help-49

[frail burrow]

Maybe it isn't tight actually, true

Yeah

[junior flower]

but you won’t be able to

[frail burrow]

Ok you're right

[frail burrow]

I made two inequality steps and they are equailities sometimes

But not for the same values

so it messes up

Welp idk then

[obsidian glen]

for the minimum, can't we consider the case where $\sum_{i=1}^{1012} x_i = 0$

[grand pond]

ginny

[junior flower]

i think i just explained why not

[obsidian glen]

oh, i thought it was for some other line of thought so i didn't read all that much, im sorry

[frail burrow]

look in mg

turns out its a special case??

[junior flower]

ok like it’s at least not clear the min is achieved in this case, and it needs proof

so you can’t just say “consider this case” and be done

[obsidian glen]

well of course you always need a proof

by that i meant consider this and then try proving that this is what you want to do

[junior flower]

and it’s not as simple as x^2 <= x when x in [0,1]

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

<@&286206848099549185>

[frozen turtle]

??

[twilit field]

this is the question

[hard umbra]

smh everyone left

set [ x = \sum_{i = 1}^{1012} x_i ] and apply QM-AM inequality

[twilit field]

I pressume QM<AM

[hard umbra]

no

[twilit field]

never used QM before tbh

[hard umbra]

QM >= AM

[twilit field]

ok

[junior flower]

all i wanted to do was raise my skepticism over people ignoring the linear part and justifying it with x^2 <= x

i had nothing else to offer

[hard umbra]

[junior flower]

other than lagrange multipliers

[hard umbra]

lagrange would be proud

[twilit field]

I don't exactly follow

[fervent ember]

max = 1 is easy

[hard umbra]

welcome to the party, you are late

[twilit field]

$s=\sum_{i=1}^{1012}x_i$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so then

[hard umbra]

[twilit field]

if x=$x=\sum_{i=1}^{1012}x_i$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

then the QM is

$x=1\sqrt{\frac{1}{1012}\sum_{i=1013}^{2024}x_i^2}$?

[grand pond]

ƒ(Why am. I here)=I don't Know

[onyx fossil]

Anyone wanna help me

[hard umbra]

no x isn't equal to that

[twilit field]

that's x^2?

[hard umbra]

x is the first 1012 terms

[twilit field]

#❓how-to-get-help

[twilit field]

yes

[hard umbra]

you're looking at the last 1012 terms

[twilit field]

but what about the next 1012 terms

[hard umbra]

you wrote it here

[twilit field]

yes

so how do I use QM>AM

[fervent ember]

min =1/1012

[twilit field]

how did you get that>

[hard umbra]

!nosols moment

[midnight plank]

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

[hard umbra]

everyone knew the min was 1/1012 we got like 3 different people saying to set the non-squared terms to 0

not adding anything new here

[hard umbra]

you apply QM-AM to the squared terms

[junior flower]

ok but most of those do not have good reasoning for it

[twilit field]

I don't follow

[hard umbra]

,, \sum_{i = 1013}^{2024} x_i^2 \ge \f {\parens {\sum_{i = 1013}^{2024} x_i}^2} {1012}

[fervent ember]

as it is asked for a proof naming the min isnt a solution. so dont claim "nosols" if you don not understand the rule.

[hard umbra]

[twilit field]

hmm, where did the 1012 come from?

[hard umbra]

blud

[twilit field]

squaring the QM?

[hard umbra]

yes

[twilit field]

ok, cool

[hard umbra]

you just rearrange for the sum of squares

[twilit field]

that was suprisingly simple

thanks!

.lcose

[hard umbra]

you're not done

[twilit field]

oh

[hard umbra]

we haven't even gotten a min

[twilit field]

right

[hard umbra]

you have to apply this to the original sum

[twilit field]

It's half past 12 here, I'd much rather continue this later today

sorry

[hard umbra]

okay

sure

[twilit field]

thanks everyone who helped

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[junior flower]

blud…

[hard umbra]

don't come in here providing the same amount of (lack of) input like everyone else here and start accusing me

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[midnight plank]

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[boreal geyser]

I know the A value is 4

I took the derivitive

but for X

is it x^4?

