help-10

[obtuse pebble]

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[timid silo]

I don’t get it

[timid silo]

Basically show that grad 1/r is -1/r^3 times vector r

r being the abs value of vector r

[obtuse pebble]

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[timid silo]

<@&286206848099549185>

Is this correct

[obtuse pebble]

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[obtuse pebble]

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[plucky temple]

can anyone help me with part 2

[obtuse pebble]

@plucky temple Has your question been resolved?

[plucky temple]

<@&286206848099549185>

[plucky temple]

<@&286206848099549185>

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@plucky temple Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

Barry found a delicious recipe for creating cinnamon rolls. However, the recipe measurements are in metric, and his scale uses ounces and pounds.

The recipe calls for 900 grams of flour to create 15 cinnamon rolls.

Rounding to one decimal place, how many ounces of flour should he use?

(I’ve tried multiplying 900 by 28.35oz but the answer was incorrect. Then I tried dividing it by 15 and it was also incorrect).

[obtuse pebble]

@timid silo Has your question been resolved?

[hexed blaze]

Best solution is Barry should go learn how to use metric
It's like stepping from a cave to a civilization

[hexed blaze]

28.35 grams - 1 oz
900 grams - ? oz

[timid silo]

💀 harsh

[timid silo]

According to Google 900 g would still be 28.35oz

31.747 if dividing I guess

[hexed blaze]

Yes

[hexed blaze]

Nah everyone in the world except majority of people living in the US use metric

So it's not harsh, but to be globalized 😎

[timid silo]

Ykw that’s a very good point

& dividing with 31.747 didn’t work either

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

The hw problem says the correct answer was 32.1 (I only get 3 tries until they give me a similar question) but idk what steps were taken to get that answer /: clearly wasn’t by multiplying or dividing those numbers tho

[upbeat island]

answer key has an error.

900 g  *  1 oz / 28.3 g ≈ 31.8 oz

,calc 900 grams to ounces

[warm shale]

Result:

31.746565754622 ounces

[timid silo]

Huh I see

Thank you, I’ll try again with a similar question

.close

[obtuse pebble]

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[timid silo]

how to determine the distance between points that lie on a parabola (the equation of the parabola is known)

[obtuse pebble]

Please don't occupy multiple help channels.

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[brazen gyro]

what do you mean by that?

you mean the length of the parabola between the 2 points

of the general distance between the two point?

[obtuse pebble]

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[red raft]

I got this far with the teacher but I honestly don't get it to begin with.

[obtuse pebble]

@red raft Has your question been resolved?

[red raft]

<@&286206848099549185>

[red raft]

.close

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[steel raft]

Determine, if they exist, an a and b so that (2,-2,1) is a possible solution to

[kind hawk]

what do you get if you plug (2, -2, 1) into the first equation?

[steel raft]

I considered that the best way to start doing this is with a matrix, in order to shorten some of the equations

[kind hawk]

you could, but not necessary

[steel raft]

2 -4a + 1 = -1

[kind hawk]

so what can we say about a?

[steel raft]

a = 1 ???

[kind hawk]

at least for the first equation to hold

what do you get when you plug (2, -2, 1) into the second equation?

[steel raft]

im sorry

im back

i was washing the dishes

that b has to equal 2

[obtuse pebble]

@steel raft Has your question been resolved?

[steel raft]

is this alright?

[obtuse pebble]

@steel raft Has your question been resolved?

[steel raft]

i will say yes just cause no ones helping

[obtuse pebble]

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[fringe hawk]

Somehow I got 6a+8b=72 but I think that's definitely wrong so I don't know what else to try, could someone help please?

[obtuse pebble]

@fringe hawk Has your question been resolved?

[fringe hawk]

<@&286206848099549185>

[raw warren]

Sorry my work is really sloppy

[fringe hawk]

no, not at all, it makes alot more sense now, thank you so much

[raw warren]

💖

You divide by the mph because each distance is in miles

This will leave each term in hours

You can cancel all of the hours, multiply the right by a common denominator of miles

And then solve for total miles

[fringe hawk]

ok, tysm

.close

[obtuse pebble]

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[bold nest]

Does the root test work for C)?

[bold nest]

lim sup $(\sqrt[n]{n}-1)^n$

[warm shale]

Mirinim

[bold nest]

I think I can show that nsqrt(n) goes to 1, so the lim sup would go to 0

does that sound right or am i way off

[obtuse pebble]

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[bold nest]

<@&286206848099549185>

[mint tendon]

yeah this would do it

[obtuse pebble]

@bold nest Has your question been resolved?

[obtuse pebble]

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[chrome bridge]

this is an estimation question, but I have no idea how to approach this problem... Can someone help me get started?

[slow fulcrum]

You could try figuring out the volume of a nickel

[oblique sage]

So basically he needs to figure out how many cells could fit inside the volume of a nickel?

[slow fulcrum]

yeah which is kinda dumb

[obtuse pebble]

@chrome bridge Has your question been resolved?

