help-13

[tropic oxide]

there is no #1 here

the numbering starts at 2

did you mean #3 (XY and YW)?

if you did, then that's angle XYW. "angle Y" in reference to the diagram on the right is ambiguous.

[tall dome]

my bad

i meant three

[tropic oxide]

then see above.

[tall dome]

i feel dumb for asking but could u dumb it down a bit

sorry

[tropic oxide]

the angle between lines XY and YW is named XYW.

saying "angle Y" in reference to the diagram that's shown is ambiguous.

there are more than two lines drawn in the picture that go through Y, and so there is more than one angle with Y as its vertex.

[tall dome]

ohhh i understand

the angle thats correct would be

angle XYW

[tropic oxide]

yes, as i said.

[tall dome]

yea idk how i dindt catch that thanks

.close

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@vocal raptor Has your question been resolved?

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[long arrow]

integration by parts

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[tribal sonnet]

sorry it was this

it looks hell complicated

[long arrow]

still IBP should hep

I'll do, but wait

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[long arrow]

now you have same integral, move it to the left, then simplify the rest

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[vernal rivet]

How many 4 digit numbers that have at least 2 same digits next to each other are there?

[solid juniper]

I guess you can count the numbers that have exactly 2, 3, and 4 digits next to each other separately

so like for exactly two digits, those look like aabc, or baac, or bcaa for each digit a and each b, c not equal to a

except for arrangements that make leading 0s

I'm guessing those don't count

[bitter reef]

I think you can just count ones that have 2 digits next to eachother (e.g. _aa_) since it says at least 2

you have 3 cases from what I can see aa__, _aa_ and __aa and for the first one you can't have a = 0

[vernal rivet]

Oh yeah my bad

Forgot 1 more line

Digits exclude 0

[solid juniper]

no 0s at all or just leading 0s?

[bitter reef]

so 1002 doesn't count?

[vernal rivet]

At all

1002 doesn't count

[bitter reef]

for each case you have 10^2 * 9 then I think, so the answer would be 2700?

wait does 1101 count?

it has a 0 but not in the same digit part

[solid juniper]

how'd you get that?

[vernal rivet]

1101 does not count

I got to 2007

[bitter reef]

oh yeah not 2700 then

[storm badger]

i think its more easier to subtract 9 permutation 4 from the number of all 4 digit numbers that exclude 0 from the digits or maybe not.. not good with probs..

[bitter reef]

so e.g. we have
aa__
We have _ -> 1,2,3,4,5,6,7,8,9 so 9 possibilities for each empty digit so 9^2 then we have a -> 1,2,3,4,5,6,7,8,9 so 9 for the same digits so we'd have 9 * 9^2 = 9^3

then we apply the same to _aa_ and __aa so we get 3 * 9^3 = 3^7?

is the answer 3^7

[solid juniper]

I think you are overcounting

for example, don't you count 1111 three times?

[bitter reef]

oh yeah

so we have to subtract that

hm

[vernal rivet]

1111 only counts once

[solid juniper]

you can also do it like this, but subtraction after will work too

[vernal rivet]

The total number of ways u can write that is 3 9!

Then subtract all repeating numbers

I didn't know how to do that

So I eyeballed it xd

[quiet sinew]

It'd be 1953.
9^4 (6561) 4 digit numbers without 0.
The number of 4 digit numbers without zero and without consecutive same numbers is 9x8^3 (4608) (you have 9 choices of your lead digit, then 8 for each following digit as you exclude your previous choice)
9^4-9*8^3=1953

[vernal rivet]

Ah I fucked up then

Gg

[bitter reef]

oh that's a nice way to do it

[solid juniper]

hmm I got something else

[vernal rivet]

Naw 1953 is the correct answer

Thnx fyriole

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[quiet sinew]

No problem

[cedar kiln]

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[royal wasp]

Ok

[royal wasp]

How this

[long arrow]

start from domain

then it's enough to solve (3x-2)(x+4) <= 0

[royal wasp]

Sorry but what do u mean by that

[long arrow]

can you plug x = -4?

[royal wasp]

No

Oh so

The domain can't be -4

[long arrow]

so we remove -4 from the possible set of solutions

now you solve inequality, can you solve (3x-2)(x+4) <= 0

[royal wasp]

I can but how did u get to that

Wasn't it division

How did u turn it into multiplication

[long arrow]

sign of the expression is same for product and quotient

look:
+/+ = + and + * + = +
+/- = - and + * (-) = -
-/+ = - and - * + = -
-/- = + and (-) * (-) = +

or simpler explanation is to multiply both sides by (x+4)^2 (squared because then we're sure we don't reverse inequality sign)

[royal wasp]

Ok

So

I just set each bracket equal to 0

Then I got -4 < x <= 2/3

HOWEVER

How do I know if it's greater than or greater than or equal to

[long arrow]

it's <= (as questions says)

[royal wasp]

Between -4 and X

[long arrow]

but we've said we remove x = -4 cause then denominator is 0, right?

[royal wasp]

OOOH

Thanks alot

U have my thanks

.close

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[royal wasp]

Ok

[royal wasp]

I'm at -2 < x <= 1

What do I do after

@long arrow

Sorry for bothering u

[long arrow]

it's your final answer?

[royal wasp]

It was but then I realized it's wrong

Idk what to do next

[long arrow]

what have you done so far

[royal wasp]

I put each bracket equal to 0

Removed the -2

[long arrow]

have you turn it into (x+2)(2x+5)(x-1)^2 >= 0?