(x+1)^4 doesn't feel right

[sage olive]

so the limit definition of f'(a) is the lim as h approaches 0 of (f(a+h) - f(a)) / h, now matching your parts, you have f(a+h) = (1+h)^4 and f(a) = 1

so if f(a+h) = (1+h)^4, then f(a) = 1^4 by just subtracting out the h inside, that matches the f(a) = 1 that you know

it's just x^4

[reef solar]

f'(a) = 4 not a itself

[boreal geyser]

got it

misread that one

a itself is 1

f'(a)=4

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[granite sphinx]

hi

[dull path]

Helo

[granite sphinx]

2

46 2

[dull path]

y(1) = 3 means y is 1 when x is 3 ?

Sorry
Other way around

Y is 3 when x is 1 ?

[granite sphinx]

I need the Y and the X on one side

This is 1 if it helps

[dull path]

Take the 1 - x to the other side and divide by x

[dull path]

How did it become x + 1

[granite sphinx]

its - i see now

This is what I have

ig its good

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[sinful cipher]

What does symmetry in this article mean? This is a subsection concerning the "two-point form" of an Article concerning linear functions (of elementary algebra) by Wikipedia.

[sinful cipher]

<@&286206848099549185> somebody please help

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[mystic mason]

I'm betting on B for this one

but the answer key doesnt have this question soooo idk

[subtle zinc]

yes B correct

alternating series test

[mystic mason]

ty

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[mystic mason]

I'm pretty sure the answer key is wrong for part a and part d

a) we arent given any f'(5), im pretty sure he meant to put f'(4) which would make it 0.1/2 or 1/20

d) second to last line of work is incorrectly simplified from what im seeing

can some1 have a second look at it and ping me if im just crazy or if im on to something

[fathom tangle]

For the first one judging by the value of “f(5)”, it looks like he meant to say f(4)

Second to last line should go like this I believe:

24-12-(15-f(6)) = 7
12-15+f(6) = 7
-3 + f(6) = 7
f(6) = 10

[mystic mason]

alright so

I think part c on the NEXT question is also wrong 😭

can u also... perhaps... maybe pls... check that too

my answer is 102

on his second line

he did 7(x-7)

i did 18(x-7)

[fathom tangle]

Yeah it looks like it should be 18

[mystic mason]

...

so uh

i think part d is also wrong

i typed it in the calclulator and i got something different

AND

it looks like he did a RIGHT riemann sum

and here it looks like he didnt even use the mvt

C' is a rate so theres no chance it can be the solution to a "total number of blah" question

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[magic cosmos]

Hi why is this wrong

[olive matrix]

show your work

[magic cosmos]

I’ll just tell u what I did since I did it all in my calculator

[tribal temple]

(you should be able to find exact values here: notice that in this one, they don't tell you to round, in contrast to our last one)

[magic cosmos]

I did the quadratic formula and got [-4+-sqrt(12)]/2
And then I put the answers into my calculator and added pi until it was between the 0,2pi range

[magic cosmos]

It does I just cropped it 😭

[tiny geyser]

from the answer you got from the quadratic equation, instead of adding pi

take the arctan of that answer

[magic cosmos]

That’s jus the textbook link bruh I already went over the textbook otherwise I wouldn’t be asking

[magic cosmos]

Wait why

[sweet reef]

i think hes saying that u found the value of tanx
so u need to find x

[tiny geyser]

yea

[magic cosmos]

@tribal temple how come for this problem we need arctan but for the previous one we didn’t

[tribal temple]

We did need arctan for the previous one
#help-45 message

[magic cosmos]

Oh I forgot

[tribal temple]

It was a long question, after all

[magic cosmos]

Lol true

Okk i think(?) I understand now lol

Thx guys :]

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[hollow sun]

the frame of the tent shown is defined by a rectangular base and two parabolic arches that connect the opposite corners of the base. the equation y=-1.8x^2+1.6x models the height y of one of the arches x along the diaonal of the base.
can a child who is 4 feet walk under one of the arches without having to bend over? show your work and explain your answer

[hollow sun]

y = -1.8(4)^2 + 1.6(4)
y = -1.8(16) + 6.4
y = -28.8 + 6.4
y = -22.4

The height of the arch at x = 4 is -22.4 feet.correct or nah

[golden cloak]

It's better to draw a simple diagram or a pic

And send here

[hollow sun]

wor dproblem

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[woeful thicket]

This screenshot shows the problem I am having trouble with.

