help-26

[vestal sigil]

they gave you the equation of the first line

that gives the y-intercept

aka the y-value of the power station

do you see that :p

[sleek haven]

um

one sec

oh yes

sorry

-3

[vestal sigil]

yesss

thats one thing done

our equation is now y = mx - 3

now, you gotta find the range of the slope m

so that our new line does not go through the mountain range 1

[sleek haven]

That I found out earlier and = 0

[vestal sigil]

what did you find

[sleek haven]

the b^2-4ac?

[vestal sigil]

what did you do exactly

[sleek haven]

that it's a tangent that kisses it

I believe

that's what it says in my notes

there is only 1 answer

[vestal sigil]

yeah, but did you find the specific value of the slope m

for which they "kiss"

[vestal sigil]

yes, its a specific value of the slope m

that you gotta find!

[sleek haven]

so
b^2-4ac
-10^2-4x1x25
-100-4x25
-100+100

=0

m = 0?

[vestal sigil]

no what :p

you calculated the discriminant of the quadratic?

[sleek haven]

lol

[sleek haven]

how do you calculate that?

[vestal sigil]

hahaa

consider our objective

we want to find the value/interval of the slope m

so that our new line and the parabola of the mountain range 1 "would kiss"

so you equate thee 2 expressions

the equation of our new power line and the parabola's

[sleek haven]

I don't really understand

[vestal sigil]

the equation of the moutain range 1 is y = (x-5)² - 1

and the new power line equation is so far y = mx - 3

if you want them to "kiss" each other

then you gotta equate those 2 expressions

[sleek haven]

equate?

[vestal sigil]

yess

(x-5)² - 1 = mx - 3

[sleek haven]

OOOOOOOOOOOOOOOOH

[vestal sigil]

and find out which value of the slope m makes them "kiss"

[sleek haven]

y = (x-5)² - 1 and y = mx - 3

[sleek haven]

make that

[vestal sigil]

yess

[sleek haven]

yes

very good

[vestal sigil]

thats the quadratic

not the given one

[sleek haven]

I'll quickly calculate that

[vestal sigil]

okay, try to write it as ... x² + ... x + ....

standard

[sleek haven]

I don't really understand that

anyway I've got x^2-10x+27=mx (haven't finished)

[vestal sigil]

what about mx

didnt finish 😛

[sleek haven]

I'm trying to find m

not x

[vestal sigil]

just write it in standard form

youll see

... x² + ... x + ... = 0

[sleek haven]

do I factorise the x^2-10x+27?

[sleek haven]

oh

[vestal sigil]

no, just follow my instructions, you will see

[sleek haven]

ax^2+bx+c
1x^2-10x+27/mx=0

[vestal sigil]

what, how :p

[sleek haven]

no

1x^2-10x+27=mx

wait

[vestal sigil]

x² - 10x + 27 = mx

x² - 10x - mx +27 = 0

carry on :p

[sleek haven]

OH silly me

[vestal sigil]

lol

[sleek haven]

now do I factorise?

[vestal sigil]

... x² + ... x + ... = 0

finish putting it in this standard form

you need it

[sleek haven]

the quadratic formula?

[vestal sigil]

x² - 10x - mx +27 = 0

[vestal sigil]

we only need the discriminant

[sleek haven]

do I do the b^2-4ac?

[vestal sigil]

yess, but i ask you

what do we do with it?

[sleek haven]

put it as m into the orignial equation?

[vestal sigil]

sorry, i was on a phone call

this is the quadratic

x² - (m+10) x + 27 = 0

now recall our goal

we want the new line and the moutain range 1 parabola to not intersect

how do you achieve that using the discriminant?

must it be > 0

or = 0

or < 0

?

[sleek haven]

yes

one moment

[vestal sigil]

we want the new line and the moutain range to not intersect

aka, we want to not have a solution

for x² - (m+10) x + 27 = 0

what should b² - 4ac be?

[sleek haven]

the difference between the two

that's what I remember

[vestal sigil]

if you want the quadratic to not have any solution

[sleek haven]

being <0

[vestal sigil]

aka, the new line and the moutain range never intersect

[vestal sigil]

then yesss

b² - 4ac must bee < 0

thats the only case where they dont intersect

good job

so write that out and find the range of the slope m

[sleek haven]

righto

[vestal sigil]

so what did you end up with :p

[sleek haven]

do you list the values

a =
b =
c =

my notes are very confusing

[vestal sigil]

a,b,c?

find the range of m!!

