help-42

[potent lotus]

riemann

[rare drum]

.close

[calm coral]

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[polar elbow]

Hi

[astral mural]

Hi

[polar elbow]

Hi. I have a trigonometry problem. Here it is: find the length of the side c of the triangle ABC when a=6ft, b=3ft and the angle A equals 3B

Can anyone help me solve this?

!status

[calm coral]

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

[polar elbow]

1

[leaden thunder]

Try drawing a picture

[polar elbow]

Smth like this?

[leaden thunder]

Try setting up some equations using law of sines and cosines

,tex .sine law

[potent lotus]

riemann

[leaden thunder]

,tex .cosine law

[potent lotus]

riemann

[calm coral]

@polar elbow Has your question been resolved?

[polar elbow]

I don't see how i can estabilish a relashionship between the 2 formulas that will help

[calm coral]

@polar elbow Has your question been resolved?

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[polar elbow]

.reopen

[calm coral]

✅

[polar elbow]

Actually i got the minus wrong on the cossine law equation but i already fixed it. Still couldn't solve it

<@&286206848099549185>

[calm coral]

@polar elbow Has your question been resolved?

[calm coral]

@polar elbow Has your question been resolved?

[calm coral]

@polar elbow Has your question been resolved?

[winter trench]

That is law of sines. Can you send a clearer picture?

If you are a bit confused, list down the given and note the things you need to find. Match the missing pieces with the given equations to find the missing pieces. It's like a puzzle where you have to figure it out yourself. If it helps, you only need law of sines most of the time, and if law of sines don't work/not applicable, then use law of cosines. Law of cosines does look too complicated to use so I suggest use law of sines FIRST if it works.

[sly trout]

Which two formulas

[calm coral]

@polar elbow Has your question been resolved?

[calm coral]

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[remote mural]

"The sculpture will be made of solid cement. Calculate the exact volume of the composite figure. Calculate the cost of the cement if it costs $32/m3. Show all relevant working out."

[fathom shuttle]

It would be nice if we can see the composite figure

[remote mural]

It says these are the figures:
Cube (length: 2.5m)
Rectangular prism (length: 3m, width: 1m, height: 1.5m)
Triangular prism (base: 2m, height: 1.5m, length: 2.5m)
Cylinder (radius: 1m, height: 1.5m)

[fathom shuttle]

Do you know how to find the volume of each shape given the dimensions?

[remote mural]

no

[fathom shuttle]

Ok lets start with the cube

Do you know the formula for volume of a cube?

[remote mural]

volume = s3?

[fathom shuttle]

s³ yes

[remote mural]

ok

[fathom shuttle]

So what is the volume of the cube?

[acoustic totem]

Better question "what is the value of side given?"

[remote mural]

well you see guys

it's 1am and my brain isn't rlly functioning

• the thing is due in the morning and i barely got started

[fathom shuttle]

Do you understand what s³ means?

[remote mural]

no

[acoustic totem]

Cube of side

Side of a cube (shape)

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[strange lichen]

not too sure how to start, any help would be appreciated

[leaden thunder]

Just integrate?

Or do you need help setting up the bounds

[strange lichen]

yea i need help with the bounds

[leaden thunder]

The inner one is dy. And y>x^2 tells you the lower bound is x^2

The upper bound is?

[strange lichen]

is it 1?

[leaden thunder]

Nah

It's in the integral already

[strange lichen]

o so +inf

howd u know to change the lower bound but not the upper one?

sorry its been a minute since my calc 3 class

o wait its cuz they dont give an upper bound for y right

[leaden thunder]

You use all the information given

[calm coral]

@strange lichen Has your question been resolved?

[strange lichen]

ah thanks riemann

lemme see what i get

[calm coral]

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[strange lichen]

.reopen

[calm coral]

✅

[strange lichen]

i got 1/2

ill send my work in a second

[fathom shuttle]

Looks good to me

[strange lichen]

Niceeee

Thanks @fathom shuttle and @leaden thunder 👍

.close

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[carmine warren]

.

[indigo gust]

y = tan (x)/ tan(3x)

[indigo gust]

I need to find range of this function

y<1/3 or y>3 - {0}

[bright light]

step 1 would be to write tan 3x in terms of tan x

[indigo gust]

Yes I did that

[bright light]

try converting that into one sort of trignometric function

[indigo gust]

Then I sub tan²x = t and then I wrote t in terms of y and then t>0

[bright light]

like maybe sin

or cos

[indigo gust]

But t≠1/3 for tan3x to be defined

[bright light]

whatever u got after the first step
try converting ur equation to either sin or cos

[indigo gust]

y=0 should not come as range?