[woeful thicket]

These are some of the formulas that I have been provided with.

I know that a full circle equals 360 degrees however the angles in the middle make this really hard

Supplementary angles tell me that Angle U_R should be 146 degrees because 180 - 34 = 146. However this would not line up with what is taught to me by the first formula.

<@&286206848099549185>

[abstract harness]

Could it be that the vertex of the angle U_R isn't ON the circle like the angles are as shown in the diagram? (I'm not sure just a thought, I don't remember much from Geometry)

[woeful thicket]

How would this change the way the problem is solved?

Different formula maybe?

[abstract harness]

Maybe

Could you use the second formula given 34=(1/2)(TR+SU) to then use TR+SU plus the 106 to find that then divide that by half to find SVT using the first formula given ?

[woeful thicket]

So 34 = 1/2(68 + X)

But then solving X = 0

[abstract harness]

The sum of SU and TR is 68 but you don't need to find what SU and TR are individually if that makes sense

Then you could do
SVT = 1/2(68+106)

87

[woeful thicket]

Why would the sum of the two be 68? The first formula tells us that only tr is 68.

[abstract harness]

For the first formula the vertex has to be on the circle

The vertex of the angle is inside the circle so you'd use the second formula

You can't use the first formula for T_R

[woeful thicket]

OHHHHHHH

[abstract harness]

Yay

[woeful thicket]

shit ive been running around trying to do this so stupidly

imma try to resolve, but im going to keep this open in case i run into any other problems

Thank you

[abstract harness]

We all do that sometimes I totally understand

I found 87 let me know what you get

You're welcome

[woeful thicket]

For this did you mean Angles S_U and T_R

@abstract harness

[abstract harness]

I meant the arcs sorry if that wasn't clear

[woeful thicket]

no worries

[woeful thicket]

I am rereading this message and it makes so much sense now istg my eyes must have been closed'

[abstract harness]

Glad I could open them for you lol

Are you starting to figure it out now

[woeful thicket]

Everything up to solving for SVT makes sense

Just being able to say that SVT = 1/2(68 + 106) doesnt compute yk

OH

THE FIRST FORMULA '

[abstract harness]

Yep!

[woeful thicket]

How would you solve for arcs TR and SU again?

[abstract harness]

Using the second formula

The angle being 34 degrees is given so you could put that in

[woeful thicket]

Got it

34 = 1/2(TR+SU)

[abstract harness]

34 = 1/2(SU + TR)

[woeful thicket]

Multiply by 2 to get rid of 1/2

68 = TR _ SU

EZ

[abstract harness]

And you already got the 106 so add it to that

Then you have the full arc you need to find SVT

Using the first formula

[woeful thicket]

Holy shit i'm planning our wedding

that crazy ]

.closer

[abstract harness]

I'm taken-

[woeful thicket]

Gang...

.close

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[forest thorn]

So like real quick why is the correct answer correct?

[forest thorn]

The one that’s circled

[ebon kettle]

you integrate 2wice or smth

/rotate

[topaz geyser]

,rotate

[ebon kettle]

/rotate

[ebon kettle]

mb

ur sure its D?

[topaz geyser]

a(t)=P''(t)

[edgy sphinx]

you forgot the +c in the first integral

[edgy sphinx]

no like fr

[ebon kettle]

this is why

rip @forest thorn

[edgy sphinx]

the +C will account for the velocity at t=0

[ebon kettle]

this

[edgy sphinx]

same for the second step

the next +C will account for the position at t=0

[forest thorn]

Ah I probably did that in my head back when I did it a few months ago and now with reviewing for the AP exam I forgot

But yeah that makes sense

Makes a lot more sense

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[steel imp]

how do we solve for B?

shoudlnt it bet 25-x

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[midnight plank]

@red flint Has your question been resolved?

[red flint]

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[grand pond]

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

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[hollow elbow]

I need help again💀

[hollow elbow]

What method do I use now😭😭

[sage helm]

factorize it

you can start by dividing through by 2

probably makes it easier to see

[hollow elbow]

OH

That makes so much sende

Sense

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[gentle fossil]

can someone explain this?

[queen ermine]

Note, this is a unit circle

it's radius is 1

[gentle fossil]

but how is tangent 2 then

@queen ermine ?