[sleek haven]

you plus the values together

plus/minus

and that gets you smth

[vestal sigil]

i dont get what you are saying

[sleek haven]

nor do I tbh

[vestal sigil]

take a picture of what you wrote

okay

look

the discriminant = b² - 4ac

we want it to be < 0

so that the new line and range 1 do not intersect (no real solutions)

you got x² - (m+10) x + 27 = 0

so

(m+10)² - 4 * 1 * 27 < 0

you see

[sleek haven]

Not really

[vestal sigil]

what do you not understand

O:

[sleek haven]

one sec

[topaz sinew]

@sleek haven Has your question been resolved?

[sleek haven]

thanks for the help, much appreciated.

.close

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[halcyon willow]

or any thoughts?

[halcyon willow]

.occupied

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[shut orchid]

I would try to substitute sin(x)

[neon iron]

solve $4y^3 + 2y^2 -2y - 1=0$ first, once you get all the roots, substitute $y=sin(x)$ and solve each individual factors

[thorny flame]

uop

[shut orchid]

yeah, this^^

you can, but not from all the terms

you have a term without y...

[neon iron]

I think you should revise basic algebra before attacking this problem.

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[round canopy]

A polar bear rests by a stack of 3000 pounds of fish he has caught. He plans to travel 1000 miles across the Arctic to bring as many fish as possible to his family. He can pull a sled that holds up to 1000 pounds of fish, but he must eat 1 pound of fish at every mile to keep his energy up.

What is the maximum amount of fish (in pounds) the polar bear can transport across the Arctic? How does he do it?

[round canopy]

ive been really conffused with this question

and dont know how to solve it

my teacher said that there was a solution to it, and it isnt just zero

if someone could help me that would be amazing

<@&286206848099549185>

[glossy wharf]

What if you bring all the fish halfway

Then you fill the sled

Wait I'll think about it a bit

It depends on whether going back also uses up fish

Because of that is so I think it would be 0

[neon iron]

A stack of 3k lbs (1.5 tons)

[glossy wharf]

I'm not sure but i got 500

[neon iron]

sled can carry half a ton

1k mi to get to location

1 lb fish per 1 mi

just carry 2k on his back

idk

🤷

bc that sled can only hold enough fish for the bear to consume on 1 journey (of 1k mi)

[glossy wharf]

If you bring everything halvway to the Destination every time

[neon iron]

Polar bears can weigh up to 1.5k lbs or so

so

he can handle carryin 2k fish on his back 👍

[round canopy]

wtf bro

[neon iron]

(this is not a real, thought out answer)

[round canopy]

💀

i think ima tell my teacher i gave up

[neon iron]

Just say you ate the other ton of fish XD

[round canopy]

the thing is ur gonna use all the fish

[glossy wharf]

Ok wait

[round canopy]

cuz if u go halfway

its 500

[neon iron]

thats 500 fish

[round canopy]

and back ur gonna need 500 too

so u use all the fish

[neon iron]

which totals to 1k

[glossy wharf]

Not if you use the fish only when you're pulling the sleigh with fish

[neon iron]

well

i have one question

whats the friction constant of the ice?

and how many downward sloping inclines are present on the journey?

[round canopy]

lmao

[neon iron]

you must try to find any loophole possible

so

just say that there was a very gradual slope, but just steep enough for gravity to move the sled and bear, allowing for the bear to return with 999 fish (he ate 1 to not starve) 👍

sicne its ice, we can assume friction is negligible

and you can make this into a physics problem

and calculate the slope of the ice there

and then, using google and stuff, confirm if there is a stretch of 1k mi where such a slope occurs 🤷

just a thought 👍

[glossy wharf]

If you do this until the total goes under the total the sleigh can bring

It's 500

[neon iron]

the sled can only carry 1k fish

no more

[glossy wharf]

Yeah I just cut the passages a bit

[neon iron]

anyways, you got this vita 👍

[glossy wharf]

3000 you bring 3 times to half and they're 1500

1500 you bring twice to ¾ and that's 750

Then you bring the last 750 to the end and that's 500 total

[glossy wharf]

I don't think there's any other possible answer

It should be this

Unless

Wait a minute

[glossy wharf]

You don't have to leave all the food there once you get to a point

Ok it's 100% at least 250

Even considering you consume fish going back

[glossy wharf]

I got to 250 and you just leave 1250 in the back

Because they're not worth taking

[glossy wharf]

Doesn't say anywhere you have to bring it all

Ok so
You go ¼ of the way bringing 1000 , you arrive with 750, leave 500 and go back with the remaining 250
Do it 1 more time

Then you are ¼ of the way with 1250

You fill it all up And go to the end

You get there with 250

Should be the solution

[topaz sinew]

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[timber schooner]

can show me where im goingwrrong

with this deriving

[deep crow]

