{(n+1)^{2}} \binom{2n}{n}$

[jolly parrot]

black_couscous

[potent hedge]

Try to show that $\frac{(2n+2)(2n+1)}{(n+1)^{2}}\ge \frac{4}{2n+2}$

[jolly parrot]

black_couscous

[potent hedge]

Which translates to $ \frac{2n+1}{n+1} \ge \frac{1}{n+1}$

Which is fairly doable

[desert forge]

Thanks! Got it

.close

[pearl pond]

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[golden ice]

These are the answers it could be

It's one of these

[ebon skiff]

ok

[golden ice]

I do not understand the question

[ebon skiff]

x is between -1 and 3

so

which tables can we instantly exclude?

[golden ice]

the 2nd and 5th

[ebon skiff]

good

now

if x is -1

what's y?

look at the equation of the line

[golden ice]

not sure

[ebon skiff]

just plug x=-1 into the function

[golden ice]

wdym

[ebon skiff]

y=4x+2

and x is -1

find y

[golden ice]

y=-4+2?

is that right?

y=-2*

alr i got the answer

[pearl pond]

@golden ice Has your question been resolved?

[pearl pond]

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[shrewd basin]

try part a again

[pearl pond]

@tardy sun Has your question been resolved?

[pearl pond]

@tardy sun Has your question been resolved?

[rose rock]

@tardy sun replace r^3 with 1/4 your answer in part a

Then it is a 👏 👏 telescoping series!

[pearl pond]

@tardy sun Has your question been resolved?

[pearl pond]

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[desert wraith]

for 10 , do I continue using l'hospital rule or does it stop there

its confusing me

[woven matrix]

what's the derivative of ln(x+1) ?

(it's not x+1)

[desert wraith]

1/x+1?

[woven matrix]

1/(x+1) yes

so now, when you use l'hospital on this

what do you get ?

[woven matrix]

yes

[desert wraith]

oh yeah , i guess used the a/b /c = a/b*c

yeah, makes sense now lol

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[pearl pond]

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[tribal thunder]

I need 18

I already know the answer just need the steps to solve it

Answer is 5

Don’t take me step to step just the general equation and the formula to solve it

<@&286206848099549185>

[pearl pond]

@tribal thunder Has your question been resolved?

[tribal thunder]

<@&286206848099549185>

[pearl pond]

@tribal thunder Has your question been resolved?

[pearl pond]

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[sacred pagoda]

can someone confirm if I did this correctly? Or am i supposed to modify proofs differently?

[sacred pagoda]

i feel like i didn't understand the question properly

[pearl pond]

@sacred pagoda Has your question been resolved?

[sacred pagoda]

<@&286206848099549185>

[pearl pond]

@sacred pagoda Has your question been resolved?

[near echo]

you started with the conclusion

the idea is to start with a similar proof as they did, only that since x>y, (x-y)^2 > 0, not >=

@sacred pagoda does that make sense?

[sacred pagoda]

so i just rewrite the whole proof, but replace the inequality with the new one and go on from there?

[near echo]

pretty much

[sacred pagoda]

but isnt that what i did up there

[near echo]

well no

you started with the conclusion

and worked back

[sacred pagoda]

oh

so do i start with (x-y)^2 >= 0?

[near echo]

no

[sacred pagoda]

ah so i need to make a whole new inequality to start off and work my way to the conclusion

[near echo]

start with (x-y)^2 > 0

since x-y > 0

[sacred pagoda]

oh right

thanks

.close

[pearl pond]

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[fast lagoon]

I am unsure how to solve this, it seems like a pretty low tier question but it has me stumped, im in algebra 1

[fast lagoon]

bro looking at these other channels makes me feel like an idiot 💀

[fast zealot]

I had to google how much a nickel and dime was worth

[fleet sky]

nah bro people have various starting points

[fast zealot]

EU problems

[fleet sky]

okay

so we know that we have 500 total coins yeah?