[lament fox]

what does the tangent represent?

relating to the line PQ

[gentle fossil]

sin/cosine

[lament fox]

well yes,

but tan also represents the slope of the line PQ

so if you show that the slope is 2...

[gentle fossil]

since y axis is 2?

[lament fox]

yeah since you go up two units for every one unit to the right

[gentle fossil]

oh alr

thank you

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[dense pagoda]

I've been watching this problem for about 5 minutes, i have most of it done, I still need the time (y axis.)

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[late roost]

What question???

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[verbal ivy]

for d) do I first find the dot product with the angle from using tan(-). Then use cosine law to find the resultant vector?

[verbal ivy]

just want to confirm that my process is correct

[tiny drum]

just do the dot product by multiplying terms, see what you get

I don't think the tangent comes into play with dot products

[verbal ivy]

in order to find the angle for cos, I've learned to use tan

[tiny drum]

you don't need to worry about angles here

[verbal ivy]

am I trippin?

[tiny drum]

is that the only definition you learned?

[verbal ivy]

yes

[tiny drum]

ok

well then calculate the dot product however you normally would

[verbal ivy]

is this correct?

for d

[tiny drum]

yes you can first calculate the dot product

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[round wharf]

if $y=\frac{e^x+1}{e^x-1}$, then $\frac{y^2}{2}+\frac{dy}{dx}=$

[grand pond]

Specific Joe

[round wharf]

quotient rule

to get the dy\dx

$\frac{dy}{dx}=\frac{(e^x-1)(e^x)-(e^x+1)(e^x)}{(e^x-1)^2}$

[grand pond]

Specific Joe

[round wharf]

then we can split the fraction

$\frac{(e^x-1)(e^x)}{(e^x-1)^2}-\frac{(e^x+1)(e^x)}{(e^x-1)^2}$

wait

not necessary

we can simplify the numerator by distributing the e^x so that it will be

$\frac{dy}{dx}=\frac{e^{2x}-e^x-e^{2x}-e^x}{(e^x-1)^2}$

[grand pond]

Specific Joe

[round wharf]

so the $\frac{dy}{dx}=\frac{-2e^x}{(e^x-1)^2}$

[grand pond]

Specific Joe

[round wharf]

so $\frac{y^2}{2}+\frac{dy}{dx} = \frac{(\frac{e^x+1}{e^x-1})^2}{2} + \frac{-2e^x}{(e^x-1)^2}$

[grand pond]

Specific Joe

[round wharf]

so here is where im stumped

we can do $\frac{1}{2} \cdot\frac{(e^x+1)^2}{(e^x-1)^2}+\frac{-2e^x}{(e^x-1)^2}$

[grand pond]

Specific Joe

[round wharf]

I dont know what to do next

[valid fiber]

Well you did the hard part

What can you do with 2 fractions basically ?

[round wharf]

add them ?

[valid fiber]

Yeah

[round wharf]

so $\frac{1}{2}\cdot\frac{(e^x+1)^2-2e^{x}}{(e^x-1)^2}$ ?

[grand pond]

Specific Joe

[valid fiber]

Ouch

Are you sure they have a common denominator ?

In the first place

[round wharf]

yes

[valid fiber]

What about the 1/2 ?

Doesn’t it contribute to the denominator ?

[round wharf]

oh

wait

mb

we need to multiply by half first

so $\frac{(e^x+1)^2}{2\cdot(e^x-1)^2}+\frac{-2e^x}{(e^x-1)^2}$

wait

[grand pond]

Specific Joe

[valid fiber]

Yup now common denominator

[round wharf]

we multiple the other fraction by 2 right ?

[valid fiber]

Up and down

[round wharf]

the denominator and numenator

ye

[valid fiber]

Ye

[round wharf]

so $\frac{(e^x+1)^2}{2\cdot(e^x-1)^2}+\frac{-4e^x}{2\cdot(e^x-1)^2}$

now we add

[valid fiber]

Wait

[round wharf]

oh

I got it'

[grand pond]

Specific Joe

[valid fiber]

Ah yes more like it

But I mean who does his homework on latex

[round wharf]

exactly 😭

$\frac{(e^x+1)^2-4e^x}{2\cdot(e^x-1)^2}$

[grand pond]

Specific Joe

[valid fiber]

All good

[round wharf]

now we distribute the 2 ?

[valid fiber]

On what ?