U can’t combine those into 1 fraction

[timber schooner]

mb i just multiplied everything by 2

wait

which bit

this is negative power

so i made it a fraction

[deep crow]

Yes but

2x(x^3+1)^1/2 cannot be combined in that fraction

No common denominator

[timber schooner]

this is the sol btw

@deep crow can u help me go through iot

[sweet shard]

Use product rule

,tex .diff rules

[thorny flame]

pizzanator

[timber schooner]

yeah im trying

but im doin smthn wrong

[sweet shard]

And then use exponent rules to simplify

,tex .exp rules

[thorny flame]

pizzanator

[timber schooner]

$\frac{3}{2}x^2(x^3+1)^{-\frac{1}{2}}(x^2+1) +2x(x^3+1)^{\frac{1}{2}}$

[thorny flame]

pixel

[timber schooner]

@sweet shard am i on the right track

[sweet shard]

Just keep simplifying

[timber schooner]

$\frac{3x^4+3x^2+2x(x^3+1)^{\frac{1}{2}}}{2\sqrt{x^3+1}}$

[thorny flame]

pixel

[sweet shard]

No

You tried to put everything over a common but didn't do it correctly for the last term

a/b + c/d = (ad+bc)/(ad)

[timber schooner]

$\frac{3x^4+3x^2}{2\sqrt{x^3+1}} + 2x(x^3+1)^{\frac{1}{2}}$

[thorny flame]

pixel

[timber schooner]

i am so

confused

😭

save me man

just walk me through it

please

anyone

[sweet shard]

Use this now

Or even simpler
a/b + c = (a+bc)/b

Identify a, b, and c first

[sweet shard]

Then use this table

[timber schooner]

$a = 3x^4+3x^2 \ b = 2\sqrt{x^3+1} \ c = 2x(x^3+1)^{\frac{1}{2}}$

[thorny flame]

pixel

[timber schooner]

$\frac{3x^4+3x^2 + 4x(x^3+1)}{2\sqrt{x^3+1}}$

[thorny flame]

pixel

[timber schooner]

holy shit its lookingh better

oh my ggod

i jjust didnt know how to put it in a proper fraction????

💀

man

im so dumb

i dont know why i struggled this much

oh my god it was just getting the same denominator

and i just realised u said the same thing

like 6 times

😭 thank u

.close

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[fervent junco]

Can some help me substitute f(-x) into 5c

[night imp]

You just need to replace x with -x right?

[fervent junco]

Yeah

[night imp]

What is the issue?

[fervent junco]

I’m just messing up with some of the steps

[night imp]

1/(-x) - (-x)

[night imp]

Show me

[fervent junco]

Is there a way to simplify it

Or is that just the answer

Cause I was changing the denominator to add the fraction

[night imp]

Well you can change it to a/b form if you like

a/b + c/d = (ad + bc)/bd```

[fervent junco]

I’m trying to find if the function is odd or even

And the way my teacher told me to was to substitute -x, then if it’s the complete opposite then it’s odd, if it’s not then it’s even

[night imp]

Complete opposite means what?
Suppose f(x) = a, so f(-x) should be 1/a?

[fervent junco]

As in f(x) the opposite is -f(-x)

[night imp]

Ohhh well you can test that easily

Find f(x) first for 5c

In terms of a/b

f(x) = 1/x - x , change this to a/b form , can you do it?

[fervent junco]

So 1/x- x squared/c

X

[night imp]

Why /c?

[fervent junco]

I meant x

[night imp]

x²/x

Yeah

Then

[fervent junco]

Sub -x?

[night imp]

This is still not a/b

You have
1/x - x²/x

[fervent junco]

-x+x cubed/xswuared

I’m lowkey lost

[night imp]

(1-x²)/x

[fervent junco]

How did you get that

[night imp]

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

Do you know this?

[fervent junco]

Nah I never learned that before

Wait no

I did

I see what you mean

[night imp]

Yes

[fervent junco]

So now I substitute negative x

[night imp]

Yes and reach something like this

[fervent junco]

Alright, thank you

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[ripe hollow]

hi i was looking for help on #23.i was able to find the bounds but im confused on how to set up the actual integral cause the way that i thought it was done was wrong

[ripe hollow]

volume by rotation

[limber wigeon]

For which value of k does the polynomial function p(x)=8x3−12x2−2x+k have two opposite zero values?

