help-36

[foggy minnow]

Looks correct

Congrats

[vocal grove]

but why dont we multiply on both

why just one

[foggy minnow]

Well it’s simpler to multiply one equation

[vocal grove]

how is it still correct?

[foggy minnow]

The equation is still correct as long as you do the same operation to both sides

[vocal grove]

wdym

[foggy minnow]

when you multiplied the equation by 4, you did it to the left and right side because that would keep the equation equal

[vocal grove]

yes i know that

But i only multiplied on the

I

not the II

why is it still correct?

[foggy minnow]

hmmm why should it not be correct

[vocal grove]

wait

i meant the other way around

[foggy minnow]

The equations are not equal

[vocal grove]

huh

i dont think you understood my question

[foggy minnow]

I know why you’re asking it’s hard to explain maybe someone with better terminology can explain it

But can you explain why we would need to do the same operation to the first equation

It is still “correct” because the second equation is equal to itself after it was multiplied

[terse dagger]

if you have a system of 2 equations

[foggy minnow]

If you graphed the equation before and after you multiplied they are still the same line

[vocal grove]

hmm i see

[terse dagger]

you can subtract one from the other and the resulting equation has equivalent information to the equations you got it from

and since you can subtract one from the other an arbitrary number of times

[foggy minnow]

So the intersect of the X and y will stay the same since the lines did not change

[terse dagger]

even a non integer number of times

you can just take equation I and subtract 4 times equation II from it

and the information about the system won't be lost

[vocal grove]

so the proportion is always the same?

not sure if thats the right english word

[foggy minnow]

the proportion of the equation is the same as long as you multiply to both sides

Not sure what you meant by proportion but maybe that helps

[vocal grove]

okay thanks

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[vivid walrus]

Hello I have a question about transformation rules for a value

[vivid walrus]

does it works vertical stretch by a factor ½ or stays a factor by 2 if we have a 2 in a?

or does it only works for k value

$$y=-2(x-3)^4+5$$

[soft zealot]

アキラ (>_<)

[vivid walrus]

here's the actual function but im just wondering about a

I keep forgetting if it's factor by -½ or just factor by -2

<@&286206848099549185>

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[vivid walrus]

<@&286206848099549185>

:(

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[gaunt meteor]

dumb question but why are there 3 solutions instead of 2, and why is it not just pi/6 and pi/2

my teacher said its bc it goes around the unit circle more times but i dont get what that means?

so if its any trig function where x has a coefficient > 1 does the coefficient determine the amount of solutions

[bold wadi]

yes, if you have a coefficient of y then the new period is 2π/y

in this case the coefficient is 3, so the period is 2π/3

so π/6 is a solution, but so is (π/6)+(2π/3) = 5π/6

[gaunt meteor]

ohhh

so if i got a question like this the last box would be all the solutions? bc the period is pi/3 instead of pi

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[wise light]

can someone help me understand how to factor this

x^4 + x^3 + 2x^2 + 4x -8

[rain compass]

i would first try to plug in some low numbers like
-2, -1, 0, 1, 2
and see if any of those give you 0

[wise light]

1 does

[rain compass]

ok good, so if 1 does

what does that mean?

[wise light]

its a zero

is that why it starts out

(x-1) because 1 equal to 0

[rain compass]

sure

since 1 is a zero, the polynomial you have has a factor of (x-1)

so you would divide the polynomial by (x-1)

[wise light]

okay got that

that just blew my mind

the plugging in numbers part

cause i was doing synthetic division and its the exact same thing 1 is a zero

so okay whats next

do i just group the rest

[rain compass]

well after you divide it out you should be left with a cubic equation

once again

plug in some numbers

to see if any are roots

[wise light]

the factoring by grouping doesnt work

[rain compass]

what do you mean

why are you factoring by grouping

[wise light]

i was just trying stuff out

the x^3 +2x^2 +4x - 8

[rain compass]

ok ill suggest you try plugging in some values for x again

[wise light]

okay

[tawny halo]

There are a lot of methods to do it

Like division

Or horner's method

Or whatever

[vital surge]

it is + 8 at end

[wise light]

ik i tried it

[wise light]

so generally use -2 , -1, 0 ,1, 2?

[tawny halo]

yes

[wise light]

why is that?

