help-19

[dapper raptor]

anyone know how they got x = (u^2-2)/3?

[mystic saffron]

they just wrote x in terms of u

[dapper raptor]

how did they get to the first term after hence, ?

I'm so confused

[faint knot]

you ought to try figuring out how to turn one line into the other instead of expecting it to be smooth 100% of the way through

theres a lot of steps involved to get that

you can treat it as its own problem

\begin{align*}&\int\frac{2x+1}{\sqrt{3x+2}}\dd{x}\
=&\int(2x+1)\cdot\frac1{\sqrt{3x+2}}\dd{x}\
=&\int(2x+1)\frac23\cdot\frac3{2\sqrt{3x+2}}\dd{x}\
=&\int\qty(2\cdot\frac{u^2-2}3+1)\frac23\cdot\frac3{2\sqrt{3x+2}}\dd{x}\
=&\int\qty(2\cdot\frac{u^2-2}3+1)\frac23\dd{u}\
=&\int\qty(\frac{2(u^2-2)}3+1)\frac23\dd{u}\\end{align*}

[clever fjord]

matt07734

[dapper raptor]

that makes sense now

damn, i couldn't even come to that conclusion lol

.stop

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[icy junco]

how can i solve a partial fraction integration problem in which the roots are complex numbers?

[faint knot]

you can still keep going with the problem

also you should post the specific problem youre stuck on

[round bone]

Dont

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[rancid compass]

someone please help me out with this?

i tried calculating it but i got the answer wrong

[faint knot]

did you try multiplying both sides by x^2

[rancid compass]

yes

i got 4/x^2 - x/x^2 = 5x^2/x^2

then i multiplied everything by x^2

so 4-x = 5x^2

[faint knot]

then 5x^2 + x - 4 = 0

[rancid compass]

bro im so stupid

now i get it

[faint knot]

knew it, be careful with those signs next time

[rancid compass]

i got a math exam tmr man

[faint knot]

Ive made that mistake too many times

[rancid compass]

and i don't feel prepared

[faint knot]

youre just making simple mistakes tho

[rancid compass]

still

[faint knot]

but you know what youre doing

check over your work again maybe

but youll get enough right at least

[faint knot]

was it the signs or was it something else?

[rancid compass]

the signs

ok im stuck again

[dark shard]

just dont sleep

[rancid compass]

bad idea

[dark shard]

just practice lol

[rancid compass]

that's what im doin man

[dark shard]

coming from me no its not a bad idea i got a 95% on a test i had no idea about the previous night

[rancid compass]

yeah but i don't wanna ruin my sleep schedule once again

i'll stay up until like 12 or something but that's it

not sleeping will make me worse off

[dark shard]

its 12 am for me wby

[rancid compass]

10.37

[dark shard]

crazy

[dark kraken]

sleep is important

[dark shard]

• 🤓

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[plain sun]

can someone please help me with b?

[plain sun]

b (i)

aren't there too many variables to solve even if we use vietas

[mystic saffron]

check if discriminant > 0

[plain sun]

b^2 - 4ac?

[mystic saffron]

yes

[plain sun]

so i can literally use any values

[mystic saffron]

no

[plain sun]

but what is a and B supposed to be then?

[mystic saffron]

you use a) equation

check if discrimnant > 0 on a) equation

[plain sun]

im not sure

oh wait its plus

hold on

yeah this should be greater than zero

so what to do from here?

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[tiny bobcat]

How to calculate the chance with math shown
What’s the chance of winning powerball if you need 7 out of 35 numbers but you get to choose 13 and you also need the powerball which is 1 in 20

[odd edge]

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[tiny bobcat]

.closr

.close

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[modest field]

P, Q and R are angles of a triangle.

show tanP + tanQ + tanR = tanP x tanQ x tanR.

[round bone]

That doesn't seem true

That's too unbelievable

[modest field]

it's a question

[nimble blaze]

are you omitting any info

[fluid tundra]

no this is a common geo proof question

[nimble blaze]

oh. seems like a nice property.

[leaden widget]

i've never seen that before either, that's pretty wild 😄

[modest field]

really

[nimble blaze]

what have you tried

[modest field]

i tried to draw a triangle

but then i realised tan90 is undefined

so it wasn't gonna work

i don't know much else

[nimble blaze]

try using a non right triangle

[modest field]

do i start with the left or the right

the addition of tangents (left)
or product of tangents (right)

[leaden widget]

i have no idea how to do this proof but my first thought is to try to write the sum as sin a/cos a + sin b / cos b + sinc / cos c, then give the fractions a common denominator (cos a * cos b * cos c), the numerator will become a sum of the sines multiplied with a bunch of cosines and then maybe you can use cos a * cos b identities or something? not sure where it goes from there

[modest field]

.close

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[stuck mountain]

Question 1 please i dont know how to approach it other than doing the limits which is
Lim f(x) = -2
x -> +- infinity
Lim f(x) = -+ infinity
x -> 1^(+-)

[lyric dust]

to me its not really care what 'study' precisely includes but im guessing they just want u to describe the graph

it seems from 0 up until vertical asymptote it is positive and as it u have described, when x approaches infinity (pos and neg) itapproaches negative 2

[stuck mountain]

But shouldnt we find the function as in question 2 they need the derivative of the function and then apply 2 in the function*

[lyric dust]

i dont think u can find the function with the information given

[stuck mountain]

I see

[lyric dust]

other than the general shape and form

[stuck mountain]

I described the graph like this
As x approaches - infinity it gives -2

As x approaces + infinity it gives -2

As x approaces 1^- itgives -infinity

As x approaches 1^+ it gives +infinity

Is that considered correct/enough?