[kindred oasis]

Findind the dimensions of a nickel should be pretty easy to do online, then if you can also find the volume of a cell you can just divide the nickel volume by the single cell volume to find the number of cells.

That's a weird question though

Doesn't give you a lot of information

[obtuse pebble]

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[slow fulcrum]

yep

if the discriminant is negative, the roots are not real

the roots are the x intercepts

when there are no x intercepts there are no real roots

this goes into the quadratic formula

do you know what that is?

do you under stand why if the discriminant is negative there are no real roots?

what is the discriminant

correct

if that value is negative why wouldn't the roots be real

if b^2 - 4ac in the quadratic formula is a negative number

what would the roots include?

ok say b = 1 and a = 2 and c = 3

find the roots using the quadratic formula

right simply the sqrt

it would be sqrt(-23)

what is sqrt(-23)?

no

a square root of a negative number is not real

there is no such thing

is has to be expressed with i

which is imaginary

thats why if the discriminant is negative then there is no real roots because they all have i in them

if there are no roots

there are no x intercepts

yes, no real roots

when the discriminant is 0 there is 1 real root

and when the discriminant is postivie there is 2 real roots

ok have fun

byeee o/

[obtuse pebble]

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[obtuse pebble]

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[woven walrus]

i'm dumb and forgot to check back last time and ask further, does something being multiplied by 2 (e.g 1 to 2) equal a 100% increase or 200% increase?

[timid silo]

100% i believe

[versed cave]

increase means that something is being added
if you have 100 of something, and then you check later and you still have 100, it didn't increase by 100%, it didn't increase at all
If you have 100 of something, and then you check later and you have 200, it increased by 100%, so that's what you had before + 100% of what you had before

[obtuse pebble]

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[obtuse pebble]

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[lethal sphinx]

if (2a+2b+1)(2a+2b-1)=63, find the value of a+b

[lofty jay]

if x = a+b, (2x+1)(2x-1) = 63

then just solve for x

[lethal sphinx]

ok thx

[timid silo]

a^2 - b^2
= (a+b)(a-b)
So ye u can just replace 2a + 2b with some variable for ur convenience

[lethal sphinx]

ah ok i see

thank you so much

.close

[obtuse pebble]

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[bitter holly]

Is there a way to prove that a simple graph is connected? Or is there a general algorithm to verify that a simple graph is connected that can be applied to all simple graphs?

[obtuse pebble]

@bitter holly Has your question been resolved?

[unique nebula]

Do you consider running an algorithm that verifies connectedness as a proof?

[obtuse pebble]

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[bitter holly]

Yes.

[bitter holly]

.reopen

[obtuse pebble]

✅

[royal basin]

as far as algorithms go you could run a spanning-tree algorithm

it will succeed and return a spanning tree iff the graph you were starting with is connected

[bitter holly]

Thanks @royal basin

.close

[obtuse pebble]

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[icy pelican]

pls explain this to me

[proven zephyr]

ok so

for the first step

you add 45 to both sides

and then for the next step, you divide both sides by 4, isolating the x^2

you still understand until this part?

@icy pelican

[icy pelican]

wait

im understanding it still

[icy pelican]

what do i divide by 4?

is the 46

45*

[proven zephyr]

both sides

[icy pelican]

im sorry im so dumb but wym by bothsides

this is why i dont understand math,the most little details i dont understand

[proven zephyr]

like this

[warm shale]

help

[proven zephyr]

ike this

like*

@icy pelican

[icy pelican]

so um after u make it x² = 45/4 what do u do

next step

[proven zephyr]

you square root it

to remove the ^2

[icy pelican]

get the square root kt 45 and 4?

am i right

[proven zephyr]

wdym?

[icy pelican]

do i get the square root of 45 and 4

[proven zephyr]

ok.. what do you get after doing that

[icy pelican]

apperantly im wrong so um,what do i do next

[proven zephyr]

can you show your working so i know what you did wrong?

[icy pelican]

im just trying to learn it rn

[icy pelican]

im so confused on how 45/4 become 5.9

[proven zephyr]

ok so first of all

you only took the square root of 45

you left the 4 as it was

-harry styles reference

second of all

that's 5 times 9

they usually use a dot for multiplication

it's just simplifying radicals

[icy pelican]

can u tell me how it become 5.9

[proven zephyr]

(square roots in this context)

[icy pelican]

ok

[proven zephyr]

45 is 5 times 9

[icy pelican]

bro dedas 💀 i dont understand it

like 45 become 5.9

[proven zephyr]

as i said, they used a dot for multiplication

it's not 5 point 9

bad font ig,

[icy pelican]

did they multiply anything?

