help-33

[vocal leaf]

1 min

[true iris]

So...?

@vocal leaf

[marsh citrus]

@true iris Has your question been resolved?

[true iris]

<@&286206848099549185>

[sleek flame]

Hello I need help rn

[marsh citrus]

@true iris Has your question been resolved?

[wet sundial]

<@&286206848099549185>

ineed help

formula to find area of equilateral triangular prism

<@&286206848099549185>

hello?

[marsh citrus]

@true iris Has your question been resolved?

[marsh citrus]

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[fading bough]

by rounding each number to it's nearest 10, estimate the value of 804/14.9

[fading bough]

i got 53.

[marsh citrus]

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

[fading bough]

i cant ss on macbook

[pale agate]

what's 804 rounded to nearest 10?

[fading bough]

800

[pale agate]

and 14.9?

[fading bough]

15

[pale agate]

how is 15 a multiple of 10?

[fading bough]

but it says round to the nearest 10

14.9 is 15 rounded to the nearest 10?

wait 14.9 is 10? right

becuase i rounded it to the nearest whole number not the nearest 10!

[pale agate]

don't forget to .close if there are no more questions

[fading bough]

thanks !

.close

[marsh citrus]

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[marsh citrus]

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[visual wolf]

The area of ​​the base of a regular quadrangular prism is S, and the diagonal of the side face forms an angle α with the plane of the base. Find the lateral surface area.
I don’t quite understand how to solve this, I need help.

[marsh citrus]

@visual wolf Has your question been resolved?

[visual wolf]

<@&286206848099549185>

[fickle coyote]

Eminem better

[spark siren]

i doubt that this is the whole example.

[visual wolf]

This is the whole example

[stiff jewel]

Is a quadrangular prism a four sided pyramid?

[visual wolf]

no

[stiff jewel]

Nevermind, then ._.

[spark siren]

should the result be a number or what else?

[visual wolf]

The result should be a formula with letters. Because no numbers were given in this problem. Letters instead.

[spark siren]

do we agree that A regular quadrangular prism is a quadrangular prism whose bases have squares, and the side faces are rectangles.

this?

[visual wolf]

yes

[spark siren]

well then draw the diagonal and the angle alpha in the sketch.

and describe the lateral surface area

[visual wolf]

but I don't quite understand how to do it correctly

[spark siren]

come on, you only need this: A regular quadrangular prism is a quadrangular prism whose bases have squares, and the side faces are rectangles.

What is the lateral surface described in words?

[visual wolf]

the area of ​​all its lateral faces.

?

[spark siren]

and what are the lateral faces?

[visual wolf]

four side faces which are rectangles.

[spark siren]

well. and what can you say about the rectangles? are the all different? are the all the same? or are some equal and some not?

[visual wolf]

Well, since the regular quadrangular prism, they are equal

[spark siren]

so, in total does this mean: the lateral surface is 4 times the area of one rectangle., isnt it?

[visual wolf]

yes

[spark siren]

so you only need the area of one siderectangle. what do you know about the rectangle?

[visual wolf]

Opposite sides of equality

[spark siren]

do you know one side?

[visual wolf]

no

[spark siren]

is the rectangle connected to the base?

[visual wolf]

yes

[spark siren]

how?

[visual wolf]

perpendicular?

[spark siren]

yes. do the rectangle and the base have something in common?

[visual wolf]

ribs

[spark siren]

what is ribs?

[visual wolf]

well, the lines that connect the rectangles to the base

this line

[spark siren]

where is the base?

[visual wolf]

?

[spark siren]

where in the sketch is the base?

[visual wolf]

this

[spark siren]

how would you call this?

[visual wolf]

maybe the top?

vertex

[spark siren]

the rectangle and the base have a side in common.

[visual wolf]

ahhh exactly.

[spark siren]

what do you know about the base?

[visual wolf]

that the base is a square

[spark siren]

yes, and something else?

[visual wolf]

it has a common side with a rectangle

[spark siren]

yes, and something else? look at your example.

[visual wolf]

Well, according to the condition, it makes an angle with a diagonal

[spark siren]

yes, but i mean something else. its area is S.

[visual wolf]

ohhh ok

[spark siren]

do you know the side of the base?

[visual wolf]

no

Using the area formula, you can imagine that this is a.

[spark siren]

you know the area is S, you know the base is a square, what is the formula for the area of a square?

[visual wolf]

but I'm not sure

S=a*a like side to side

[spark siren]

well, you know S, can you calc a?

[visual wolf]

a=S:a?

[spark siren]

$a=\sqrt{S}$

[elfin berry]

ThM

[visual wolf]

Well, that's possible

[spark siren]

what do you mean "thats possible"?

[visual wolf]

It's hard to explain, but in general you're right.

[visual wolf]

with this

[spark siren]

hmm, i am not sure what i should do with this answer. I am beginning to wonder if it makes sense to continue here.

[visual wolf]

at what plan?

<@&286206848099549185>

[spark siren]

you have one side of an rectangle, you know an angle between the diagonale and one side in the rectangle, so it should be a simple calculation to get the area of the rectangle. You should be able to do this for your own.