@lyric dust

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@stuck mountain Has your question been resolved?

[stuck mountain]

Yo

[odd edge]

@stuck mountain Has your question been resolved?

[stuck mountain]

<@&286206848099549185>

[stuck mountain]

.close

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[gusty blaze]

what is c_1 and c_2?

[gusty blaze]

i mean like

are they a part of a formula?

idk where c_1 and c_2 came from

[fluid tundra]

these are called weights, and they represent non-vector constants

the main thing you need to know is that the set of vectors v1 and v2 are considered linearly dependent if there exist weights c1 and c2 (not all zero) such that c1 * v1 + c2 * v2 = 0

[fluid tundra]

that's also essentially what these answer choices are communicating

[gusty blaze]

so i basically need to find the weights that make v_1 and v_2 linearly dependent?

[fluid tundra]

to be more precise, you'd say "that prove v1 and v2 are linearly dependent" not "that make v1 and v2 linearly dependent"

but yes i guess so

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[signal oar]

We have a linear DE if y only appears linearly
Though here, it doesn't appear at all

[spiral basalt]

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

[signal oar]

We had it like this

[spiral basalt]

german handwriting
That's 2 problems

[signal oar]

Hmm

The linear polynomial is b(x), right?

[spiral basalt]

"b(x) an arbitrary differentiable function"

[signal oar]

"that is defined by a linear polynomial in the unknown function"

What's the unknown function

[spiral basalt]

y

[signal oar]

"in the unknown function"

How do we know if y is a linear function or not..

Maybe the solution tot the DE is y = x^2

Then it's not linear?

[spiral basalt]

it is

[spiral basalt]

not y

[signal oar]

then what?

[spiral basalt]

this is of the form f(x, y(x)) = b(x)

f needs to be linear in the second argument

[signal oar]

Ok, and if we had (y'(x))^2 in there, that'd be not linear, right?

[spiral basalt]

no

[signal oar]

Even though that's not y(x)

[signal oar]

no as in: no it would not be linear?

[spiral basalt]

it would not

[signal oar]

Ok, now in our case, y doesn't appear at all, but y'' does

And y'' appears linearly

So it's linear

Right?

[spiral basalt]

because it's linear in y

[signal oar]

What do you mean?

[spiral basalt]

the same linear as the one you know from linalg

[signal oar]

y, y', y'', y''', ... are all linear (most of which are 0)

So our DE is linear

That's how I'd argue

[spiral basalt]

because in this case you can prove this

(though technically it also needs to only involve y's derivatives to be called a diff eq, hence the definition on wikipedia)

[signal oar]

Yep. Thank you

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[woeful rivet]

excuse me

[woeful rivet]

where does the 4a x b come from

i just got this

oh wait

do i need to do crab arms

[weary pelican]

FOIL, whatever you call it

[woeful rivet]

oh

ok

[weary pelican]

the inner and outer terms

[woeful rivet]

i got confused by this stuff lol

[leaden widget]

it's just FOIL-ing a polynomial, I'm very confused why they are putting absolute values around a and b in the provided solution though

[weary pelican]

a*2b + 2b*a

[leaden widget]

oh those are vectors

[woeful rivet]

do i have to write it as

4a x b

why cant it just be 4ab

[leaden widget]

write it as $4a\cdot b$, it's a dot product, a x b (which denotes cross product) would be wrong

[woeful rivet]

ooh i see

ok

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[clever fjord]

Soosh

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[eager sigil]

hi how do i find the coordinates in a line graph where both the x and y are equal eg y=3x-2 passes through (b,b) find the value of b

[viscid flint]

if you have that x=y then you can just replace all of the y's with x's in that line equation

[eager sigil]

dont get it

oh nvm

now ik what you mean

.close

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[mystic saffron]

Confused

I thought it was 4 hours

Guess I’m wrong

Is it 3?