[trim ermine]

! occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[neon iron]

bro you already have like 5-6 channels

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[ripe hollow]

<@&286206848099549185>

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[topaz sinew]

@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

[vestal sigil]

i ask you, what do we conclude when the determinant of a matrix is non-zero?

yes, we conclude that the matrix is invertible

but we also conclude that the columns of that matrix are linearly independent

right?

yes

sooo, if you want to check whether or not the vectors of the Basis B' are LI or LD

construct a matrix using them

and calculate its determinant

its a quicker method

better than doing a1 v1 + a2 v2 + a3 v3 = 0

yes

if the matrix constructed by the basis vectors is square, only

some will be rectangular ofc

[gleaming reef]

well otherwise our kernel would not be trivial

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[vestal sigil]

only one, not at least one

$A , \vec{X]=\vec{b}$

[thorny flame]

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

[topaz sinew]

✅

[gleaming reef]

$A , \vec{X}=\vec{b}$

[thorny flame]

nosqldb

[gleaming reef]

note that if we have multiple solutions

Ax = b and Ax' = b

then A(x-x') = b

where x \neq x'

then x-x' \in ker(A)

basically whatever that maps to 0

the 0 vector is in the kernel

so like

if we have more than that

we are in trouble for invertibility purposes

of course

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[hot flame]

hello

[hot flame]

how do i do 82

i have x+2 as denominator

and x-2 as part of numerator

how do i make the slant asymptote happen

[drowsy coyote]

Do you remember how to find a slant asymptote?

[hot flame]

yeah

divide

and the quotien is x+1

[drowsy coyote]

Right

So just do the opposite

We really just want x+1+1/some polynomial

[hot flame]

x+1+1?

[drowsy coyote]

$$x+1+\frac{1}{p(x)}$$

[thorny flame]

Cycadellic

[drowsy coyote]

Where p is a polynomial term

[hot flame]

i think ur onto something

[drowsy coyote]

Then we use cross multiplication to write it as a fraction

[hot flame]

i have been so focused to try to get x+1 exactly

with no remainder

which quite frankly i realize now isnt possible 😔

[drowsy coyote]

If we did that, wed just have a line

Yeah

Remember that we can have polynomials of degree 0, its not exactly 0 but a number

Polynomial division is a little different than normal division, but this is a huge sidebar

we can get the other asymptotes from the p(x)

[drowsy coyote]

Dont let me confuse you this is irrelevant

[hot flame]

wait would x^2-x-6 world

i worked polynomial division backwards kinda and realized i need x^2-x in the front

[drowsy coyote]

We dont want this to be more complicated than it needs

We just want to make an error when we divide by p at x=2

So anything which satisfies p(2)=0

But make it suuper simple

Like x-2

[hot flame]

we already have x-2

as [art of the denominator

[hot flame]

also im pretty sure p(x) should be on top

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[drowsy coyote]

No it must be in the denominator or we cant get the pole

We need the division by 0 for it to make an asymptote like you want

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[lime delta]

How to simplify
3x²y² -8xy + 7y²+5xy²-11y²+5yx

[lime delta]

How do I know what to group together

Thoughts

Grouping these

[topaz sinew]

@lime delta Has your question been resolved?

[uneven ginkgo]

cant u take out y first

[lime delta]

Wym

<@&286206848099549185> please 😔

Idk how to find out what are different like terms

[twin heath]

so

okay

is this a circle equation?

[lime delta]

what

[twin heath]

like

[lime delta]

Idk I just gotta simplify so I don't have to solve for anything

[twin heath]

yeah grouping those is fine

go ahead and do that

[lime delta]

Is there a better way

[twin heath]

add like terms first

and then we'll move on

[lime delta]

Yes, is there any like terms I missed

[twin heath]

unless i have suddenly morphed into a bat, im pretty sure that's all the terms 😭

[lime delta]

Okay so I'm answer is 3x²y² - 3xy -4y² +5xy²

Am I cooking with gas

Wait on skibidi 😭

Okay I fixed

[twin heath]

nice

so

i'm not sure if this next part is what they want you to do

but

do you see how all of the terms have a y?

[lime delta]

So xy² is a different variable than xy and x²y² right?

[twin heath]

yeah

[lime delta]

I think that's all I had to do but I'm intrigued

[twin heath]

because it kind of looks like you can simplify it more beyond that

only kind of tho

[lime delta]

What part

[twin heath]

all of the terms have a y

[lime delta]

Yuh

Idk how I would simply that though

[twin heath]

update: it is not simplifiable i lied 😔

[lime delta]

W

Alright thank you my squire 😼

And good night am so sleepy 😴

.close 😼 mic drop

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[timber schooner]

First batch starts brightness at 15 units and loses 5% yearly

Second batch starts at 20 and loses 6% yearly

Guy buys carpet from batch 1, 1 year passes and he buys new one from 2nd batch

Find how many years from there until brightness of both carpets become the same

[timber schooner]

I had this question in the exam today

And I was wondering if my answer is correct

I got 32

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[eager hatch]

can someone check this pls

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[eager hatch]

<@&286206848099549185>

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[raw sphinx]

hello!