[visual wolf]

4a x (tgalpha x sqrtS)?

@spark siren

[visual wolf]

but i need to find the lateral surface area.

[spark siren]

we had this discussed and solved a long time ago. #help-33 message

[visual wolf]

I forgot, I'm sorry

[spark siren]

i have to go, but anyway it is all said from my side. you have everything you need to solve your example.

[visual wolf]

4a x (tgalpha x sqrtS)? but is this a solution?

<@&286206848099549185>

.close

[marsh citrus]

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[still temple]

this does not feel right bruh

d = 1 btw

[pale agate]

if 2x+1=1.7/7 it is fine

[still temple]

i utilized this method and the answer was wrong

the one i showed at the start of this chat was an example someone showed

doing that math with different values i got like 1.8 for this practice problem ^

when its 1.34

oh shit

wtf

i keep practicing other problems and getting the right answer but not the one im trying to solve

because im silly and using 7 instead of 6.4 :))

.close

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[ancient bronze]

Hii, can someone help me?
The problem translated is the next one
1- For each system given you need to
A) Express it as a matrix and ¿scale it? (dont know how to translate that haha) using Gauss method and clasify it using the values of each parameter (K in S1, A in S2) using rouche-frobenius theorem
B) Find all the solutions for the case that the system is a ¿Indeterminate copatible system? (the range(A) = range(A*) ≠ number of variables )

[ancient bronze]

https://media.discordapp.net/attachments/387447715312828416/1153033998776406146/image.png?width=1423&height=424

[marsh citrus]

@ancient bronze Has your question been resolved?

[ancient bronze]

<@&286206848099549185>

[marsh citrus]

@ancient bronze Has your question been resolved?

[marsh citrus]

@ancient bronze Has your question been resolved?

[marsh citrus]

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[terse folio]

idk what to do with the square root

[somber current]

did u learn ablut l'hopital rule?

[terse folio]

no

[somber current]

about*

[terse folio]

what is it

[bright jay]

Try multiplying by the conjugate of the numerator

[terse folio]

conjugate is just same thing - right

[somber current]

then try what dldh06 mentioned

[bright jay]

Not quite

You don't flip the sign in the sqrt

You flip the sign that separates the two terms

[terse folio]

so x-7

[bright jay]

You don't flip the sign in the sqrt
You flip the sign that separates the two terms

[bright jay]

No

[wise lance]

Yeah as he said

[terse folio]

which two terms?

OHHH

nvm

so (x+7) + 3

multiply top and bottom byt that

[bright jay]

Yes but don't forget it's sqrt(x + 7)

[terse folio]

yah

[bright jay]

Yes

[terse folio]

just didnt know how to do sqrt on keyboard

[bright jay]

You can do what I did here

sqrt(x + 7)

[terse folio]

whilke your herer

for this one do i just simplify the numerator

and take out the 1/4

[bright jay]

Yes

[terse folio]

did i do som,ething wrong

cause when i plugged in 2 i got 0

[marsh citrus]

@terse folio Has your question been resolved?

[terse folio]

@he

[marsh citrus]

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[humble river]

is

one half percent

x+0.005

or

cuz ik wat to do here but the translation part got me

[dry prawn]

that's how I'd interpret it

[humble river]

Wat am i doing wrong

im using this setup

i keep getting wrong answer

i think 0.005 is getting me wrong answer idk]

thx in advance

and the question is above^^^ (word problem

[dry prawn]

that setup looks fine

what's the answer you're ultimately getting?

[humble river]

0.5

and then i plug the 0.5 into x

so i gotta do 0.5+0.005

i think

0.05*** i get

[dry prawn]

x = 0.05?

[humble river]

ya

[dry prawn]

,w solve 4000x + 8000(x + 0.005) = 640

[elfin berry]

Results provided by WolframAlpha

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

Requested by steakanator

[dry prawn]

yeah looks good

so 0.05 is what in percent?

[humble river]

5%?

[dry prawn]

5%

[humble river]

so thats the answer?

i thought we had to find

the interest rate for the $4000

[dry prawn]

we do. That's x

[humble river]

Oh

oh

Thnx u

.close

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[misty kettle]

I need help solving this. How would I go about showing the work?

[marsh citrus]

@misty kettle Has your question been resolved?

[misty kettle]

.close

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[still temple]

.close

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[humble river]

hi

[humble river]

WHere did the Variable (R) go here?

It just went from 6(r)(1.20r) to 55.12r

da heck

[bright jay]

Did you read the rest of the line?

[humble river]

Yea

[bright jay]

Because notice how there's an r in each of the terms on both sides so it got canceled out

[humble river]

OH

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[iron pond]

May someone help me with this question

I don’t understand it

<@&286206848099549185>

[marsh citrus]

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.

[winter lava]

!status

[marsh citrus]

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

@iron pond Has your question been resolved?

[marsh citrus]

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[umbral finch]

For what value of x, does f(x) = 2 ?