Lol

Sorry

Dumb question

Just miscounted by 1 lol

.close

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[leaden widget]

so m = 0 and 1
and n = 3^n and 2^n.

confused, how does this fulfill the original equation? right side of equation n! + 3 >= 3

is n! something other than n factorial?

so how does 1 = n! + 3 ?

n! is at least 1

so right side is >= 4

i'm just in general confused about everything you are doing in the proof, but maybe i just dont understand whats happening

i see that but i dont see the logic for it

m^2 = 3mod(n)
like what does the first line follow from

nope

but i think youre using mod notation wrong for one

m^2 = 3mod(n) are you trying to write that m^2 when divided by n has remainder 3?

which is true because m^2 = n (d) + 3 where d = (n-1)...etc

you would write: m^2 mod n = 3

its like the same order as when you write division:
5/4 = 1.25
5 mod 4 = 1
but instead of the result you have the remainder

as far as when you "checked stuff" i have absolutely no clue what is going on there

m^2 = 3mod(3)
m^2 = 3mod(2)
m^2 = 3mod(1)
also when you get down to these well you cant have remainder of 3 when dividing by 3, 2, or 1

(even if the notation was corrected)

anyway that's all i've got to say, good luck

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[merry beacon]

can someone walk me thru this

[hushed island]

Ok so first solve for theta

Sin(x) = sin(3x)

One of the answers is 0 but I gtg

[merry beacon]

rip

im thinking

but there's a minus sign between the two sin's

(bing chilling rn ping me)

be back in 5

[dense sonnet]

You don’t need to do that idt, just consider the periodicity of sine

I.e sin(x) = sin(x+2pik) for anyinteger k

[merry beacon]

dont get it

<@&286206848099549185>

https://tenor.com/view/idk-gif-1156795358012555738

.close

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[white stirrup]

Can somone help me with this

My answer is not matching

[viscid flint]

show your work

[white stirrup]

This is the diagram I drew

The page above used for rough work

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@white stirrup Has your question been resolved?

[white stirrup]

<@&286206848099549185>

Hello

[long smelt]

note that the question states the force acts at an angle of 30° above the slope

not sure if that's your issue, but i wanted to point it out

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@white stirrup Has your question been resolved?

[white stirrup]

Tried this but still it doesn't match

Ohh

Wait

It still doesnt match

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[sly laurel]

I had an answer but it was wrong.

I need help/answers. I will be asking my teacher tomorrow. Deadline for this is 1hr 55.

<@&286206848099549185>

[odd edge]

@sly laurel Has your question been resolved?

[obsidian yarrow]

you want to first find the equations of the lines

[sly laurel]

i know

but when I find the answers, they are apparently wrong

[obsidian yarrow]

what did you get for the equations of the lines?

[sly laurel]

here

[obsidian yarrow]

those are inequalities not lines

[sly laurel]

that is how/what I can remember I was taught

[sly laurel]

i got 22 mins so help is desperate

please

[sly laurel]

bruddah

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@sly laurel Has your question been resolved?

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[odd edge]

@viral inlet Has your question been resolved?

[viral inlet]

<@&286206848099549185>

[lavish knot]

do you know partial fractions?

[viral inlet]

Yh I got the pmf as c/x - c/x+1

[lavish knot]

nice

[viral inlet]

I'm stuck on how to get the value of c

[lavish knot]

if i sum up values of f across all values of x, what should i get?

[viral inlet]

c-c/x+1?

the sum of the pmf is 1

[lavish knot]

correct

so $\sum_{x=1}^\infty\left(\frac{c}{x}-\frac{c}{x+1}\right)=1$

[clever fjord]

emphatic_wax

[viral inlet]

Yh and that leaves you with c-c/x+1 over x values?

[lavish knot]

what’s the limit of c/x+1 as x goes to infinity?

[viral inlet]

0

[lavish knot]

good

so what is c?

[viral inlet]

x?

[lavish knot]

how did you get x from c - c/x+1?

[viral inlet]

1 nvm

[lavish knot]

great!!

[viral inlet]

Do i just integrate to get the cdf?

[lavish knot]

yes

[viral inlet]

ok thank you

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[foggy epoch]

\

[foggy epoch]

How do I do this question

I did 2x+3-(x+1)=x^2-5-(2x+3)

and got x^2-4x-6

But the anwser is x=5

[forest sky]

it should come out to be a quadratic equation, which you then can solve using the normal methods for quadratic equations

i think you made a mistake somewhere in simplifying

[foggy epoch]

.close

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[foggy epoch]

How do I do this one?

[odd edge]

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[hollow carbon]

ik im supposed to use the identity sinx/x = 1 but how do i apply it on when 2sin(3x)^10 is nested in it?

[hollow carbon]

(this is before the lhopitals rule unit so im not allowed to use that)

idk if it even helps here tho

[hushed island]

uhhh I'm thinking you can start with sin^10(3*0) = 0

[hollow carbon]

ok

[hushed island]

wait nvm if you want to use that identity

I'm not the best at this stuff but we could see?

nvm that leads to weird places I'm thinking of using sin(2x) rule give a moment cuz I'm probably gonna lead you somewhere wrong

[hollow carbon]

i found a way to reduce it a little

[hushed island]

what

[hollow carbon]

separate bottom into x^10 * x^10

and multiply by sin(3x)^10/sin(3x)^10

[hushed island]

x^10 * x^10 is x^20

[hollow carbon]

you are right

1 sec

[hushed island]

could this help?

lim x > 0

[hollow carbon]

oh yeah whole thing is ^10

ok i think i got it

thx

.cose

.close

[hushed island]

can I see the solution @hollow carbon

[hollow carbon]

https://math.stackexchange.com/questions/3538058/evaluating-lim-x-to-0-frac-sin102-sin103xx100-without-l

[hushed island]

ty

[hollow carbon]

.close

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[lofty lagoon]

If seb has x credits, and john has x + 2.6 credits, and they can each play 3[0.5(x-7)]+1 games ([] is the floor function), how many more games can john play than Seb?