[proven zephyr]

no

they just changed 45 to 5 x 9

(x is not a variable here)

[icy pelican]

where did they get 9

[proven zephyr]

it's a method for simplifying square roots

https://www.youtube.com/watch?v=wSBxW7LW3DA

this guy explains it pretty well

[icy pelican]

ok wait me

[proven zephyr]

it's at minute 3 btw

[obtuse pebble]

@icy pelican Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[wanton dagger]

Is it c) I’m a bit unsure

I dont think its b)

[high lily]

explain your reasoning

[wanton dagger]

Well for c) it looks like the time increases as speed increases as time decreases

[high lily]

whut

[wanton dagger]

So speed increases and time slows down and then speed decreases as time ok i feel confused

Wait i need time to think

[high lily]

consider what's happening to the speed at each stage of the journey

before the hill

going up the hill

going down the hill

after the hill

[wanton dagger]

Ok

Before hill the speed is constant

Going up the hill the speed would decrease I suppose

Now going down the hill the speed would increase

And after the hill the speed would return to its constant

[high lily]

yes, and which graph beat represents that

[wanton dagger]

Looks like b) ?

[high lily]

yes

[wanton dagger]

I see

[wanton dagger]

I guess this made it easier to interpret from the graph

Ill probably do that for further questions

Thanks

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[obtuse pebble]

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[obtuse pebble]

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[olive jungle]

I got this wrong and I don’t understand why/how

[high lily]

didn't i walk you through on how to set up bearing diagrams properly?

[olive jungle]

You did but I keep messing up

[high lily]

i repeatedly mentioned drawing compass axis

i.e. you should have a compass axis at the point 3km west of the starting position

and the 150° angle would be the angle measured clockwise from the north axis

[olive jungle]

Ohh u mean like drawing more than one compass?

[high lily]

yes

[olive jungle]

Ok one sec

Like this ?

[high lily]

no,

angle measured clockwise from the north axis

that clearly acute angle is NOT 150°

[olive jungle]

Wait oops I I meant it like this

Like this right ?

[high lily]

yes, diagram is decent now

redo your calculations accordingly

[olive jungle]

I keep getting this

[high lily]

that clearly acute angle is NOT 150°

[olive jungle]

Yeah Ik but I csnr be bothered drawing it again

[high lily]

well the bearing diagram from earlier was correct

the angle from the north axis to the path you're walking is 150°

that is NOT the same as the angle in your right triangle

that WHOLE RED angle is 150°,

if you plan to use right triangle trig,
you want the GREEN angle

(which isn't 150°)

[olive jungle]

How do I find the green out

[high lily]

angle addition/subtraction

like before

how many degrees in a quadrant etc.

this has all been explained to you over 10 times now

(by me)

you should have notes

[olive jungle]

I’ll remember it this time I promise

Anyways thank you so much I got it now

.close

[obtuse pebble]

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[obtuse pebble]

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[plain grove]

Hi guys can you please check if my answer is correct

[plain grove]

Part A

Part b

[high lily]

why'd you close the other one

[plain grove]

Because someone interfered in it

[high lily]

tell them to move

[plain grove]

Im sorry if that was not ok, my bad

[high lily]

part a) is incorrect

[plain grove]

Ok

[high lily]

your use of t for the derivative function is also questionable

[plain grove]

So how can I fix it

[high lily]

for the aroc,
you'd just use the original function, not its derivative

[plain grove]

Make it f(3)?

[high lily]

and also do it from 0 to 3, instead of 1 to 3

[plain grove]

For part a?

[high lily]

(f(3) - f(0))/(3-0)

first 3 hours, not from the end of the first hour to the third hour

[plain grove]

True

[plain grove]

I don’t get the 3-0

[high lily]

the interval you're measuring

is from t=0 to t=3

[plain grove]

Oh ✅

[high lily]

$AROC_{AB} = \frac{f(b)-f(a)}{b-a}$

[warm shale]

ℝamonov

[plain grove]

b is 3?

[high lily]

yes

[plain grove]

a is 0?

[high lily]

yes

[plain grove]

I’ll try that, can I show u after?

[high lily]

yes

as for the notation for the derivative, you should use
$$\blue{f'(t)} = 3t^2 -2$$

[warm shale]

ℝamonov

[plain grove]

Is that for part b?

[high lily]

yes

[plain grove]

This is what I got so far for part a

[high lily]

no

f(3) which has a value of 22,
NOT f(22)

and b is just 3, not f(3)

and a is just 0, not f(0)

[plain grove]

Ok I’ll try that

[high lily]

and just f(0) which has a value of 1, not f(1)

[plain grove]

Ok

Is that better?

[high lily]

still has problems

with the denominator

denom is b-a
where b=3
and a=0

[plain grove]

Like that?

[high lily]

yes

[plain grove]

What’s the next step?

[high lily]

plug in the values for f(3), f(0)
and simplify

[plain grove]

22-1/3-0?

[high lily]

yes

missing () in your plain text

(22-1)/(3-0)

[plain grove]

Oh ok I’ll do it

So is 7 the answer to part a?

[high lily]

7kg / hour
if you were to include units

[plain grove]

Oh sweet

How’s my part b, is it correct?

