help-10

[slim cove]

great! now there's one thing you need to be careful about, did you check to make sure r is oriented correctly around the circle?

[prisma glacier]

No

[slim cove]

Okay keep in mind the hemisphere is oriented towards the positive y-axis

So that induces a certain orientation on its boundary circle

You can figure out this orientation using the right-hand rule

Do you know how?

[prisma glacier]

Yes

[slim cove]

okay cool

so is your current orientation for r good or no?

[prisma glacier]

No

[slim cove]

Yeah, so how should you fix it?

[prisma glacier]

Switch the orientation

[slim cove]

yeah, so what's your new equation for r?

[prisma glacier]

Is it any of these

[slim cove]

wait, where am I supposed to be looking

isn't that the same equation for r you already had

[slim cove]

yeah that equation for r works fine, the picture you drew is the wrong orientation though

the arrow should be going clockwise

[slim cove]

yeah

but it looks like you got the right answer do you have any other questions

[prisma glacier]

If it’s possible to do so, use the divergence theorem to find the result

[obtuse pebble]

@prisma glacier Has your question been resolved?

[slim cove]

The surface is not closed

However if you wanted, you could use the divergence theorem to conclude that

integral over the curved surface + integral over the flat circular base = volume integral of 0 = 0,

with everything being outwardly oriented

[obtuse pebble]

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[prisma glacier]

Is this right

[prisma glacier]

I’m in lecture for class

Hopefully I got what my professor was trying to say for this specific example

<@&286206848099549185>

[obtuse pebble]

@prisma glacier Has your question been resolved?

[timid silo]

wow i learned that like 3 weeks ago but never used it cuz i had other ways

[prisma glacier]

Really

???

[obtuse pebble]

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[obtuse pebble]

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[hidden sand]

Did I solve for the right thing I haven’t done this since last year

[fiery flower]

yes

you did it right

[obtuse pebble]

@hidden sand Has your question been resolved?

[obtuse pebble]

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[halcyon hull]

snap

[halcyon hull]

i thought i had things under control, but i need help with #2

[timid silo]

Which one?

Aye

You know loagrithm?

Logarithm

[halcyon hull]

yeah. this is a log worksheet

so i assume we gotta use logs for this

[timid silo]

Oops

Yeah

So

You want 1-x to not be the exponent right?

[halcyon hull]

that property?

1-xlog3 = log2?

i have a bad feeling about that

[timid silo]

Yeah that but different way

So you can take log both sides right?

[halcyon hull]

yeah

but wait

[timid silo]

What would you keep as the base?

[halcyon hull]

how do i remove the -1 part of the exp?

[timid silo]

When you get that in multiplication with log, you can easily do basic arithmetics

Like I'm sure you can find x when I give you (1-x)(7) = 8 or something like that right

[halcyon hull]

yeah

[timid silo]

Yeah

So coming back

You would take log both sides

With what base?

[halcyon hull]

3?

[timid silo]

Remeber , you need to bring down the exponent , and also if possible , use some property of log such that the 3 isn't there anymore

[timid silo]

Yes good

Write it down , tell me what you'll get

[halcyon hull]

x-1(1)

x-1 = 2

[timid silo]

Yeah and on the RHS?

[timid silo]

No

[halcyon hull]

ohh wait

[timid silo]

You take log both sides with base 3

[halcyon hull]

lkog on bith sies

[timid silo]

Yeah

Yeah

[halcyon hull]

x-1 = log 2

x = log 2 + 1

[timid silo]

1-x

You don't change the sign, when you bring the exponent down it'll stay as it is

1-x = log 2 base 3

[timid silo]

And log 2 wouldn't be the correct notation. As log2 would be assuming your base to be 10

[halcyon hull]

true

i can calculate log 2 base 3

with change of base property

[timid silo]

Yeah

Log 2 divided by log 3

And we know the values of log2 and log3

Good

So you'll get x = 1 - (whatever that division is)

Got it?

[halcyon hull]

1-x = 0.63
1- 0.63 = x

[timid silo]

Yep

[halcyon hull]

got it

[timid silo]

Good

[halcyon hull]

appreciated

[halcyon hull]

could i do the same thing for #3? because i used a more complex way and got the answer. but if this is a better way, ill do this way

[timid silo]

Yes you can

Ummm Lemme just try it

[halcyon hull]

same

[timid silo]

Yeah

[halcyon hull]

wait how

[timid silo]

It'll be x = log(5 4) / [ 1 - log (5 4) ]

It will be a long calculation unless you have a calculator

Okay so

You there?

[halcyon hull]

yeah

im guessing i use log base 4 on both sides?

[timid silo]

Taking log with which base would be easier?