[royal wasp]

Yes

[long arrow]

now u can check if x = 1 satisfies this

[royal wasp]

It does

[long arrow]

so now u can divide by (x-1)^2 both sides because it's always positive (except x = 1)

and x = 1 is a solution

so in fact (x+2)(2x+5) >= 0

and x = 1

[royal wasp]

Ok

[long arrow]

obviously x ≠ -2

now just do what u did in the previous one

[royal wasp]

On the previous one

I just set everything to 0

Then solved

This is what I did here

Now I'm at -2 < x <= 1

But don't know what to do next

[long arrow]

why do you think -2 < x <= 1?

x + 2 = 0 --> x = -2

2x + 5 = 0 --> x = -5/2

[royal wasp]

Isn't that all in the range

Of my answer

Wait so can X be anything greater than -2 then

Wait

It cant

[long arrow]

I don't know your method of solving, but I prefer doing schematic sketch of a parabola

then just read solution

[royal wasp]

Idk tbh, just tell me how u do it

Preferably without a graph tho

[long arrow]

so how have you determined answer to the previous one having only zeros?

[royal wasp]

I just put each bracket equal to 0

So x +4 =0

3x - 2 = 0

Solved

Got the range where X can be

Removed the zero

Which is 4

[long arrow]

yes, but then what? why u did -4 < x <= 2/3

[royal wasp]

-4*

Because I got

2/3

And -4

I figured that was my range

Or domain

Sorry

[long arrow]

u know it's not the rule to do a < x < b?

where a, b are zeros

[royal wasp]

What is the rule

Listen I wouldn't be here if I had an idea of what I was doing 😭😭😭

[long arrow]

u can check signs by doing intervals but easier method is to do sketch

[royal wasp]

Wdym intervals

[long arrow]

https://www.youtube.com/watch?v=_gWjLKsFOPE&ab_channel=TheOrganicChemistryTutor

[royal wasp]

Ok

[long arrow]

let's watch, I guess it will help you

u can skip to solving having brackets

(factored form)

[royal wasp]

Is there one

With division

But that doesn't change much tbh

Just zeroes

Open circle

I think I understand tho

Ima reattempt

[long arrow]

ok

[royal wasp]

Iz the answer just

X is greater than-2

[long arrow]

nope

Maybe I'll show you my method?

u know when the parabola is opening downwards and opening upwards?

[royal wasp]

X <= -5/2

[long arrow]

o

ye

now it's ok

[royal wasp]

But why is that an answer

Like

If I just said

X greater than-2

That covers that answer

As well

[long arrow]

$$x \leq -\frac{5}{2} \vee x > -2$$

[wraith dagger]

Modus

[royal wasp]

Ikik

But if X less than or equal to

WAIT

[long arrow]

how? -5/2 is less than -2

[royal wasp]

Nononk

I get it

Now

That answer is also there cuz

If X is -3 for example

Then top would be negative as well

Making it positive

Right

Or am I just spouting nonsense

@long arrow 😭

[long arrow]

reasoning is ok, if we sub x = -3 then both the numerator and denominator are negative --> expression is positive

[royal wasp]

Ok I think I get it now

[long arrow]

but bro, doing sketch gives you an answer in idk 10 seconds if you're experienced

[royal wasp]

Oh

Well its ok

What happened it

Is*

I'm taking calc bc

But never rly learned how to solve inequalities

And it's in this summer homework packet

So as long as ik how to do it it should be ok

Thanks again

.close

[cedar kiln]

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[keen drum]

In a satellite launch operation, the probability of the failure of launch is: 0.01 when for
the system both the software and the hardware are approved, 0.25 when the software
is approved but the hardware is not, 0.3 when the hardware is approved but the
software is not, and 0.6 when neither the software nor the hardware is approved. The
probabilities are 0.5, 0.3 and 0.75 that respectively the software is approved, the
hardware is approved, at least one of the software or hardware is approved. Find the
probability that (a) the satellite launch fails, (b) both the software and the hardware
were approved given that the satellite launch succeeded.

[grizzled dome]

Hey, sorry about earlier -- had to step away for a while. Got a couple minutes now though.

Do you know how to translate the things in this problem into statements of the form "P(some event | some other event) = something"?

[keen drum]

yes

but I still get confused sometimes

[grizzled dome]

Ok, try it for the probabilities of failure-to-launch in the various situations described above.

(We've got a few bits of info to collect, so may as well start there)

[keen drum]

I did
P(F|(S∩H))=0.01
P(F|S)=0.25
P(F|H)=0.3

[grizzled dome]

Great -- F is failure-to-launch, S software-approved and H hardware-approved.

[keen drum]

P(F|(S∩H)')=0.6
P(S)=0.5
P(H)=0.3
P(SUH)=0.75
P(S∩H)=0.05

This is what I got

[keen drum]

Yes

[grizzled dome]

Right on, but I disagree with one bit of notation... by $P(F|(S \cap H)')$, do you mean "the probability of F given neither S nor H"? (I'm assuming, since that's what's in the problem?) Because that should be written as $P(F | (S' \cap H'))$.

[wraith dagger]

daveamayombo

[keen drum]

yes

is it wrong

[grizzled dome]

Yeah... $S'$ means not-S, and $H'$ means not-H, right? One of those weird quirks with set operations is that $S' \cap H'$ is not the same as $(S \cap H)'$.

[wraith dagger]

daveamayombo

[keen drum]

oh

[grizzled dome]

$S' \cap H'$ means neither event $S$ nor event $H$ happened. But $(S \cap H)'$ means they didn't both happen -- it's "not(S and H)".