[tawny halo]

Because teachers maybe give us equations that have simple roots to simple our life

[vital surge]

not "generally"

1 and -1 are always good idea to try but there is specific reason why 2 and -2 are good idea in this case

[tawny halo]

In school i mean

[vital surge]

still no

[tawny halo]

Ok

If you can't find root, try useing factorization

[wise light]

then how can u tell how to pick which numbers are good without going through it all

[vital surge]

in general what u try is number of this form

[wise light]

like how did u determing that 2 and -2 is good

[tawny halo]

Ah yeah

Forgor this thing

[wise light]

yeah im doing this

[vital surge]

so

because ur last number is 8

it would be good idea to try -8, -4, -2, -1, 1, 2, 4, 8

(but you start with -2, -1, 1, 2 since these numbers are easier to try)

[wise light]

im at the part where u factor it cause i just did 1 and i got a zero then i got

x^4+x^3 + 2x^2 +4x - 8

[wise light]

so in my notes the -2 did get a 0

but after it it goes to 1

then it also goes to zero

[vital surge]

is correct

[wise light]

how come -1 wasnt tried

[vital surge]

i mean

u can try it if u want

(u will see -1 is not a solution)

[wise light]

im just unsure how it went from -2 to 1 (maybe my teacher did try to do -1 but i did not write it down

[wise light]

see

[vital surge]

wdym

[wise light]

is this why or is there some skill that helps you determine what numbers are good

[vital surge]

not rly

i assume they either knew the answer beforehand or just mentally checked

(mentally checking becomes a quick process after some practice or time doing math i guess)

[wise light]

ic

so i just keep doing the synthetic division

[vital surge]

ye ur teacher did it twice

[wise light]

until it becomes standard form

[wise light]

the ax^2 + bx + c is that right>?

[vital surge]

idk what u r referring to

[wise light]

this last part of synth.

[vital surge]

ur teacher just divided the original polynomial by (x + 2), then divided that result by (x - 1)

if that's what ur talking about then ig answer is "yes"

[wise light]

the 1 0 4

if i write it out it would become x^2 +x +4

[vital surge]

no

0x = 0

[wise light]

ohhhhh right

forgot about that\

[vital surge]

👍

[wise light]

why do u not write it out?

[vital surge]

write what out

[wise light]

the 0x\

[vital surge]

u can think of it as

"why would you?"

[wise light]

ic, so its 0 so dont bother

[vital surge]

after u get x^2 + 0x + 4. u r done because this expression cannot be factored

therefore there is no further need to apply synthetic division and you can remove the 0

if u needed to apply division again then u can keep the 0

[wise light]

you stop doing synthetic when u can factor the expression?

like how x^2 + 4 is finally factorable in this last part

[vital surge]

u stop when the end result is no longer factorable

x^2 + 4 is the last thing u get after using division, and it cannot be factored

[wise light]

ok my b only x^2 - 4 is factorable

[vital surge]

ye

u can think of it as

x^2 + 4 is always above the x-axis

because u r taking x^2 and moving it up by 4

so it has no real roots and can't be factored using real numbers

[wise light]

u mean no zeroes cause it has to be on x axis?

does it have to be on the x axis to have zeroes?

[wise light]

is this a valid statement ^

[vital surge]

for purposes of an introductory alg class, a polynomial has a zeroes where it touches the x-axis

[wise light]

safe to say for pre calc too>

?

[vital surge]

for pre-calc u'd probably know what complex numbers r by then

so no

[wise light]

i dont think we have that part yet in the semester

[wise light]

but good to know this either way

@tawny halo @vital surge ok thank you for your help, i understand now. have a good day, im gonna study graphing rational functions now and ill come ask again here hope u guys can help again.

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[violet zodiac]

I was thinking it would be a set of sequences of length z? from 1 to n?

or is there a easier more mathematical way of putting it

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[chrome hemlock]

is there an easier way to do this

[chrome hemlock]

like faster

[barren hound]

easier way would be to just kinda list them out

[chrome hemlock]

cuz in test i only have like 15 mins to complete the non calc

[chrome hemlock]

o

so like

trial and error?

or like

the triangle

[barren hound]

essentially? triangle is quick

you can also skip writing out some steps once you've practiced this sort of thing

like a lot of the factorial stuff

[chrome hemlock]

ohh ok thanks

i dont understand

what this means

oh

do the brackets mean

cobination

combination

nCk

[barren hound]

yes

[chrome hemlock]

okok thx

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[chrome hemlock]

.reopen

WAIT

[final saddle]

✅

[chrome hemlock]

ONE MORE QUESTION

how would i do this?

is this one the pigeon hole principle

[deft ravine]

Yes

[chrome hemlock]

oh naurrrrrrrrrrrrrrrrrrrrrrrr

do u think

its worth 5 marks

because i havent started the chapter yet

and my test is tomorrow

on last years test it had one question on the pigeon hole principle

and it was only 2 marks

this is too complicated

[deft ravine]

Say we have a two digits number. What is the maximum sum of the digits?

[chrome hemlock]

hm

9 + 9?

18

[deft ravine]

Yes

So each number has a digit sum between 1 and 18

And you have 19 numbers

[chrome hemlock]

if the max sum is 18 and u have 19 numbers

that means tehre is one that is repeated?

[deft ravine]

Yes 🙂

[chrome hemlock]

wow

is that all the pigeon hole principle is

[deft ravine]

Yes, it all comes down to you "running out of possibilities". Like here we have 19 numbers, but 18 possible sums of digits

[chrome hemlock]

how come

this one is different

why did they remove one from each pigeon hole

[deft ravine]

Here they firstly want to determine the number of crayons you can have at maximum without the condition being wrong

So you can have 1 green less and 1 red less than asked for

[chrome hemlock]

oh

so there is a definiute chance of at least 6 red or 5 green

if we put some in

when theres 5 red or 4 green

because minimum being is like 1

crayon

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[prisma lantern]

How do I answer this?