[timid silo]

Oh I did that but hear me out

[halcyon hull]

xlog4(4) = x+1log4(5)

[timid silo]

You can ofc do that

But from my experience, here as you can see

Exponent of 5 is harder than exponent of 4, to solve

So take base 5

[halcyon hull]

ok

[timid silo]

You'll get the same answer with base 4, but easier with base 5

[halcyon hull]

the way how i did it at first was to remove the '+1' by making the eq 4^x = 5^x+5

since '1' exp of 5 is 5

then did a whole bunch of exp properties

[timid silo]

Yeah got you

[halcyon hull]

thats a longer method though. i like this one more

[timid silo]

That must have been a very long way

[halcyon hull]

yeah it was

[timid silo]

Ye

[halcyon hull]

anyways, thanks for the help. ill try the other questions

[timid silo]

So you take log both sides with base 5 , what you getting?

[timid silo]

Oh okay np

Best of luck

[halcyon hull]

🫡

[timid silo]

The last one is good

[halcyon hull]

i saw this a lil too late

[timid silo]

No worries man

[halcyon hull]

but just to clarify, is x - 2(0.63) = x - 1.26?

[timid silo]

And the second last is also of the same type

[timid silo]

Yeah

[halcyon hull]

alright

im actually doing #5 right now

[timid silo]

Good

[halcyon hull]

i got it incorrect for some reason

lemme type working

3^2x+1 = 2^x-2
2x+1log3(3) = x-2log3(2)
2x + 1 = x - 2(0.63)
2x + 2.26 = x
2x-x = -2.26
x = -2.26

but this ones incorrect

[timid silo]

Okay wait llemmee

[halcyon hull]

ill try doing log 2 on the other side instead in the meantime

[timid silo]

Yeah I got this

It's uploading

[halcyon hull]

analyzing

wait

[timid silo]

Waiting

[halcyon hull]

the numerator on the last step is -1-2log2(3) right?

just to confirm

[timid silo]

Yes

[halcyon hull]

ok

-4.18/0.41?

since 2log2(3) is 3.14

[timid silo]

Ohhh

Wait lkemme

[halcyon hull]

this is the hardest log worksheet ive ever done tbh

but its good for learning

since im training for calc

[timid silo]

No it shows 1.26 in my calculator

[halcyon hull]

i see

lemme retype it then

in the calculator

thats strange

i do log2(3) first and get 1.59(to 2 decimal places)

then multiply 2 on it

and since its -1 - log2(3), i do -1- 3.18

then i get -4.18

as for the bottom, i get 0.41

[halcyon hull]

whatd you get for the denominator?

[timid silo]

Lemme just get the final answer wait

[halcyon hull]

ok

[timid silo]

Numerator = -2.26

Denominator = 1.37

Answer = -1.64

[halcyon hull]

thats basically the answer on the worksheet

[timid silo]

Log 2 base 3 is 0.63

[halcyon hull]

i kept doing log 3 base 2

without knowing

[timid silo]

No worries

Happens

[halcyon hull]

i think thatll be it

ill try to solve the rest

[timid silo]

Ye

Yep

Good questions

Tbh

Like not the first 6

Just the last two

[halcyon hull]

very

i thought i was decent at logs until i saw this worksheet

its hard but im definitely learning alot

[timid silo]

Lmao 😂

But there sure are harder worksheets tbh

[halcyon hull]

it has basically everything on it. log graphing, basic logs, compound interest, expanding, exponentials, etc

[timid silo]

This is basics

[halcyon hull]

💀

[timid silo]

Umhmm

This would be the starting 2-3 day level worksheet you would be getting

Like except those last two questions

[halcyon hull]

im glad im making sure to get good at logs during the break then

[timid silo]

Ye

[halcyon hull]

then eventually there will be log derivatives

well, im gonna hop off now

have a good one

[timid silo]

Yeah

Night

@halcyon hull close the channel if you're done

[halcyon hull]

.close

[obtuse pebble]

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[obtuse pebble]

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[cobalt pendant]

I don’t understand how to find its porportion

[obtuse pebble]

@cobalt pendant Has your question been resolved?

[spare rain]

Areas have to be divisible by 3 as the other two sides (a and b) are integers

[delicate sentinel]

You simply have to find the set of numbers with a value that is not divisible by 3.

[cobalt pendant]

Oh ok

Thanks guys I got it

[obtuse pebble]

@cobalt pendant Has your question been resolved?

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[rotund vector]

How would I solve for x?

[rotund vector]

$$-\frac{2x}{\sqrt{36-x^2}}=-\frac{2\sqrt{36-x^2}}{8-x}$$

[warm shale]

Lex1729

[rotund vector]

I know I can cross multiply but do the sqrt(36-x^2) cancel out when I do so?

[nocturne minnow]

They won't but recall that $\sqrt{x} \cdot \sqrt{x} = x$

[warm shale]

dldh06

[rotund vector]

$$-\frac{1}{\sqrt{36-x^2}}\cdot \sqrt{36-x^2}$$

[warm shale]

Lex1729

[candid yarrow]

no, that’s not cross multiplying

[nocturne minnow]

That's not cross multiplying

This is

[rotund vector]

Oh so it's like this

$\sqrt{36-x^2}\cdot -2\sqrt{36-x^2}$

[warm shale]

Lex1729

[nocturne minnow]

Yes

[rotund vector]

Oh ok my bad

[obtuse pebble]

@rotund vector Has your question been resolved?