But Wolfram alpha shows that 0 is in range

[bright light]

i am getting y <= 1/3 or y => 3

[calm coral]

@indigo gust Has your question been resolved?

[bright light]

@indigo gust sry for late reply + ping

[calm coral]

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[calm coral]

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[turbid dirge]

any help

[visual nimbus]

simplify by t^{-3} I guess

[frosty sparrow]

consider the limit of the highest degrees in the numerator and denominator only

[turbid dirge]

so -t^-2 and t^-1?

[frosty sparrow]

yeah

[turbid dirge]

would u not plug in inf which is inf/inf or 0/0? which means more work is required

[frosty sparrow]

no need

if you get rid of the expressions with smaller degrees and keep everything else as is, what are you left with?

(aka. if you remove the t^-3's, what do you get)

[turbid dirge]

-t^-2/t^-1 right

[frosty sparrow]

yes

what then can you do to simplify the expression even more

[turbid dirge]

would you need to convert them to fractions cus theyre negative exponents?

[frosty sparrow]

that can work if that helps you but i would prefer doing exponent subtraction

[turbid dirge]

simplifies to -1/t?

-t^-2/t^-1 -> -t/t^2 = -1/t

[frosty sparrow]

yes

[turbid dirge]

then plug in inf -1/inf or basically -1/0 = 0

[frosty sparrow]

yep!

[turbid dirge]

is there a name to the highest power method of the numerator and denominator?

and ignoring everything else

.close

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[radiant latch]

Can someone please help me with this math question I've been stuck on it for the last hour

[balmy blade]

how far have you progressed the past hour?

[storm stirrup]

would be nice if you gave us the drop-down options as well

[radiant latch]

first one states linear or exponetial 2nd one is asking mutiplicativity or additivetily 3rd one is asking common diffrence/slop or common ratio/mutiplyer/base

[balmy blade]

alright

the easiest way is to find the difference of f(x) for each year

if the increase is the same each time then it's linear

if it isn't then it's exponential

[radiant latch]

its 170 for the first 2 then its 169 if it were linear so would that be considerd linear?

[balmy blade]

a linear function must have the same difference

so if the difference is +170 it has to be +170

but it isnt exponential either like

its probably like $x^n - ax$ for $x < 0$

[potent lotus]

Snookums

[calm coral]

@radiant latch Has your question been resolved?

[radiant latch]

But it has to be linear or expodiential

.close

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[jade bison]

help

[jade bison]

FIND L how much rise will be there in 8 packets of mass 5whole1/2 kilograms

.close

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[calm coral]

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[north ice]

What is a semi filled dot? (Seen at (2,1)

[polar jetty]

I think that just shows the graph at x=2.

[north ice]

What you mean

[polar jetty]

That f(2) = 1

[north ice]

Ah ok

I was just confused why it’s not filled in all the way

Maybe an error

.close

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[austere remnant]

I can’t even start the problem I don’t know how to get to dx/dt =-y^dy/dt /x

[calm coral]

@austere remnant Has your question been resolved?

[austere remnant]

<@&286206848099549185>

[sharp dagger]

which question are you working on?

i'm kind of confused by the structure ngl

[austere remnant]

circle centered at the origin with the radius of 4...

[sharp dagger]

finding dy/dx would essentially mean the first derivative. for here, i'd say you need to differentiate x^2+y^2=16, specifically with implicit differentiation.

you should give implicit differentiation a revision since you didn't recognise it, khan academy has great resources on it

[austere remnant]

ok hold on.

so it would be 2x time x prime + 2y =0?

or would multiply by Y prime again

[sharp dagger]

try writing y' in terms of dy/dx, it'll be make a bit more sense

but uhh hold on

so the reason for using the form dy/dx is essentially because later you'll need to sort like terms

https://www.mathsisfun.com/calculus/implicit-differentiation.html

[sharp dagger]

also i'm not sure why you crossed or cancelled out 2x and 2y because you can't do that

[austere remnant]

ok...

[sharp dagger]

you need to multiply each term by d/dx to show they are being differentiated

and write what's left after that

and sort the like terms

[austere remnant]

oh ok

so since x=2 and dy/dt = -3 and y= square root of 12

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[scarlet sentinel]

Hey is this correct? People keep telling me it’s 1

[clear delta]

for the first one we'd just say the limit is infinity

second and third are right

[silk merlin]

x(x+1)/(x+1) [eliminate (x+1)]
We will have x which is equal to infinity

[calm coral]

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[hearty osprey]

I’m not really sure what goes where, could y’all help?