[round wharf]

wait no we can't

we do the brackets then ?

[valid fiber]

Do you got any paper to try things out ?

Or do you really have only latex ?

[round wharf]

wait

alright I got some papers

[valid fiber]

Bc it’s kinda hard to try ideas on latex

For you

[round wharf]

yeah

[valid fiber]

So try stuff with your fraction, work it up a bit

And come back if you need help

But I doubt you’ll need help

[round wharf]

alright

[round wharf]

bet

[valid fiber]

Tag me it’ll be easier if you need help

What you got @round wharf ?

[round wharf]

figured it out

$\frac{e^2x+2e^x+2-4e^x}{2\cdot(e^2x-2e^x+2)}$

I expanded the brackets

then

[grand pond]

Specific Joe

[round wharf]

$\frac{e^2x-2e^x+2}{2e^2x-4e^x+4}$

[grand pond]

Specific Joe

[round wharf]

and subtracted the 4e^x in the numerator and multipled by 2 in the denominator

[round wharf]

and ended up with that

$\frac{(e^2x-2e^x+2)}{(2e^2x-4e^x+4)} = \frac{1}{2}$

[grand pond]

Specific Joe

[valid fiber]

Good job !

[round wharf]

Thank you

[valid fiber]

But you do realise you did extra work for the denominator ?

[round wharf]

what do you mean ?

[valid fiber]

A denominator and a nominator won’t interfere unless they have a common divisor

[round wharf]

maybe I shouldn't have multiplied by 2

[round wharf]

yeah

$\frac{(e^2x-2e^x+2)}{2\cdot(e^2x-2e^x+2)}$

[grand pond]

Specific Joe

[valid fiber]

So by extending the denominator, you should have in mind that you did basically nothing since at the end to simplify a fraction you need to get a common divisor

[round wharf]

so the brackets will cancel each other and we'll end up with 1/2

[valid fiber]

Even for the square in the denominator

[round wharf]

Yes I see what I did wrong

Thank you sm

[valid fiber]

But anyway

You got 1/2 so good job

[round wharf]

Thank you so much for your help

[valid fiber]

Don’t do exercice on latex ever again tho

[round wharf]

ye I learned my lesson, way easier on paper

[valid fiber]

That’s like terrifying

[round wharf]

true

thank you thoo

have a nice day

[valid fiber]

You’re welcome

You too

[round wharf]

.close

[midnight plank]

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[midnight plank]

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[shut axle]

not to tell the answer but could someone just explain how to find x&y

[shut axle]

i want to help my friend but i just forgot how to do these

[wise brook]

It's an inscribed quadrilateral so I'm pretty sure that the opposite angles are supplementary so they add up to 180 degrees.

[late roost]

Then it's just a system of equations

[late roost]

Had this saved since someone asked the same thing like an hour ago

[shut axle]

I see thank you so much

[late roost]

Not the same thing
Just similar

[shut axle]

hmm sorry one more

there is two variables when you add both up so how would i solve for one of them?

[normal river]

substitution

[shut axle]

OHH

i got it now tysm

[midnight plank]

@shut axle Has your question been resolved?

[midnight plank]

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[clear dew]

I have 0 clue on how to solve this exercise: (Learning Curve) An individual's efficiency in performing a routine task improves with practice. Let t be the time spent learning the task and y be a measure of the individual's performance. (For example, and could be the number of times, per hour, the task can be performed.) So an equation that is often used to relate y to t is:
y=A(1-e^(-kt) )

Where A and k are constants. After an hour of practice, a person on an assembly line can tighten 10 nuts in 5 minutes. After 2 hours, the person can learn 15 nuts in 5 minutes. Determine the constants A and k. How many nuts can the person tighten after 4 hours of practice?

[clear dew]

If k is a constant it shouldn't give me 2 diferent results, and when I try to get k from one of those results I'm getting 0 so I'm doing something wrong but I have no idea what

[tiny drum]

from the first equation, you can stop when you get e^{-k} = 1 - (10 / A)

then square both sides to get e^{-2k}, and plug it into the second equation. then you can solve for A

plug that value into the expression for e^{-k}, and you will find k

[midnight plank]

@clear dew Has your question been resolved?

[clear dew]

Not finished but something like this?

Just noticed a mistake but still, something among these lines?