[astral moss]

What did you try

Do you have working out

[prisma lantern]

I haven't tried anything

Idk where to start

[astral moss]

Do you know that the definite integral between a and b of a function gives you the area under the curve between a and b for that function?

[prisma lantern]

Oh I figured it out thank you

[serene wagon]

Area is always positive

[prisma lantern]

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[limpid quiver]

Hello! I have to find the first four nonzero terms in the Maclaurin Series of $f(x) = \frac{2x}{1+x^2}$

[soft zealot]

dabbingpotato

[limpid quiver]

It seems pretty impossible to brute force it and i'm pretty sure theres a neat little trick thats supposed to be used. Could anyone enlighten me?

[rain compass]

note that this is the derivative of ln(1+x^2)

[limpid quiver]

Okay so we just do the maclaurin for ln(x) and then substitute the terms with x for 1+x^2 and then take the derivative?

[rain compass]

yes

[limpid quiver]

okay thanks :D

[rain compass]

well

you should know the maclaurin series for ln(1+x)

and just sub x^2

[limpid quiver]

ohhh

[rain compass]

then differentiate, yes

[limpid quiver]

right

thts super helpful

thank you so much

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[tranquil pine]

This is what I did so far

I’m so lost when trying to find yp

Idek if I found the unprimed yp correctly

The Ax + B + Ex^x

Ee^x ****

If someone could help me I would appreciate it

Btw it’s method of undetermined coefficients

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[runic void]

sorry, im not a ready

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.reopen

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[steel kite]

wait so please do not give me the answer

or any sort of hint

so

the question asks me to find the parabola that shows the shape of this cable

so

to find the equation of a parabola

we can either have 2 points: vertex and any other point

OR

3 different coordinates

so uh

i think i can identify two points right now

one is y-intercept which is

(0, 75)

the other is (180, 75)

i cant seem to identify the third coordinate

am i on the right track anyway?

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[steel kite]

oh wait

oo

isnt it literally just

(5, 3)
(6, 9)
(5, 13)

so t = hours = x-axis

r = rate of rainfall in mm = y-axis

[oblique mica]

yup

[steel kite]

easy lol

now just 3 equations

simultaneous equations

woo!

@oblique mica is there a way to like identify which equation goes with which

in terms of simultaneous equation

my 3 equations are

$3=9a+3b+c \ 9=36a+6b+c \ 13=25a+5b+c$

[soft zealot]

pixel

[oblique mica]

what do you mean

[steel kite]

you see

how we have to use like

eliminitation method

and use those 3 equations

[oblique mica]

also your first point should be (3,5)

[steel kite]

to solve for 'a', 'b', 'c'

oh

💀

thank u for noticing that

let me fix

[oblique mica]

your points are just backward in general

[steel kite]

wym

[oblique mica]

t is the x-axis, r is the y-axis

so a point would be of the form (t,r)

[steel kite]

ohh right right

i even draw a diagram to do that

but silly me wrote the wrong coordinates

💀

[oblique mica]

this is really helpful because you get two different x values that equal the same y value

you can find the axis of symmetry that way

[steel kite]

wait

hold on

my coordinates should be

(5,3)
(9, 6)
(13, 5)

(t, r)

right

which two different x values have the same y value

they're all different

[oblique mica]

oh yeah

you cant do the axis of symmetry thing then

so if you have a set of equations and you want to do elimination, you want to eliminate a variable from two equations using one equation

that probably doesnt really make sense, so let me make an example

4 = 3a + 7b + 3c
6 = 2a + 5b + 3c
7 = a + 4b + 5c

you can eliminate the a in the top two equations using multiples of the bottom equation

you would get
-17 = -5b -12c
-8 = -3b -7c

then you would do the same thing you did as before

[final saddle]

@steel kite Has your question been resolved?

[steel kite]

oh yeah im so sorry i solved it

im now on similar questions

except i have coordinate

which is easy

thank you for your help

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[gleaming bluff]

can someone help me with upper and lower bounds theorem

[gleaming bluff]

basically, the statement says

if we divide p(x) by x-b with (b>0) using synethtic division and if the row that contains the quotient and remainder has no negative entry, then b is an upper bound

and if we divide p(x) by x-a with (a<0) using synt. divison and if the row that contains the quotient and reminder has entries that are alternately nonpositive or nonnegative then a is a lower bound for zeros of p

well in this example i got im left with

2 10 14 6 0

in a synthetic divison

this is quotient and remainder

so this has no negative things

oh okay wait

okay i got it

i was gonna say why is it upper bc it has negative and it also could be lower but i divided it by 3/2 soo its upper

anyway lol im closing the chat

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[main nexus]

ive substitued it in and like

it does equal 1

but i dont understand how to find the point that is missing

[west berry]

What did you sub

[main nexus]

well

by elimination

doesnt it mean i just do

x^2+y^2

and like

put the t values in

well anyways when i did it and simplify it all you end up w (t^2+1)^2/(t^2+1)^2

which is equal to one

but i dont get how to find the missing point

[final tangle]

consider the range of x

doing long division on (t^2-1)/(t^2+1) may make the missing value clearer

[main nexus]

im unfamiliar w how to find

the range of x

can u explain please

[final tangle]

doing long division on (t^2-1)/(t^2+1) may make the missing value clearer

[main nexus]

how do i

do that

[final tangle]

have you ever done polynomial long division before?