[austere remnant]

You’re proving why measurement of angle 2 is equal to the measurement of angle to 4

So it’s basically asking what did you do to go from step 5 to step 6

Or step 3 to step 4

[calm coral]

@hearty osprey Has your question been resolved?

[hearty osprey]

I’m kinda just guessing at the blanks since I’m not really sure what exactly they all men’s

[austere remnant]

Well for step 4

[austere remnant]

Look at angle 1 2 3 and 4

What’s something they have in common?

[hearty osprey]

They’re complementary

[austere remnant]

And that means

?

[hearty osprey]

They’re both 90°, they’re congruent

[austere remnant]

Ok so

Angles 1 and 2 make a 90 degree angle right

So do 3 and 4

[hearty osprey]

Yeah

[austere remnant]

Read step 4 and try to explain it

Why is angles 1+2 = angles 3+4

[hearty osprey]

Both of them when added together make a right angle, so even if the angles were different, they’d still be congruent

[austere remnant]

Yes so

Angles 1 and 2 two have perpendicular line separating them making them right angles

Same thing for angles three and 4

Right

So if all 4 angles are 90 degrees or right angles which reason makes the most sense?

[hearty osprey]

I figured it would it would be right angle congruence theorem pretty much right off the bat, but I’m not too sure about the rest

[austere remnant]

Oh I c

You’re right on everything you’ve written down, at least so I think

[calm coral]

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[golden crater]

hi

[golden crater]

I need help guys with domain

[cloud grove]

post question

!status

[calm coral]

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

[golden crater]

question b

[cloud grove]

ok do you know what p(m(x)) means?

[golden crater]

yes

it is equal to this

ill send pic

[cloud grove]

ok

send it

[cloud grove]

ok

simplify it

[golden crater]

Ok

[remote mural]

help mee

[golden crater]

Thanks bro

U=the man

[cloud grove]

u cant do that

sqrt(x^2-25) != x-5

sqrt(x^2-10x+25) = x-5

[golden crater]

What should I do then

Wait I dont understand the letters bro

[cloud grove]

well whats another thing you can do to remove the square root

sqrt = square root

[golden crater]

not sure bro

[cloud grove]

!= is not equals

[golden crater]

what can I do to remove the sqaure root

[cloud grove]

ok find its easier if you know

how to find ha and va

[golden crater]

ha? va?

[cloud grove]

horizontal and verical asymptotes

vertical*

[golden crater]

yeah but we havent got there yet

that happens later in the course

near the end

[cloud grove]

ok basically if numerator has a lesser degree than denom the horizontal is 0

in this case thats true right?

[golden crater]

wait please dont teach me that yet

my foundations are not built yet

if u build a tower on a weak foundation it will collapse

[cloud grove]

tbh thats like how id do it

[golden crater]

oh ok

well thanks for the help anyway bro

[cloud grove]

npnp

[golden crater]

I appreciate all u guys helping

im gonna close it now

.close

[calm coral]

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[cloud grove]

gl in ur endavors

[golden crater]

.close

[cloud grove]

its closed

[golden crater]

oh ok

u too bro

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

[prisma mango]

.close

[calm coral]

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[calm coral]

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[summer crow]

Hi, I used the formula I learned in class

(1/m)(v1*m1+...+vk*mk)

and ended up with the following, but it seems to be incorrect. I simplified the z coordinate previously but it was still marked wrong. Wondering where I went wrong.

[calm coral]

@summer crow Has your question been resolved?

[calm coral]

@summer crow Has your question been resolved?

[calm coral]

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[floral marten]

Simplifying expressions

[floral marten]

i know that i’m suppose to get the common denominator but don’t know how to work it out

[calm coral]

@floral marten Has your question been resolved?

[floral marten]

<@&286206848099549185> pls

[calm coral]

@floral marten Has your question been resolved?

[willow gyro]

Is $abc=acb$

[potent lotus]

Kai The Doge

[willow gyro]

You can just multiply the square roots together ignoring the 2, then multiply the 2 in

[calm coral]

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[floral marten]

.reopen

[calm coral]

✅

[floral marten]

oh

so if i multiply the square roots

does it cancel out

[willow gyro]

What

It removes the square roots

[floral marten]

ohh

[willow gyro]

You get x³+1

[floral marten]

so it’ll be 2(x^3+1)

[willow gyro]

Yup

[floral marten]

ok

so when you multiply square roots

it’ll take it out

as a rule

[willow gyro]

Yup

[floral marten]

thank you

[willow gyro]

Assuming the base is equal

[floral marten]

ok so just when it’s equal

[willow gyro]

So if u have sqrt(x)*sqrt(x) it becomes x

But if it's sqrt(x)*sqrt(y) it becomes sqrt(xy)

[floral marten]

okok

thank you

[willow gyro]

👍

[floral marten]

one more question

this is from the answer key

how did they do the denominator

[willow gyro]

(x³+1)=(x³+1)¹

[floral marten]

so 2sqrt x^3+1

is equal to (x^3+1)^2

then i add the powers to make (x^3+1)^3/2

?

wait

oh i’m suppose ignore the 2 first

so sqrt(x^3+1)*(x^3+1) is equal to (x^3+1)^3/2

[calm coral]

@floral marten Has your question been resolved?