[lofty lagoon]

(Please show your work when answering)

[elfin zodiac]

Why don’t you show your work first to see what’s happening with your answer?

[hushed island]

if seb has x credits he can play 3[0.5(x-7)]+1 games

[lofty lagoon]

Well I have a lot of difficulty removing the floor function

[hushed island]

is x an integer?

[lofty lagoon]

Not necessarily, that's where I get stuck

I think I know how to solve but not how to show work

[hushed island]

uhhhhhhhh well I think when you divide x by 2

actually idk then I would just keep it in floored form and take out the [0.5*-7]

I'm assuming you can do that

[lofty lagoon]

No, 0.5 is steplength of 2

3 is step height

3[0.5(x+2.6-7)]+1 - 3[0.5(x-7)]+1 is what I would do,

[hushed island]

is this a graph?

[lofty lagoon]

No

But it's a step function

[hushed island]

ok idk step length and step height

[lofty lagoon]

So it can be graphed

[hushed island]

OHHhhh

[lofty lagoon]

3([0.5(x+2.6-7)] - [0.5(x-7)])

How would I simplify something like this

[hushed island]

is 3[0.5(x-7)]+1 = 3[.5x] + [0.5*-7] +1

[lofty lagoon]

Is that allowed?

[hushed island]

I'm asking you cuz Idk but it seems right

[lofty lagoon]

I don't think so

[hushed island]

and I think that's the only way to go because i don't see another way of simplifying

[lofty lagoon]

Cuz it ain't a distributive function

[hushed island]

yeah

[lofty lagoon]

Well... u have to take multiple answers

Like, if u have

[hushed island]

yeah it's not distributive a

[lofty lagoon]

[2.5 + a]

[hushed island]

= 3([0.5(x-4.4)] - [0.5(x-7)])

ig distribute the halves

= 3([0.5x-2.2)] - [0.5x-3.5)])

[lofty lagoon]

If a - [a] =< 0.5, [2.5 + a] = 2 + [a]

If a - [a] > 0.5, [2.5 + a] = 3 + [a]

Is there a way to rigorously prove this?

[hushed island]

nvm 🤩

is x an integer? do they give that

[lofty lagoon]

Nope, x is anything

It'd be a lot easier if it was tho

[hushed island]

can you send the original problem

wait

can't we just put in a value for x

cuz it should be constant right

x = 1

= 3([0.5-2.2)] - [0.5-3.5)]) +2

= 3([-1.7)] - [-3)])+2

-6+9+2

so 5 games????
@lofty lagoon

I graphed it on desmos and there is 1 more solution 😱 try to find it

[lofty lagoon]

Ya lmao, I know how to do that

My question is, how do you solve ALL solutions without plugging

[hushed island]

this is a way to solve it actually

[lofty lagoon]

Ya but it ain't rigorous

I want a formula

[hushed island]

ok so if you have a constant and a variable

actually yeah

[lofty lagoon]

If a - [a] > x, [am + n] = z + [a]

[hushed island]

i ain't making a formula but that's basically it you solve for 2 cases

[lofty lagoon]

How do I do this

[hushed island]

let me check that formula

if a-[a] > 1-x, x = b-[b], [a+b] = a+b

[lofty lagoon]

????

[hushed island]

wait actually im dead

[lofty lagoon]

How would I apply [2a + 3.7] to that

Like show me the steps to do that kinda thing

[hushed island]

btw this doesn't work

[hushed island]

no that is wrong also 😭

[lofty lagoon]

How so

[hushed island]

if you input .5 for a, then it equals 3 instead of 2

[lofty lagoon]

And how would you solve it then

[lofty lagoon]

Oh so remove the =

[hushed island]

ye

[lofty lagoon]

Idk man

[hushed island]

uhhhh

let's see

[lofty lagoon]

Try finding the solutions for this algebraically

[hushed island]

wait actually this works I think

does it make sense

[lofty lagoon]

No

Lol, what's the comma and what's b

[hushed island]

but like floored

ACTUALYL IDK

[lofty lagoon]

I'm gonna post a question so this doesn't get shut

[hushed island]

3([0.5x-2.2)] - [0.5x-3.5)]) +2

what value x makes [0.5x-2.2] jump up

I think it's .4 right

0.20 - 2.2 = -2 which is a jump from .19-2.2 = -2.01 which floors to -3

@lofty lagoon wait so I think you can just evaluate for when x =.2 and x = .19 then you get both solutions

[lofty lagoon]

I don't really care about the original, I think I will get more info from this, since there is less number clutter

[hushed island]

just find what value a increases the value

[lofty lagoon]

Ya but u don't know always which number to evaluate at

That's what my question js

[hushed island]

so to jump [3.7] to 4

you need to add .3

so 2a = .3

a = .15

so a = .14 is a different value then you evaluate

[lofty lagoon]

Ok, so basically for
[ax + b], test at (b - [b])/a ± ε?