[main nexus]

i thought it just equals to 1 no?

[main nexus]

wiat

ohhh

wait no 1 was for the big one

ok lemme try

oh ok

so i got

the remainder as -2

and the quotient is 1

[final tangle]

yes, so
$$x = 1 - \frac{2}{t^2 + 1}$$

[soft zealot]

ℝαμΩℕωⅤ

[final tangle]

can you now see what value(s) x can't attain

[main nexus]

wait how did u do that?

[final tangle]

do what

[main nexus]

wait oh i get it

[final tangle]

that's what the result of long division gets you

[main nexus]

Ohh ok

ohhh i see

um

um i still cant see what x cant attain

sorry

[final tangle]

think for a few minutes

[main nexus]

wait

is it

x cannot equal to 1>

?

[final tangle]

yes, x can't be 1

[main nexus]

and then what does

y =

do i just like

use

x^2 + y^2 =1

and of x cannot be 1

then y cannot be 0?

hi

i have a bit of a question

why cant y not equal to 0?

<@&286206848099549185>

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[faint locust]

#1195800467880558672 message

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[formal gazelle]

hii, got a question about an induction case in n^k < 2^n where i've defined n = k^2 + 1

[formal gazelle]

i want to make induction for this term, but i need to factor out (k^2 + 1)^k in my right side, and idk how

this

if i can factor out that, i can remove the factors out, and just reapply induction on the factors that has left

this is my hypothesis

thats why i need in the right side (k^2 +1)^k as a factor in the right side

any help on the way to factor out this crap? 😦

oh btw it was the left side XDD

not the right

<@&286206848099549185>

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[abstract pelican]

I've made these two equations both equal y^2, now I need to solve them simultaneously, how would I go about doing this? It's getting late and I've been doing this all day, I don't have it left in me to figure this out myself lol

[ornate knot]

where is there a xx in first equation?

are they different or is that x^2

[abstract pelican]

its x^2

that is residual from earlier

[ornate knot]

also in second equation u could take square root on both sides and substitute the value of y into first

get value of x and then use that value of x to get y

[abstract pelican]

I don't have the value of y

ohhhhhh

silly me

they equal the same

right, that is why I just did all of that

[ornate knot]

there is a y in the first equation

that would lead u no where

[abstract pelican]

wdym?

[ornate knot]

$$-(x^2 + 15 - y(8.5)) = (\frac{(x+0.5)(x)(x-0.43)}{0.4}+5)$$

[soft zealot]

JustToPro

[ornate knot]

u still have both x and y in there

[abstract pelican]

oh, whoops

uhh, how do I get rid of that? I am so far gone

[ornate knot]

what i would do is like i did in the picture
$$(\frac{(x+0.5)(x)(x-0.43)}{0.4}+5)^2 = y^2$$
square root both sides
$$(\frac{(x+0.5)(x)(x-0.43)}{0.4}+5) = y$$

[soft zealot]

JustToPro

[abstract pelican]

yeah, that is what I had earlier, but that equation (cubic polynomial) does not equal the other equation

[ornate knot]

what u mean?

[abstract pelican]

I am trying to get both equations to equal each other, so that I can solve them simultaneously

[ornate knot]

ik

[abstract pelican]

the first equation equals y^2
but that second equation does not

[ornate knot]

thats 1 of the methods known as substitution method

[abstract pelican]

yes, but can

wait

[ornate knot]

where u substitute the value of 1 variable into the other equation

[abstract pelican]

ohhhh

I substitute that whole equation into every other y in the first one

ok, that is going to be extremely confusing but I can do it

I will report back in a few minutes

[ornate knot]

ok

[abstract pelican]

ah, this is awkward

your little substitution trick has just invalidated all the math work I've done today

eh, its fine

[ornate knot]

can u show ur question?

maybe there is a simpler way ( i doubt there is tho)

[abstract pelican]

well the equation I started with has 1 Y in it

[abstract pelican]

there isn't really a question, I am trying to find where two relationships intercept

it's for an assignment, but it doesn't explicitly tell you to do it

[ornate knot]

oh ok

[ornate knot]

which equation is that?

the one below?

[abstract pelican]

the bottom one

it has 1 Y in it to substitute for, rather than a few

[ornate knot]

here its different equations

do u have more than 2 equations ??