[calm coral]

@floral marten Has your question been resolved?

[calm coral]

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[fervent sky]

y=2x^2; {non-positive reals}, would this mean the domain is non positive?

[fervent sky]

.close

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[rapid ivy]

can i get some more help with my multi step equation homework

[clear delta]

!show

[calm coral]

Show your work, and if possible, explain where you are stuck.

[rapid ivy]

i dont have work i dont know how to do the first one

[clear delta]

okay, solving for x is like unwrapping an onion

[rapid ivy]

ok

[clear delta]

x is at the core

what's surrounding it?

$5(x-1) = 23$

[potent lotus]

hayley!

[rapid ivy]

the parentheses

[clear delta]

yep, and outside of that, there's a 5

so 5 is the outer layer here

[rapid ivy]

yea

[clear delta]

5 is being multiplied, so we need to do something to both sides to cancel that out

[rapid ivy]

i use distributive property

?

[clear delta]

sure, let's use that

[rapid ivy]

$5x-5=23$

[potent lotus]

sk8ull

[clear delta]

okay, and what do you want to do next to both sides?

[rapid ivy]

add 5 to both sides

[clear delta]

the outer layer here is that addition

yep

so $5x - 5 \orange{+5} = 23 \orange{+5}$

[potent lotus]

hayley!

[rapid ivy]

now heres where im stuck

with the 5x=28

how do i divide that

[clear delta]

you won't get an integer, that's for sure

but you can still divide it

if we list out the multiples of 5

5 * 1 = 5
5 * 2 = 10
...

5 * 5 = 25
5 * 6 = 30

so our answer should be somewhere between 5 and 6

actually, do you know long division?

[rapid ivy]

i forgot how to do it

[clear delta]

this is our setup

[rapid ivy]

yea

[clear delta]

5 doesn't go into 2, so we look at 28

how many times does 5 go into 28? maybe with a remainder

[rapid ivy]

5 and 3/5

[clear delta]

great so we write something like this

and then we pull that first 0 down and look at 30

[rapid ivy]

ok

[clear delta]

so how many times does 5 go into 30?

[rapid ivy]

6

[clear delta]

great so our next digit is 6

do you think we're done?

[rapid ivy]

no

[clear delta]

okay, what next?

[rapid ivy]

drag 0

[clear delta]

sure, then we get, well, 00

but 5 doesn't go into 00 at all

[rapid ivy]

yea

[clear delta]

do you see how we're kind of in the same place again?

[rapid ivy]

yea kinda

[clear delta]

if we keep going we'll just keep putting more 0s on the end

[rapid ivy]

yea

[clear delta]

and that means we're done :)
5.6 is the same thing as 5.6000000

[rapid ivy]

o so its 5.6

or 5.60

ok

[clear delta]

those are the same number, yes

[rapid ivy]

ok

Thank u

[clear delta]

you will want to practice long division

[rapid ivy]

i think i have more

ok

ill come back in here to ask

i have like 5 questions left

[clear delta]

okay, um if it's not soon then would you mind closing the channel and opening a new one when you do? we're running out of channels haha

[rapid ivy]

i have another one question

the question on the bottom

the x-2/3+9=10

@clear delta

[clear delta]

so this is how i see that equation

$\boxed{\frac{\boxed{\boxed{x}-2}}{3}} - 9 = 10$

[potent lotus]

hayley!

[clear delta]

what can you do to peel back the layers?

[rapid ivy]

i think maybe multiply 3 to -9 and itslef

[prisma mango]

Cry \s

[clear delta]

the -9 is the outermost layer, just take care of that first

[rapid ivy]

ok

so we subtract 9 on both sides?

[clear delta]

yep

[rapid ivy]

ok

[clear delta]

so now you should have

$\frac{\boxed{\boxed{x}-2}}{3} = 19$

[potent lotus]

hayley!

[clear delta]

maybe with fewer boxes :)

[rapid ivy]

?

now we multiply 3 on both sides?

[clear delta]

yep!