[high lily]

final value is alright, just very bad notation/choice to use t for the derivative function which was already established to be the variable for time

[plain grove]

How can I make it better?

[high lily]

https://discordapp.com/channels/268882317391429632/903430332324405288/1013059372609445889

and then evaluate
$$f'(3)$$
for the instantaneous rate of change at t=3

[warm shale]

ℝamonov

[plain grove]

Saw that on top so I should write f’(etc

F’(3)=3t^2 - 2 for the instantaneous rate of change?

[high lily]

no

f'(t) = 3t^2 - 2

and when finding the iroc at 3, you replace all the ts with 3

[plain grove]

Oh ok now I got what u meant

Or maybe

[high lily]

no

that just gives you the amount at 3, not the iroc

which you already had earlier for part a

[plain grove]

Can you please go to another channel?

[high lily]

now its good

[plain grove]

So the Answer to part a is 7kg/h. And answer to part b is just 25?

[high lily]

same units

[plain grove]

Same units right?

[high lily]

that's what i said

[plain grove]

Thanks so much for reaching out, I really appreciate your effort with me ✅

@high lily

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[strange umbra]

A little confused as to what the phi(t) stands for here? Does it have a specific meaning (i.e., is it an actual function or is it simply a stand-in variable?).

[weary delta]

It’s saying if some function of t we’ll call \phi is a solution to eq(2)

S.t. d/dt phi(t) = f(y)

[strange umbra]

Ah I see, thanks!

.close

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[fallow ember]

Hey can someone help me find out what I did wrong here

[fallow ember]

Maybe it’s to do with splitting the integral

[timid silo]

maybe multiply and divide by rooty and the sub lny =t then ibp

ig

[kindred oasis]

You can't divide an integral like you did between your second and third step, since it's a product between two functions, not a sum.

[kindred oasis]

@fallow ember

[warm shale]

Andrea276

[obtuse pebble]

@fallow ember Has your question been resolved?

[fallow ember]

ahhh iget it, tysm

[obtuse pebble]

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[violet compass]

when working with eigenvalues and eigenvectors, is it (A-tI) or (tI-A)? for the characteristic polynomial

[kind hawk]

depends

[violet compass]

hmm how?

[kind hawk]

some people use the first, some the second

essentially only changes the sign

[violet compass]

oh so it don't really matter then

[kind hawk]

not really, no

will still have same roots etc

so all the relevant stuff is the same

[violet compass]

ahh thank you, because I keep reading different notes and they all use either one.

and got confused whether one was to find evalues and evectors

i got it now thanks.

.close

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[hard fable]

Hi I got one of these problem solving questions

[hard fable]

The product of three numbers: a, b and c = 0. The number b is two times bigger than a +1 and the number c = b - 4

How do I solve for the numbers?

[topaz bridge]

The product of three numbers: a, b and c = 0
The number b is two times bigger than a +1
c = b - 4
Make system of equations

@hard fable

[hard fable]

sorry I am back

so then one number can be 4

as 4-4 = 0

what about b= 2(a+1)

a= -1

ah sorry I now understand what you mean

brb

[obtuse pebble]

@hard fable Has your question been resolved?

[hard fable]

.close

[obtuse pebble]

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[hard fable]

@topaz bridge helped me you have

thank you

[obtuse pebble]

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[fallow ember]

how do you do this? i have no idea where to start or wat to do

[summer spire]

Im guessing youd have to apply the formula recursively and see if theres some sort of pattern?

[proven breach]

just evaluate from 0 to pi/2

[summer spire]

or yeah that actually

[proven breach]

actually you can see directly...

it's multiple choice cheese but

only one of the four even appears in the original formula...
now ofc you can evaluate and cos(pi/2)sin^(n-1)(pi/2)-cos(0)sin^(n-1)(0)=0

[fallow ember]

oh so the second one

looks like the second part of the formula

[proven breach]

yeah

when you evaluate you can also find that the first term drops off

so

[fallow ember]

ohhh i see thanks

where should i learn reductino formulas if my teacher didnt teach it?

[proven breach]

this is something you can prove by induction
but I don't think they'll ask you to memorize this

[fallow ember]

so we just have to memorize the formula and learn to evaluate it?

[proven breach]

they'll most likely give you the formula when they think you need to use it in a specific problem

stuff you need to memorize is basic trig identities

[fallow ember]

Did I do smth wrong in evaluating this? I’m not getting the right answer

I fixed some stuff but the answer is 7/9 I think

[proven breach]

cos pi/2 = 0

you put -0.5

[fallow ember]

is the process right? i did it again and got 1/3 😭

[proven breach]

so you only have the last term

• -2/3cos(0)

which is 2/3

[fallow ember]

Like this? :,),

[proven breach]

yeah

[fallow ember]

Yay thank you :)))

[obtuse pebble]

@fallow ember Has your question been resolved?