My work so far: The values of x where f(x) = 2 are 0 and 5.2. Using the points (4, 6) and (6, 0) we can find a slope of -5/2 for the equation on the interval (4, 6). From there we can use a point to define the lines equation as y= -(5/2)(x-6) or y = -(5/2)x + 15. To find the value of x for that f(x) = 2, we can plug 2 in for y and solve for x. The solution is 5.2. The other value for x is obvious from the graph.

[umbral finch]

this question is coming from my college algebra teacher who I don't think meant for us have to do all that to solve this

but I'm like 99% sure I'm right on this

I'm worried I'm overthinking it and I'm wrong

thank you for reading 🙂

[cunning basin]

<@&268886789983436800> is this ok

[umbral finch]

bro

<@&286206848099549185> would appreciate your guys' thoughts

thank you @quaint hill

[cunning basin]

I don't see any mistak

e

[umbral finch]

aight

i hope my professor says the same

😭

ty

.close

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[radiant sluice]

where does the e disappear here? How did they manage to remove it, or better, is there an easier explanation then the text lol

[stoic saddle]

they consider just what the limit of the exponent is

at the end they'll take e^(that) to get the answer

[radiant sluice]

thank you!

.close

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[sudden mauve]

I was wondering what i did wrong because the denominator will end up as 0 which will make it undefined

[cunning basin]

multiplied wrong

[stoic saddle]

4th step

should have been (4a+4h+8)(a+5) and not 4a+4h+8(a+5)

likewise (4a+8)(a+h+8) and not 4a+8(a+h+8)

i have a suggestion for making your own life a little easier tho...

[sudden mauve]

Okay. Thank you 😊

[stoic saddle]

write $\frac{4t+8}{t+5}$ as $4 - \frac{12}{t+5}$

[elfin berry]

Ann

[sudden mauve]

.close

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[wise oxide]

How am I supposed to sketch the graph

[still temple]

whats the qn?

[wise oxide]

Sketch the graph of the equation

[still temple]

write the eqn

[wise oxide]

How

[still temple]

ok send me a pic of the eqn

then

[wise oxide]

It’s up there

In the pic

4x+5y=30

[still temple]

4x+5y=30?

ok

so its simple really

u can take random values of x and find the corresponding values for y and just plot them on the graph sheet

but to reduce the hassle i got a trick

u can write y=(30-5y)/4 right?

[wise oxide]

I already did

It’s in the picture

[still temple]

then join the points and u got the answer

[wise oxide]

But y only goes to 16

And my answers r above that

Ok well it seems I was wrong

[still temple]

is it there in the qn that y less than 16?

[wise oxide]

I got my answers wrong

Mb

[still temple]

ok

[marsh citrus]

@wise oxide Has your question been resolved?

[marsh citrus]

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[gloomy salmon]

just need to make sure i have this right. Given (x, y) = (x0, y0)... x0 = f(x) and y0 = f(a)?

[gloomy salmon]

or do i have it flipped

.close

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[sudden mauve]

I was just wondering if i did this right

[sudden mauve]

<@&286206848099549185>

[boreal rose]

Your formula is right and your handwriting is so good that it makes me wanna kms

But

It says between each pairs of years

[marsh citrus]

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[violet prairie]

Could I get assistance of showing proof of log(u/v) = log(u) - log(v)

i tried it myself using a similar example to show proof of log(uv) = log(u) + log(v) and didn't really get very far myself

word doc so poor formatting lol

but this is what i thought to do but honestly i have zero clue how to get u/v and x - y isolated on both sides

[abstract knot]

wait

dude

this shows that u/v=a^(x-y)

which means taht by definition

[violet prairie]

does it?

[abstract knot]

yes

you mixed up v and y though

[violet prairie]

oh...

[abstract knot]

it should be a^x/a^y

[violet prairie]

i forgot u can subtract exponents that are divided like that

[abstract knot]

yeah

ok do you see how to do it now

[violet prairie]

yep

thanks for the help

also amazing pfp

love that game

[abstract knot]

oh

same lol

[violet prairie]

.close

[marsh citrus]

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[sleek crater]

Wuestion

[cloud iron]

[sleek crater]

When you root

does it become | q |

Or no

Which of these are not the inverse

Hello??

[wise lance]

Can you state the question clearly

[sleek crater]

Right here

Root means ^2

In the picture

The root is around q + 3

[wise lance]

That is called square actually

[sleek crater]

[wise lance]

Root means ^(1/2) or √

[sleek crater]

Ok

Is it | | ?

Which of them is not an inverse

[wise lance]

Yes

[sleek crater]

f composed of g or g composed of f

[wise lance]

It can be
|q + 3| - 3

[sleek crater]

So what is absolute value -3

[sleek crater]

This is the problem right

it comes down to |q|

Which is not q, so not inverse

[wise lance]

No
It doesn't come to |q| you can't just minus that 3 out of the absolute value to the absolute value

[sleek crater]

FUCKKKK

[sleek crater]

So is that the answer

|q + 3| - 3

You cant go any further

[wise lance]

Yeah that's fog

You can simplfie it making it like this
fog = q if q is bigger than or equal -3
fog = -q - 6 if q is smaller than -3

@sleek crater what do you mean by inverse
I mean Inverse to what ?