[raw sphinx]

can someone help me with this

f(x)=∣x−3∣+4

[fallow igloo]

What’s the original question?

[royal sky]

are you trying to find x

or graph it

or both

[raw sphinx]

im trying to look for

type, domain, range, vertex (if any), graph (descripition only)

<@&286206848099549185>

okay nvm i got it but i need help w the other one

f\ \left(x\right)\ =\sqrt{3x-5}

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[fresh adder]

i understand 5i but how did 10tsquarej become 20tj

[sudden temple]

they differentiated the 10t^2

[fresh adder]

they havent taught that yet in maths

[sudden temple]

But clearly they are differentiating the expression of vec(x)

Maybe consult your teacher about it or learn about differential calculus

[sudden temple]

just the basic knowledge of it might be enough, till they don't teach you in your curriculum

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[deft valve]

I don't know how i screwed this up
Teacher's answers are 3 and 2

[deft valve]

factoring via completing the square

[restive inlet]

you didn't root the 0.25

[deft valve]

ugh
that was really stupid, sorry for wasting your time

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[sweet shard]

which part is confusing?

the $\pm$ ?

[thorny flame]

pizzanator

[loud oasis]

whether or not you should use a calculator to get a decimal approximation for your answer is a matter of teacher's preference

[deft valve]

they want it, yeah.

Let me elaborate, sorry.
i can simplify sqrt 4425 to 5 * sqrt 177

Then what do i do to get the answers x = 8.9 & -7.7? (These came from the teacher)
5 sqrt 177 equals approx. 66.5

[prisma mesa]

this looks like xy prob

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[prisma mesa]

What was the original question?

[deft valve]

factor 4x^2 - 5x - 275 = 0

Answers should be 8.9 and -7.7 per the teacher.
I've currently simplified it to (5 + sqrt 4425)/8

[prisma mesa]

I see

can you use calculator?

or not?

[deft valve]

yes

but i've got to show how i got there, which i'm not sure how he got the two decimal answers

i don't know if i miscalculated or not

[prisma mesa]

What do you mean by this?

You can get the answer by plugging in (5 + sqrt(4425)) / 8 and (5 - sqrt(4425)) / 8 in your calculator

which as you said is allowed

[sly pasture]

,w solve 4x^2 - 5x - 275 = 0

[thorny flame]

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[sly pasture]

,w calc (-5/8)*(sqrt(177) - 1)

[thorny flame]

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[sly pasture]

,w calc (-5/8)*(sqrt(177) + 1)

[thorny flame]

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[sly pasture]

looks like those are approximations

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[still loom]

what is the proof that: Pr(A & B) = Pr(A & B | B)?

Or what is the proof that Pr(A & B) ≤ Pr(A & B | B)?

[topaz sinew]

@still loom Has your question been resolved?

[prisma mesa]

P(A & B | B) ||= P(A | B)|| ||= P(A & B) / P(B)||

[neon iron]

☝️ Case P(B)≠0. Or ||P(A & B | B)=0|| otherwise

[prisma mesa]

hmm, isnt it undefined if P(B) = 0?

depends on convention probably

[neon iron]

yeah, I guess so

[still loom]

i dont follow the whole proof there

where do we go from P(A & B | B) = P(A & B) / P(B)?

[prisma mesa]

Then you only need to prove that P(A & B) <= P(A & B) / P(B)

it's not whole proof btw, i intentionally left out the details

[prisma mesa]

do you understand that it's sufficient to prove that?

[still loom]

yeah, P(A & B) <= P(A & B) / P(B) = P(A & B | B)

oh i see

P(A & B) <= P(A & B) / P(B) since 0 <= Pr(B) <= 1

[prisma mesa]

yes

the idea behind this approach was remembering that definition of P(A | B) includes P(A & B), and then chasing the definitions until I eventually got to that inequality where some stuff cancels nicely

so the motivation was getting P(A & B) on both sides

[still loom]

okay yeah i didnt think about it like that

then when you have P(A & B) on both sides, it should be straightforward from there

[prisma mesa]

indeed

[still loom]

thanks for the help!

[prisma mesa]

np

[still loom]

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[vestal hatch]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[vestal hatch]

Number four

my work: basically i just identifide the points (0,2) and (1,3.6) and subbed the x=0, f(0)-2 in equation , simplified and solved, to get b=1

for a I got a=2.6

then i computed that with f(3) = 18.58

[outer delta]

What is your question though?