[obtuse pebble]

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[timid silo]

Can someone who is good at proofs check my proof that the average of two numbers is bounded between the two numbers? I'm not good at proofs

[timid silo]

Let there be two real numbers a, b such that |a|≥|b|. Let there be a third real number, c, such that c=(a+b)/2.
Suppose that |c|>|a|:
For the case a=b=0, c=a=b=(a+b)/2 = 0, thus the statement |c| > max(|a|,|b|) is incorrect in this case.
For the case |a|> 0 and |b|< 0, as well as the case |a| > 0 and |b|<0 the contradiction is apparent. As |a+b| ≤ |a|. Following this, |(a+b)/2| > |a| is impossible. Proving that the average, c is bounded within the range (b,a) or the range (a,b).
For the a≥0 and b≥0, as well as the case a≤0 and b≤0, let us rewrite the average function.
Average = (a+a-x))/2, where x is the difference between a and b. Since we are proving that the statement |c|>|a| is never true, let us take the norm of the average function
|Average| = (|a|+|a-x|))/2
Clearly, this function is maximized when x is equal to zero. Which gives
|Average| = |2a|/2 = |a|.
Thus, max(|c|) = |a|
This proves that the average two real numbers a,b is bounded between the closed interval [a,b]

[obtuse pebble]

@timid silo Has your question been resolved?

[zenith raft]

this looks more complicated than needed...?

[timid silo]

Yes.

[tardy epoch]

try to simplify your proof

it looks fine, but could probably make some generalization

[timid silo]

I could simplify it to
|Average| = (|a|+|a-x|))/2
Clearly, this function is maximized when x is equal to zero. Which gives
|Average| = |2a|/2 = |a|.
Thus, max(|c|) = |a|

couldnt i

[zenith raft]

can I suggest...$$a<b \implies\frac{a}{2}<\frac{b}{2} \implies \frac{a}{2}+\frac{b}{2}=\frac{a+b}{2}<b$$

[warm shale]

layla💜

[zenith raft]

that's like half the proof, just that

[obtuse pebble]

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[timid silo]

I’m genuinely confused on how to solve this question. I also looked at the answer key my teacher posted online and I’m still at a loss.

[timid silo]

Is the form PQ the distance from P to Q?

[timid silo]

I’d think so, I don’t know what else it could be. Answer key says it’s 5.

that is correct

But the p is negative?

in general to find the distance from AB you would do B-A

Ohh

in this case we have PQ where P=-2 and Q=3 to 3-(-2)=5=PQ

[obtuse pebble]

@timid silo Has your question been resolved?

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[tame valve]

So I know how to find some points like if I put in x=2 I get (2, -5), how would I graph this?

[tame valve]

It's a logarithm ik that

I mean exponential equation mb

[nocturne minnow]

Go to (2, -5), put a point there

[tame valve]

Alright

I did that

[nocturne minnow]

You need more points

[tardy epoch]

simplify first

[nocturne minnow]

Do like 5 points, then connect the dots

[tame valve]

Is there a way I should do this

[tame valve]

how can I simplify this?

-4 is a negative

[nocturne minnow]

Not really, just plot points, and connect the dots

[tame valve]

I can't do 2^2

[tame valve]

Oh ok

[nocturne minnow]

It's a joke

[tardy epoch]

-4 = (-1)(4)

[tame valve]

Oh

So -2^2

[tardy epoch]

$2^{x-2} = 2^{-2}2^x$

[tame valve]

That becomes

[warm shale]

riemann (eric tao for honorable)

[tame valve]

2^x?

[tame valve]

Mhm

So

2^x - 1

[tardy epoch]

yea sorry i don't know why i did strikeout haha

[tame valve]

Uh

I'm confused

So

Get the points right?

[tame valve]

And is this correct?

[tardy epoch]

not quite. check your algebra again

[tame valve]

That's what I thought

U can't make -4 = -2^2

Right?

[tardy epoch]

yes you can

you did $2^2 \cdot 2^{-2} = 1$ correctly, but there was more you left out

[warm shale]

riemann (eric tao for honorable)

[tame valve]

Oh

-2^2 * 2^2 * 2^x = 2^x

So 2^x - 1

I don't get it

How is that wrong

[tardy epoch]

try plugging in x=1 to both sides

[tame valve]

Yeah

They're the same

😭

[tardy epoch]

Nah you're plugging numbers into your calculator incorrectly

,calc -2^2

[warm shale]

Result:

-4

[tardy epoch]

,calc (-2)^2

[warm shale]

Result:

4

[tardy epoch]

See how they're different?

[tame valve]

Oh-

I'm dumb

-2^2 * 2^x * 2^-2

Uhhhh

So -2^x - 1

I see

[tardy epoch]

now you can continue with

[tame valve]

Oh lol ty

.close

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[royal light]

hello, i'm having trouble solving this PDE

i've solved the steady state solution (i think) but im not sure how to solve the transient sub-problem

[slate zephyr]

What tools do you ahve available?