[sleek crater]

If function g is an inverse to f

f(g(q) must = q

[wise lance]

Oh yes

You can get the inverse directly from f(q) right ?

[sleek crater]

Yes

[wise lance]

I think the only way to get the inverse of f
Is to get signum function @sleek crater

[marsh citrus]

@sleek crater Has your question been resolved?

[marsh citrus]

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[still temple]

hi i dont know how to approach solving Find sets X and Y such that ((1, 2] × [0, 2]) ∪ ([0, 2] × (1, 2]) = X^2 \Y^2

[still temple]

where do i even start?

<@&286206848099549185>

[marsh citrus]

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[frigid fulcrum]

im going over my image and kernel knowledge and i made up this function

f : $\mathbb{R}{\leq 4} \rightarrow \mathbb{R}{\leq 5}$ with $f : a + bt + ct^2 + dt^3 + et^4 \rightarrow (b-c)t + t^5 + a$

So the rank is 5 right since its the dimension of the image... so what would the dimension of the kernel be?

I was gonna approach it with this formula

dim(V) = dimimf + dimkerf

But what exactly would the vector space here be?

[elfin berry]

Levens

[sharp mortar]

i don't think f is a linear map

[frigid fulcrum]

why

[sharp mortar]

consider scaling

[marsh citrus]

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[still temple]

what is this formula again

.close

[marsh citrus]

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[barren ermine]

Is it true that an integral which comes up with ln(x) is always actually ln|x| ?

[barren ermine]

So int du/u is always actually ln|u| + C ?

[quartz jetty]

yup

a lot of ppl omit it because it might not be a relevant restriction to the problem

like it wouldn't make a diff

but abs is the more correct way

[barren ermine]

What would make it relevant?

Since it's indefinite integral, couldn't x be anything after integration?

[quartz jetty]

u have a point

my profs have always just hand-waved it though, not totally sure maybe someone else has a better answer

[marsh citrus]

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[pine ravine]

Hi, why do we add an x if there is only one solution for a second order differential equation? Ex:

[main idol]

There isn't only one solution

Or I'm not sure what you're counting as "one"

[pine ravine]

Sorry one r

As the geniral solution is e^(rx)

[main idol]

exp(3x) and x * exp(3x) are two linearly independent sols

That's not the general form for repeated roots

[pine ravine]

So im learning second order with costant coifisants where the geniral solution is y=e^(rx)

you plug that in and solve it throught the equation for r

And there are these rules where:

and I was just wondering why they exist as such

[marsh citrus]

@pine ravine Has your question been resolved?

[marsh citrus]

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[arctic crow]

how would i use l'hopital on this/ solve it

[crystal lintel]

$\ln\left(\lim_{x\to 0} (2x)^x\right) = \lim_{x\to 0} \ln((2x)^x)$

[elfin berry]

layla

[arctic crow]

then i would bring the x to the front, but after that what would i do

hello

[crystal lintel]

i might have a way with lhopital's rule but it seems convoluted, trying to simplify

ok i think you can just lhopital $\lim_{x\to 0} \frac{\ln(2x)}{1/x}$

[elfin berry]

layla

[arctic crow]

how do you get ln(2x)/(1/x) tho

[crystal lintel]

$\lim_{x\to 0} \ln((2x)^x) = \lim_{x\to 0} x\ln(2x) = \lim_{x\to 0} \frac{\ln(2x)}{1/x}$

[elfin berry]

layla

[crystal lintel]

just x = 1/(1/x) is all it is

[arctic crow]

ohhh youre right

thank you

.close

[marsh citrus]

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[crystal lintel]

^_^

[marsh citrus]

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[thorn wharf]

Can someone pls help me with part c

[thorn wharf]

No idea how I would solve this

[marsh citrus]

@thorn wharf Has your question been resolved?

[thorn wharf]

<@&286206848099549185>

[marsh citrus]

@thorn wharf Has your question been resolved?

[thorn wharf]

.close

[marsh citrus]

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[still temple]

Ive tried to substitute but it doesnt lead me anywhere

[paper raptor]

maybe try converting everything into terms of sine and cosine, usually helps with these problems :)

[still temple]

oh my god

wow that fucking worked

[static quarry]

probably you know a useful trig identity for sin(2theta)

[still temple]

thank you

[still temple]

i used it just had to convert all the tans after

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[still temple]

help

[still temple]

<@&286206848099549185> any trigonamatry masters?

[marsh citrus]

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.

[paper raptor]

no unfortunately I'm only well versed in trigonometry idk much about trigonamatry

what's your question?

[native frigate]

trigonamatry is the astrology of math

[abstract knot]

Lmao

[still temple]

lmaoo

um

i have a few

hold on

what does coterminal mean?>

sorry this stuff is so confusing its hard to even ask good questions

[quick moth]

An angle that starts and end in the same place

[paper raptor]

no that's a valid question coterminal angles are angles that yeah a propus said ends up in the same place

i.e. 110º and 470º are coterminal

[quick moth]

Basically they “terminate” in the same place

[still temple]

so does that mean one of the rays are the same?