[wraith dagger]

daveamayombo

[grizzled dome]

Anyways, don't get bogged down in this... $S' \cap H'$ definitely means neither S nor H.

[wraith dagger]

daveamayombo

[keen drum]

Oh oh oh

I get it now

[grizzled dome]

Ok cool!

So collecting it all, we've got:

P(F|(S∩H))=0.01
P(F|S)=0.25
P(F|H)=0.3
P(F|(S'∩H')=0.6
P(S)=0.5
P(H)=0.3
P(SUH)=0.75
P(S∩H)=0.05

[keen drum]

yeah

[grizzled dome]

Ok. Now there's two approaches. We'll do the better one here, but on the previous problem I was going to suggest an alternative: simply make up a pretend population that fits the numbers you've been given. Like imagine 1000 launches... P(S) = 0.5, so of those 1000, there must have been 500 occurences of event S. You keep going like that and think directly in terms of the population you've made up. But let's not do that in this problem.

Instead, there are a few rules, like Bayes' rule (or theorem or whatever you call it), that can answer the questions we've got.

[keen drum]

Yeah, I was doing this question to get good a Bayes theorem.

[grizzled dome]

Sorry, gimme a moment.

[keen drum]

ok np

[cedar kiln]

@keen drum Has your question been resolved?

[grizzled dome]

Hey, I'm really really sorry -- our 6-month-old woke up, and my window is closed way sooner than I hoped! When I have time I'll check back, but really hope someone else steps in to help meanwhile.

[keen drum]

Np

[grizzled dome]

Heh, got him back to sleep. Let's try and knock this thing out.

So, to find $P(F)$... in general $P(A) = P(A|B) + P(A|B')$. And we've been given $P(F | S' \cap H')$. If we can figure out $P(F | S \cup H)$ we'll be golden.

[wraith dagger]

daveamayombo

[grizzled dome]

That's based on the same thing we were talking about before: $(S \cup H)' = S' \cap H'$. In words, the opposite of "S or H" is "not S and not H", aka "neither S nor H".

[wraith dagger]

daveamayombo

[grizzled dome]

Haha, oh man. I just realized I made a big mistake earlier -- I didn't read the problem closely enough, and OK'd two of those original P(foo|bar) = blah things that were actually wrong.

The problem stated the probability of failure-to-launch when software-approved-but-hardware-not-approved was 0.25.

So that should have been translated into $P(F|S \cap H') = 0.25$.

[wraith dagger]

daveamayombo

[grizzled dome]

And the one after that should have been $P(F | S' \cap H) = 0.3$.

[wraith dagger]

daveamayombo

[grizzled dome]

So all together, they should have been:
P(F|(S∩H))=0.01
P(F|S∩H')=0.25
P(F|S'∩H)=0.3
P(F|S'∩H')=0.6
P(S)=0.5
P(H)=0.3
P(SUH)=0.75
P(S∩H)=0.05

[cedar kiln]

@keen drum Has your question been resolved?

[fair geyser]

black is like, data, blue is inferences

[cedar kiln]

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[crimson sedge]

Hello, I need help with Algebra 1 work. I don’t quite understand how how to solve it. I’m capable of completing the question and such, but to solving it is the problem itself.

[cold briar]

• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.

_ _:)

[crimson sedge]

The picture is sending but my internet is rather slow

[cold briar]

oh ok

[crimson sedge]

I apologize

[cold briar]

np

[crimson sedge]

This is the problem, it’s already been solved and such but I need help understanding how to solve it.

I’m well aware on how to complete the equation and such, but to find the remaining amount is always such a hassle for me. I’ve had my friend help me out but I’m always so slow to comprehend it and even so AU still don’t.

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

<@&286206848099549185>

[crimson sedge]

What do I do? Do I multiply the the 150 with the 0.5? Or do I multiple the 0.5 with the 4?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

.close

[cedar kiln]

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[cedar kiln]

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[harsh yarrow]

Is g(x) = 0 the lower/upper limit?

[muted bear]

It depends, it could even be both!

When you graph it, you are looking for the area between curves

Seen as this is a cubic equation, it might touch the x axis more than twice

[muted bear]

Ah so it both

[harsh yarrow]

I’m trying to use a calculator to help me verify the right answer as the one I have right now is wrong

But i’m not sure how to input the equation

[mighty drift]

the unsigned area is like the integral of the absolute value

[muted bear]

Yes

[harsh yarrow]

Could you type it out as an integral for me

[muted bear]

Ah um i dont know the texing so im just gonna write it rq

Unless mateo is doing it rn

[mighty drift]

yes

[muted bear]

Ah ok

[mighty drift]

$\int_{-4}^1 |f(x)| dx$

[wraith dagger]

themateo713

[mighty drift]

with f(x) your cubic function

[harsh yarrow]

hmm i’m trying to use symbolab to help but i’m getting an input error

thanks for all the help this far btw

[mighty drift]

you put the 1 in the wrong place. It is always useful to read the error when it pops up

be it in programming or elsewhere

[harsh yarrow]

When i tried to move i still get the same error

It was put there automatically

Better yet would you be able to point me in the right direction of a calculator I could use for this problem? This is my last chance for this question and I really can’t afford to miss it

[mighty drift]

I don't know that software but I doubt it

,w \int_{-4}^1 |12x - x² - x^3| dx

[wraith dagger]

Results provided by WolframAlpha

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

Requested by themateo713#5303

[harsh yarrow]

Thank you! But it’s still wrong lmao

You’re not allowed to give me the answer and explain how you got it are you? 😳

[mighty drift]

that was to check. To actually compute it yourself, break it up into the negative and positive parts and compute the absolute value appropriately

[harsh yarrow]

this is the equation i’m trying to solve for reference

yeah i have tried that but i’m still stumped either bc somehow i’m imputing it in wrong or maybe i just don’t understand. this software is very picky and the answers have to be pretty exact

[mighty drift]

so what's your answer and what are you putting in ?