[obtuse pebble]

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[sleek stag]

Help with vector calculus

[coarse ibex]

,rotate

[sleek stag]

How to solve this

[slim cove]

Do you know how to parameterize the line y=x?

[sleek stag]

No

[slim cove]

Okay so the first step is to do that, do you know what parameterization means?

[sleek stag]

No

[slim cove]

Basically you want to imagine that you're a bug and you're walking along the line y=x from (0,0) to (1,1)

So at time t=0 you're at (0,0)

And then at time t=1 you're at (1,1)

Does that make sense so far?

[sleek stag]

Yes

[slim cove]

So for every value of t between 0 and 1, you have a point r(t) which describes your current position

[sleek stag]

Okay

[slim cove]

for example, what would your position be at t=0.5?

[sleek stag]

Between 0 to 0.5

[slim cove]

but what would your coordinates be

like your x position and y position

at the time t=0.5

[sleek stag]

I am not understanding

[slim cove]

Did you understand the part about the bug walking from (0,0) to (1,1)?

Along the line y=x

[sleek stag]

Yeah

[slim cove]

Okay so at time t=0 it's at (0,0), and at time t=1 it's at (1,1), so now I'm asking where is it at if t=0.5? right in the middle of t=0 and t=1?

Assume it just walks straight from (0,0) to (1,1) at a constant speed

[sleek stag]

okay

[slim cove]

So where is it at when t=0.5?

What coordinates?

[sleek stag]

0.5 ,0.5

[slim cove]

yes!

exactly

[sleek stag]

ohhh okay

[slim cove]

so in general, at any time t, the bug is at the coordinates (t, t)

This is called a parameterization: we put the coordinates x and y in terms of another parameter t

[sleek stag]

okay

[slim cove]

So we can write r(t) = (t, t) for 0 ≤ t ≤ 1, and that's a parameterization of the line segment from (0, 0) to (1, 1)

And this will help us evaluate the line integral

Because [\int_C \vec F\cdot \dd \vec r = \int_0^1 \vec F\cdot \frac{\dd \vec r}{\dd t},\dd t]

[warm shale]

Molothrusater

[sleek stag]

okay

[slim cove]

So what is dr/dt equal to?

[sleek stag]

I dont know?

[slim cove]

Well we just said r(t) is (t, t)

What is the derivative of that?

[sleek stag]

is it 0.5,0,5?

[slim cove]

Not quite

What's the derivative of t with respect to t?

[sleek stag]

I am not understanding what you are saying

[slim cove]

I'm saying to calculate the derivative of t

[\frac{\dd}{\dd t}[t]]

[warm shale]

Molothrusater

[sleek stag]

X^2i+xyj?

[slim cove]

Maybe it would help if I asked you instead: what's the derivative of x with respect to x?

[sleek stag]

Okay
2x+y

[slim cove]

How did you get that?

[sleek stag]

I differentiate it with respect to x

[slim cove]

Like imagine you're in your first calculus class, and I just ask you: what's the derivative of x?

What would you say

[sleek stag]

1

[slim cove]

yup, because you're taking the derivative of x with respect to itself

So if I change the letter name to t

And ask you what the derivative of t with respect to t is

Changing the letter name doesn't actually change the math behind it right?

It should still be 1

Does that make sense?

[sleek stag]

Yeah

[slim cove]

Okay cool

So if r(t) = (t, t)

Then what is the derivative of r with respect to t?

Well it's just

The derivative of the first part

1

and then the derivative of the second part

1

so dr/dt = (1, 1)

Does that make sense?

[sleek stag]

Yea

[slim cove]

okay cool

[slim cove]

So now do you think you can plug everything into this equation here

and solve?

[sleek stag]

I don't know

[slim cove]

Try it

[sleek stag]

Not getting it

[slim cove]

Show me what you've tried

[fierce lagoon]

Are you confused on the $F \cdot \dv{r}{t}$ part?

[warm shale]

Umbraleviathan

[sleek stag]

Yes

[fierce lagoon]

Here I'll give you hopefully a more detailed definition (at least, for 2D):

Let $C$ be a curve defined parametrically by $\vecb{x(t), y(t)}, t \in [a, b]$. Then we have a vector field $F(x,y)$ (note this is the same as $F$). Then:

$$\int_C F \cdot \dd r = $\int_a^b F(x(t), y(t)) \cdot \vecb{x'(t), y'(t)} \dd t$$

[warm shale]

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

[slim cove]

Don't post in the help channel if you're not helping.

[fierce lagoon]

@sleek stag hopefully this helps a bit more

It's not as compact as the other definition

In your case, $F(x,y) = \vecb{x, xy}$

[warm shale]

Umbraleviathan

[gilded needle]

why

[coarse ibex]

<@&268886789983436800>

[obtuse pebble]

@sleek stag Has your question been resolved?