[quick moth]

Uh

[paper raptor]

because 470º-360º=110º

[quick moth]

It’s like if you add or subtract 2pi or 360 degrees from angle

You can add or subtract 4pi or 720 degrees etc

[still temple]

idk guys ima just quit college

lmaoo i want too so bad

3rd week btw

taking a math was a mistake

u guys helped i finished one assigntment

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

kinda

ima have more soon prolly

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[nova crow]

how can i set up this equation to solve?

idk what to do with the (a) at the end, but i can multiply G and H

[gray shoal]

This is a composite function its basically a function inside of a function
Here its asking you to do g(h(a))
For reference g(h(a)) is basically the same thing as (g . h) (a)

[nova crow]

oh they are the same?

g(h(a)) and (g . h) (a) ?

[gray shoal]

To do g(h(a)) you just substitute the h(a) for a+3 since it states that h(a) = a+3

[gray shoal]

Yeah they're the same thing

[nova crow]

oh thats all i needed thanks

ill try it again rq befero i close this

[gray shoal]

Ok np

[nova crow]

OHHH YEAH i remember idk how i forgot

but whats the difference between open circle and closed circle?

cuz i know thats what u do for open circle... but whats different about when its closed

[gray shoal]

Not sure what you mean by that

oh

[floral cypress]

Is the closed circle not multiplication?

Thats how i read the question

To me it seems like g(a)*h(a) rather than g(h(a))

[nova crow]

yeah the multiplication thing, when its open ik u just convert it to (x(x(x)) but wb closed

[gray shoal]

I think its just generally supposed to be a open circle for a composite function

typically a composite function would have a open circle

Not sure why this one has a closed circle?

[nova crow]

yeah we wnet over it in class today but i forgot lol

ok let me solve it agian fr this time

it was right les goo

[gray shoal]

To be honest I've always been taught that a composite function would have a open circle

[gray shoal]

nice

[nova crow]

alr thank u

.close

[marsh citrus]

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[nova crow]

@gray shoal sry for ping but do u know how to set this up lol

[still vortex]

(h/g)(t) is basically just h(t)/g(t) isn't it?

[nova crow]

uhhhh

yes

[still vortex]

So which step are you stuck on exactly?

[nova crow]

here

not sure how i hsould divivde

[still vortex]

You can leave it like that or you can distribute the 1/3t out if you want to

Note that (a+b)/c=a/c+b/c

[nova crow]

i tried leaving it like that, said it was wrong let me try agin

[still vortex]

Ok

[nova crow]

ok my bad

yeah i just put it in wrong 💀

thanks anyway tho

.close

[marsh citrus]

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[nova crow]

.reopen

[marsh citrus]

✅

[nova crow]

what does this mean 😭 i was absent when we learned ab this

[still vortex]

So it's asking for the domain of the function (h/g)(t)=(2t-1)/3t

[nova crow]

how can i find that

[still vortex]

You can find that by finding all values of t such that (h/g)(t) is defined

[nova crow]

can u explain it again but for an idiot to understand 😭 😭

[still vortex]

So the question is asking for the restrictions that needs to be put on t so that (h/g)(t) is defined

For example t is not equal to 2 is a restriction

[nova crow]

ok i think i see

how would i write it tho

[still vortex]

Can you write a not equal sign in the space?

[nova crow]

yes

[still vortex]

Oh wait you're asking for how to find the restriction right?

[nova crow]

yeah

and how to write it as an answer

[still vortex]

Let's first find it

A good way to do that is to ask: for what values of t is (h/g)(t) not defined?

[nova crow]

2t and 3t

[still vortex]

Hint: division by zero is not defined

[nova crow]

i dont understnad

what am i trying to find

[still vortex]

The values of t for which the function (h/g)(t) (which you found last question) is not defined

[nova crow]

2 and 3?

[still vortex]

Division by zero is undefined so when t=0, (h/g)(t) has a denominator of zero and is therefore undefined

[nova crow]

can u just tell me what it is

cus im lost and need to move on

[still vortex]

Well it's just 0

[nova crow]

oh

so at what point does the equation equal 0!?!?

[still vortex]

Which equation?

[nova crow]

oh no im restating the question

ok can u just tell me in dumbass terms what a restriction is

[still vortex]

A rule that limits what t is

For example t≠0

[nova crow]

ok how do i find the DOMIAN from esomtihjgn liek this:

where is the domain

[still vortex]

I actually said the wrong thing it's asking for the restriction not the domain

[nova crow]

"What would be the restriction on the domain from #12?"

i know this is a super easy question i need you to just explain it in extermemly simple terms

the last oneee

[still vortex]

That's the simplest I can get

Ok let's go back to the drawing board

The domain is the values of t for which a function g is defined (i.e. g(t) is defined)

The restrictions on a domain (or more generally on a set) is a rule that can uniquely determine the domain

[nova crow]

so does the -1 on h(t) make it undefined?

[still vortex]

If you substitute -1 for t in h(t) you get 2(-1)-1=-3 which is defined

[nova crow]

ok

so then why is 3t defined too

[still vortex]

3(-1)=-3 is defined so 3t (or g(t)) is defined for t=-1

[nova crow]

where TF DID 3(-1)=-3 COME FROM

[still vortex]

Wait so you're talking about when t=-1 right?