[rapid ivy]

how do i multiply it on the left

[clear delta]

well, the whole point of multiplying by 3 is that the /3 completely goes away

[rapid ivy]

o

i got 57 from 19x3

[clear delta]

yep

so we have

$\boxed x - 2 = 57$

[potent lotus]

hayley!

[rapid ivy]

now do i add the 2 on both

[clear delta]

yep! very simple eqn at this point :)

a lot of math is reducing complicated problems down to ones you've solved before

[rapid ivy]

oh

thank you

ima see if i can do the last one

gimme a sec

wait but the answer cant be 59

@clear delta

[clear delta]

oh whoooooops i typod

[rapid ivy]

o

[clear delta]

i wrote it down as ... - 9 = 10

[rapid ivy]

oh

[clear delta]

but it's +9

this gives you a chance to retrace my work but with the right equation ;)

the procedure is the same

[rapid ivy]

and this is the last question and i got stuck and dont think i did it right

@clear delta

sorry for pinging you so much

i just need to finish this within 20 minutes

[clear delta]

which problem is this?

oh

,calc 11 + 4

[potent lotus]

Result:

15

[clear delta]

you said 14

[rapid ivy]

im on the last one now

[rapid ivy]

oh

i got it now thank you

you can close the channel now

thank you for helping me im very grateful @clear delta

[clear delta]

oh okay! glad you got it figured out :)

btw you can close your channels by typing .close

but i'll get this one for you

.close

[calm coral]

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[old heath]

id like to know how this became?

[still marlin]

first statement is wrong

should be n^2

[velvet osprey]

^

[old heath]

thanks!

.close

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[calm coral]

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[pulsar yoke]

can anyone tell me what should I do after replacing the roots in the equations?

[calm coral]

@pulsar yoke Has your question been resolved?

[calm coral]

@pulsar yoke Has your question been resolved?

[calm coral]

@pulsar yoke Has your question been resolved?

[pulsar yoke]

.close

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[eager acorn]

Hint for a short method

[clear delta]

$\frac{z^2-1}{z^2+1} = \frac{z-\f1z}{z+\f1z}$

[potent lotus]

hayley!

[eager acorn]

It seems like a line

Which is (i)

i tan thita

Yoo got it

. close

.close

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[high drift]

Can I get some help please

[polar pebble]

Sure which problem

[high drift]

2 problem please

[polar pebble]

Problem 2?

[high drift]

Yes

[polar pebble]

Ok

Which part r u stuck on rn

[high drift]

I need help setting it up

[polar pebble]

Ok so after you rewrite the question, you're going to have to get rid of the x from one side.

In my opinion you should get rid of it from the right side in this case

To do this, divide x from both sides

[high drift]

So like that

[polar pebble]

Yes but now you need to make sure x is subtracted from the other side

It looks like u kept it the same

[high drift]

So it would be -3 1/2 then?

On the right hand side

[polar pebble]

Since the 3 is negative and the 1/2 is positive, it would be -2 1/2

[high drift]

Ok

[polar pebble]

I gtg sorrrry

[high drift]

Ok

[calm coral]

@high drift Has your question been resolved?

[calm coral]

Channel closed

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[calm coral]

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[spare swallow]

how can we be sure that B can exist?

[spare swallow]

for example let's imagine A is finite, and that $f: A \rightarrow R$

[potent lotus]

Olomej

[spare swallow]

$R = { { B } \cup S }$

$S \subset P(A) \space such that \space \abs{A} = \abs{S} + 1$

[potent lotus]

Olomej

Olomej

[mortal orbit]

The set B is a subset of A that exists thanks to the axiom schema of specification

If you let "H" be the element-wise proposition for all elements of $A$, $H(x)$:"$x\notin f(x)$", which is a valid true or false proposition, then $B = {x\in A|H(x)}$ is well-defined

[potent lotus]

rafilou2003

[spare swallow]

the set A and R have the same cardinality, so there exist a bijection, but one of the elements of R is B

[mortal orbit]

this isn't possible as you can't construct R while f is defined this way

as we saw, B is never in the image of f

You can't both define R = f(A) and R = {B}u S because as we just saw this would mean there exists x such that f(x) = B

And that's impossible by definition of B

[spare swallow]

i don't really understand

for example

you're saying that it's impossible to construct R this way?

[mortal orbit]

by definition of what you wrote, f(a4) = B

but B = {x in A | x not in f(x)}

So if a4 is in B, then a4 is not in f(a4) = B

And if a4 is not in B, then a4 is in f(a4) = B

Contradiction

[spare swallow]

yeah

like in Cantor's argument

right?

[mortal orbit]

yep

[spare swallow]

so how is that different

[mortal orbit]

?