[fleet sky]

so let's say we have x nickels and y dimes. Then we have the relation x + y = 500. Does that make sense?

we also know that nickels are worth $0.05 and that dimes are $0.10. So we know that we have a total of $29.45 in the jar, so would it be fair to make another equation for the value, say, 0.05x + 0.10y = 29.45.

so now we have 2 equations

think you got it from there? 🙂

[fast lagoon]

sorry i wasnt checking my phone

yes thank you for your help 🙏🏻

.close

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[pearl pond]

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[shrewd widget]

Two objects differ in density and mass. object A has a mass of 8 times that of object B. The density of object A is 4 times the density of object B. how do their volumes compare

[shrewd widget]

I’m stupid and don’t understand

[unreal pewter]

You know how density is mass/volume right? Try writing each line of information down in terms of the variables

like, mA = 8 * mB

$m_A = 8 m_B$

[jolly parrot]

void

[unreal pewter]

Then the density of A is 8 times density of B, but you can write the densities as mass divided by volume

[shrewd widget]

Yeah

[unreal pewter]

so how would you write that down

[shrewd widget]

Idk I’m slow one sec

8/4

Nope

I’m given density

Yeah

I’m actually dumb

[unreal pewter]

$\frac{m_A}{v_A} = 4 \frac{m_B}{v_B}$

[jolly parrot]

void

[unreal pewter]

did you get to this?

I actuall don't know what the answer is yet I haven't bothered to work it out let me see

[shrewd widget]

It’s just asking how their volumes compare

[unreal pewter]

yeah

so from the equation for the densities

you just need to get rid of the masses and write it in terms of the volumes

which you can do because you know that $m_A = 8m_B$

[jolly parrot]

void

[shrewd widget]

I don’t have volume I have mass and density

Yeah

[unreal pewter]

so if you look at that equation for the density

you can substitute the masses

$\frac{8m_B}{v_A} = 4 \frac{m_B}{v_B}$

[jolly parrot]

void

[unreal pewter]

did you see how I got that

[shrewd widget]

Yeah 8(4)= 4(4)

Divided by volume

8(8) I mean

4(8)

I think

[unreal pewter]

wdym

[shrewd widget]

You just put the value of the mass where m is

Right?

Yeah

Idk I’m out of it lol

[unreal pewter]

okay so you don't need to know what the mass actually is here

[shrewd widget]

What

It’s given to me

Oh

8 times

[unreal pewter]

you know what the relationship is

[shrewd widget]

Nvm

Yeah

[unreal pewter]

but it doesn't tell you what the mass catually is

yeah

but it doesnt really matter because you can just write mass of A as

8 times mass of B

[unreal pewter]

which is what I did here

[shrewd widget]

Yeah I see that

[unreal pewter]

then you can divide both sides by the mass

[shrewd widget]

I still don’t understand how their volumes compare

[unreal pewter]

and the mass doesn't matter anymore

since there's a mB on both sides

right

[shrewd widget]

So we got rid of mass

[unreal pewter]

so what equation do you get now

[shrewd widget]

8/VA = 4/VB

[unreal pewter]

mmhm

and then can you now get a ratio between VA and VB?

[shrewd widget]

4=2

Somehow

[unreal pewter]

huh

what did you do there

[shrewd widget]

I have no idea

This is what happens when you don’t sleep 😂

[unreal pewter]

$\frac{8}{v_A} = \frac{4}{v_B}$

[jolly parrot]

void

[unreal pewter]

you got to this equation right?