[nova crow]

eb567 n89sd79rvx tb56489ybep4895 y689o45n68ygoles yg6io48l5y674io893 yb89p2456h7n 5wtyv4um8vterysuip ytersvup hgtsvop htgsdhuiopsgvuhgv dfiojgsdfvhjk sdfgu tjkhui yjghv/kb jkxcmv,hn

[nova crow]

ill ask my teacher how to do this tmrw in class

im not computing bro

[still vortex]

Uh computing what?

[nova crow]

my brain

brain

not work

[still vortex]

Ok good luck

[nova crow]

i KNOW this is an easy question but its getting way overcomlicated for me

.close

[marsh citrus]

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[rapid jacinth]

need help plz

[rapid jacinth]

<@&286206848099549185>

idk how to graph it

is it right?

what about my equations

[marsh citrus]

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[marsh citrus]

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[jolly sigil]

I do not follow the last step here

[main idol]

cropping is your friend

[unreal oxide]

well the original integral turned up on the right hand side

so they moved it to the left

[jolly sigil]

I'm not sure I follow

[unreal oxide]

are u okay with the step just above the last line?

[jolly sigil]

No

[unreal oxide]

upto which line are u okay

[jolly sigil]

I'd say when doing the second u sub

[jolly sigil]

There's something about where we ended up that looked like what we started with?

[unreal oxide]

they have integrated by parts ig

$\int e^{2t}\cos t,dt=e^{2t}\cdot\sin t-\int e^{2t}\cdot2\cdot\sin t,dt=e^{2t}\sin t-2\int e^{2t}\sin t,dt$

[jolly sigil]

Yes

[elfin berry]

SilverSoldier

[jolly sigil]

Twice

Mb

Wrong word

[unreal oxide]

okay so this is clear ig?

then $2\int e^{2t}\sin t,dt=-e^{2t}\cos t+2\int e^{2t}\cos t$

[elfin berry]

SilverSoldier

[unreal oxide]

put that back in here

[unreal oxide]

but notice that the integral u have on the right here is the same one u are evaluating

[jolly sigil]

Hold on

Let me re do the problem by hand

[jolly sigil]

For the first integration by parts

should I continue to doing it again?

or should I see what you're talking about here?

[unreal oxide]

yes integrate the second one again by parts

[jolly sigil]

btw, I used cos(t) for my u

[unreal oxide]

and that plus inside the integral is not an addition right?

[jolly sigil]

I pulled the negative out

along with the 2

[unreal oxide]

oh right okay

[jolly sigil]

because of integration rules

[unreal oxide]

so u can integrate this again

by parts

what does that give u?

[jolly sigil]

what we started with

[unreal oxide]

okay so put that back into the first equation that u got

[jolly sigil]

are you talking about where I did the second integration by parts or

[unreal oxide]

you got an expression for the integral in the first expression (integral e^2t sin t), by integrating it by parts again

replace the integral in the first expression with that new expression u got (sin t*2e^2t - 2 int e^2tcost)

[jolly sigil]

hmmm

I gave each step numbers. Maybe that will help

if you refer to the numbers

are you saying replace the integral I got in 2* with 3?

[jolly sigil]

^

[unreal oxide]

so in 2 thers a part u underlined in red

[jolly sigil]

yes

[unreal oxide]

replace that with 3

[jolly sigil]

that inegral

okay

the whole thing of 3?

[unreal oxide]

only replace the entire underlined-in-red part with 3

[jolly sigil]

okay

[unreal oxide]

coz its just that underlined part that u showed is the same thing as 3

[jolly sigil]

look good?

[unreal oxide]

theres a 2 just before the red part

[jolly sigil]

yes

it's there

[unreal oxide]

u must multiply every term in 3 by that 2

[jolly sigil]

right next to the sin

[unreal oxide]

u havent multiplied the last one

[jolly sigil]

oooo

I see what you're saying

let me add this

[unreal oxide]

right

so u showed that the integral u have is the same thing as 4

(1) = (4)

consider writing $I=\int e^{2t}\cos t,dt$

[elfin berry]

SilverSoldier

[unreal oxide]

and replace every occurence of that integral with I

[jolly sigil]

wait

in 4?

or like

I'm following what you're saying so far

just not where to replace it

I showed that 1 = 4 huh?

I guess this will take me a few to see it

[unreal oxide]

u showed that the green thing on top equals the green thing at the bottom right?

if it helps write both of them in one line

[jolly sigil]

Should I sub the integrals out for I

hmm

na

I guess I just need to think about it

I'll try to look at it again later

thank you for walking through that with me

[unreal oxide]

well how did u get a 16?

[unreal oxide]

yes u can do that

and then u get an equation in I

[jolly sigil]

I pulled out the 2's

using integral rules

[unreal oxide]

there arent that many twos

[jolly sigil]

Yeah then it's just I = all that stuff on the rhs (except the integral) * I

[unreal oxide]

I = all that stuff on RHS - 4I

5I = all that stuff

so I = 1/5(all the stuff)

[jolly sigil]

There'd a 4 there, times 2 that's 8. Times the second 2, that's 16.