[tardy epoch]

<@&268886789983436800>

[sleek stag]

How to solve the problem

[obtuse pebble]

@sleek stag Has your question been resolved?

[obtuse pebble]

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[fathom flicker]

@wooden cipher

[obtuse pebble]

Please don't occupy multiple help channels.

[fathom flicker]

oops

.lcose

.close

[obtuse pebble]

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Use .reopen if this was a mistake.

[wooden cipher]

stop pinging me

[urban patrol]

@fathom flicker

[fathom flicker]

@zenith raft

[compact shadow]

What is the question asked in this channel

[obtuse pebble]

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[south whale]

A knife has a blade that is 33.7 mm wide and 2.40 mm thick. If it takes 1.25 lb of force to cut into a ribeye steak and you apply 0.239 lb of force on the knife, what is the efficiency of the blade? Answer in percent, and round your answer to 2 decimal places.

[obtuse pebble]

Please don't occupy multiple help channels.

[south whale]

hello

[tired sierra]

don't open more than 1 channel my guy

you have 3 (as of now)

[south whale]

how do i get the other one closed??

sorry im new man

[tired sierra]

.close

[south whale]

i did but it still shows /:

[tired sierra]

give it a second, its like a temporary buffer

[south whale]

okay sounds good

[tired sierra]

just in case you decide to reopen it

[south whale]

.close

[obtuse pebble]

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[sleek stag]

help

[sleek stag]

with vector

[candid yarrow]

[sleek stag]

this

[candid yarrow]

can you show your work

[sleek stag]

I am not able to slove

solve

[candid yarrow]

what have you tried?

[sleek stag]

yeah

[pulsar quarry]

he asked

what have you tried

wdym by yeah

[sleek stag]

yeah I tried

[trim portal]

what have you tried

[pulsar quarry]

emilyisalwaysright

[obtuse pebble]

@sleek stag Has your question been resolved?

[sleek stag]

integration

I got the answer

[obtuse pebble]

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[astral aurora]

20

[astral aurora]

I didn't understand the question number 20 solution

[royal basin]

do you know what a random variable is?

[astral aurora]

Nunber of possibilities

[royal basin]

do you know what a probability density function is?

[astral aurora]

X=0,1

....

We make this type of possibilities for a function

Binomial distribution

[royal basin]

sounds like you have a lot to learn. it's not that you don't understand the solution, you don't understand the question!

you are missing basically this entire topic.

[astral aurora]

I see

Can you help me so that i can learn it? If i see beyond statistics it is a function

[royal basin]

i don't have the energy to do that

[astral aurora]

.close

[obtuse pebble]

Channel closed

Closed by @astral aurora

Use .reopen if this was a mistake.

[astral aurora]

.reopen

[obtuse pebble]

✅

[astral aurora]

I understand the solution of integration and limits

.close

[obtuse pebble]

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[timid silo]

not rlly sure how to answer these questions

[heavy shore]

first one can be solved by either drawing graphs or by solving the trigonometric equations. second image is not clear

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[heavy shore]

ah okay

[obtuse pebble]

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[obtuse pebble]

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[winter pewter]

Last 2 digits of 11^11^11^11

[winter pewter]

My answer is 11

[novel grotto]

show your work

[winter pewter]

11 mod 100 = 11

11^(11^(11^11)) = 11^(11^11) = 11^11 = 11

[novel grotto]

you meant 11^11 mod 100 \cong 11 right

yeah anyway it is correct

you can close the channel

[compact shadow]

Which is wrong

11^4=121^2=21^2=441=41 mod 100

So it’s 41

[novel grotto]

?

[novel grotto]

its exponent not multiplication

[compact shadow]

Oh my bad. Never mind sorry

[winter pewter]

Thanks

.close

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[static furnace]

sir?

[trim portal]

Do you have a question

[royal basin]

don't spam.

[obtuse musk]

.close

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[edgy needle]

can someone help me draw dis diagram

[quaint acorn]

yh sure

[edgy needle]

Oke

Hey u still there?

[edgy needle]

??

[quaint acorn]

Sorry Hi

[edgy needle]

Hi

[quaint acorn]

Yeah...

[edgy needle]

I drew it like this but its wrong

[quaint acorn]

good for u

[edgy needle]

what..

[quaint acorn]

just shoot ur teacher an emial, Hey Ms/Sir blah blah blah, Im sure they can help u out

[edgy needle]

Lol its 9 pm rn haha

[quaint acorn]

Ikr teachers are lazy

[edgy needle]

Do you know how to do it?

[quaint acorn]

yeah but I can't upload it

my phone is dead

[edgy needle]

Rip

Can u do it on like paint

[quaint acorn]

how

[edgy needle]

So

U go on ur pc and search paint

And then u draw it :D

[quaint acorn]

voice call?