[harsh yarrow]

Well i can only get half credit now so i’m not sure if the answer but I thought the answer was roughly 78 sq units

937/12

[mighty drift]

I can't tell you anything about what you might have done wrong without seeing how you got there

[cedar kiln]

@harsh yarrow Has your question been resolved?

[harsh yarrow]

nah but

/close

[mighty drift]

.close it is

[cedar kiln]

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[versed tinsel]

hello i havent a clue as to how rate of change is done because i was not taught so may i please get a lesson 😄

[muted bear]

Average rate of change is (endpoint-startpoint)/length of interval

[versed tinsel]

okay so for the first one it would be (17 - (-3))/2

is the rate of change 10?

[muted bear]

Looks like it

[versed tinsel]

alright awesome ik how to do that thanks

one more thing

[muted bear]

You take the derivative and plug in your x value

[versed tinsel]

derivative is [f(x + h) - f(x)]/h right

[muted bear]

Do you know power rule?

[versed tinsel]

i do not

[muted bear]

Wow

[versed tinsel]

i just finished taking precalc

[muted bear]

Um

Oh

Ok

[haughty cloud]

can any help me find the derivative of ln(4 − x)^2

[muted bear]

#❓how-to-get-help

So yeah you can limit definition

[versed tinsel]

yeah that was the only way i was taught lmao

[muted bear]

Power rule states if you have y=x^n, then the dy/dx=nx^(n-1)

[versed tinsel]

i dont know any of that differential stuff is yet

i think i'll stick to limit 😄

[muted bear]

Sounds good to me

[haughty cloud]

why cant i talk here?

[muted bear]

Help channels are usually reserved to one person so that epople get helped faster

[versed tinsel]

thanks a lot:)

[muted bear]

Np

[versed tinsel]

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[junior palm]

ok hi

um

idk how to do it

[glossy canopy]

4, maybe?

[junior palm]

why

because of bd?

[glossy canopy]

The angles are the same, so the distance probably should.

[junior palm]

i thought so at first

then i noticed the "Round your answer" part

and

everytime they say it

you have to round

lmao

so

[glossy canopy]

Can you find out the angle?

[junior palm]

well

[glossy canopy]

I fink it's 88.694°.

[junior palm]

no

[junior palm]

how

the hell

[glossy canopy]

https://www.calculator.net/triangle-calculator.html?vc=60&vx=6&vy=&va=&vz=10&vb=&angleunits=d&x=74&y=19

[junior palm]

i dont think it will help us

[glossy canopy]

Half of that would be 44.347°.

[junior palm]

still

if we needed the measurement of an angle

they wouldve given it

[glossy canopy]

Welp, I still think it's 4. When you round that to the nearest wholly number, the answer is "4."

[junior palm]

well thanks for the help

<@&286206848099549185>

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[vocal girder]

im utterly useless when it comes to solving systems of equations. With this one in particular, I solved for "a" (2nd equation), and plugged it into the first which was long. Any faster way?

[ocean badger]

You can make the top a little shorter by multiplying by 2 and then the first term is the bottom equation and turn it into 1 +...

[cedar kiln]

@vocal girder Has your question been resolved?

[vocal girder]

idrk what you mean...?

[ocean badger]

Multiple the top equation by 2

To get 0=a2e^4b + 4ab(e^(4b))*4

Then a2e^4b = 1

[vocal girder]

idk how but we got diff answers

i got sqrt(e) / 2 = b

[ocean badger]

Could you show your work?

Does my work make sense to you?

[cedar kiln]

@vocal girder Has your question been resolved?

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[swift compass]

Is this a remotely acceptable proof

[hazy minnow]

The proof is correct in the sense of the logic.

[high coyote]

Yes, you should write it formally logically correctly with cuantifiers and all, but the idea of the proof is exactly what you've done

[hazy minnow]

It’s more like scratch work for a formal proof.

[cedar kiln]

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[versed bone]

,tex Given h(x) = $\frac{8}{x-k}$ + 2, find k if h$^{-1}$(8) = 3, then state the equation for h(x)

[wraith dagger]

tye

[versed bone]

<@&286206848099549185>

[cedar kiln]

@versed bone Has your question been resolved?

[versed bone]

<@&286206848099549185>

[ocean badger]

Do you know how to find the inverse?

@versed bone

[versed bone]

i tried but nothing matched up

[versed bone]

so no not in this case

[ocean badger]

Turn h(x) into y to make it look nicer. Then get x by itself on one side to get the inverse.

[versed bone]

yeah still a bit difficult

<@&286206848099549185>

[dire geode]

What methodology did you try

[versed bone]

essentially what i got to was

,tex y = $\frac{8}{x-2}$ + k

[wraith dagger]

tye

[versed bone]

,tex so if h$^{-1}$(8) = 3, that means that my equation will be 3 = $\frac{8}{8-2}$ + k, right?

[wraith dagger]

tye

[versed bone]

if so

what i got was k = 1.66

[ocean badger]

Leave K as 10/6

[versed bone]

why?