ε is an infinitesimal btw

[hushed island]

lets see

[hushed island]

1-(b-[b])

1-(3.7-3) = .3

[lofty lagoon]

y, its floor not ciel

[hushed island]

that's how i did this ^

[lofty lagoon]

Ya but y u add .3 not .7

Round to 4 is ceil func

[hushed island]

you need to add .3 to 3.7 to have it floor to 4

so I'm asking how much ax is required to increase [b] in [ax+b]

[lofty lagoon]

Ok, so (1 + [b] - b)/a ±ε

I'll try it

[hushed island]

infinitessimal is just an extremely small number right

does it work

[lofty lagoon]

Ya

[hushed island]

ok

and you need to do it to both floor equations to find where the jump at

UNLESS

you just want the 2 values

[lofty lagoon]

I already know where the jump at lol

[hushed island]

very nice I learned something today

[hushed island]

but do you know how long for tho?

this is what the graph looks like

[lofty lagoon]

1/b 💀

[lofty lagoon]

Tf

[hushed island]

the place I examined was this

it gets weirder cuz if you change one of the .5x into .4x then it becomes infinite values

math classes don't teach floor i think but this is interesting

[lofty lagoon]

Ya my math class spent a week on floors like it was nothing then spent the next 4 months explaining the easiest 2nd degree shit ever, and all topics are on the midterm 💀

He didn't even show us how to do a problem like this, he said "as long as you test 3 random values you'll get correct, even if your final answer is wrong" XDDD so not mathematical

Why

[hushed island]

what is 2nd degree ?

quadratics?

[lofty lagoon]

x²

Ya

[hushed island]

lol

yeah floors are confusing but hope that you learned something today

I did haha

[lofty lagoon]

Mmm I guess, I learned I'm fucked if they are on a long answer question tmrw

Anyways gn

Thx for ur help

.resolve

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[upper grove]

Hey! So in this piecewise function, is "a" just 2? f(2) = a, and a is 2?

[upper grove]

If not, how do I go about finding it?

[gritty holly]

this function is said to be cont. on x=2
so
limx->2- [g(x)]= g(2)=limx->2+[g(x)]

[lavish osprey]

from 5x - 2, we know that in order to be continuous they need to "connect" at x = 2.

Aka when x = 2, the line a and 5x-2 should align. So we can plug in x = 2 into a = 5x-2 and solve for a

same logic connecting a to ax^2 + bx

[upper grove]

Thanks for the info! Can you expand on the "from 5x-2..." part? Why do we know that they connect at 2 from 5x-2?

[lavish osprey]

basically in order for this piecewise function to be continuous at x = 2, each part needs to "meet up" at where x = 2.

So we know that at x = 2, we have the point (2, a). We need that to be the endpoint of 5x-2 as x approaches 2

hopefully this make sense

https://www.desmos.com/calculator/maxoyspnmw

try playing around with this part and and move the a part around until it lines up with your endpoint of 5x-2

[upper grove]

I appreciate the help! I'm still trying to really understand it, and am working the problem on paper.

[lavish osprey]

basically it just comes down to what something means when it's continuous

in a function it means there's not any jumps like you can see happening in piecewise function or something like tan(x)

so in order to find a, it has to connect with the graph. So you need to connect a to 5x-2 somewhere, so we can set them equal and plug in x = 2

[upper grove]

So, given that f(2)=5x-2 = 8, I would just set a = 5(2)-2 = 8? Would I then just plug a and 2 into ax^2 + bx, so f(2) = (8)(2)^2 + b(2) and solve for b?

[lavish osprey]

you got it! 🙂

nice intuition

but also for that f(2), it's also equal to a

so you do ax^2 + bx = a for the second part

but you know x and a 🙂

so you just gotta solve for b

[upper grove]

So all of these functions have to equal each other, then? Are we setting a = ax^2 + bx because we both know a and x, leaving us with b? Is that why I couldn't set 5x-2 = ax^2 + bx, because that would only elminate the x and not the a or b?

That's also why we started with 5x-2, because it gives us a direct output value?

Also, this is somehow related to calculus but there wasn't any calculus in this process. Is there a method to do this using calc... stuff?

[lavish osprey]

we can technically use the approach of connecting ax^2 + bx = 5x - 2, but it's much easier to use the a to solve for a 🙂 Otherwise you'd have to use the quadratic equation with 2 unknowns.

[lavish osprey]

oh in that case I would follow this part that 崖から投げ落とす said

[lavish osprey]

this part right here

[upper grove]

Awesome, thanks for the help! I'm going to try and wrap my head around the calc approach, as well.

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✅

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[tight venture]

suppose that f(x) is a function with f(2x+4)=6x+3. Find f(5-3x)+f(5x-1)

[spark cargo]

is $f(2(\frac{x-4}{2}) +4))$= f(x) legal?