[jolly sigil]

:0

I see that

Yeah

[unreal oxide]

theres a 2 just in front of the red underline, and by simpliying the red underline u get a 2 in front of that integral

[jolly sigil]

Wait jk

It should be 8

[unreal oxide]

those are the only twos

when u replace the red underline with 3, the 2 in front of it mutliplies the 2 in front of the integral sign in 3, giving a 4 and thats all

[unreal oxide]

there arent any more twos

[jolly sigil]

hmmm

maybe I did something wrong then

[unreal oxide]

if u multiply the two inside the with the two outside, u dont need those 2s inside the integral again

[jolly sigil]

oh

okay

noted

ty

[jolly sigil]

But to be clear, I should be seeing that (1) = (4) here right?

I do get that with the replacing the integral with I you pretty much show this to be true

[unreal oxide]

yeah u have shown that 1 is equal to 4

[jolly sigil]

Right

Okay

I just gotta see that better

Its not exactly clicking

Like it does when I replace the integral with i

[marsh citrus]

@jolly sigil Has your question been resolved?

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[crimson frost]

What have you tried?

[paper relic]

I tried Adding

I dont understand the number before the letter

[stoic saddle]

4C means 4 times C

multiplying a matrix by a number means multiplying each element of the matrix by that number

(different from matrix by matrix multiplication)

[paper relic]

oh okay

thank you

.close

[marsh citrus]

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[marsh citrus]

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[main idol]

why so much caps

go through this lesson
https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations-inequalities/x2f8bb11595b61c86:multistep-inequalities/v/multi-step-inequalities-3

[main idol]

these are resources

...

there are multiple videos

you see that right?

no

watch the videos and try the problem on your own

then show your work

[marsh citrus]

@stiff wave Has your question been resolved?

[marsh citrus]

@stiff wave Has your question been resolved?

[marsh citrus]

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[alpine tendon]

For this ntegration by partial fractions problem can someone please help me understand how to get B i don't get where A^2 + B^2 = 0 is coming from

[alpine tendon]

I understand getting A and C values I did it through setting 10 = A(x^2+9) + (Bx+C)(x-1) and letting x = 1 which gives you 10A = 10 therefore A = 1 and for C I let x = 0 so 10 = 9A + -1C so therefore C = -1 I don't understand how to get B though

Is it through the coefficient thing i dont rlly get that

[main idol]

where does it say A^2 + B^2 = 0 ?

[alpine tendon]

very bottom

[main idol]

this?

[alpine tendon]

yeah

[main idol]

Do you see the X^2?

AX^2 means A times X^2

Similarly BX^2 = B * X^2

[alpine tendon]

yes i get that from when u distribute the factors getting 10 = Ax^2 + 9A + Bx^2 - Bx + Cx - C i just don't understand like how they're making that equation Ax^2+Bx^2=0 from that

[main idol]

they factored out x^2 from Ax^2 + Bx^2

and since the coefficient of x^2 = 0 on the left side A + B = 0

[alpine tendon]

okay i understand that now

im still confused how they solved for B though

[main idol]

Solve for B in this equation

[alpine tendon]

wait how did it go from Ax^2 + Bx^2 = 0 to A + B = 0 you're allowed to just take away the x^2

okay that makes sense tho solving that

[main idol]

0 / x^2 = 0 for x not equal to 0

[alpine tendon]

okay

makes sense

thanks

[marsh citrus]

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[wind light]

So how do I know if a plane contains two lines?

[wind light]

I thought just taking the cross product of the two vectors from the line to get the normal vector, then get one of the points of the line, and use the plane equation

[marsh citrus]

@wind light Has your question been resolved?

[wind light]

Eh, <@&286206848099549185> So if I am given equations of 2 lines, I can just take the direction vector from the lines eg (x=3+5t, y=4+4t, z=5+6t), and ( x=2+t, y=9+8t, z=4+9t), I can just do <5,4,6> x <1,8,9> correct? and get the normal vector and just plug in using the formula where N is the normal vector and <A,B,C> Na(x-3 or 2) + Nb(y-4 or 9) + Nc(z-5 or 4)

[fast pumice]

Yeap

[wind light]

So if I I said x=3+pt, what values of t would still be okay

if I didn't know the value of one of the direction vectors

I can still find the normal vector in terms of p

[fast pumice]

If $\mathbf{d}_1 = \langle p, 4, 6 \rangle$ and $\mathbf{d}_2 = \langle 1, 8, 9 \rangle$, you still have to continue the same steps. Find the normal vector using the cross product and write the plane equation using a direction vector (it is useful to use the direction vector with unknown here), then substitute the other direction vector into the equation and solve for $p$.

[elfin berry]

adzetto

[wind light]

Can you elaborate a bit?

so find the cross product

and write the plane equation using a direction vector?