[abstract pelican]

that top equation is the circle

and so is the 2nd one in that most recent image

just I spent all of today trying to make it equal Y

[ornate knot]

im confused btw , the recent image is that ur original equation which u transformed into the old image (the one u posted first)?

[abstract pelican]

no, old image came first

ok, let me organize my thoughts

this was the circle

but I turned it into this to try to make it equal y

that is the exact same equation

this is a cubic polynomial

I am trying to solve for X

because X is where they intercept

[ornate knot]

ok

i get it , so u are trying to find where the cubic polynomial and cricle meet?

[abstract pelican]

yes

this might help visualize it

[ornate knot]

why is the circle like pac man?

[abstract pelican]

doesn't matter

it's two functions

the mouth is an inverse reciprocal

would I be able to just find the square root of both sides to get rid of the brackets?

[ornate knot]

nop

$$\sqrt{a^2 + b^2} \neq a + b$$

[soft zealot]

JustToPro

[abstract pelican]

so how would i remove the brackets on the left?

would I have to expand them?

because that is going to take a very long time

[ornate knot]

yeah i dont see another way

[abstract pelican]

ugh, ok

\left(\left(x+0.5\right)\left(x\right)\left(x-0.43\right)+0.75\right)^{2}

oh

[ornate knot]

where did the 0.4 go from the denominator ?

[abstract pelican]

can't even put it in here

[ornate knot]

use dollar signs

[abstract pelican]

I just got rid of it to make it easier

[ornate knot]

$\left(\left(x+0.5\right)\left(x\right)\left(x-0.43\right)+0.75\right)^{2}$

[soft zealot]

JustToPro

[abstract pelican]

I was hoping to put it here because I don't have anywhere to write this out while expanding it

[ornate knot]

well u got rid of it wrong way

[abstract pelican]

no I mean I literally got rid of it

[ornate knot]

oh like u skipped it?

[abstract pelican]

I made these equations to begin with

if I was smart I would have made it a quadratic

[ornate knot]

btw what u would get is a 6th polynomial equation which u would have to solve

[abstract pelican]

ah man

[ornate knot]

idk how to solve 1 of those

[abstract pelican]

you know what, I am just going to make it easier and get a lower mark

but actually a higher mark cause I can do it

thanks for the help anyway, I've learnt some stuff even if I can't use it

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[faint monolith]

part ii) how do i do

AB=27
BC=18
CD=12

[west berry]

You can use the vector equation of a line

Assume the parameter as lambda or smth

And you know the distance CD, so use the distance formula to find possible values of lambda

[faint monolith]

D can be lying between B and C or outside B and C right

[west berry]

Yup

Points B,C,D should be collinear, that's all

[faint monolith]

wait i draw

[west berry]

Ok nvm one will lie between B and C and one outside, yeah

Idk why I deleted that

[west berry]

Yeah

Oh section formula works too then

[faint monolith]

so i can take vector direction of BC x 6/12 right

[west berry]

Distance formula isn't required

Yup

[faint monolith]

doesnt get the right answer tho

no idea what this marking sheme yapping

[west berry]

You mean unit vector?

[faint monolith]

uh no?

oh wait

thats ony BD

not positionn vector of D

[west berry]

You can find unit vector of BC

And then multiply it with 6 to find vector BD

[faint monolith]

why cant i juz do this

i still get the wrong answer

[west berry]

I don't get which vector you're multiplying with 6/12

[faint monolith]

according to here, BC direction vector is (12,-6, 12)

so if i take this BC direction vector multiply by 6/12, wont i get BD direction vector?

[west berry]

BC length is 18 though

So divide by 18 to find unit vector

And then multiply with 6

[faint monolith]

o-

[west berry]

Yeah

[faint monolith]

BC divide 18 times 6

[west berry]

Yeah

Now note that it's BD and you've to find OD

[faint monolith]

hmmm alright

i will try and figure

thanks alot

[west berry]

Np

[faint monolith]

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[vernal vortex]

how is the blue part true?

[vernal vortex]

it should be false right?

[compact breach]

Why do you think it's false?

[vernal vortex]

{ a } is not a subet of { <1, a>, 1, a }

oops wait

ah wait

nvm lol

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[hexed bison]

Someone told me that having delta <1 is a stricter bound than being >0, so that for this specific problem the inequality being true for delta <1 means it should automatically hold true for all delta >0, and by extension all epsilon >0. Is this correct, and if so, how do I describe it in the informal proof?

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[faint locust]

#1195800467880558672 message

[faint locust]

Maybe I should post it here too:

Is that proof missing some steps or would you say it's good?

[wintry kindle]

hm it's an interesting approach

arguing via the availability of pivots

we proved it in a completely different way :D

[wintry kindle]

have you been taught the elimination/division manipulation before

or rather are you allowed to use them?

if so then I'd say yes, not sure if it doesn't cover edge cases

the common approach is to say that if the only solution is the zero vector, then ker(A) is {O_n} and from that follows Injectivity

because if any two vectors would then be mapped onto the same vector

due to linearity dim(ker(A)) would have to be > 0

which would defy injectivity

[faint locust]

Would you say it's missing some steps?