You should also consider asking in #odes-and-pdes

[obtuse pebble]

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[obtuse pebble]

@mossy lotus Has your question been resolved?

[timid silo]

yep! @mossy lotus

[obtuse pebble]

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[eager knot]

I simplified it down to sinx-tanx but not sure where to go next

[eager knot]

I think I have to plug x into the equation

[wooden cipher]

The table is your friend

[eager knot]

still don't know what to do

[tardy epoch]

You're asked to find the limit as x goes to zero

The table has values of x that goes from 0.3, 0.2, all the way down to 0.0001

[eager knot]

I understand that because of the + next to x--> 0

[tardy epoch]

What does f approach in the table

[eager knot]

y approaches .50

-.50

[tardy epoch]

That's all you have to do

You're done

[eager knot]

but I just don't get the sin and tan
am I supposed to ignore them?

[tardy epoch]

You're not ignoring them. They're encoded in f(x) in the table

[eager knot]

oh I get it
you think I'll ever see that same problem without a table?

.close

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[wanton dagger]

The sum of the areas of two circles is 106pi cm^2 and the radius of the larger circle is 4cm longer than the radius of the smaller circle. Find the length of the radii.

[wanton dagger]

I can’t seem to form an equation here

[prime scarab]

A=pir^2

106-A

[high lily]

let r represent the radius of the smaller circle

[wanton dagger]

Ok

[high lily]

what will be the radius of the larger circle? (in terms of r)

[wanton dagger]

(r+4)

[high lily]

yes

[wanton dagger]

Wait so now its just sum of the areas

And we know what its equal to

And we also know radius

So we could solve for r

[high lily]

you mean you know the area

[wanton dagger]

Ya

So its just a straight forward equation now

Ok, thanks

.close

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[prisma ravine]

this gets me
(tan^2x)(csc^x)-1 / (tan^2x)

[ (sin^2x / cos^2x)(1/sin^2x) - 1 ] / (sin^2x / cos^2x)

[fierce lagoon]

tan^2(x) + 1 = sec^2(x)

Use that

[prisma ravine]

ooo forgot about that one

[limber mason]

the numerator tan was multiplied by csc

[prisma ravine]

which is [ (sec^2x) - 1 ] / (sin^2x / cos^2x)

which is tan^2x / tan^2x

=1

.close

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[limber mason]

welp

[fierce lagoon]

Gamer

[prisma ravine]

mhm

this I got cosx = 1

.reopen

[obtuse pebble]

✅

[prisma ravine]

but how do I put it as an answer? cos^-1(1)?

[tardy epoch]

You should know these numbers from trig

[fierce lagoon]

arccos(1) doesn't exist

[fierce lagoon]

Also no, cos(x) ≠ 1

Use quadratic formula

Let u = cos(x)

Then $u^2 = u+2$

[warm shale]

Umbraleviathan

[mint tendon]

wait what isn't it 0?

[fierce lagoon]

No

[mint tendon]

if you plug that back into the equation it doesnt work

,w arccos(1)

[warm shale]

Results provided by WolframAlpha

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

Requested by S̨̥̫#8871

[fierce lagoon]

Wait no it's 0

Lmao whoops

[mint tendon]

no worries happens to the best of us

[fierce lagoon]

I was thinking about something else

Uh

Brain moment

My brain went into cos(1) isn't 0

Which is like

A brain moment

[mint tendon]

https://tenor.com/view/big-brain-lateralus-parabol-third-eye-tool-gif-16499016

[prisma ravine]

whatd I miss

I Was gone

ah its 0

wait

now im confused

[fierce lagoon]

My brain cells charging up to be the biggest dumbass in my classes

[fierce lagoon]

cos(x) ≠ 1

Look at your problem

[prisma ravine]

yea its -1

not 1

messed up factoring

[fierce lagoon]

Yeah

[prisma ravine]

so its pi then

the answer

[fierce lagoon]

Yeah

[prisma ravine]

.close

[obtuse pebble]

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[deft citrus]

hello I have 2 questions

[timid silo]

Okay what are they

[deft citrus]

these 2

[crimson berry]

is this a quiz

[astral ivy]

Bruh

[deft citrus]

it is not a quiz

why?

[crimson berry]

it looks like a quiz

[deft citrus]

oh alright

yeah no

[astral ivy]

Let’s look at the first one

How do you know whether a point is on the graph of a function?

[deft citrus]

Im not sure

[timid silo]

Here is the hint y=f(x)

[astral ivy]

Do you know how functions work?

[deft citrus]

this is a algebra review I put together. I forgot these two lol

I do know how functions work

[astral ivy]

Ok

Well for example

If f(1) = 2, then the point (1,2) is on the graph of the function

[deft citrus]

yea

[astral ivy]

So with (a,b), the input is a and the output is b

Following?

[deft citrus]

yes

[astral ivy]

Ok

So they give you a bunch of points and ask which one is on the graph

How do you think you could find out whether it’s on the graph

[deft citrus]

by plugging in points for x

[astral ivy]

Yeah exactly

Try (1,3)

Is that on the graph?