[fast pumice]

So $\mathbf{N} = \langle -12, 6 - 9p, 8p - 4 \rangle$. Write plane eq. $-12(x-3) - (6 - 9p)(y-4) + (8p-4)(z-5) = 0$

[elfin berry]

adzetto

[wind light]

haha so

get the plane equation with p

subsitute the 2nd point from the line into it

to get the value of p

[fast pumice]

correct

[wind light]

I did that and erased it

😭

now

[fast pumice]

:/

[wind light]

if I substitute the first point into the plane equation

I should get 0

is 0 a valid value of p

or no

[fast pumice]

of course

[wind light]

alright ty

.close

[marsh citrus]

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[jade wedge]

Got the first part but I've been struggling with the second, someone plz explain it to me

[wraith cobalt]

Try solving the equation you obtained in the first part for x/1 (or in other words, x)

[jade wedge]

So solving (x-9)/-2 for x, right?

actually, doesn't make sense

if you can explain more I'd appreciate it

[wraith cobalt]

In 2d, you might have the equation of a line being $\frac{x-2}{-2}=\frac{y-7}{6}$. You could solve this for $x$: $$x-2=-2\frac{y-7}{6}$$ $$x=2+\frac{y-7}{-3}$$ $$\frac{x}{1}=x=\frac{y-13}{-6}$$

[elfin berry]

Damian

[wraith cobalt]

I'm suggeting you do the same thing here, except that you would have to do the same thing to the y and z sides as you do this process of solving for xx

does that make sense?

[jade wedge]

So in my case, multiply the y and z sizes by -2 and add 9, I see

[wraith cobalt]

Yeah exactly

[jade wedge]

this way, it's the same equation rewritten differently

I'll try that

[wraith cobalt]

mhm

[jade wedge]

hold up, is the denominator supposed to be -6 or -3 at the end here? @wraith cobalt

[wraith cobalt]

What have you got so far?

or are you talking about in my working?

[jade wedge]

asking about your work

I also got y + 74 / 9

I mean (y+74)/9

[jade wedge]

at the end you had -6 as the denominator in this image, that's why I asked

[wraith cobalt]

oh lmfao typo!

should just be -3

[jade wedge]

I got the y one correct! The z one probably just a miscalculation

[jade wedge]

yeah I forgot the *-2 for the z

@wraith cobalt I got it right, thanks bro, much appreciated

.close

[marsh citrus]

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[wraith cobalt]

No worries :))

[marsh citrus]

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[limber eagle]

can somoene help me with my math

[limber eagle]

I will send it once somoene replys

because no one has been repleying to me for the whole day

[wet holly]

no

there u got a reply now u have to send even if I'm not the help u have

[limber eagle]

oh

is the answer

7c^6

[jovial bear]

huh

[wet holly]

no

Needs same exponent to add together

can't combine terms with diff exponents

[limber eagle]

o

is this the answer?

@wet holly

[wet holly]

ye

[limber eagle]

o

ok

[marsh citrus]

@limber eagle Has your question been resolved?

[marsh citrus]

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[stoic saddle]

!status

[marsh citrus]

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

[keen briar]

3

[stoic saddle]

show work & answer

$10 says it's an arithmetic fuckup

[keen briar]

i did it in two ways

[stoic saddle]

oh boy...

[keen briar]

one way i just found the magnitude of each vector then multiplied

the other i multiplied the vectors first then found the magnitude

[stoic saddle]

... that's not how the dot product works.

at all.

[keen briar]

yea idk then

[stoic saddle]

$\ang{x_1,y_1,z_1} \cdot \ang{x_2, y_2, z_2} = x_1x_2 + y_1y_2 + z_1z_2$

[elfin berry]

Ann

[stoic saddle]

this should have appeared in your notes and in your textbook.

[keen briar]

is this further maths?

[stoic saddle]

it's not any "further" than what you came with.....

[keen briar]

what are you talking about?

[stoic saddle]

what are YOU talking about?

[keen briar]

do u know what further maths is?

[stoic saddle]

no

[cloud iron]

If you’re being asked to do dot products then you’ve probably been shown that dot product computation that Ann sent

I can’t imagine you’d be asked to do them without being taught how first

[keen briar]

i havent this is my first time

[cloud iron]

The dot product of a set of vectors is the sum of the products of corresponding coordinates. Take for example the vectors {1, 2} and {3, 5}

The dot product would be 1(3) + 2(5) = 13

The product of the x-values added to the product of the y-values

[keen briar]

oh that makes sense

thank you alot

[cloud iron]

The thing that Ann sent is the formula for the dot product of a three dimensional vector (same idea, but with a z-value too)

[stoic saddle]

yeah but because i couldn't categorize it as "further maths" or "not further maths" it is in fact worthless

[cloud iron]

I don’t think it’s further maths at that level. Maybe for someone that hasn’t seen dot products before yeah but the assumption I made was that someone doing dot products has been taught how to lol

[keen briar]

.close

[marsh citrus]

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[chrome thistle]

Where in the world does the sqare root and sqware come from??????

[sharp vessel]

They take the fraction, and apply both sqrt and square it.

Therefore it's still equal to the same value, because it cancels out.

[chrome thistle]

not to be rude but

i was suppose to pull that out my ass?😭