[tiny drum]

I just left it as 1 - (10/A) and squared that

[clear dew]

You lost me

im replacing e^{-2k} for 1 - (10 / A)^2 since thats its value

then operating

[tiny drum]

instead of just calculating (1 - 10/A)^2, you did ((A - 10)/A)^2

just be careful w/ your arithmetic

[clear dew]

ah

just saw it

turned a 1 into an a

I'm re doing it and it's just the same thing

I don't get it

[tiny drum]

you need to solve for A

keep going and see what you get

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[clear dew]

.reopen

[midnight plank]

✅

[clear dew]

I'm running in circles

[stable owl]

What are you trying to solve

[lavish venture]

you squared a negative and got a negative

[clear dew]

Before I got like a2+5a+101 and putting that in a cuadratic equation gives an error on both sides

[clear dew]

This

[lavish venture]

line 5

[tiny drum]

be careful w/ your arithmetic as I said

and why did you square root both sides?

[clear dew]

its an equality, if i do something in one side i have to do it in both

[tiny drum]

ok... why are you taking the square root at all

[clear dew]

throwing anything agaisnt the wall

Not doing it leads me to this

And it's marked as an error Ina cuaddratic equation on both the positive and negative

[tiny drum]

you're not doing the arithmetic correctly still

you had (15 / A) - 1, you multiply both sides by A^2

go back and do things carefully. you will not end up with a quadratic

start from $15 = A \left ( 1 - \left (1 - \frac{10}{A} \right )^2 \right )$

[grand pond]

cwatson

[tiny drum]

you don't even need to divide by A. just work from the inside out

[clear dew]

This is my logic

[tiny drum]

that is not correct

[clear dew]

Is it because I move the 1?

[tiny drum]

yes. either way, you should not be ending up with a quadratic

[clear dew]

?

[tiny drum]

you need a minus sign out front on the right

like I said, go back to the start and work through it carefully

[clear dew]

This?

[tiny drum]

yes

[clear dew]

Ok so my train of tought is this i need to get rid of ^2 somehow the only two ways i see of doing it is using square root on both sides or to spred the ^2 to (-10+a)/a making it (10+a)^2/a^2 and (10+a)^2 is a^2+20a+100

[tiny drum]

why would you take the square root? then you'd have a complicated square root expression on the left

you need to solve for A

this is just an algebra problem

[clear dew]

becuase √(a)^2=a

they cancel eachother out

[tiny drum]

no, it equals the absolute value of A

[clear dew]

I mean i know it does that with some numbers but when ive solved equations in the past using that method its been fine

[tiny drum]

you are free to try it if you want

[clear dew]

i mean im not trying to be diffcult

its just i dont get the miskate im making

Like this is what I mean

so thats where im coming from

Or this

I think I did it

A=20

[tiny drum]

yup

[clear dew]

thank you, sorry for being frustrating

.close

[midnight plank]

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[last slate]

Hi, is this correct?

[midnight plank]

@last slate Has your question been resolved?

[last slate]

.close

[midnight plank]

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[midnight plank]

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[wooden knoll]

question

[wooden knoll]

in a triangle

yk the formulas $\ \ tan\theta = \frac{opp}{adj} \ cos\theta = \frac{adj}{hyp} \ sin\theta = \frac{opp}{hyp}$

[grand pond]

pixel

[cobalt swan]

yes

[stable owl]

Also, its for right angled triangles

[wooden knoll]

yes sorry

right angled triangles

if i were solving for theta

[cobalt swan]

yes

[wooden knoll]

which angle am i finding

[stable owl]

Yes, then its true.

[wooden knoll]

if that makes sense

[stable owl]

It makes perfect sense.

[wooden knoll]

so you see how we have the 90 degree angle

and the other 2 angles

say the other 2 angles are both missing

when solving for "theta"

[cobalt swan]

u know all the sides right?

[wooden knoll]

how do I know which angle im solving for

yes

[cobalt swan]

ok so look

can u name the triangle and send pic again?