Or is it a fine proof

[faint locust]

oh

[wintry kindle]

I'd put it slightly more explicitly

Why does from "we can manipulate A into the unit matrix" follow "A is invertible"

and also

Why are there no "free variables" if A has a single solution

[faint locust]

Well if there were, then we'd have infinitely many solutions

So I'd write it down like that?

[wintry kindle]

Hm maybe more like "Since the H.S. of A has only the zero-solution, there are no zero-rows if A is in RREF form."

which you can then use to express that there are no variables in the set of solutions

[faint locust]

Alright, thanks

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[shut spade]

It says not all series will need the nth term test, does that apply for #1? Can I assume its divergent?

[formal trail]

you can apply the nth term test here, although there should be more justification

[shut spade]

I could, but is it really necessary to add up 6/7 +7/12 +8/17+9/22?

Sounds like a quagmire

[formal trail]

the nth term test says that a series diverges if its associated sequence doesn't converge to 0

[shut spade]

Yea so i have to find the sum right

Or is there another way??

[formal trail]

that's not what the nth term test asks for

it's not the sequence of partial sums, just the sequence of terms

[shut spade]

What does that even mean?

Terms of what?

[formal trail]

https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-3/v/divergence-test

[shut spade]

So i just find the limit of (n+5)/(5n+2) as it goes to infinity, and if its not 0 then it diverges?

[formal trail]

yes

[shut spade]

Ok thanks for posting this

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[heavy laurel]

Can someone check to see if my work makes sense for these problems

?

[heavy laurel]

work:

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[paper lion]

hey

can anyone help wit the chain rule in calculus?

[gritty drift]

!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).

[gritty drift]

I think you would have better luck asking it in the appropriate early-uni/advanced-math channel.

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[mental root]

struggling a bit with this optimzation problem

[mental root]

I formulated $L(x) = \sqrt{x^2+36}+\sqrt{(20-x)^2+196}$

[soft zealot]

The Great D

[mental root]

and after differentiating I find its minimum at x=6

but the book says the correct answer is 5.71

Did i construct a wrong expression for L(x)?

[silver roost]

Looks good to me idk

[mental root]

Yes, it seems that the book is not very right.

Thanks

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[peak ice]

https://cdn.discordapp.com/attachments/905285721886187580/1220129863487324170/image.png?ex=660dd17c&is=65fb5c7c&hm=8092e14376f34f1772b96a31bd5b13d6a255ef9c2bffdc4b8f9420f38143e9db&

[peak ice]

is 3 true because

https://cdn.discordapp.com/attachments/905285721886187580/1220129880813998080/image.png?ex=660dd180&is=65fb5c80&hm=9791dd9ad579ecd7e7bca46bbabed3f5843f32ec84959b80298b986825038dfc&

?

[marble agate]

what is your justification that this equality is true?

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[lapis wedge]

my answer is wrong, where did I go wrong?

[lapis wedge]

oh

took the integral wrong again

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[peak estuary]

how do i simplify this

im not sure what to do

😭

[mossy bridge]

Just use Newton’s third law of motion

[gentle barn]

Remember n! is defined $$1 \cdot 2 \cdot 3 \cdot ... \cdot (n-1) \cdot n$$

[peak estuary]

haha i wish i could

[peak estuary]

ohh so do i think of it as a number

[soft zealot]

Good

[gentle barn]

Kind of.

$$\frac{n!}{(n-1)!}$$ $$=\frac{1 \cdot 2 \cdot 3 \cdot ... \cdot (n-1) \cdot n}{1 \cdot 2 \cdot 3 \cdot ... \cdot (n-1)}$$

[soft zealot]

Good

[peak estuary]

ohh thank you so much

[gentle barn]

Welcome.

[peak estuary]

that looked like a nightmare to type 😭

[peak estuary]

🤗

just curious how do you know when to stop though

like when do i stop subtracting 1

[gentle barn]

Could you give more information? I'm kind of not following. Do you mean when do one stop to type numbers until it reaches n, or n-1?

[peak estuary]

cause you know for factorials its like in decreasing order e.g 7! is 7 times 6 times 5 times 4... 1

how do i know when to stop for n

[formal trail]

it's still down to 1... n is just some integer

[gentle barn]

$1 \cdot 2 \cdot 3 \cdot ... \cdot (n-1) \cdot n = n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1$

[soft zealot]

Good

[peak estuary]

OH WAIT

I GOT IT NOW

thank you so much

🥹 🤗

[peak estuary]

thank you as well!

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[gray acorn]

how do we perfect square this

[gray acorn]

when the third is negative

[coral prawn]

What do you mean by "perfect square" ?