[deft citrus]

no

(1.3) doesnt work for either of them

[astral ivy]

Either of what?

[deft citrus]

either functions

[astral ivy]

There’s only one function written

Wait do u know how piecewise functions work

[deft citrus]

i remember learning them but i do not currently remember how they work

[astral ivy]

Ok

You know how most functions are like a recipe

“if your input is x, the output is 5 times x”

This has only one formula you need to write, namely f(x) = 5x

But what if you want a function that squares positive numbers and cubes negative numbers?

There’s no single formula that can contain both cases

[deft citrus]

ok

i got it now

how about question 2?

[astral ivy]

Ok lol

Do you know what an “invertible” function is

[deft citrus]

i remembered how it goes

no

i dont

[astral ivy]

Again with functions as recipes

Consider a function f that doubles the input, namely f(x) = 2x

You can make an “inverse” function that undoes whatever f did

[deft citrus]

yes

[astral ivy]

So in this case, the inverse function would halve the input

g(x) = (1/2)x

[deft citrus]

why 1/2x?

[astral ivy]

The fundamental idea here is that plugging the output of f into the input of g will give you the original number back

[astral ivy]

Because that would halve the input

Undoing the doubling

[deft citrus]

K

[astral ivy]

Does that make sense?

[deft citrus]

yes

[astral ivy]

So this property, written mathematically, means
g(f(x)) = x

following?

[deft citrus]

yes

[astral ivy]

Cool

[astral ivy]

So for this problem, they’re asking what f^-1(6) is

[deft citrus]

yeah

[astral ivy]

In other words, what number when plugged into f will give you 6

And you can simply look at the table to see that it’s -6

[deft citrus]

oh

so your saying the answer is -6?

not 4?

@astral ivy

oh i get it

[astral ivy]

Yeah

[deft citrus]

i think

yeah i got it

[astral ivy]

Nice

[deft citrus]

id like to go back to the piecewise functions question

[astral ivy]

Alright

Wait you said u got it?

[deft citrus]

Ive graphed it but I do not understand how to find what point is on the graph. it seems that there are multiple points

[astral ivy]

Multiple points? Wdym

You mean multiple of the answer choices are on the graph?

Or that the graph has multiple points (generally) on it

[deft citrus]

yes

[astral ivy]

Which ones

[deft citrus]

(3,8) and (2,3)

[astral ivy]

Let’s see

(3,8)

So 3 >= 2

So we plug it into 3x+4

That yields 3(3) + 4 = 13

Which is not 8

@deft citrus did that make sense

[deft citrus]

yes

but it fits 5x-7

[astral ivy]

But that’s not the right recipe

5x-7 is only used if x < 2

[deft citrus]

none of those ordered pairs match for 3x+4 and 5x-7

yeah

[astral ivy]

They don’t need to match for both

They need to match for the corresponding one

[deft citrus]

so what do they need to do to be correct?

but there is only 1 answer

[deft citrus]

(4,16) works in 3x+4

oh

so thats the answer then

[astral ivy]

Yeah lol

[deft citrus]

wow

math can make you feel so dumb

[astral ivy]

Hahaha

Everything’s obvious after you realize it

[deft citrus]

yeah

thanks for the help

[obtuse pebble]

@deft citrus Has your question been resolved?

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[obtuse pebble]

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• Show your work, and if possible, explain where you are stuck.
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[raven ridge]

hello

[timid silo]

hi

#❓how-to-get-help

[kindred oasis]

Do you have a question to ask?

[raven ridge]

YESSSS

[kindred oasis]

Then you should probably ask it?

[raven ridge]

Can i message you?

[astral ivy]

No

[kindred oasis]

No, ask it here

That's why these channels exist...

[raven ridge]

i know..! but

[kindred oasis]

I don't even know what your question is, I might not be able to solve it

[raven ridge]

im not used to it yet

its calculus 3

Or multivariable calculus

[astral ivy]

Just post the screenshot in this channel

[kindred oasis]

Wdym? If you have a question ask it

[astral ivy]

Lol

[kindred oasis]

Here

[raven ridge]

here it is

i dont understand anything

i watch vids all over again and it's frustrating

And its 1 out of 11 problems

[obtuse pebble]

@raven ridge Has your question been resolved?

[stable rain]

for Q1, u can try graphing the values of temperature when y=5

so u just pick the approriate temperature values as y axis and x axis as the current x axis and u should get a normal x,y graph

[obtuse pebble]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[tranquil prairie]

.open

.reopen

[deep dirge]

.reopen

[buoyant apex]

.reopen

[slow sonnet]

what did you try?

[quiet wadi]

Typing this out for my sanity:
$$f(x) = \sqrt{x+40}, g(x)=x^2$$ $$(f \circ g)(3)$$

[warm shale]

nHail

[slow sonnet]

what do you think (fog)(3) means?