[shrewd widget]

Yeah that’s what I have

Yeah

[unreal pewter]

then you just want to find

[shrewd widget]

Volume

[unreal pewter]

well not the volumes themselves

[shrewd widget]

The relationship

[unreal pewter]

the ratio between the volumes

yep

which would be

$\frac{v_A}{v_B}$

[jolly parrot]

void

[shrewd widget]

What happened to 8 and 4

[unreal pewter]

oh I'm just telling you what you want to find

so how would you rearrange the equation

to get something that looks like that

[unreal pewter]

to this

[shrewd widget]

Yeah

[unreal pewter]

you'd multiply both sides by vA

[shrewd widget]

Yeah

I was gonna say that

[unreal pewter]

$8 = \frac{4v_A}{v_B}$

[jolly parrot]

void

[unreal pewter]

and then you know what to do from here

[shrewd widget]

So now I have 8 = 4Va/Vb

Wouldn’t it be 4/Vb x Va

[unreal pewter]

that's the same thing

[shrewd widget]

I multiply both sides by 4

[unreal pewter]

what would that give?

[shrewd widget]

32 = Va/Vb

[unreal pewter]

on the right side, you multiplied a 4 to a 4

$84 = 4\frac{4v_A}{v_B}$

[jolly parrot]

void

[shrewd widget]

Yeah

Wait that is a mistake because that makes 4 on the right side bigger

No

Man what am I on right now I’d like to know

[unreal pewter]

yeah, you'd want to get rid of the 4 on the right side

[shrewd widget]

4x4Va/Vb

Does that get rid of 4

[unreal pewter]

nope

if something is multiplied by a number

what operation undos that multiplication?

[shrewd widget]

Division

[unreal pewter]

yep

[shrewd widget]

Can’t remember how to divide a fraction

[unreal pewter]

you have

$8 = \frac{4v_A}{v_B}$

[jolly parrot]

void

[unreal pewter]

I wrote it as a fraction, but that's the same thing as

$8 = 4 * \frac{v_A}{v_B}$

[jolly parrot]

void

[shrewd widget]

Oh

So wouldn’t that just be 32 = Va/Vb

[unreal pewter]

you said division would undo multiplication

so on the right the ratio you want is being multiplied by 4

you'd want to divide both sides by 4

[shrewd widget]

Oh

Right

So it would be

2 = Va/Vb!

[unreal pewter]

yep!

[shrewd widget]

I still don’t see how they compare though lol

[unreal pewter]

you now have the ratio between Va and Vb

[shrewd widget]

Yeah

[unreal pewter]

oh maybe I shouldn't have put in as a fraction

well you have

$2 = \frac{v_A}{v_B}$

[jolly parrot]

void

[unreal pewter]

which just becomes

$2v_B = v_A$

[jolly parrot]

void

[unreal pewter]

(do you see how?)

[shrewd widget]

Volume b will always be 2 times the volume of volume a

[unreal pewter]

mm the other way around!

the equation says

2 times volume of B is equal to volume of A

[shrewd widget]

But object A has a mass 8 times and density 4 times

So

[unreal pewter]

yeah, and object A has a volume 2 times greater than object B

[shrewd widget]

Wouldn’t it be object A will always be 2 times volume of object A than object B

But the equation is saying that it will be 2 times volume B than object A

[pearl pond]

@shrewd widget Has your question been resolved?

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[tardy path]

Can anybody show me how to get the formula for the sum of all odd numbers squared e.g.(1²+3²+5²+7²...+n²)?

[foggy mason]

$\sum_{k=0}^n (2k+1)^2$

[jolly parrot]

Ｈｅｒｅｌｓ

[foggy mason]

?

$(2k+1)^2 = 4k^2 + 4k + 1$

[jolly parrot]

Ｈｅｒｅｌｓ

[foggy mason]

$$\sum_{k=0}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

$$\sum_{k=0}^n k = \frac{n(n+1)}{2}$$

[jolly parrot]

Ｈｅｒｅｌｓ

[tardy path]

Thanks

But, i still dont see the sum of all odd numbers squared

Anybody?

Can anybody show me how to get the formula for the sum of all odd numbers squared e.g.(1²+3²+5²+7²...+n²)?

<@&286206848099549185>

Oooh i think i got it

Thanks again @foggy mason

[pearl pond]

@tardy path Has your question been resolved?

[pearl pond]

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[midnight haven]

need hep

[midnight haven]

fill in the other angles then solve two equations for a and b

;-;

2a + b = 180 ?