[edgy needle]

Um... its my bed time im supposed to be asleep lol

Its ok ill figure it out myself

[quaint acorn]

Alr all good, it just hard to explain by typing

I beileve in u

Actaully go to eddie woo drawing bearing diagrams, he'll help u out

[pulsar quarry]

it this correct?

@edgy needle

[obtuse pebble]

@edgy needle Has your question been resolved?

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[obtuse pebble]

@fossil spoke Has your question been resolved?

[fossil spoke]

<@&286206848099549185>

.close

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[violet cosmos]

<@&286206848099549185>

[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.

[violet cosmos]

pls help i dont know how to start

.close

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[graceful marten]

Hello

[graceful marten]

I do not understand how to do Part B, first time doin such a worded question

I know two distinct roots is B^2-4ac > 0

I know k^2 +16 > 0

But it makes k^2 > -16

Which is a negative and u cant root a negative

So I dont know what to do

[latent walrus]

Youre doing things in the wrong order, youre trying to show that k^2+16>0, and the crux of that is that k^2>0 since k is real

[graceful marten]

Crux?

[latent walrus]

just means the main point

[graceful marten]

ok

Ive seen a lot of questions which have the same way of answering

Like the exact same words/method

Is it important to mention K^2 >_ 0

Equal than or greater than 0

[latent walrus]

in this case yes, thats how you can then say that k^2+16>0

since if you add 16 then it must be strictly greater than 0 now

[graceful marten]

Oh ok Yeah

Bc Square any number is posotive

but can be zero

So +16 makes it bigger than zero

Ok

Thank you

Anyone know how to attempt this, I though about just puttin in integers of a and c

and happy guessing

didnt work

Part A

[latent walrus]

start from what it means for there to be distinct real roots with the discriminant and try go from there

[graceful marten]

Distinct real roots I thouight is

B^2-4ac

=0

But it said >0

I dont get it

[latent walrus]

=0 is a repeated root

[graceful marten]

Oh ok

What do they mean Always possible to choose B

[latent walrus]

it means that for any choice of a and c, there can always be some value of b such that there are 2 roots

youre just trying to prove or disprove that such a b can exist for all cases

[hidden compass]

,rotate

[graceful marten]

Not sure what to do for the next two lines

But at the same time I know somehow b^2 -8>0

And that somehow makes 2 roots

Because of somehow k^2 _> 0

Should I send the markscheme

[latent walrus]

i would recommend not using specific numbers, but you could say suppose a,c>0 or a>0,c<0 or whatnot. b^2>-8 is fine, for example, b^2 could be 256>-8 then b=16 or b=-16

[graceful marten]

Ok I think I get what you are saying

So by making both a and c pos/negative it always makes
b> root 4ac

and by one pos or negative always makes 4ac<0 ----- I dont get that

Oh i got it

Alright alrighty

Thank you Azo

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[prisma flax]

the probability of recovering a disease with a medicine is 90%, if 2 people take the medicine, what is the probability that 2 will not recover?

[royal basin]

!status

[obtuse pebble]

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

[prisma flax]

4

I got 1/100

and its most likely not correct

[royal basin]

why is it most likely not correct?

[pearl bison]

why

[prisma flax]

I think so at least, is it true?

[royal basin]

well you got it as 1/10 * 1/10 yes?

[prisma flax]

indeeed

[royal basin]

yeah

that is correct

[prisma flax]

oh okay

ty

[timber island]

@prisma flax close the channel if u have no further questions

[prisma flax]

.close

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[obtuse pebble]

@opaque pier Has your question been resolved?

[hexed gull]

I'm not that familiar with uncertainty propagation, but I think you'd take the root of all uncertainties divided by their value squared

sqrt((dC1/C1)²+(dC2/C2)²+(dC3/C3)²+...)

I wish I could help but I'm really uncertain when it comes uncertainties 🐣

Just hoped I could potentially spark a thought since nobody with knowledge about it answered yet

[polar fossil]

things i'm seeing online say RMS is correct

root mean square

like lunatic described

[obtuse pebble]

@opaque pier Has your question been resolved?

[obtuse pebble]

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[warm shale]

Şêro

[polar fossil]

\begin{tikzpicture}[rotate=180]
    \begin{scope}
     \clip (-2,0) rectangle (2,1);
    \draw (0,0) circle(2 and 0.35);
    \end{scope}
    \begin{scope}
     \clip (-2,0) rectangle (2,-1);
    \draw (0,0) circle(2 and 0.35);
    \end{scope}
    \draw[densely dashed] (0,4) -- node[right,font=\footnotesize] {$h$}coordinate[pos=0.95] (aa)(0,0)
                            -- node[above,font=\footnotesize] {$r$}coordinate[pos=0.1] (bb) (2,0);
    \draw (aa) -| (bb);
    \draw (-2,0) -- (0,4) -- (2,0);
  \end{tikzpicture}

i'm assuming the containe looks like this

[warm shale]