[golden blade]

looks good your work

[vestal hatch]

status = 4

[vestal hatch]

👍 thanks

.close

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[tawdry ice]

Kinda stuck

[worthy storm]

well first you should probably make the sum go from k=0 to k=n-1, to match your binomial coefficient

then you might want to add a clever factor of the form 1^(some power) where (some power) is chosen so you can apply the binomial theorem

[tawdry ice]

like this

[worthy storm]

no, leave the binomial coefficient alone, but do a change of variable

say j = k-1

[tawdry ice]

Like this

or wait

it should be n-1

right

like this

[worthy storm]

where did the j factor in the middle come from?

[tawdry ice]

From the original function

k(n Chose k)

[worthy storm]

start with this form:

and do a change of variable j = k-1 here

[tawdry ice]

Like that

[worthy storm]

couple of issues

[tawdry ice]

my brain is not braining rn

[worthy storm]

j = k-1 implies k = j+1 so you should have an extra factor of (-1)

and (n-1)! in the numerator shouldn't change, it doesn't depend on k

[tawdry ice]

but it should be from j=0 to n-1

[worthy storm]

yes that part is right

[tawdry ice]

does that look better

forgot 1 bracket

but yh

[worthy storm]

wait no, the two factorials in the denominator were correct before but now they're not

you should have the following:
$$n \sum_{j=0}^{n-1} (-1)^{j+1} \frac{(n-1)!}{(n - 1 - j)! j!}$$

[thorny flame]

Bungo

[worthy storm]

yep

[tawdry ice]

alright

[worthy storm]

now you can recognize the fraction as a new binomial coefficient

[tawdry ice]

n-1 chose j

[worthy storm]

yep

and you can pull out a factor of (-1)

so you'll have (-1)^j inside the sum

[worthy storm]

yep

now you need a clever trick

you know the binomial theorem tells you that:
$$(x+y)^{n-1} = \sum_{j=0}^{n-1} x^j y^{n-1-j} {n-1 \choose j}$$

[thorny flame]

Bungo

[worthy storm]

comparing with the sum you have, you can take x = -1

but you seem to be missing y

unless you cleverly take y=1

[tawdry ice]

so like this

[worthy storm]

yes!

and -1 + 1 is also known as ...

[tawdry ice]

alright

3 billion

haha

[worthy storm]

haha indeed

[tawdry ice]

just messing with u

0

[worthy storm]

yep so that's why the sum is 0

[tawdry ice]

and the restriction i guess is just not to end up with the 0 to the 0 case

Thank you a lot @worthy storm very helpful!

[worthy storm]

yea, and in fact if you directly evaluate your original sum with n=1, you get (-1)^1 (1) (1 choose 1), which is -1, not 0

[tawdry ice]

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@elder hare Has your question been resolved?

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[obtuse cliff]

hello

[obtuse cliff]

i am asking for help

Question #7: No response

Let ( C={1,2,3} ). Which of the following sentences is false?

From the 5 options, select all that apply

If ( A={1,3} ) then ( A \times A \subseteq C \times C )

( (3,1) \in C \times C )

( C \times C ) has 9 elements

( \emptyset \subseteq C \times C )

( C \times C^{c}=\emptyset ) (where the complement is taken in ( \mathbb{N} ) ).

[thorny flame]

Arlix

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@obtuse cliff Has your question been resolved?

[opal vault]

Have you found what the set C x C is?

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[misty sand]

@unborn mason

[misty sand]

i forgot to ask

why is there a constant K

[unborn mason]

When you write any polynomial as a product of it's roots there is a constant along

It's all the same curves just scaled

You'd need a third point to find the exact value of k

[obtuse cliff]

.repon

/reopen

.reopen

[misty sand]

got it ty

[misty sand]

you can reopen now

.close

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[obtuse cliff]

.reopen

.reopen

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[neon iron]

What does the [[ symbol thing mean

[neon iron]

I already claimed this

[acoustic pecan]

is there more context

[neon iron]

Np bro

I can show the whole question:

65c

[acoustic pecan]

it could be the floor function?

[neon iron]

Not familiar

Also I think it looks a lil diff

[acoustic pecan]

it returns the greatest integer less than or equal to the input

im not familiar with the specific notation used here but my little search made it come up a lot

[neon iron]

Hm

I’ll try it ig

How annoying though

Lol

[acoustic pecan]

has it not been used anywhere previously in your course for something?

[neon iron]

Never

Lit like first week ap calc never seen it in my life lol

Thx for trying tho

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[tawdry lake]

why can we not integrate xtan(x)?