[ocean badger]

Then you can combine the denominator a little better

[versed bone]

ah i wasn’t in fraction mode

so what my steps then were

[ocean badger]

You did it correct

[versed bone]

,tex 3 = 1.33 + k; k = 3 - 1.33; k = 1.66

[wraith dagger]

tye

[ocean badger]

But you can plug 10/6 into k, turn x into 6x/6 to add 10/6 and combine

[versed bone]

or 10/6

[ocean badger]

Then just state h(x)

Sounds like you did it just fine.

[versed bone]

okay

thanks

.close

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[sturdy bay]

$\int\frac{dx}{\sqrt{x^{2}+16}}$

[wraith dagger]

homme

[sturdy bay]

Need help with this integral

[tropic oxide]

any progress so far?

[sturdy bay]

I don’t know how to start

[cosmic steppe]

Have you done trig subs

[tropic oxide]

there are many ways to do this

one of them is trig subs

[sturdy bay]

I lied

I did try x =4tan theta sub

and i got to integral of sec theta

[tropic oxide]

yeah 1/cos(θ) is one of those functions that has an antiderivative but it's kind of annoying lol

[cosmic steppe]

It should be one of your identities to memorize

[tropic oxide]

you could also do something sneaky

and do what my highschool math teacher called the ``long substitution'': $u := x + \sqrt{x^2+16}$

[wraith dagger]

Ann

[tropic oxide]

it's very nonobvious

[sturdy bay]

Right

[cosmic steppe]

the antiderivative of sec(x) is ln|tan(x)| + C

Long substitution sounds like pain

I guess I can see where it's going

[sturdy bay]

How do you work that out

[cosmic steppe]

It's one of the identities I had to memorize

[sturdy bay]

And how does this work

[tropic oxide]

wait but is it tho

im getting sus vibes from that one

shouldn't it be log|sec(x) + tan(x)|, that one

[cosmic steppe]

that's for sec(x)

Oh wait

HA whoops

[sturdy bay]

Lol

[tropic oxide]

well you will have $\dv{u}{x} = 1 + \frac{x}{\sqrt{x^2+16}} = \frac{u}{\sqrt{x^2+16}}$

[wraith dagger]

Ann

[cosmic steppe]

Short brain short circuit never happened

[tropic oxide]

so in fact $\frac{\dd{x}}{\sqrt{x^2+16}} = \frac{\dd{u}}{u}$

[wraith dagger]

Ann

[tropic oxide]

and you get log(u)

[sturdy bay]

How do you know to do that

[tropic oxide]

it's highly nonobvious

so i don't think you do

[cosmic steppe]

In my defense it's almost midnight for me.

Lemme give you the stuff from my notes in case you run into stuff like the antiderivatives of cot, tan, etc

Ignore the stuff on the side that was for notes lol

And as you see I did do a brain dum dum moment

[sturdy bay]

I have “worked solutions” for odd questions in this textbook and they’ve done this integral in 1 line so that’s probably what they did

[cosmic steppe]

Just have this memorized and then it should make trig subs a lot easier. Just make to not have a smooth brain moment lol

Unless your teacher asks you to prove them, then uh

[sturdy bay]

I worked it out using that anti derivative for sec

They don’t seem so bad to memorise but ye proving that doesn’t look very fun

Ty for the help anyways

.close

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[sweet quiver]

How does the underlined term equal "a"?

[tropic oxide]

let a = xi + yj + zk

[sweet quiver]

I see it now; Thank you for your time and assistance.

.close

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[supple drum]

I'm back with this question. I'm able to get the GCF but what do I do after that.

[supple drum]

Like for the first number the GCF is 3x. What do I do after that

[jovial snow]

Short answer: You factor out the GCF.
Slightly longer answer: You will use the distribution property in reverse i.e. a(x+y)=(ax+ay). Essentially we multiply the entire expression by the GFC. However, to compensate we have to divide the original expression (what will be in the parenthesis) by the GFC.

[supple drum]

Ok I'm going to try one

[buoyant latch]

there are faster methods

for example, the first one i look at x^7 + x^3

only the 3rd one on the right has x^3 term

faster methods are helpful when doing a test/exam when time is of the essence

[supple drum]

Ah ok. I will try to utilize that.

I chose the 15x^7+10y^2

Am I dividing at this step now.

Or would I have to divide at 5(15x^7+10y^2)

[jovial snow]

which cleans up to answer (2)

so if we factor out the GFC of 5, it means that we divide 5x^7 and 10y^2 by 5 respectively.

Notice that if we distribute we get back the original expression.

[supple drum]

Ok I think I understand now.

I would like to attemp one of the other to make sure I have it down

Actually nevermind I have to go now

But thank you very much for the help

.close

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[rigid current]

I'm still confused on how to get the angle between a vector and x, y and z axis

[rigid current]

problem is the angles gotten using dot product is always in range of 0 to 180

but I want the angles from 0 to 360

[dire geode]

Do you have a specific problem to work out

[rigid current]

I want to rotate an object in 3d space but I just know the direction vector along which it has to be placed

so I need to calculate the angles this direction vector makes with x, y and z axis

https://stackoverflow.com/questions/5188561/signed-angle-between-two-3d-vectors-with-same-origin-within-the-same-plane/5190354

I found this method but it uses an up direction vector which I don't have

[cedar kiln]

@rigid current Has your question been resolved?

[rigid current]

no

[cedar kiln]

@rigid current Has your question been resolved?

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[brittle plume]

My question is : A friend plans to purchase an 85 inch TV at the cost of 2500.The store is offering 30% off any one item. He also has an internet coupon for an additional 10% off any discounted price. How much will your friend save (a) in dollar amount and (b) percent?