Hailey

[polar fossil]

i think it's just open on top

like a paper cup

well, no, i think it looks like how i drew it

with the point at the bottom

like this

yeah

so the cool thing about this is that if you look at the water in that cup

it also forms a cone

with a radius and a height

for cones, the radius and the height are in proportion so like if the radius is half, the height is half

we're trying to relate volume and height right? so we want an expression for the volume of the cone (of water) in terms of the height

volume of a cone is given by $V = \f13\pi r^2h$

[warm shale]

Hailey

[polar fossil]

can you convert that into an equation for volume given just the height? for the cone we're working with

yep

[warm shale]

Şêro

Şêro

[polar fossil]

there's no t

but with respect to h

exactly this yes

[warm shale]

Şêro

[polar fossil]

yeah

oh right you do actually need time

yeah so you probably want to take that $V = \frac1{12}\pi h^3$ and differentiate both sides with respect to $t$

[warm shale]

Hailey

[polar fossil]

the question is asking for dh/dt given dV/dt right?

look at the original question

then what is the rate of change of the height of the liquid in the cone

it asks about the rate of change of the height of the liquid

which is dh/dt

yeah

(when i said not to do that earlier, i had forgotten the original problem)

no it's fine, i forget which channel i'm helping sometimes

[obtuse pebble]

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[ivory pawn]

How would i approach part c

i've done a and b, i will need my answers from those parts

i got the the coordinates for (a) as (53/17, 22/17, 75/17) and for part (b) angle as 1.100 rounded to 4s.f.

[obtuse pebble]

@ivory pawn Has your question been resolved?

[obtuse pebble]

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[novel pewter]

dont mind that I clicked exponential it was a misclick

alg 2 btw

[timid silo]

Soo do the numbers seem to have a consistent pattern?

[novel pewter]

yes

+1 for the x side

[timid silo]

and what direction do they seem to be going

[novel pewter]

I really dont know sorry im new to this

[timid silo]

acc ignore direction

[novel pewter]

yes

does that mean its linear?

sure np

[timid silo]

Okay so the x value seems linear

But the f(x) isn't

[novel pewter]

I see

[timid silo]

because it's not increasing by a constant amount, so it can't be linear

1 -> 3 -> 7

[random rose]

how to I do spoiler lol

[novel pewter]

yea

[novel pewter]

not sure

[timid silo]

||<text>||

shift + \ key (above enter)

[random rose]

||quadratice. e^x is never negative and its not linear||

[novel pewter]

alright

[timid silo]

The amount is gradually increasing

[novel pewter]

so I know it isnt linear

[timid silo]

Yep

[novel pewter]

nor exponential

so is it quadratic?

how can i prove that tho

[timid silo]

Well we proved it's not linear

Now just prove it's not exponential

[novel pewter]

so u said linear can never be negative right?

[novel pewter]

how do you do that

[timid silo]

@random rose any idea

it can't be negative u said?

[random rose]

the exponential function e^x is always positive

providing that is what is meant by exponential function

[timid silo]

Ah

[random rose]

if it means any base then not necasserily

[novel pewter]

I see

so exponential is the one that cant be negative

[random rose]

If it is referring to e^x then correct. However something like -2^x could be

[novel pewter]

ah okay

[timid silo]

I see

[novel pewter]

so

its quadratic

right?

[random rose]

Im pretty sure but hold on a sec

[novel pewter]

alr

[random rose]

Yes I would say exp

I mean I tried forming a quadratic and couldnt

[timid silo]

sorry for cutting in but couldn’t we just find the first and second differences?

[novel pewter]

hi again haudi

[random rose]

Ye lol I guess so

I was just working off properties of functions

[novel pewter]

and yea actually I was thinking ab that

[timid silo]

that’s what’s normally taught in alg 2

[timid silo]

hi

[random rose]

Im from uk so there is no alg 2 here lol do whatever method ur taught I guess, but as I said if the function exponential is referring to e^x is can't be negative no matter the input so that should make it clear...

[novel pewter]

rightt

so now what

[random rose]

then again if it was a negative base it would always be negative

so that should answer the question

just basic properties of functions

[novel pewter]

so its quadratic or exp?

[random rose]

quadratic

[novel pewter]

I see

I have to go afk for 10 min

I will close this then come back

[random rose]

its ok im off now anyway if u need anymore help im sure someone will come

[novel pewter]

thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[safe reef]

which steps are wrongs?
Derivating z

[tardy epoch]

,w diff (x^2-1)^(-1)

[warm shale]

Results provided by WolframAlpha

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

Requested by rie.mann

[tardy epoch]

Why do you think it's wrong

[safe reef]

hmm then the other part of the equation is wrong

Show that the functions: 𝑦 = 𝑥^2 / (𝑥^2 - 1) and 𝑧 = 1 / (𝑥^2 - 1) have the same derivative

[pulsar quarry]

you subtract the power by 1

take a look

[tardy epoch]

Show the entire problem and the entire work

[safe reef]

Hopefully it doesn't look awful

For y = 𝑥^2 / (𝑥^2 - 1)
x^2 / x^2−1 = x^2−1+1 / x²−1 = x^2−1 / x^2−1 + 1 /x^2 -1 =
1 + 1/x^2-1

[safe reef]

The entire problems

Show that the functions: 𝑦 = 𝑥^2 / (𝑥^2 - 1) and 𝑧 = 1 / (𝑥^2 - 1) have the same derivative

[tardy epoch]

What's wrong with this?