[spare swallow]

P(A) exist and contains B

[mortal orbit]

yes

[spare swallow]

isn't the only difference between P(A) and R that R has less ellements?

[mortal orbit]

R does have less elements because it can't contain B

[spare swallow]

oke

ty

.close

[calm coral]

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[empty gale]

Hi we need to find the graphs of each of them

I know that one of them is a x^3 graph

And the other two are interconnected through a 1/x graph

But don’t know how to find it

[supple needle]

This looks like a rational function if you've worked with them much

It'll be one function, not piecewise

[empty gale]

I know the denominator but am getting confused in the numerator

so x(x-4) is the denominator

But idk how to get numerator in 1 function

[supple needle]

Play around with functions of x in the numerator and see if you can get something that looks similar

[empty gale]

We have 1 attempt once we put in smth we can’t delete it

OH I GOT IT

FIRST TRY

.close

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[shrewd dawn]

need help\ on 15 B

[shrewd dawn]

idn what just sent

[calm coral]

@shrewd dawn Has your question been resolved?

[shrewd dawn]

,help

[potent lotus]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[calm coral]

@shrewd dawn Has your question been resolved?

[shrewd dawn]

.closee

.close

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[cloud grove]

how does it not span r^3

original matrix is shown

<@&286206848099549185>

Thought pivot points = span not sure though please help 😄

<@&286206848099549185> if anyone can help me with lin alg 😄

still dont get it?

can anyone help

<@&286206848099549185>

damn im stuck in the abyss

<@&286206848099549185>

[calm coral]

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[sand trench]

im a bit confused about how the blue line is plotted

why does it extend to the left? not to the right? how is it decided which side i should draw the line

well ig, ray, not line

[sand trench]

from the point (2, 2root3). Its the line for P3 in the question

.close

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[wooden wharf]

I need help with a DERIVATE APLICATION PROBLEM with (IMPLICIT DERIVATES)

[wooden wharf]

this is what i tried to do (I think its wrong)
TRASLATION: An airplane flies at an altitude of 5 miles and at a speed of 600 miles per hour toward a point located exactly vertically to an observer (see figure). How quickly is the angle of elevation of THETA changing when the angle is 30, 60, and 75 degrees?

[prisma mango]

Please do not close and open new channels if you have not gotten help, just be patient and wait

[wooden wharf]

😦

[wooden wharf]

help me pls

[calm coral]

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[calm coral]

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[noble plover]

Hey can anyone help me w geometry

[noble plover]

I need to say how big an angle is without a protractor

[wanton wave]

do you have a specific question/problem?

[noble plover]

A yeah

I

Wait nvm

.close

[calm coral]

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[prisma mango]

Why do you keep opening and closing channels? Like #help-26

You literally said hi then closed @noble plover

[calm coral]

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[safe niche]

Hello, I was able to find this conditional probability but am pretty stuck on the mean question.

A p.m.f is a dicreet distribution so would that mean it has a mean? or is my logic off.. pretty stuck here

[safe niche]

I get a divergent series when I calculate the sigma x*f(x)

or maby I am just ovethining it

wow yeah its a cauchy distro lol

all good

[calm coral]

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[latent zinc]

hi

[latent zinc]

really I do not understand this one

given to complex numbers w1 and w2, calculate the length of below:

i got sqrt(185)

but its wrong

[pure kayak]

Could you show your working?

[mortal orbit]

Remember, the module of a quotient is the QUOTIENT of the modules, not the product...

[calm coral]

@latent zinc Has your question been resolved?

[latent zinc]

got it

.close

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[torpid canopy]

How do I prove this without sounding like, "Oh yeah, $AI_n=A$ because when you multiply $A$ by $I_n$, you get $A$ again?"

[potent lotus]

Ari

[analog heart]

Do the explicit multiplication then

[torpid canopy]

what would that look like exactly?

[analog heart]

Consider a_ij where = 1,...,m and j= 1,...,n to be the elements of A

Then do the multiplication AI_n and I_mA

you'll get A in both cases

[torpid canopy]

K thanks

.close

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[noble cedar]

This question asks me to find the area using basic geometry

[fallen dragon]

Interpret the areas geometrically

[noble cedar]

How am I supposed to know the height of the right triangle?

[fallen dragon]

You know the endpoint

So you know the y-value at that endpoint

[noble cedar]

Plug in 9 for x to find y?

36-1 = 35

Well I’ll be damned

It worked

Now I just gotta divide by 2 and subtract 2

17.5 - 2

15.5 + 4

Area should be 19.5

If I did that correctly

Noope

Oh I forgot to multiply 35 * 9

[noble cedar]

Getting closer but Not sure where I went wrong

Answer should be 153.25

(8.75*35) / 2

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[calm coral]

@steady nymph Has your question been resolved?