[gray acorn]

how do make it ( )^2

[coral prawn]

You can't

,w 9x^2+6x-8

[soft zealot]

Results provided by WolframAlpha

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

Requested by stipendi

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[viral pivot]

how do i find the kernel and range of this?

[viral pivot]

diff channel pls

[warped summit]

Ok

[tranquil pine]

@warped summit #help-46

[tulip coyote]

For the kernel, you wanna find the matrices M such that T(M) = 0

[viral pivot]

when c and b =0 then ?

[tulip coyote]

The range, basically what each of the outputs "look like"

[tulip coyote]

Yep, all matrices $\pmqty{a & b \ c & d}$ with $b = c = 0$, so stuff that looks like $\pmqty{a & 0 \ 0 & d}$

[soft zealot]

@tulip coyote

[viral pivot]

here @fiery trail

[viral pivot]

ook

how do i express all those possible matrices?

like

will it be a matrix with variables

[tulip coyote]

You can, you could leave them as per how I did, or you can split that matrix into linear combinations of linearly independent matrices and say the kernel is spanned by those

Hopefully it's clear how you could break down that $\pmqty{a & 0 \ 0 & d}$

[soft zealot]

@tulip coyote

[fiery trail]

and what would the range be for this question?

[viral pivot]

can i say $\pmqty{a & 0 \ 0 & d}$ , a,d in R?

[soft zealot]

risa★

[viral pivot]

LOL

$\pmqty{a & 0 \ 0 & d}$

[tulip coyote]

two \

[soft zealot]

risa★

[viral pivot]

OH YEAH

$\pmqty{a & 0 \ 0 & d}$

[soft zealot]

risa★

[tulip coyote]

And yea, that should be fine, or you could also notice that
[
\pmqty{a & 0 \ 0 & d} = a\pmqty{1 & 0 \ 0 & 0} + d\pmqty{0 & 0 \ 0 & 1}
]
so the latter two matrices span the kernel

[soft zealot]

@tulip coyote

[viral pivot]

yes

[tulip coyote]

It's a similar idea for the image, you can either say it's all the matrices of the form $\pmqty{3c & 0 \ 0 & 2b}$ for $b, c\in \bR$, or you can similarly split up that matrix if you want

[soft zealot]

@tulip coyote

[tulip coyote]

(it's worth also noticing that the kernel has dimension 2, the image also has dimension 2, and as a real vector space, the 2x2 real matrices have dimension 4, so that's all great when it comes to rank-nullity)

[viral pivot]

for kernel we were looking for matrices M such that T(M) = 0, but whats the range basically?

[tulip coyote]

Everything that you can "get out of" T, what all the outputs of T look like

[viral pivot]

im just confused cuz the $\pmqty{3c & 0 \ 0 & 2b}$ is just the question

[soft zealot]

risa★

[tulip coyote]

Do they ask you to represent the range in a particular way or anything for the question?

[viral pivot]

they didnt specify

for example this one was in the form of a quadratic

both the kernel and range

as the answer

but i dont think u can rly do taht for the matrix?

[tulip coyote]

a quadratic for the range of that one?

And for the matrix, you can give them as matrices (in fact, the kernel is the same space as the image here(!))

[viral pivot]

oops sorry, it was proven using a quadratic but the answer was "any polynomial in R^2"

FOR KERNEL

oops

wait

[tulip coyote]

Ohhhh I was gonna say

[viral pivot]

kernel = {c(x-5)(x-7) | c in R}

[tulip coyote]

Ah, I see, yea, cause 5 and 7 would be roots of your polynomial so (x - 5) and (x - 7) are factors

[viral pivot]

okk, wait but kernel was (a 0 0 b) and range was (3c 0 0 2b) tho?

[tulip coyote]

Yep, but like you can replace the $3c$ with another variable, as with the $2b$, so you see that all the matrices are gonna look like something like $\pmqty{x & 0 \ 0 & y}$ with some real $x,y$

[soft zealot]

@tulip coyote

[tulip coyote]

Which matches with what the kernel looks like

[viral pivot]

okk i get it!

[tulip coyote]

Perfect

[viral pivot]

thank u chartbit !!!

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[tulip coyote]

Always a pleasure

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[summer dragon]

for this theorem isnt it also necessarily that vectors must not be the zero vector?

[tiny gorge]

if one of them is the zero vector, then there will be a nontrivial solution

you don't need to make a special exception for that case

[summer dragon]

thanks

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[cyan kayak]

[vital surge]

i despise these problems

[final saddle]

@tawdry minnow Has your question been resolved?

[cyan kayak]

Let x * y = f(x, y) and x + y = g(x, y)

Now we have f(7, 3) = 158, g(3, 9) = 45, g(f(2, 3), 4) = 377

Now you can see that this is clearly not nearly enough information to peg down the general form of f(x, y) or g(x, y). We have no restrictions on these functions at all. Are they commutative? Are they even continuous?

All we have is that a finite number of points have been defined. Functions from R^2 -> R have cardinality of (R^2)^R = R^R = beth_2. That's a lot of possible functions to choose from.