120 = 3

b = 40 ?

a = 70 ?

so answer is 110 ?

no

;-;

a and b are exterior angles

yes

yes but

aint

third exterior

a

too

sure

so a + a + b = 180 ?

or wait

360 ?

the exterior angles of a triangle do not add up to 180

360

then

xd

[midnight haven]

yes

ok

then

360 = 2 times 30

300

then 300 divided by 3

[tardy path]

Yes

[midnight haven]

;-;

angles are a = 130 b = 100 ?

so

a + b = 230 ?

,calc 2*(180-130)

[jolly parrot]

Result:

100

[midnight haven]

yup

;-;l

ok

.close

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[tardy path]

Hey this doesent work

[tardy path]

It only works for zero

[plush bramble]

expand your numerator

[tardy path]

Okay

Still the same dud
If you plug in 1 you get 9
Where the answer should have been 10

@plush bramble

[plush bramble]

why should it be 10?

oh cuz k = 0?

1^2 + 3^2 = 10

[tardy path]

@plush bramble

See?

[plush bramble]

right. you just verified k=1

you need to add k = 0 to get to n=1

[tardy path]

Ooooh

[plush bramble]

no wait you did the sum. just left it out of your work. i found your error

you forgot to sum 1

[tardy path]

No i did add it

4+4+1=9

[plush bramble]

how'd you get from the top to the bottom ?

[tardy path]

I put the values for k² and k in there

[plush bramble]

you're leaving out the summation

[tardy path]

Where?

[plush bramble]

do you know what $\sum_{k=0}^n k^2$ means?

[jolly parrot]

riemann

[tardy path]

Ya it means the sum of all natural numbers squared

[plush bramble]

close enough

[tardy path]

Thenk what exactly?

Then^

[plush bramble]

when you did this substitution, you need a sum over k=0 to n

[plush bramble]

first n natural numbers. the sum stops at n

[tardy path]

Oh yah

So how do i procceed to get the right answer?

[plush bramble]

the way you're describing this makes it sound like you think k^2 EQUALS n(n+1)(2n+1)/6

put a sum k=0 to n on the left side here

then distribute the summation using these

and also figure out what $\sum_{k=0}^n 1$ equals

[jolly parrot]

riemann

[plush bramble]

the result will ALMOST look like this

[plush bramble]

it's not 1

[tardy path]

So basically first i find out the sum of the first n natural numbers squared
Then i find sum of first n natural numbers.. then put it here?

[tardy path]

I mean first i find the values for these two?

Anyway thanks man i think i can solve it now @plush bramble

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[glad oar]

We are using rectangles to estimate area under a curve

I'm confused about how to do this as there is no function. and I don't know how to get middle points.

answer is .70

.C

.c

.close

[pearl pond]

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[copper stratus]

hlp

pls

my brain aint braining today

thats it

,calc 0.000001^0.000001

[jolly parrot]

Result:

0.99998618458488

[copper stratus]

had to do that

[pearl pond]

@copper stratus Has your question been resolved?

[copper stratus]

<@&286206848099549185>

[copper stratus]

…

[steel coral]

ok so its in the linear equation form y=mx+b

you know that b is your y-intercept

what is the y-intercept on that graph?

it appears to be 4, so you can say that your b value is 4

next, you can apply the rule (y2-y1)/(x2-x1)

for gradient

so we are given two points where it intersects, so it lies on the line

(0,4) and (10,0)

assign it as (x1,y1) and (x2,y2)

and sub into gradient eqn

so that should be (0-4)/(10-)

which gives your gradient as m=-2/5

remember it is decreasing, so it has to be negative

so finally create your linear equation by subbing in your known values

so since b is 4, m=-2/5

it will be y=-2/5x+4

i hope it helps, I have to go

[copper stratus]

so whats this one

@steel coral

oh its m=-4

right

[steel coral]

that the y-intersect, as it cross the y line

and remember how y intersect is the value of b, if it’s y=mx+b

You can see one grid above it is positive 5

[copper stratus]

so am i right or wrong

im right

right

[steel coral]