[steady nymph]

.close

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[clear cypress]

im unsure as to how to express this

[mortal orbit]

What does it mean for x to be in the interval (-1,3]?

[clear cypress]

?

that x exists within that plane

[mortal orbit]

In terms of inequalities

[clear cypress]

im not sure

[mortal orbit]

Maybe something that looks like ... < x < ...

[clear cypress]

oh

-1<or = x < 3

[mortal orbit]

Ah, wrong way

parentheses () are for exclusion, [] are for inclusion

[clear cypress]

ohh okay

[mortal orbit]

So in (-1,3], -1 is excluded

[clear cypress]

right

[mortal orbit]

While 3 is included

[clear cypress]

what is this

[mortal orbit]

It means intersection

So "only take elements that are in BOTH"

[clear cypress]

would that be that points between the two though

so like from 3, 6

and -1, 0

[mortal orbit]

5 doesn't belong to BOTH of them for example, so "from 3 to 6" is not true

(5 doesn't belong to (-1,3] here)

[clear cypress]

okay

but then what would i write

because 3,6 is all above -1,0

[mortal orbit]

Can you first try to write what it means for x to be in (-1,3] again in terms of inequalities?

[clear cypress]

-1<x<or=3

[mortal orbit]

Correct

[clear cypress]

thats my final answer?

[mortal orbit]

No not yet

Let's do the same for [0,6)

[clear cypress]

0<or=x<6

now what

[mortal orbit]

Yes, where getting closer

So now, if x is in both sets...

[mortal orbit]

It means that this

[mortal orbit]

AND this

Are both true

So $-1<x\leq 3$ AND $0\leq x < 6$

[potent lotus]

rafilou2003

[mortal orbit]

Let's try to simplify that

[clear cypress]

how would i do that

[mortal orbit]

Let's take all the "left" inequalities together

$-1<x$ AND $0\leq x$

[potent lotus]

rafilou2003

[mortal orbit]

This simplifies into?

[clear cypress]

-1>x

?

[mortal orbit]

First of, the inequality is on the wrong side

[clear cypress]

what does it simplify into

[mortal orbit]

like you used > instead of "<" or "< or ="

[clear cypress]

right

i understand

but im confused on what it simplifies into

[mortal orbit]

One of those statements implies the other, can you see which one?

[clear cypress]

yes -1<x implies that 0< or equal to x

oH

wait

[mortal orbit]

Uh the other way around

If x is non-negative, then x is bigger than -1

[clear cypress]

right

so then 0<=x

[mortal orbit]

Exactly, this is the simplification

[clear cypress]

then the right side

x<=3

[mortal orbit]

Exactly let's do that

[mortal orbit]

Yep correct

[clear cypress]

so final answer is 0<=x<=3

[mortal orbit]

And in interval notation, this is...?

[clear cypress]

so in interval its [0,3]

[mortal orbit]

Correct!

[clear cypress]

thank u so much

can u help me with one more question?

i just want some confirmation

im doing homework

would this be correct?

ive plotted it on desmos to visualize it

[mortal orbit]

Yep, correct

[clear cypress]

okay awesome thansk

can i message you if i need some help later?

[mortal orbit]

Sorry, I usually keep help confined in this discord

It is always more useful, as I won't always be around so others can take my place when you need help and I'm not there

Don't worry, there are always a lot of very competent people online ready to help!

[clear cypress]

okay thanks

this is still correct right?

@mortal orbit

[calm coral]

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[astral mural]

For this question I used the equation v = v_o + at

And got the answer a = 30.49

Did I use the correct equation?

Anyone here?

[calm coral]

@astral mural Has your question been resolved?

[astral mural]

.close

[calm coral]

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[remote mural]

How do I do this

[trim rose]

calculate the y for 2 and 6

subtract g(6) from g(2) and divide that by (6 - 2)

[remote mural]

I don’t understand

-6?

[remote mural]

What is g

Do I plug 2 and 6 in the place of x for equation?

[calm coral]

@remote mural Has your question been resolved?

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[cloud grove]

what does this mean

idk where to start a-5i3 and 5i3a

idk what i is

[civic dirge]

I is the identity matrix

[cloud grove]

what does that mean

[civic dirge]

They're using I3 to mean the 3×3 identity matrix

[cloud grove]

like i is a?

or

sorry if im not understanding I am new to the class

so its hard for me

yo

hello

[prisma mango]

No

The identity matrix is a square matrix where the diagonals are 1

$I_3$ means it's a 3x3 matrix with 1s along the diagonal

[potent lotus]

dldh06

[cloud grove]

like r1 = 1 0 1 r2 = 0 1 0 r3 = 1 0 1?

or

[prisma mango]

That is not the diagonal

[cloud grove]

theres 2 diagonals right?

or

[prisma mango]

The main diagonal people refer to is top left to bottom right

Like that

[calm coral]

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[outer dawn]

Is there any trick for me find the trial and error answer in first try

[polar jetty]

Rational roots theorem.