@tawdry minnow ^

[cyan kayak]

I mean, ok

That answers nothing and doesn't narrow it down at all

Now we are getting somewhere.

We have a model.

Sorta

There are still infinite solutions

But we managed to reduce the size of the infinity

From beth_2 to beth_1

So he said it's a binomial, I suppose both f and g, which means that it's two terms.

[vital surge]

this sounds like overanalysis

[cyan kayak]

Not really overanalyzing, because I'm not really analyzing it at all, just explaining why it's a broken question

[vital surge]

the way these questions work is that ur given some numbers & symbols denoting obscure operations with the goal of figuring out what those operations are

there's plenty of these online and this problem is just another one of those

it's true there's "infinitely many solutions" or smth similar but the creators usually intend for there to be one reasonable way to proceed

... not that i have a high opinion of this kind of question, but i wouldn't say it's inherently broken

it's just pure guesswork ig

[cyan kayak]

Well, even if we assume that this is as simple as f = ax + by and g = cx + dy we still have one too few equations to determine the coefficients.

,w Solve[{7a + 3b == 158, 3c + 9d == 45, (2a + 3b)c + 4d == 377, 3a + 2b == 8.5, 4c + 5d == 20.5}, {a,b,c,d}]

[soft zealot]

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[vital surge]

(i assume) that's enough info to find both functions

[cyan kayak]

So the model is more complicated than ax + by

[austere peak]

whoa

what is this?

looks like a tricky problem

[cyan kayak]

It's not, I can find many different solutions. The problem is finding the intended solution.

[austere peak]

which makes it tricky

[cyan kayak]

I mean... tricky is not the word I would use.

But sure

Let's change the model to something more complicated: ax + by + c that's 6 unknowns

Total

[vital surge]

that third line is correct right

[cyan kayak]

,w Solve[{7a + 3b + e == 158, 3c + 9d + f == 45, (2a + 3b + e)c + 4d + f == 377, 3a + 2b + e == 8.5, 4c + 5d + f== 20.5}, {a,b,c,d, e, f}]

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[cyan kayak]

Pick any value for a.

And the rest of the values are determined.

Except for a = 123/4, apparently

c(ax + by + e) + dy + f, where a is a free variable not equal to 123/4, and b, c, d, e, and f are defined as above.

Simplification left as an exercise to the reader.

Or hell, I'll type it in to WA and get it to do it.

,w simplify 2901/(1230 - 40a) * (ax + (299/2 - 4a)y + 5a - 581/2) + (8259 - 245a)y/(1230-40a) + (27684-405a)/(40a - 1230)

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[cyan kayak]

Any value of a that you select will make the function hold

And of course, you can have more complicated models, like quadratic models

Oh sorry my bad

That's (x * y) + y

Not (x + y) * y

,w simplify a * (2901x/(1230 - 40a) + (8259 - 245a)y/(1230-40a) + (27684-405a)/(40a - 1230)) + (299/2 - 4a) y + 5a - 581/2

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[cyan kayak]

@tawdry minnow ^

Generally, there's some "clever trick" that the person posing the question thinks is obvious, but the puzzle is typically utterly lacking in conveyance.

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[inland kettle]

the fu is this

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[mortal horizon]

I’m having trouble proving this, I’m trying to do something with x*m_2 = m_1 but I can’t get a rigorous argument

[mortal horizon]

Wait I have something

Let me get it written

[grizzled tusk]

a = b mod m1 so a-b = 0 mod m1 ie m1 | a-b

now, if m2 | m1 ...

[mortal horizon]

This is my argument

Excuse the handwriting

[grizzled tusk]

yep that's it

[mortal horizon]

Hell yeah

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[next ermine]

anyone able to do this?

[frank nebula]

sure

[next ermine]

consider the region in the first quadrant bounded by the graphs of y = x^2, y = 1, and x = 0. Use the shell method to calculate the volume of the solid formed by revolving this region about the line y = 2

[frank nebula]

umm

[next ermine]

just encase it's hard to see

[frank nebula]

whats shell method

nvm

i got it

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[hollow hollow]

Not sure where to go from here

[frank nebula]

ok so

SA = 4pi 4^2

[hollow hollow]

oh ok

So i find the value for SA

then take the derivative of the SA equation?

[frank nebula]

bingo

[hollow hollow]

thank you

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[forest pagoda]

I tried putting it in the calculator and all that fun stuff. I’m guessing it’s pretty simple but I have no idea what to do 😭

[frank nebula]

um

-f(x)

is just

add a - sign

in front of above f

F(x)

[forest pagoda]

like in the calculator?

[frank nebula]

calc is not required

[forest pagoda]

sorry im like sleep deprived rn so just do the opposite of f(x)

For the -f(x)

[frank nebula]

for eg..

-f(x) of -6 is 6

[forest pagoda]

oh no I knew that I was referring more to bottem row

f(2x) +4