[ivory sorrel]

,w integrate xtan(x)

[thorny flame]

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[cursive patrol]

not in terms of elementary functions

[ivory sorrel]

oops

[tawdry lake]

or at least like using integration of partrs

parts

[ivory sorrel]

yeah, you can't.

my bad

[cursive patrol]

you can use integration by parts to integrate it

the issue is that you eventually have to integrate ln(cosx) at some point and that does not have an elementary antiderivative

here is a derivation of the indefinite integral of x * tanx

as you can see, it is possible to integrate... the result just cannot be expressed using elementary functions

[tawdry lake]

elementary function is that like in terms of trig, logs?

basic operation?

[cursive patrol]

here is typically how they are defined

[tawdry lake]

what about this?

isnt this possible to integrate?

[cursive patrol]

same issue as before

using ibp, you have to integrate ln(cosx) and that cannot be done in terms of elementary functions

[tawdry lake]

what did i do wrong?

[cursive patrol]

the derivative of x^2 is not 2

[tawdry lake]

2x

ohh

i see what you mean

it goes back to almost the original problem

since id have to integrate 2xtan(x)

[topaz sinew]

@tawdry lake Has your question been resolved?

[tawdry lake]

im integrating this again how should i apporach thius

<@&286206848099549185>

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@tawdry lake Has your question been resolved?

[untold rapids]

You're on the right track

Just do integral by parts again (where you derive cos(mx) and integrate sin(nx))

It will get tricky after that

Since you would return to the original problem

Now then assume that the original problem equals y

then combine y in the right side of the equation with y that is on the left side (idk if I'm saying this correctly) but yeah

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[graceful leaf]

how does he know its north of west though

the a and b are positive

like i know its north for sure

[loud oasis]

that's how he draws it on the diagram

[worn aurora]

the drawing

[graceful leaf]

ohh

lol

ok thnxs

.close

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[cold wasp]

Guys for some reason when I evaluate at the bottom I get the wrong answer idk where I go wrong

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[quartz smelt]

how do I do this Im in physics 1

[quartz smelt]

I tried writing the vectors and calculating the angle then finding the velocity of him going at an angle but it was an incorrect answer

chat gpt and the quizlet Ai thing both cannot get this question right and I am completely stumped on what I am actually supposed to do

the answer is 1.16

hours

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[dapper sparrow]

Im having a hard time wrapping my head aroung proofs, I just finished a question and I was wondering if anyone can help me understand what/if I did something wrong? Im lost about what parts I have to state and what I am allowed to use within the proof without being too close to the actual assosiative property:

[left juniper]

C is complex number, right?

Yeah, it's complex number

Yeah, the problem is that we don't know yet if (a + bi) + c = a + (bi + c)

It's better to use the pair notation to prove the associative property first. Only after we establish complex number as a vector space you can use the notation a + bi

[keen matrix]

essentially you know that real numbers are communtative that, is (4+3)+5 = 4 + (3 + 5)

but how do we know that holds for complex numbers?

well we compare both sides of the equation using what we know about real numbers (that they are commutative)

to prove that complex numbers are also in fact commutative under addition

[topaz sinew]

@dapper sparrow Has your question been resolved?

[dapper sparrow]

@left juniper yes

Wait so that’s what I tried to do @keen matrix

But I’m lost in if I didn’t right or not

[keen matrix]

looks sound to me :)

[dapper sparrow]

Thank you :) lmao

Proofs are weird

But I think I’m getting it

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[keen matrix]

[keen matrix]

keep at it, you'll get it eventually

I got faith

[left juniper]

I disagree. It's better to rewrite your proofs in pair notation.

[dapper sparrow]

What does that mean @left juniper

[left juniper]

Maybe say .reopen first?

[keen matrix]

.reopen

[topaz sinew]

✅

[left juniper]

Basically, like
$(\alpha + \beta) + \gamma = ((a, b) + (c, d))+(x,y)$

[thorny flame]

Xwtek

[left juniper]

Don't use $\alpha = a + bi$ first.

[thorny flame]

Xwtek

[left juniper]

That's later when you found out that the complex number is a vector space.

[keen matrix]

well yeah but for such an elementary introduction to complex number which I assume this is, assuming all the complex number in their a+bi form should suffice here?

[left juniper]

I still think it's circular.

[keen matrix]

well yeah you're using the commutative property of addition in reals to prove it in the complex numbers, is that your gripe?

[left juniper]

in a step $(a+bi)+(c+di)=(a+bi)+c+di$, both $(a+bi)$ and $di$ are not a real number

[thorny flame]

Xwtek

[left juniper]

In order to make this step even legit, you need to prove associativity, which you're doing right now.

Unless he wrote something like $(a+c) + (b+d)i$, which is kinda correct, but feels like it should have been done in a pair notation anyway.