[buoyant apex]

Dam I don’t know math isn’t my strong suit but if I had to get the sum of the missing number

<@&286206848099549185>

[oblique sage]

So your first step is to read the rules

!15min

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[oblique sage]

Do not prematurely ping, and do not ping again.

[buoyant apex]

My fault

[obtuse pebble]

@buoyant apex Has your question been resolved?

[buoyant apex]

!15min

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[short zodiac]

$$(f \circ g)(x)=\sqrt{x^2+40}$$

[warm shale]

KD

[short zodiac]

Now substitute 3 in this expression

[rigid pine]

To help you understand the notation it is read "f of g". Normally you might see f(x) which is read as "f of x" so in place of x in the function f(x) you place g(x).

[obtuse pebble]

@buoyant apex Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[finite portal]

is there any way to prove that triangle AMC isn't isoceles other than using pythagorean theorum or side ratios

[finite portal]

like im looking for multiple solutions

[cinder badge]

getting that sunday night homework finished 🤝 respect

[finite portal]

ty man, wanted to do it at a friends house but ended up not doing shit

ive already got one solution but i cannot physically find another one for the life of me

[obtuse pebble]

@finite portal Has your question been resolved?

[finite portal]

<@&286206848099549185>

[obtuse pebble]

@finite portal Has your question been resolved?

[timid silo]

area of △AMP = (AP)(PM)/2 and area of △CMP = (CP)(PM)/2 in that case let us do area of △AMC = area of △CMP + area of △AMP = (AP)(PM)/2 + (CP)(PM)/2 = 2PM+AP+CP/2 = PM+(AP+CP/2) which is PM the height and AP+CP/2 is the base and that is true when both △AMP + △CMP = △AMC so △AMC is asosceles

[finite portal]

hold on wait

for "2PM+AP+CP/2", is it addition or multiplication thats needed?

i know it sounds stupid but wouldnt you need to multiply them

[timid silo]

2PM = PM+PM

[finite portal]

yeye i mean the AP+CP part

[timid silo]

it is like PM/2 + PM/2 + AP/2 + CP/2 so 2PM/2 + AP+CP/2 so PM + AP+CP/2

hold on

[finite portal]

alr

[timid silo]

I did wrong

sorry

[finite portal]

oh thats ok

[timid silo]

it is multiply not addition

[finite portal]

right so its 2PM * AP * CP / 2

[timid silo]

no

$\frac{(AP * PM)+(CP * PM)}{2}$

[warm shale]

流星宗教

[finite portal]

alright

where would you go from there then

[timid silo]

So $\frac{(AP)(CP) + (AP)(PM) + (PM)(CP)+(PM^{2})}{2}$

[warm shale]

流星宗教

[finite portal]

yeah

hold on

you'd only expand it if it were multiplication right

but its addition

[timid silo]

omg

why i'm doing multiply

[finite portal]

rip

[timid silo]

so $\frac{(AP)(PM)}{2} + \frac{(CP)(PM)}{2}$

[warm shale]

流星宗教

[finite portal]

mhm

what next

[timid silo]

we should prov that CM the same as MA

[finite portal]

right

[timid silo]

AM = √PM² + AP²
CM = √PM² + CP²

so we can see X in triangle △CMP the same as △AMP

[finite portal]

ok so we've proved that they're congruent

so how does that go onto proving they're isosceles

[timid silo]

when we prove that the triangles △CMP and △AMP are the same we also prove that target triangle is isosceles

[finite portal]

ah right

cause the base would be different

[timid silo]

yes

[finite portal]

one last thing

[finite portal]

can you walk me through how you got that

[timid silo]

from r = √x² + y²

[finite portal]

what is that

[timid silo]

second

[finite portal]

ah pythagorus right

[timid silo]

yeah, idk its name in english

[finite portal]

alr thanks man

[timid silo]

np

[finite portal]

.close

[obtuse pebble]

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[timid silo]

Shouldn't the line be the span of R2?

[timid silo]

If it is then why the professor is writing like this?

[rich sparrow]

"span" means all possible combinations of the vectors

the lines are in the span of those two vectors, but the span itself means the whole plane, AKA R2

[timid silo]

I see.

Thank you 👍

.close

[obtuse pebble]

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[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[timid silo]

i don’t understand my mistake

i hope everyone can read me

the answer says it’s (q-3) at the end and not (q+3)

[hexed agate]

What did you do in the third line?

I can understand upto the 2nd line and seems correct

[timid silo]

since (p-2) and (p+2) repeated themselves

i did (p-2)(p+2) multiplied by the terms outside of the brackets

like 12+q+q^2 were outside of the brackets

and since it’s a remarkable identity i used (a+b)^2

[hexed agate]

(2-p)(2+p) and (p-2)(p+2) are the same?

[timid silo]

ah shit

got it

-12

thanks

[hexed agate]

Wait

Nvmd yeah -12

[timid silo]

yeah irs correct

thanks pal

.close

[obtuse pebble]

Channel closed

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