[cobalt swan]

ABC also

[cobalt swan]

ha yes

so if u consider angle A is thetha

u already know wht hypotenuse is

the side which is common to angle A that is AC is called adjacent

and BC is called perpendicular

[wooden knoll]

oh right right

ok yes sorry I was just a bit overthinking it

[cobalt swan]

so for angle A

mhm

[wooden knoll]

depending on which angle I want to calculate, the adjacent and opposite swap

if that makes sense

so if I were solving for angle B, the "opposite" would be 1.6 and adjacent is "1.2"

whereas solving for angle A, the "opposite" is 1.2 and adjacent is '1.6"

[cobalt swan]

yes

[wooden knoll]

hypotenuse stays the same

[cobalt swan]

yes exacty

[wooden knoll]

yeah thank you for clearing that, I'm just a bit tired and overthinking some questions

😭

are you good with vectors by any chance

cause im solving a question for that rn

[cobalt swan]

yep

[wooden knoll]

and theres no answer sheet

[cobalt swan]

np

[wooden knoll]

is it fine if i send you a question and have you check my answer

[cobalt swan]

yes thts fine

[wooden knoll]

mm ok

let me finish writing smthn up

also it asks

"Describe the position at which she arrives"

thats the angle

right?

[cobalt swan]

idk put the question so i can understand the question

[wooden knoll]

ok

easy

one sec

[cobalt swan]

ok

[cobalt swan]

its just another right angle triangle

[wooden knoll]

well yeah idk if thats clear

[cobalt swan]

woah wht is this

[wooden knoll]

but I've done most of the question

[cobalt swan]

the position is not thetha

[wooden knoll]

that is a horrible image

[cobalt swan]

lol

[wooden knoll]

oh

what would it be

would it be a vector then?

so in terms of 'i' and 'j'?

[cobalt swan]

yes

the length of the vector

[cobalt swan]

or ya u can consider this too

[cobalt swan]

this is distance

[wooden knoll]

essentially we're finding the vector that takes us to point B, correct?

can it be 1.6i and 1.2j?

1.6i + 1.2j

[cobalt swan]

that would be the velocity vector

[wooden knoll]

oh

[cobalt swan]

he gave length of river

[wooden knoll]

right

[cobalt swan]

use that

[wooden knoll]

oh yeah

i mean in terms of length of river

in the triangle for that

we only have one side

[cobalt swan]

but u have the angle

[wooden knoll]

oh 90?

[cobalt swan]

from the velocity triangle

[wooden knoll]

what angle do I have

unless ur telling me to calculate them

which I can do

using what we mentioned earlier, the cos/tan/sin thing

oh yes

thats right

and since its parallel banks

i can find the alternate angles or whatever

and find the angles of that other triangle

that makes a lot of sense

[cobalt swan]

u got both angles

[wooden knoll]

I already have them?

oh

45

wait no

[cobalt swan]

no

[wooden knoll]

okay im confused, how do I have the angles?

I can calculate them yeah

unless that's what your saying

[cobalt swan]

from this

triangle

u can calculate them

[wooden knoll]

😭 yeah thats what I said I was gonna do, sorry i misunderstood you

i thought you were saying I already had them

so i was like huh?

[cobalt swan]

oh ok lol

[wooden knoll]

ok i calculated angles

now

since the banks are parallel

i am able to use some sort of rules

to find the angles in the other triangle

[cobalt swan]

wht?

[wooden knoll]

https://cdn.discordapp.com/attachments/1018703621841494076/1239825113126277151/rn_image_picker_lib_temp_68375d2b-3ac7-40a4-a081-2c49743c5bf7.jpg?ex=6644549d&is=6643031d&hm=631906c61e51c594027582bba7ef4bdd563a101bcf9d30646db6f4735c901cd7&

the bnaks are parallel

[cobalt swan]

yea just draw the position triangle

[wooden knoll]

since we know what alpha is on the position triangle

we can find the last angle easy

[cobalt swan]

yea tht works

[wooden knoll]

let last angle = theta

theta = 180 - (90+ alpha)

[cobalt swan]

but the one u marked is not 48

the other side is 48

he said width of river

[wooden knoll]

its the same thing

1.6 just shows how fast the swimmer went

48 is the width tho

remember the swimmer tried to swim across the river

so that means like a horizontal line pretty sure

https://cdn.discordapp.com/attachments/1018703621841494076/1239825113126277151/rn_image_picker_lib_temp_68375d2b-3ac7-40a4-a081-2c49743c5bf7.jpg?ex=6644549d&is=6643031d&hm=631906c61e51c594027582bba7ef4bdd563a101bcf9d30646db6f4735c901cd7&

[midnight plank]

@wooden knoll Has your question been resolved?