[tight venture]

what???

this is my help channel

go find your own

[spark cargo]

jeez, im trying to help you

[mystic saffron]

He's trying to help u

[tight venture]

what are you saying tho

doesn't make sense

i did not ask tat

i said User
suppose that f(x) is a function with f(2x+4)=6x+3. Find f(5-3x)+f(5x-1)

suppose that f(x) is a function with f(2x+4)=6x+3. Find f(5-3x)+f(5x-1)

[clever fjord]

Dubs

[nimble blaze]

consider doing a sub like
u = 2x + 4
to get f(u)

[tight venture]

and then

[nimble blaze]

after you have f(u)
you can get
f(5-3x) and f(5x-1)

[tight venture]

how

[nimble blaze]

start with the first step

[tight venture]

show me pls

[nimble blaze]

consider doing a sub like
u = 2x + 4

substitude that into your equation

[tight venture]

ok

[nimble blaze]

also isolate x, as you'd want the whole thing (including the right side) in terms of u

[tight venture]

ok

how

[nimble blaze]

how what

[tight venture]

how do you isolate x if the whole thing is u

[nimble blaze]

have you ever solved any sort of equation before

[tight venture]

yea

but it's a function

[nimble blaze]

u = 2x + 4
solve that for x, isolate or however you want to word it

[tight venture]

you can't just isolate stuff

what is your age

[nimble blaze]

treat u like you would any other number

[tight venture]

how old are you

[nimble blaze]

old enough

[tight venture]

i dont think you understand

my teacher did it without sub

[nimble blaze]

its doable without sub
using sub makes it easier to process

[tight venture]

can you do it without sub pls

<@&286206848099549185>

[nimble blaze]

i "could", but i won't.
it involves skipping steps, doing stuff in your head which will likely lead to mistakes
and I'm not going to encourage that.

you could try to go through this with my recommendation of sub
or you can wait for someone that will try and do it the way you want

[tight venture]

thx anyway

suppose that f(x) is a function with f(2x+4)=6x+3. Find f(5-3x)+f(5x-1)

<@&286206848099549185>

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[fast frost]

struggling with this limit, i tried a few different ways but i always end up with an indeterminate form

[odd edge]

@fast frost Has your question been resolved?

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@fast frost Has your question been resolved?

[toxic monolith]

look and analyse it:

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[fast frost]

thank you!

[toxic monolith]

yvw:)

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[oblique rapids]

Is this correct?

[amber schooner]

yes

[autumn bolt]

no

you marked the angle between the floor and the ladder as 18

It's the other way

So you should get 22cos18

Not 22sin18

just that

[fathom ivy]

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[wild rune]

If anyone happens to know economics, which of these shaded regions represents welfare loss and why?

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[wild rune]

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[fringe jungle]

which is the odd number out, 13,14,17,23,31,37,53

[minor moat]

Do you see any pattern?

Anything sticking out?

[wild rune]

think about divisibility

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[viral frost]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[viral frost]

Do you know derivatives? Or just the maximum/minimum value of a quadratic?

[midnight ore]

Max height is 350 metres, at 5 second

[odd edge]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[viral frost]

Oops

[midnight ore]

😳

[viral frost]

But don't hand out answers.

[midnight ore]

but he asked that th-

[viral frost]

Give him hints, not the answer.

Look up rules before you're helping.

[midnight ore]

Why can't I give an answer and afterward explain how I arrived at that answer?

[viral frost]

Because many people wouldn't care whatever the heck you're saying once they have the answer.

And isn't it better if the op arrives at the answer on his own?

Rather than having to look at a readymade solution.

In any case, this isn't the place to discuss this. I'd take this discussion elsewhere, just do not give answers directly here, that's how it works here.

[midnight ore]

So, as you can see, it is the graph of a quadratic function. Since the coefficient at t^2 is negative, its branches will point downward. We need to find its highest point.

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@primal atlas Has your question been resolved?

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[mortal hinge]

hi can someone help me w this ?

[mystic saffron]

alright elaborate what you are struggling with

[mortal hinge]

i put what i thought was the answer but it says it's wrong. i feel like the amplitude and vertical shift are correct so maybe it's the period

[mystic saffron]

period is fine

why -2 tho

[mortal hinge]

i thought it was -2 cause that's between the two parabolas

wait a second

why'd i put 2 for the amplitude

nvm i got it the amplitude was 3

[spring knot]

yah correct

[mortal hinge]

ty

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[worn drum]

Hello, I was just wondering how I would integrate this. I thought originally it would just be -cos((2πnt)/T) but looking at the answer key it tells me that the answer is instead
(-T/(2πn))cos((2πnt)/T)

[spring knot]

take the u substitution u=kt, du= kdt, dt = 1/k du

k is a cst

[worn drum]

oh okay I see
sorry would I be able to clarify how you got dt?

wait no nevermind

okay that makes sense thank you

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[mystic saffron]

Question 36

[hoary marsh]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[mystic saffron]

1

[hoary marsh]

We need to calculate the height of the side of the triangle, right?

[mystic saffron]

yes

U can use pythagoras theorem to figure out the height

[hoary marsh]

Given 10 as the height of the pyramid and 8 as the base, what is the height we are trying to solve for?