Oh you missed parentheses and messed up algebra

(a-1)^2 = ?

[safe reef]

where's this

[tardy epoch]

(x^2-1)^2

[tardy epoch]

So if you can answer this, then you can fix your mistake

[safe reef]

(a-1)^2 = a^2 - 2a +1

(x^2-1)^2 = x^4 - 2x^2 +1

Feels like I am doing something wrong

[tardy epoch]

You messed up the numerator too

You should have a(b-c) but you did ab-c

[safe reef]

Ok gotcha (2x (x^2-1)) - (x^2*2x) / (x^2-1)^2 = 2x^3 - 2x - 2x^3 = -2x

tysm

btw do I need to expand the denominator into x^4 - 2x^2 +1 for both of them or not necessary for the answer?

[safe reef]

spotted a mistake here as well, shouldn' the exponent be -2? @tardy epoch

[tardy epoch]

actually it's probably a bad idea to expand

looking at the other work, your numerator should factor so that it'll cancel

something like [ ab - ac ] / [a^2] = [b-c]/a

[safe reef]

This is what I have done

.close

[obtuse pebble]

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[safe reef]

Thanks for the assistance

[obtuse pebble]

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[zealous path]

hey

[zealous path]

I have Distributing Negatives

and Ive been at this for 20 minutes and I cant get it right

can someone walk me through this?

[worthy cargo]

sure what have you tried

[zealous path]

Thats the thing Ive been struggling on how to even start it since Im not usually familiar with a negative sign starting instead of a number.

[worthy cargo]

ok

[lethal talon]

rewrite it as -1 since its the same thing. might make it easier for you

[zealous path]

alright Ill try that right now

thanks man you are a lifesaver.

.close

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[wary elm]

I dont understand what im supposed to do in part B

[wary elm]

I got the first part right

[worthy cargo]

i think for this you are supposed to just talk about the graph and see if it makes sense with the theoretical probability you obtained

it depends really how far they want you to explain

i see that it says algebra 2

so it cant be too deep

not sure what they taught you for this type of stuff

[wary elm]

Confidence intervals

[wary elm]

Yes

But is it possible to find the standard deviation from the gigen information?

It looks like its in the 95% confidence interval which basically means its fair

Bc it looks like a normal bell graph

Its not skewed

Its symmetrical

95% confidence interval means that if the thing im looking for is within 2 SD’s of the mean, its fair

[wary elm]

Exactly thats wht im confused about

Im not sure if they want me to prove that .1 is in the confidence interval or not

I checked the answer sheet

This is so vague tho

How is it clearly?

Why are we allowed to assume?

[worthy cargo]

ok so just like compare that to your dot plot

[wary elm]

ok ik that

[worthy cargo]

yeah

[wary elm]

But like

[worthy cargo]

we are looking at the value 0.10

[wary elm]

Why are we allowed to assume

[wary elm]

Yes

[worthy cargo]

essentially, the reason they make u assume is because they make it glaringly obvious

the test makers wouldn't make you assume if it was even the slightest bit close

i took the regents lol

[wary elm]

I see

Yo deadass?

Algebra 2?

[worthy cargo]

well 2 yearsd ago

so not the same one

but yeah ik how these work

[wary elm]

yooo W

[worthy cargo]

like they wouldnt ask if 0.40 is within 95%

yes

W algebra 2

[wary elm]

I have another question then

[worthy cargo]

ok

[wary elm]

This right

[worthy cargo]

yep

[wary elm]

It wants to calculate it in months

But it should be the same thing

But like

If we are calculating it in months

Wouldnt it make more sense if the exponent is x/12

Bc there are 12 months in a year

[worthy cargo]

it gives you the yearly rate right and gives you x in terms of years

the exponent would only be 1 if it was over the course of 12 years

[wary elm]

Yes

[worthy cargo]

that doesnt make sense

if we have 12x

[wary elm]

Ohhh hmmm

Youre right

[worthy cargo]

then if x was 1/12 of a year then its one month

[wary elm]

We can balance the equation

[worthy cargo]

or the exponent is 1

[wary elm]

Thats why we do 1/12 insude

Inside

[worthy cargo]

yea

[wary elm]

So the 12 exponent and 1/12 cancel out?