I know the answer to A which is he gets a total of 925$ off of 2500 , I'm confused as to how I find out the percentage. The only thing I can think of is adding both 30% and 10%.

[tropic oxide]

it depends on whether the coupon stacks with the special offer

but if it does, then the price will be 2500 * (1-0.30) * (1-0.10)

[brittle plume]

So if I multiply (1-0.30) * (1-0.10) I get 63%, and that would be the total amount discounted?

[tropic oxide]

no

that's what percent of the original price you would be paying

the effective discount is 100% minus that

ie 37%

[sturdy inlet]

The answer for b is to just simply converst the 925 to % with reference to 2500

[brittle plume]

Oh okay, thank you guys I understand it a lot better now

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[rigid current]

If I have an equation of an ellipsoid given by 1st answer https://math.stackexchange.com/questions/1403126/what-is-the-general-equation-for-rotated-ellipsoid which needs an angle theta from x, y and z axis. If I have a direction vector whose angles I need to use what can be used to calculate these angles?

[cedar kiln]

@rigid current Has your question been resolved?

[cedar kiln]

@rigid current Has your question been resolved?

[cedar kiln]

@rigid current Has your question been resolved?

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[tribal bobcat]

question 8 part 2 and 3

[frigid canyon]

part two would be

difference in position at t=5 and 0

[tribal bobcat]

could we call or do i type first

[frigid canyon]

just type

calling is really rare here

[tribal bobcat]

i see

[frigid canyon]

ping to high and not everyone has good quality mic

typing must work, it always does

now, back to your problem

[tribal bobcat]

ive already calculated displacement of the particle at t=5 and t=0

using the above equation

[frigid canyon]

hmm

[tribal bobcat]

but the answer is different from the answer key

[frigid canyon]

umm

lemme see

it should be 50 i suppose

[tribal bobcat]

answer is 58

[frigid canyon]

hmm

[tribal bobcat]

also displacement of the particle at t=3 is 0

[frigid canyon]

ohh god why

[tribal bobcat]

i mean

wait

wait no no

not displacement

velocity

my bad

[frigid canyon]

they have asked distance

[tribal bobcat]

yeah but by knowing displacement we can find distance right

[frigid canyon]

no we cant

i mean if displacement is fixed distance can be anything

it depends on path

[tribal bobcat]

ok but the particle changes direction during the 5 seconds

[frigid canyon]

exactly why distance and displacement are different

[cedar kiln]

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[hardy sable]

where did the green car come from

[cold briar]

what?

[fair geyser]

it's D because it's the largest

[cold briar]

ok so @hardy sable

[hardy sable]

yes

[cold briar]

white cars = 50
red cars = 30
silver cars = 20

how many cars are left

[cosmic steppe]

Big D

[hardy sable]

100

[cosmic steppe]

Yeah

[cold briar]

what is 60% of 200?

[hardy sable]

120/200?

[cold briar]

yes

so can there be 120 green cars?

[foggy tendon]

The green car is the rest of those cars other than white red and silver I believe

[hardy sable]

no

[cold briar]

ok it's D then

[hardy sable]

thanks all

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[upper flame]

I'm doing math from openstax math books I don't have any idea that they are effective or not can anyone suggest me please.

[foggy tendon]

Ask it in #book-recommendations

[upper flame]

Ok

[foggy tendon]

Math problems only here

[foggy tendon]

Remember to close this channel if you have no other questions

[upper flame]

.close

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[feral isle]

tell me the thinking process someone or how do i approach it ?

[quartz frost]

i would use the scalar product

it defines an angle

[feral isle]

like how?

i didnt get it exactly

[quartz frost]

xy=|x||y|*cos alpha

you can calcuöate y

and you can calculate |x|

and then solve for alpha

also since you have to show that it is a Rotation
compare x and y

if it is a rotation they should also have the same length no matter what alpha id

[feral isle]

:/

y is A*x

and we have given A

[quartz frost]

yes

[feral isle]

so y is A matrix * x?

[quartz frost]

yes

[crimson sedge]

could you help me in #help-17 pls

I am having trouble identifying what skey lines are.

[feral isle]

.close

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[leaden apex]

How do I find the derivative of this without using the chain rule?

[trim sentinel]

Send the question

[formal nest]

Use the limit (definition) but it may be near impossible

[trim sentinel]

We still dont know what this is

[formal nest]

yea lol

[formal nest]

Why don’t you want to use the chain rule ? Its pretty simple

[dire geode]

show the problem.

[leaden apex]

Sorry! Forgot to post it.

[leaden apex]

Idk what the chain rule is

I tried using Photomath but the only solution used the chain rule

I’ll look it up if it’s easy to learn

[formal nest]

its really easy check out my chemistry tutor on YouTube

https://m.youtube.com/watch?v=HaHsqDjWMLU

[leaden apex]

Thanks!

[formal nest]

np

[leaden apex]

.close

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[limpid gull]

Let $f,g:\mathbb{R}\to\mathbb{R}$ be continuous. Show that $A={(x,y)\in\mathbb{R}^2:f(x)<g(y)\land f(y)<g(x)}$ is open.

[wraith dagger]

Felix R.

[limpid gull]

So my first try was to show that $\mathb{R}\setminus A$ is closed, i.e. that every convergent sequence in $A$ converges in $A$, but that wasn't that helpful. Should I try to find a neighbourhood of every point in $A$ that is still in $A$?

[wraith dagger]

Felix R.
Compile Error! Click the reaction for more information.
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[cedar kiln]

@limpid gull Has your question been resolved?