That point you circled would be b=4

[copper stratus]

ok\

.close

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[fervent pawn]

How can you tell that this function is concave or convex? Its not the main point of the question and its not even in the answer sheet, they just simply state its concave

[fervent pawn]

But im not sure how to tell

[plush bramble]

logarithms are concave

you should know that from single variable calculus

[fervent pawn]

ahh right, thanks

[plush bramble]

if you forgot, take a second derivative of the natural logarithm and show the second derivative is always negative

[fervent pawn]

yeah I just wasnt sure because this is a multi-variate function

[real jungle]

it's not as obvious as this

because there is a -log

which is convex

[fervent pawn]

can the Kuhn-Tucker theorem only be applied when the function in concave btw? what should I use if it's convex?

[plush bramble]

g is not a function of Q

so Q is a constant

[real jungle]

oh ok mb

[pearl pond]

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[gusty bluff]

heya

[gusty bluff]

this isnt specifically a problem i have, rather a method

i know how to find the scalars for linear combinations of vectors

but im not sure how to do it for matrices of size nxn and so on, and neither the book nor the lecture covered it. could someone explain how to set up a problem to find these scalars? im assuming something with rref

[plush bramble]

is that really the whole problem?

seems like you should be getting optional choices

[gusty bluff]

Yes, we have b matrices

But I just want to know how to set it up

[plush bramble]

show what you mean

[plush bramble]

my guess is write a generic matrix M = xA + yB + zC and set those equal to each a), b), c) and see if any solutions (x,y,z) exist for each of them

[gusty bluff]

yes, i know that, but im just confused on how the written thing would look

i could just rewrite it as a system of equations but i dont think thats the way prof wants it done

since i know for vectors you can just make an augmented matrix with column vectors and the thing youre checking. im just not sure how to structure that with 2x2 matrices

[plush bramble]

what do you mean?

do you know how to multiply a matrix by a number?

e.g. xA = [4x 0; -2x -2x]

[gusty bluff]

Here let me draw something rq

[plush bramble]

then what way do you think it's supposed to be done?

[gusty bluff]

Alright, as a preface, there were problems previous which ask the same thing but involved column vectors rather than larger matrices

i solved one like so

and rref the augmented matrix you get (and the reduced values are the scalars)

so i was wondering how youd set up an augmented matrix for the 2x2 matrices of the other problem

[plush bramble]

you can rref M here

M is just a 2x2 matrix

[gusty bluff]

Let me write what I think it is

is this what youre suggesting?

[plush bramble]

yes

[gusty bluff]

Ok, I've done it. But it raises another question

The book answers suggest option b is a combination

But the only coeffs I can seem to get is 0 for x y and z

It thought they had to be nonzero?

[plush bramble]

then there's something wrong with your algebra

[gusty bluff]

i checked with a tool, it cant be

[plush bramble]

show it then

[gusty bluff]

if visibility is poor i can resend

hmmm actually

could it be infinite solutions for coefficients?

[plush bramble]

that's clearly a 4x4 matrix

[plush bramble]

this is a 2x2 matrix

[gusty bluff]

yes, but the matrix insinuates that 4x+y+0z = 0, 0x-y+2z = 0, -2x+2y+z = 0, and -2x+3y+4z = 0

right?

[plush bramble]

is this just for b?