[worthy cargo]

they dont really cancel its just that

1.0325 is the yearly percentage right

so in one year you get that much

but in one month your money wouldn't go up by a factor of 1.0325

so we divide by 12 to get the correct approximate percentage which it would go up

this is an approximation because really, the amount of money you get in the first month is not equal to the amount of money you would get in the 11th month but that is why the problem states 'an approximately equivalent function' so we can assume the rate of growth is linear over the months

[wary elm]

Hmmmm

Interesting

I could just use process of elimination tbh

[worthy cargo]

yes

you wouldnt use the yearly rate for the monthly rate

so

B is incorrect

and from what we said about the exponent

a is left

[obtuse pebble]

@wary elm Has your question been resolved?

[obtuse pebble]

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[honest flicker]

Wat is this asking me😭

[fathom flicker]

It is saying there are two questions with two options

#1
(T)
(F)

#2
(T)
(F)

it is asking you, what is the chance that you get BOTH questions right by guessing

so, that is getting question #1 right, and, getting question #2 right

and btw the answer is not 50%

@honest flicker

[timid silo]

(b)

Two questions

true or false. so 50% chance to get right or wrong if you choose random answer

What is the probability you get the first one right but not the second?
What is the probability you get the second one right but not the first?

[timid silo]

Also, i'm 74% sure you got (a) wrong

[fathom flicker]

you can be 100% sure of that

[honest flicker]

Is it 25%

[fathom flicker]

why do you think it might be 25%

[honest flicker]

Uhh cuz like 50% chance twice idk how to say it

[fathom flicker]

.5 * .5 = .25

[honest flicker]

Fr

[fathom flicker]

the probability of A and B is the same as the probability of A times the probability of B

agreed?

you can write this as

[honest flicker]

Idk my teacher didn't really teach it she just gave us it😭

[fathom flicker]

P(A and B)=P(A) * P(B)

well now you know

that is the formula to use

A is answering the first question right

B is answering the second question right

P(A) is 50%, it is true or false so 50% chance

P(B) is the same

so P(A and B) represents the probability of getting A and B, and using our formula is equal to .5 * .5 = .25

does that make sense?

[honest flicker]

Wat is P

[timid silo]

yes

for (a)

[fathom flicker]

probability of

P(A) = Probability of A

A being an event

[honest flicker]

Wat the answer for question B

[fathom flicker]

We can't give you the answer

[honest flicker]

Oh

[fathom flicker]

!show

[obtuse pebble]

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

[honest flicker]

I still don't get what B is asking me😭

[fathom flicker]

it is asking you for the probability that you get exactly one right

so whether you get #1 right and #2 wrong

[honest flicker]

Out of both questions

[fathom flicker]

or #2 right and #1 wrong

but not both right, or both wrong

[honest flicker]

Uhh

50%??

[fathom flicker]

why

[timid silo]

Why

[honest flicker]

Um

Idk😭

[timid silo]

Are you able to split the question into sub problems?

[honest flicker]

Waht

[timid silo]

What is the probability taht you get exactly one of them right by guessing?

[fathom flicker]

[timid silo]

Are you able to create sub-problems or sub questions from this?

[honest flicker]

Uhh 50% idk there is two options

[timid silo]

that explains nothing

[honest flicker]

I have no clue

[timid silo]

You should

learn dynamic programming

[honest flicker]

Idk wat that is

[timid silo]

So basically

The question (b)

is asking you

what is the probability

that you get only one answer right

so that means

what?

[honest flicker]

Uhh it's like asking if u get one right out of 2 questions w 2 options each

Idk

😭

[timid silo]

It means that you have to get one wrong

[honest flicker]

Fr

[timid silo]

ong

[honest flicker]

So like 50% because one is wrong out of 2 questions

[timid silo]

Wrong because you didn't exaplain why

[honest flicker]

😭

[timid silo]

Wait

[honest flicker]

YES I DID

[timid silo]

because there is two questions is good an explanation

[honest flicker]

Is this answer good enough

[timid silo]

They most likely will not give you points for that

[honest flicker]

WHAT

Idk

Plz give hints

[timid silo]

so look at this horrible drawing

mainly at the 2 guess portion

What is the chance you get a question right and then get it wrong? ( two guesses in a row)

Then the next question
What is the chance you get a question wrong and then get it right? ( two guesses in a row)

What is the chance you get either of those outcomes is what (b) is asking

[honest flicker]

Uhh

25

[timid silo]

25 for what?

[honest flicker]

Percent

[timid silo]

which question are you answering

[honest flicker]

B

[timid silo]

wrong

[honest flicker]

😭

Oh wait I had the right answer earlier didn't i

I just didn't explain it well

[civic zealot]

There's only 4 ways your exam can turn out
✅ ✅
✅ ❌
❌✅
❌❌

How many of those do you get exactly one correct?

[honest flicker]

3/6

Wait

E

[timid silo]

3/6

where did the six come from he said there were only four

[honest flicker]

2

[civic zealot]

yes. How many different ways are there for the exam to turn out?

[honest flicker]

I thought I saw 6 total😭 😭 😭 😭

[honest flicker]

4

[civic zealot]

yes, so what is the probability of getting exactly one correct answer?

[honest flicker]

50%

[civic zealot]

yes. Now can you use everything we just said to justify that answer?