[mighty drift]

my first idea was to say it's the intersection of 2 open sets. If you can take for granted that sets defined with strict inequalities are open, then you're good. Otherwise it's probably better to just do a direct proof

[limpid gull]

Ah ok thanks!

.close

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[tulip river]

A space dwarf doubles his height every day. On the ninth day he begins to reduce his height twice as fast. After how many days from the beginning he will be back to his former height?

[tulip river]

,rotate

[tulip river]

i tried to do this in this pretty primitive way, but i still got it wrong

[crimson sedge]

Lyuka do you know how to determine paralell lines?

[tulip river]

yes

Its y=mx+q, the m on both equations has to be the same

i dont know its name in english

[tough aurora]

18th day, no?

[tulip river]

no

Possible answers:
15 days
12 days
10 days
13 days
14 days

[cedar kiln]

@tulip river Has your question been resolved?

[tulip river]

<@&286206848099549185>

[dire geode]

Did you get 13 or 14 from this?

[tulip river]

personally, i got 13

[cedar kiln]

@tulip river Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[cedar kiln]

@plain bridge Has your question been resolved?

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[buoyant latch]

she uses 3/5 of her salary to repay the house loan

how do you express "2/3 of the remaining money she saves"

what does "the remainder" point to

If I say I have 5 apples

I ate 2 apples

Then I used 2/3 of the remaining apples to make a pie

How many apples did I use for the pie

👍

[cedar kiln]

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[frozen tendon]

https://math.stackexchange.com/questions/475251/explanation-of-the-binomial-theorem-and-the-associated-big-o-notation

[frozen tendon]

Guys I have the same doubt so can anyone explain me whats happening with the theorem in there?

[cedar kiln]

@frozen tendon Has your question been resolved?

[dire geode]

what about it

[cedar kiln]

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[shy wind]

what is the average rate of change and how do i solve these kinds of questions

[dire geode]

Do you know the formula for average rate of change

[shy wind]

I dont actually

[crimson sedge]

$\frac{f(b)-f(a)}{b-a}$

[wraith dagger]

∆y/∆x=πy+π^2x

[crimson sedge]

where b,a are the bounds

[shy wind]

So f(3)-f(1) over 3-1 ?

[crimson sedge]

ye

[crimson sedge]

this can be derived by integration btw

[shy wind]

what is that?

[crimson sedge]

ig u havent learned calculus yet?

then u can ignore it

[shy wind]

yeah i have an upcoming SAT exam in 1 month and i am at pretty low level so I dont know anything really

[crimson sedge]

oh

[shy wind]

but thank you both you helped me greatly

.close

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[shy wind]

.reopen

[cedar kiln]

✅

[shy wind]

If i have a question where there are no equation just the inequality and a table like this one how do i solve it

[tropic oxide]

the average rate of change of a function f over the interval [a,b] is (f(b)-f(a))/(b-a)

all you need to know is the values of f at the endpoints of your interval

[noble ice]

The question should really specify the rate of change of y with respect to x

[shy wind]

i didnt quite get what you both are saying tbh

[noble ice]

The average rate of change is just the slope of the line connecting the endpoints

[shy wind]

ah ok

[noble ice]

So you draw a line connecting (-3,5) and (5,4)

[shy wind]

oh ok thank you

[tropic oxide]

you know f(-3) = 5 and f(5) = 4 from the table (or you should be able to read these off)

nothing else is required to know about the function

just these

[shy wind]

yes i figured that out thank you both

[cedar kiln]

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[undone star]

very simple question(at least it is for most ppl but i’m kinda dense)

[undone star]

i’m just not really sure what to do

if that y’= 18(6-a)^2=12

(i think that’s what it is saying)

[tropic oxide]

9a?

[undone star]

yeah

[tropic oxide]

the derivative of (x-a)^3 is 3(x-a)^2, not 18(x-a)^2

3(6-a)^2 = 12 is the equation you want to solve here

[undone star]

oh i see

that makes more sense

so it’s 4 and 8?

i think

[tropic oxide]

sounds good to me.

[undone star]

aight tysm

.close

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[meager ermine]

hi guys

so far i have

suppose the graph of y=X^2 and y = x-5 intersects

then there exists an x in both graphs where y= x^2 and y = x-5

[tropic oxide]

@meager ermine suggest going to an already open channel. this one will close on you soon.

[meager ermine]

this x^2 = x_5

[meager ermine]

ohhh

okay

[cedar kiln]

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[grand umbra]

Hi, I would like ro know if I am doing modulos arthemerics correct:

[crimson delta]

Yes

[grand umbra]

The exercize I am doing is this one, but we don't have a facit for it. I am practicing for an exam in august.

I am not so good in math, especially discrete-math. However, I am trying to learn the different moments like modulo arithmetic.

So I am doing the [7]^-1 class right, by using extended euclidean?

[tropic oxide]

let's see

,w 7^-1 mod 181

[wraith dagger]

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[tropic oxide]

you calculated it correctly

[grand umbra]

and 26 * 10 should be 260 and 260 mod 181 gave me a rest of 79

which means that should be the solution? 🙂

Thank you so much for the confirmation. Now I know how to tacle this kind of problems

and can move on 🙂

.close

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[frozen tendon]

Guys what is the meaning of big O notation in the binomial theorem in this ? https://math.stackexchange.com/questions/475251/explanation-of-the-binomial-theorem-and-the-associated-big-o-notation

[frozen tendon]

<@&286206848099549185>

[crimson delta]

well I could tell you what the answers say

[frozen tendon]

Then plz

[crimson delta]

but I guess that isn't helpful?

also wait 15 minutes before you ping helpers

I'm not gonna waste my time typing something up that someone else already wrote down

read it

and then tell me why that isn't helping you

[frozen tendon]

What does it mean by omit?