[gusty bluff]

yes

so it was my belief that these system of equations could be made into an augmented matrix

such as this

and if it was a linear recombinations, the values of x y and z would be presented as a unique solution in the rref form of it

[plush bramble]

the way you input it makes it look like there's a 4th variable

but you only have 3 variables

[gusty bluff]

yes, the 4th is the b matrix

it should draw a line between them but it sadly doesnt

[thorn cobalt]

Hey

can someone welp me

I need to solve thsi

I got -0.5

I am just not sure

this assignment cost ALOt of my grade

so plz

[plush bramble]

Read #❓how-to-get-help

[thorn cobalt]

thx

[plush bramble]

I don't get what this means

[gusty bluff]

to show that it is an augmented matrix

and not a matrix of four unknowns

[plush bramble]

You're augmented matrix should be number of equations by number of variables

4x3

[gusty bluff]

thats just [A] though

[plush bramble]

Unless you learned rref differently

[gusty bluff]

we were taught that with Ax = b

you can do [A|b]

and rrefA while performing the same ops on b

to find x

[gusty bluff]

sorry about this, turns out i dropped the ball hard. the coefficient CAN be 0

[pearl pond]

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[novel sequoia]

need help 3 6 and 9

[twin lake]

what don't you understand about 3?

[novel sequoia]

I dont know to solve it

i cant get the common difference

[twin lake]

why?

[smoky musk]

then it's not an arithmetic sequence, then

[novel sequoia]

🤔

[novel sequoia]

i cant make it work

[twin lake]

there's a common difference

just subtract first term from the second one

[novel sequoia]

can you show me the process

knowing that will help me do the 6 and 9 too

<@&286206848099549185> i need help getting the common difference for 3

[twin lake]

d = -17/6 - -1/3

[pearl pond]

@novel sequoia Has your question been resolved?

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[kind brook]

hi

[kind brook]

hum

i cant find it anywhere, so wath's the exponential rule

on google and everything it wont say

[twin lake]

what's your quesiton?

[kind brook]

wath is the exponential rule

like an affine is y = ax+b

but wath is an exponential

a quadratic is y=ax2

but wath is an exponential

<@&286206848099549185>

please ?

bro

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[peak viper]

I'm having trouble finding the value of k in the expansion (x+k)^5 where k is all rational #'s and the coefficient of the term in x^5 is 1512

[old geyser]

Binomial theorem or Pascal's triangle

[fiery vortex]

general term

then

it’ll be easy

form there

from*

[peak viper]

binomial theorem

[fiery vortex]

yeye

$n\choose\r$

[jolly parrot]

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

[fiery vortex]

aye

nice

$n\choose{r}$

[jolly parrot]

Springsskateboard

[fiery vortex]

nice

$n\choose{r}$$(x)^{5-r}(k)^r$

Bruh

[jolly parrot]

Springsskateboard

[peak viper]

where n = 5? and r = 8?

or did i misinterpret

[fiery vortex]

n is 5 ye

how’d u get

r=8?

$5\choose{r}$$(x)^{5-r}(k)^r$

[jolly parrot]

Springsskateboard

[fiery vortex]

they gave

coefficient

of x^5

ye

look at the

exponent of

x

$5\choose{r}$$(x)^{5-r}(k)^r$

[jolly parrot]

Springsskateboard

[fiery vortex]

it’s 5-r

what’ll make

5-r be 5

[peak viper]

how does the 1512 get affected?

[fiery vortex]

okay first

look at this

$5\choose{r}$$(x)^{5-r}(k)^r$

[jolly parrot]

Springsskateboard

[fiery vortex]

what’s the

exponent

of x

here

[peak viper]

5-r

[fiery vortex]

yeah

and in the qn

u know the coeff of

x^5

what value of r makes 5-r be 5

[peak viper]

0

[fiery vortex]

uhhh wait

did u copy the

qn wrongly

or smth

cuz this ain’t adding up

[peak viper]

The question is "In the expansion of (x+k)^8, where k is all rational numbers, the coefficient of the term in x^5 is 1512. Find the possible values of k."