[mystic saffron]

wdym

[hoary marsh]

We need to calculate the area of the triangle on the surface, right?

[mystic saffron]

yes

the 4 triangles

[hoary marsh]

The general area of a triangle is?

[mystic saffron]

wdym

like how to get the area of a triangle

[hoary marsh]

What's the area of the base of that pyramid?

[mystic saffron]

64

[hoary marsh]

Nice

What are the lengths for the side of the base?

8, right?

We want to cut the pyramid in half

[mystic saffron]

yes its a square

[hoary marsh]

We need the length of the base divided by 2

[mystic saffron]

32

[hoary marsh]

That will help us calculate for the surface triangle

Nono

"Length"/2

8/2 = 4

[mystic saffron]

4

[hoary marsh]

Right?

[mystic saffron]

yeye

[hoary marsh]

Great

We now have 4 and 10

Base and height

[mystic saffron]

yeah

o

[hoary marsh]

Do you know what a hypotenuse is?

[mystic saffron]

yeah

[hoary marsh]

Calculate the hypotenuse using the Pythag theorem

[mystic saffron]

hypotonusa of what

[hoary marsh]

This triangle

[mystic saffron]

10.7

is the hypotonusa

or maybe idk

[hoary marsh]

a^2 + b^2 = c^2

c = √(a^2 + b^2)

[mystic saffron]

thats what i did

[hoary marsh]

Did you use this formula

Nice

[mystic saffron]

10x10+4x4

then square root

[hoary marsh]

We now have the height of the surface triangle

[mystic saffron]

really how tf did we get there

[hoary marsh]

Hold on

Let me draw a diagram for you

[mystic saffron]

thanks would be helpful

[mystic saffron]

,w sqrt116

[clever fjord]

Results provided by WolframAlpha

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

Requested by valeldianko

[hoary marsh]

Here

Let me know if you have questions

[mystic saffron]

oh i see it now

so height of triangle times base divided by 2

thats my answer right

[hoary marsh]

Yes, correct

Good job

[mystic saffron]

thanks for the help

[hoary marsh]

Anytime 🙂

[mystic saffron]

Remember there are 4 of them

.close

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[mystic saffron]

how tall is the tree

[desert marlin]

, rotate

[desert marlin]

is this a test?

[mystic saffron]

nope

its a old paper my teacher gave me

but i am havign a test tomorrow

[desert marlin]

you know the angle, and the adjacent side, and you want to find the height of the opposite side

do you know SOHCAHTOA

[mystic saffron]

do u mind teaching me a lil bit about sin and cos and trigomentry

just the basics

[desert marlin]

You can learn the basics online

search youtube

[mystic saffron]

alr its ok u can just help me with the question

[desert marlin]

Okay I just told you how to do it

[remote vine]

I love ur pfp btw

[zinc glacier]

https://www.khanacademy.org/math/geometry-home/right-triangles-topic/intro-to-the-trig-ratios-geo/v/basic-trigonometry

[mystic saffron]

the hypotunsea?

[desert marlin]

Do you know SOHCAHTOA?

[mystic saffron]

something with sin and cos

[zinc glacier]

watch the video i sent

[mystic saffron]

sin = o/h

[desert marlin]

Yup that is what it is

you have the adjacent side in your problem

and you want to find the opposite side

so which function do you use

[mystic saffron]

yes the tree

[mystic saffron]

cos?

[desert marlin]

what is cosine

based on SOHCAHTOA

[mystic saffron]

cos = a/h

[desert marlin]

okay so is that what you want to use

[mystic saffron]

50/x= opposite

[desert marlin]

Stop guessing

That isn't the right function

[mystic saffron]

what

[desert marlin]

You have the adjacent side, and you want the opposite side

[mystic saffron]

yes which is the trees height

[desert marlin]

cosine is adjacent/hypotenuse

so cosine is the wrong one

[mystic saffron]

what

[desert marlin]

What is confusing about what I just said

[desert marlin]

Where do you see the word opposite in this?

[mystic saffron]

i dont

[desert marlin]

Okay so why would it be cosine then

[mystic saffron]

i dont know

[desert marlin]

Okay so it isn't cosine

SOHCAHTOA

Which one has opposite and adjacent

[mystic saffron]

t= o/a

[desert marlin]

Great

use that one

[mystic saffron]

what is t tho

[desert marlin]

tan(angle)=opposite/adjacent

[mystic saffron]

what is o

opposite

yes

[desert marlin]

the opposite side length

(opposite of the angle)

[mystic saffron]

which is the tree?

so what 50/x

no what am i sayign

[mystic saffron]

ok so o/a = tan

but i dont know what o is tho

oh op is 32

degrees

[desert marlin]

Are you trolling?

[mystic saffron]

no i swear to god

[desert marlin]

32 is the angle

[mystic saffron]

im just stupid

[desert marlin]

tan(angle) = opposite side length / adjacent side length

tan(32) = h / 50

does that help?

[mystic saffron]

what? is the answer 32?