[keen matrix]

wait yes there are a few errors now that I look at his logic

[thorny flame]

Xwtek

[dapper sparrow]

.reopen

oh

ok

Wait im back let me read through it if you guys are still here @keen matrix @left juniper

[dapper sparrow]

its absed off of Axlers LADR

[keen matrix]

wait shit there isn't 😭

[dapper sparrow]

First ever proof based class im taking

[keen matrix]

damn that's my bad bro, it's almost 12:30 AM over here I really should be getting to bed

you can use the pair notation if you want

[dapper sparrow]

wait is what I did ok?

[keen matrix]

but I think this is introductory

if it is, then it should be fine

[dapper sparrow]

Yeah it is

Ok perfect thank you :D

[keen matrix]

I thought (b+d)i was wrong 💀

I forgot both bi and di were both imaginary

[left juniper]

I made a typo too.

[dapper sparrow]

Yeah, making b and d real

so R rules apply

Ok wait im gonna try a few more then

thank you @keen matrix @left juniper

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[dense plover]

Let $R$ be a ring, $M$ and $N$ be $R$-modules. Let $i, j : M \rightarrow N$ be isomorphisms. Then, suppose $j \circ i^{-1}(n) = rn$ for some invertible $r \in R$. Does $j^{-1} \circ i(m) = rm$?

[thorny flame]

DavidL1450

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[onyx tendon]

I have this big giant area problem that I genunley have no idea where to even start

[onyx tendon]

If i could just be steered in the right direction that would be amazing

I need area of property, house, driveway, and yard

[halcyon willow]

u know to calculate area of square and rectangle?

that should do

divide the areas into simpler to compute rectangles and add iff necessary or substract

[onyx tendon]

okay wait

give me an example of like one thing youd break down

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@onyx tendon Has your question been resolved?

[urban barn]

start with the area of the driveway?

[onyx tendon]

okay so stop me if im wrong

would you add the 64 and 8 subtract from 95

multiply by 45

?

[urban barn]

sure

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[crisp quiver]

Sry

[ebon carbon]

Can anyone help with the notations I don’t understand it 😭

[ebon carbon]

It’s ok

[ebon carbon]

Like I know that fog is f(g(x)) and the intervals

But what do I gotta find here

Why is the input of the function an interval

So like the output is the range?

Of the set after applying the function?

So do I represent it on a graph or sth?

I’m sorry I still don’t know what to write as the output 😭🙏🏿

Oh I get it now

Ty

What do I do when the set include value out of range 🫠

Oh ok thx so much 🙏🏿

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[pine lake]

can someone just check if this solution is correct?

[sudden temple]

the series you're given is different

from the series you are calculating the sum of

[pine lake]

yes pls explain 🙏

[sudden temple]

for first term = 8 and r=1/10
the series will be 8 + 0.8 + 0.08 +...

that's different

[pine lake]

wait shit im stupid

let me calc this again

:)

[night imp]

@pine lake find a pattern first before jumping to conclusions that it is a GP or AP

[pine lake]

it is a g.p

can't be a.p

[sudden temple]

not completely

[night imp]

How it is a GP? What's r ?

[pine lake]

8.8/8 = 8.88/8.8

you get r

thats the formula for r

[night imp]

What's 8.88/8.8

,calc 8.88/8.8

[thorny flame]

Result:

1.0090909090909

[pine lake]

11/10

[night imp]

,calc 8.8/8

[thorny flame]

Result:

1.1

[night imp]

111/110

[pine lake]

i thought you were calc. 8.8/8 but oh well

nevermind, thanks again

i get it now :)

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[lusty ridge]

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[lusty ridge]

for (a) would f(x)=-1 and g(x)=1 or g(x)=2?

[neon iron]

no

[lusty ridge]

how come

[neon iron]

lim x -> 2 of f(X) is not -1

[lusty ridge]

would it be 0?

[torpid sparrow]

-1 is correct for f(x)

But the exercise was not asking for that

Right?

[lusty ridge]

yeah

i shouldve worded that better

i think it would be lim x->2 =(-1)+2

then lim x->2=1 ?

im confused reading the g(x) graph

[topaz sinew]

@lusty ridge Has your question been resolved?

[golden blade]

what is your doubt

from the right it's undefined at x = 0

but from the left it is defined

[lusty ridge]

so from x->0 would it be 2?

[golden blade]

no the limit doesn't exist

[lusty ridge]

oh

[golden blade]

from the left it approaches 3

from the right 1

[lusty ridge]

oh wait im thinking with question (a)

okay that makes sense

so then for (b)

the limit wouldnt exist because