[fiery vortex]

bruhhh

(x+k)^8

[fiery vortex]

u said (x+k)^5 here HAHA

[peak viper]

ohhh

my bad

[fiery vortex]

yeye

issok

so

same thing

find general term

$8\choose{r}$$(x)^{8-r}(k)^r$

[jolly parrot]

Springsskateboard

[fiery vortex]

so

look at

exponent

of x

what value of r

makes it 5

[peak viper]

3

[fiery vortex]

ye

so r is 3

substitute and evaluate

and u have it

ur answer

and solve*

ew

[jolly parrot]

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

[fiery vortex]

brah

ok yk what I mean

the coeff of x^5 is $8\choose{3}$$(k)^3$

[jolly parrot]

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

[fiery vortex]

BRUH

the coefficient of x^5

is

[peak viper]

after subbing is it 56k^3(x)^5

[fiery vortex]

YAS

yes

nice

so what’s the coefficient of x^5 here?

in terms

of k

it’s 56k^3

yeye

[peak viper]

1512

[fiery vortex]

^

and we know

that coefficient is 1512

so 56k^3 = 1512

[peak viper]

so 56k^3(1512

[fiery vortex]

yeye and u solve

for k

[peak viper]

root3 of 27?

so k =3

dude nice

only harder from here though 🙂

thanks for your help ill be back if i need it lmao

.close

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[hot niche]

I got this problem, im not sure how to solve it, is the diagram and equation right?

Wait how did you get 4x?

Ohhh I put 8x instead of x+8

thank you

I have a couple more if you dont mind?

For this diagram

I cant remember what my thoughts were give me a second

Ok so the first thing I did was find X

Im not sure if its right

By using 180-x=z

Because of vertical angles theorem

How did I get x? Cant remember tbh

But you only get one angle for c

X*

To find the third angle you need 2 others do you not?

All there is is 80°, and x

Thats 1 angle

Can I just share another way that I think makes sense and isnt just making magical concepts up

I think where I would start looking at this now is the triangle in the middle

Taking 35 and 36 together to get 71

Ohhhh

but i would still like to share the alternative way

so with 71

we can subtract 180-71 to find y

Which is 109

And because y is a linesr pair with the unnamed angle in the topleft triangle

180-109 is also 71

meaning 80+71=151

180 subtracted by that to get x

which still comes out to 29

Third angles theory i think

doesnt matter

theorem*

Ok, I have two more, one which is basically the same as the first one and another one that is weird

alternate exterior angle theorem?

Oh well

M is congruent to N

O is 5 times more than N

I had a trainwreck of thoughts on paper

N is 35?

Aw man I had this one right lol

i just didnt add right 💀

Yeah I did it worked

So this is the directions

Not the problem

This is the one i dont know about

P is congruent to q

S is congruent to R

I dont know man

My diagrams are all messed up

Like, the postulates?

We have only learned SAS and SSS postulates

But i dont think we can do this without my brain exploding

we knoe Segment SR is congruent to segment SR

So there is a side

but it doesnt follow sas

Didnt learn it so it shouldnt be on here

But i dont think it is

Ill put it down

Thanks so much

you a G fr

.close

[pearl pond]

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[pure depot]

How do i set up the integral for this? I need to find the total mass of the following solids by setting up appropriate
triple integral(s) with specified order of integration. Assume the density function is
uniformly δ(x,y,z) = 1.

[jolly parrot]

classic truedough

[pure depot]

<@&286206848099549185>

.close

[pearl pond]

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[red harbor]

b

[red harbor]

i try it but only got 2 for an answer which is wrong

[pearl pond]

@red harbor Has your question been resolved?

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[pearl pond]

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[pearl pond]

@graceful dragon Has your question been resolved?

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[mortal sparrow]

Hi I've a quick question. In a group G, is the subset S containing all elements of the form g**2 for g in G always a subgroup of G

[mortal sparrow]

I think the answer is no I was just hoping for a second opinion

[pearl pond]

@mortal sparrow Has your question been resolved?

[plush bramble]

do you have a proof?

it should just be checking properties of a subgroup

[mortal sparrow]

Yes see I don't think it's always closed under the operation

[plush bramble]

what's your counterexample?

[mortal sparrow]

In GL(2,Z) I can provide two elements whose product of their squares doesn't have an inverse with integral entries

Is that enough to say its not true?

[plush bramble]

do each of the elements have inverses in your subgroup?

this is all you need to check