[desert marlin]

What did I say that makes you think that

[mystic saffron]

a

[desert marlin]

What is confusing about this to you?

[mystic saffron]

i tried that on caluclator i got 0.74 so theres soemthing i dont understand

[desert marlin]

tried what on the calculator

there's nothing to calculate yet

[mystic saffron]

oh what

[desert marlin]

stop doing random things

what do you think you are calculating

what's the whole point of this problem

[mystic saffron]

the tree

[desert marlin]

to find what

[mystic saffron]

tree

[desert marlin]

to find tree?

[mystic saffron]

height

tree height

[desert marlin]

To find the height of the tree

which we called h

right?

so if tan(32)=h/50

how do we find h

[mystic saffron]

hmm

32x50?

[desert marlin]

Why

why did you just remove the tangent

it says tan(32)

[mystic saffron]

tan(32) x 50

[desert marlin]

is what

[mystic saffron]

33

[desert marlin]

no

[mystic saffron]

yes

[desert marlin]

why are you saying this

what is it

what does it represent

[mystic saffron]

32 is the angle

[desert marlin]

not my point

[mystic saffron]

what

[desert marlin]

What is the entire goal of the question

[mystic saffron]

tree height

[desert marlin]

The goal is to find the height of the tree

[mystic saffron]

yes

[desert marlin]

So what is the height of the tree

[mystic saffron]

i dont know that what im trying to figure otu

out

[desert marlin]

I just told you how to do it like 10 times

and you just solved it

[mystic saffron]

when?

33?

u said no tho

[desert marlin]

Because you keep saying random things

instead of going step by step and being specific

you're blurting

[mystic saffron]

what

apperantely the answer is 31 not 33

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

hi guys, just a question, wouldn't sin(x)/-sin(x) = -1?

[dim thicket]

almost

[nimble blaze]

as long as sin(x) isn't 0 (or undefined)

[dapper raptor]

i'm learning integrals for trigs now

[nimble blaze]

who's telling you otherwise

[dapper raptor]

and one of the question was cos^5(x)sin(x) \

and one of the solutions was sin(x)/-sin(x) cancelling out

why is that the case?

[nimble blaze]

pretty much the same way/reason you'd simplify something like 5/(-5)

[dapper raptor]

wouldn't that just equate to one?

[nimble blaze]

no

you're ignoring the -sign there

[dim thicket]

$\frac{5}{-5} = 1 \implies 5 = -5$

[clever fjord]

LF

[dapper raptor]

this one makes sense

[ember oak]

🤔

[dapper raptor]

wait

[mystic saffron]

But 5/-5 is not equal to 1

[dapper raptor]

wouldn't it be =-1?

[sonic nova]

If you could show the solution you're confused about that might help us to clarify

[nimble blaze]

yes, it will

[dapper raptor]

so 5 = 5

[dim thicket]

yes

[nimble blaze]

yeh, exact same principle

[dapper raptor]

i'm just confused on how sin(x)/-sin(x) cancels out

i always thought that it would be -1

[sonic nova]

It is -1

[nimble blaze]

it is

the work shows that

[sonic nova]

They take the -1 and move it to the front

[dim thicket]

the -1 is the negative sign in front of u^5

notice it wasnt there before

[dapper raptor]

ohhh right

[sonic nova]

You started with a positive U^5

[dapper raptor]

i probs got something wrong on mine, as I used a different method

[nimble blaze]

,rotate

[dapper raptor]

Where did I go wrong?

[nimble blaze]

order of operations when subbing your u= -cos(x) back in

you should end up having
$$= -\frac{(-\cos(x))^6}{6} + c$$

[clever fjord]

ℝαμΩℕωⅤ

[nimble blaze]

you effectively had
$$= -\frac{-(\cos(x))^6}{6} + c$$

[clever fjord]

ℝαμΩℕωⅤ

[dapper raptor]

i did end up having -(-cos(x))^6

wait, what's the difference?

wait, so what's the right answer?

[nimble blaze]

in one, the power is applied with the negative
in the other, the negative is applied afterwards

you should've done the first as u=-cos(x) is being raised to the power of 6, thus you should've had (-cos(x))^6 there

what you wrote was not that

[dapper raptor]

ohh makes sense

[nimble blaze]

you wrote
$$\red{-}\cos^6(x)$$
representing
$$\red{-}(\cos(x))^6)$$

[clever fjord]

ℝαμΩℕωⅤ

[nimble blaze]

for simplicity it's the same crime as
-5^2 vs (-5)^2
when apply the quadratic formula with b=-5

[dapper raptor]

thanks!!

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[mystic saffron]

what is sin, cos, tan

[lavish osprey]

trig functions, they represent the ratio of an angle to the sides of a triangle.

https://www.mathsisfun.com/sine-cosine-tangent.html

^ this is a good read about it!

[mystic saffron]

ty

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@mystic saffron Has your question been resolved?

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[cobalt vapor]

Hi. Can someone walk me through this, thanks.

[odd edge]

@cobalt vapor Has your question been resolved?

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@cobalt vapor Has your question been resolved?

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