[crimson delta]

"leave out"

where do you even see "omit". that word is nowhere on that stackexchange post

[frozen tendon]

$\frac{(x+\Delta x)^n - (x^n + nx^{n-1} \Delta x)}{(\Delta x)^2}$

[wraith dagger]

Night Wølf

[frozen tendon]

First of all I don't understand how this came

[crimson delta]

that is the definition of big O notation

[frozen tendon]

That's a dsa thing right?

[crimson delta]

dsa?

[frozen tendon]

Ok leave that, I just want to know whats this Big O notation in maths and how $$\frac{1}{\Delta x}((x+\Delta x )^n - x^n)$$ is equal to $$\frac{1}{\Delta x}(x^n+nx^{n-1} \Delta x + O((\Delta x)^2)-x^n)$$?

[wraith dagger]

Night Wølf

[crimson delta]

https://en.wikipedia.org/wiki/Big_O_notation

$f=O(g)$ if $|f(x)|\leq M g(x)$ for some constant $M$ near $x=a$ or for $x\to \infty$ (depending on context)

[wraith dagger]

Denascite

[crimson delta]

here $f(x)=(x+\Delta x)^n - x^n - nx^{n-1}\Delta x)$ and $g(x)=(\Delta x)^2$ and $a=0$

[wraith dagger]

Denascite

[crimson delta]

so $f=O((\Delta x)^2)$

[wraith dagger]

Denascite

[crimson delta]

written out $(x+\Delta x)^n -x^n-nx^{n-1}\Delta x = O((\Delta x)^2)$

[wraith dagger]

Denascite

[crimson delta]

or $(x+\Delta x)^n = x^n+nx^{n-1}\Delta x + O((\Delta x)^2)$

[wraith dagger]

Denascite

[cedar kiln]

@frozen tendon Has your question been resolved?

[frozen tendon]

,graph f(x) = x^n

[cedar kiln]

@frozen tendon Has your question been resolved?

[cedar kiln]

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[crimson sedge]

f(x):cubic
f(x)=0 has 2 different roots
f(2x-5f(x))=0 has 3 different roots
f(2)=3,f'(1)=2,f'(0)>2
f(0)=b/a
b+a=?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

y

how do you know

[crimson sedge]

why do u think its impossible

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[proper dove]

is this rational or irrational

[proper dove]

this shi so hard

[cold briar]

is sqrt 30 rational?

[proper dove]

irrational

[cold briar]

ok so it is irrational

just ignore the number beside the square root

irrational x rational is usually still irrational

[proper dove]

ohh ty

[cold briar]

cmiiw

[proper dove]

so it cannot or can be written as the ratio of two integers

[muted bear]

Yes

For surds usually if it does not simplify into an integer its irrational

[cold briar]

sqrt you mean?

[proper dove]

so yes or no

[cold briar]

^

[proper dove]

marvel this is we’re i get confused bro

[muted bear]

Ok sorry for introducing a new term a surd is another name for the sqrt thingy

[cold briar]

well that's ab-surd _ _:)

[muted bear]

Ah i love that joke

[proper dove]

the blanks confuse me

[muted bear]

Irrational is the opposite of ratio-nal

Ir-ratio, so not a ratio

And if its not a ratio its does not terminste nor repeat

[proper dove]

so wha is it

and can it repeat and terminate

[muted bear]

Its irrational, it does not repeat or terminate

[proper dove]

okay

ty

i added 3√2 + 3√15 i got 3√2 + 15

is it rational or na

2 is a irrational number

[muted bear]

Ah thats not how it works

[muted bear]

If the numbers in the squareroots are different, you cannot add them

[muted bear]

This is already in its most simple form

[proper dove]

so is it rational or na

[cold briar]

is sqrt 2 rational

[proper dove]

yes

[cold briar]

what

it isnt

[proper dove]

what’s sqrt

[cold briar]

square root

$\sqrt$

[wraith dagger]

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

[proper dove]

ohhh

mb it’s not rational

@muted bear

[muted bear]

Yes?

[proper dove]

what is this simplified

[muted bear]

That is simplified

The square has no other square factors

[proper dove]

oh

[muted bear]

Squareroot

[proper dove]

then i right it as that ?

[muted bear]

Yes you write it as that

[proper dove]

ty

[proper dove]

is this irrational

[muted bear]

Yes

Because sqrt3 is irrational

irrational*rational is always irrational

[proper dove]

ight ty

[cedar kiln]

@proper dove Has your question been resolved?

[proper dove]

i have to multiply the radicands i forgot

[velvet mortar]

Is suggest you go back to the basics

Review the laws of radicals and such

[proper dove]

ight but can u help me whit problem

[velvet mortar]

Solve √(2*2) first

[proper dove]

that ain’t even n the problem

[velvet mortar]

Do as I ask please

[proper dove]

i’m 2 stupid to answer it

[velvet mortar]

Solve √(2^2)

[proper dove]

how am i supposed to solve it

[velvet mortar]

How am I supposed to help you then?

[proper dove]

idk

[velvet mortar]

What is √4?

[proper dove]

2

?

[velvet mortar]

Yes

How did you do it.

[proper dove]

i jus typed it in my calculator

🚶‍♂️

[velvet mortar]

Well then type your original question also in the calculator

[proper dove]

ight

[proper dove]

wait how do i exactly right this in a calculator 😅

[velvet mortar]

I am not a calculator master