[outer dawn]

what is that

How long does it take for me to find the trial and error answer

[dull wagon]

depends

[outer dawn]

Ok but how do i use it

[dull wagon]

usually they give you something nice, so not too long

[outer dawn]

how does it work

[sour tide]

Try going for 123 and negative 123

[dull wagon]

do you know what the theorem states?

[outer dawn]

but i got 0.5

and i can't spend too much time on that question

0.5 is the trial and error answer

i can spend approximately 5 minutes on that question from what my teacher said

[sour tide]

which grade are you in

[alpine stone]

If some ratio p/q in its lowest terms is a root of a polynomial, then p should divide the free constant and q should divide the leading term

[outer dawn]

Guess i will say higher or lower

[outer dawn]

higher 10

I am in the start of year 10

[sour tide]

well

[outer dawn]

can u give an example

[sour tide]

then theorem is not needed

[dull wagon]

you can pretty do -1, 1 in your head. based on coefficients you can disregard those immediately without having to explicitly write the work for those
and start with stuff like -3,3,-5,5

[outer dawn]

from my question

[sour tide]

they give nicest options all the time

[outer dawn]

but i can't use calculator therefore i can't spend the a lot of time on the same question

[outer dawn]

its true but the answer i got is 0.5 and can i just get a direct way i don't mind even if its a formula

[dull wagon]

multiplying small numbers shouldnt take up that much time

[alpine stone]

x^2 + 3x - 2, to find the rational roots for this polynomial, you could go through 1, -1, 2, -2

[outer dawn]

but it is a 0.5

[dull wagon]

thinking of ways to try and avoid that will

[outer dawn]

The trial and error answer is 0.5

[dull wagon]

,W factor 2x^3 +3x^2 -32x +15

there are 3 rational roots here

based on what's nice to test, you'll find one of the others before 1/2

[outer dawn]

ok

[dull wagon]

once you have one, you can do long division (unless you're testing with synthetic division which hits two birds)

[outer dawn]

yeah that's what i do but the trial and error answer just gives me a hard time. Well it is easy but time consuming

[dull wagon]

and then you'll have a quadratic factor that's much easier to factor

practice more

[outer dawn]

that's what i do but it is just the trial and error answer which is also the first part

Okay thanks man. Bye

.close

[calm coral]

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[eager acorn]

suppose there a curve if I draw any slope near the curve not touching/crossing it
will I say this is slope of the curve?

[polar jetty]

What do you mean by "draw slope"?

If you mean that you draw a line which doesn't touch or cut the curve, then also - a curve might not have a single slope so - that would only be true for straight lines.

[calm coral]

@eager acorn Has your question been resolved?

[eager acorn]

.close

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[remote mural]

So

I'm supposed to determine if this converges or not

but the explanation just says

it can't be determined by direct comparison test

oh i see

they did limit comparison

and it returned a finite number

so if one converges the other does as well

okay

.close

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[remote mural]

Did i do this right?

[remote mural]

The terms that matter the most

$$\frac{k^4}{k^5} = \frac{1}{k}$$ Diverges

[potent lotus]

What should I do

[velvet osprey]

yes

$\sum \frac{1}{k}$ diverges

[potent lotus]

Ann

[remote mural]

$$\frac{1}{k} < \frac{k^4 + 2k^3 +2}{k^5+k^2+k}$$

[potent lotus]

What should I do

[remote mural]

so both diverge?

[velvet osprey]

yours has k^4 - 2k^3 + 2 on the numerator tho.

and im not sure whether that inequality actually holds the way we need it to

[remote mural]

oh

I got it

.close

[calm coral]

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[remote mural]

I need to stop using comparison test

and use

limit comparision

comparison test too much

[calm coral]

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[remote mural]

so

$$\frac{1}{2} \sum_{n=1}^\infty \frac{n^3-4}{n^5-3n}$$

[potent lotus]

What should I do

[remote mural]

$$ \frac{n^3-4}{n^5-3n} \approx \frac{1}{n^2}$$

[potent lotus]

What should I do

[hushed sequoia]

intuition says it converges

[hushed sequoia]

yeah

[remote mural]

converges

so

$$ \frac{n^3-4}{n^5-3n} < \frac{1}{n^2}$$