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[obtuse wing]

May be a stupid question, but are the equations the same? Is the default for calculating similar derivatives to use the quotient of the derivatives?

[stable owl]

What do you mean by the quotient of the derivatives?

[obtuse wing]

I mean using this formula

[stable owl]

Right, what are you confused about?

[obtuse wing]

Im wondering if it matters if I use this formula or the formula for product of two derivatives, as the results are a bit different

[stable owl]

Have you tried graphing them out?

Sometimes they may look different but are the same functions

[obtuse wing]

I have not

[stable owl]

Try it out

[obtuse wing]

I will, but if I calculated it correctly, they should be the same?

.close

[midnight plank]

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[midnight plank]

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[alpine matrix]

Idk how to do in letter B

[cedar zodiac]

which one do you need help with

[alpine matrix]

Letter B

[cedar zodiac]

Angle A = Angle E

Angle E = 32

Angle A = 9x +14

9x+14=32

solving this gives x=2

[alpine matrix]

Should I write that all?

[cedar zodiac]

that should be enough

angle A = angle E because of Z angles

What's meant with question b

what is m<A

[alpine matrix]

Angle?

Angle A and angle E

@cedar zodiac

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[normal river]

are you doing B1?

[alpine matrix]

Ye

I'm doing B1B

I'm done in 1A

[normal river]

so you have x right

already

[alpine matrix]

Yes

[normal river]

so B1B is asking you to find the measure of angle A, which basically means what's the angle

so all you have to do is plug in x for the angle A

[alpine matrix]

Wdym

[normal river]

do you see where the A is at on the triangle

[alpine matrix]

Ye

[normal river]

theres an equation near there 9x + 14

that represents the value of the angle

[alpine matrix]

Ok

[normal river]

plug in what you got for x into that equation

and that gives you the measure of angle A

[alpine matrix]

X is 2

[normal river]

yes

[alpine matrix]

So how do I put solution

[normal river]

$9x + 14 => (9 * 2) + 14$

[grand pond]

MerryTacoGamerCat

[normal river]

solve that

and you get your answer

[alpine matrix]

X=2

[normal river]

do you understand what i mean when i say plug in what you got for x

[alpine matrix]

Yes

[normal river]

so plug in 2 into the equation to get your final answer for the question

[grand pond]

Marcus

( x = 2 \), \( 9x + 14 = 32 \).
```Compilation error:```! LaTeX Error: Bad math environment delimiter.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 ( x = 2 \)
               , \( 9x + 14 = 32 \).
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

[alpine matrix]

Uh

<@&286206848099549185>

[normal river]

what are you trying to do

[alpine matrix]

Idk

[normal river]

do you still need help

[alpine matrix]

Ye

[normal river]

with what

[alpine matrix]

B1B

[normal river]

ok

what does it say

repeat it

[alpine matrix]

Find the m<A

[normal river]

do you understand what that means

[alpine matrix]

Can u explain?

[normal river]

m<A means measure of angle A

which basically means the number angle of A

[alpine matrix]

How do I get the measure

[normal river]

there is an equation near A on the diagram

labeled (9x + 14)

that represents the angle measure

you told me that x = 2

so now you just need to plug in 2

into the equation 9x + 14

to get your answer

[alpine matrix]

Wdym by plug in

[normal river]

https://youtu.be/tuVd355R-OQ

watch this to better understand

[alpine matrix]

I don't have time, I need to do this quick

[normal river]

are you just looking for answers

[alpine matrix]

Idk how to do the plug thing

[normal river]

replace x with 2

and solve the equation

[alpine matrix]

Then it's 106

[normal river]

no

[alpine matrix]

What

92+14

[normal river]

what grade are you in out of curiousity

[alpine matrix]

Bro

[alpine matrix]

I'll say after I'm done in this thing

[normal river]

9x means 9 times x so when you replace x with 2 you have 9 times 2 plus 14

[alpine matrix]

32

[arctic turtle]

if you have 680$ and thats 34% less than the original price whats is the original price then? and how do you solve it

[normal river]

yes

[normal river]

!help

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[alpine matrix]

So is that how u answer it

[normal river]

yes

[alpine matrix]

K

What to do in B2A

[normal river]

i have to go sorry but i suggest watching the video it might help

[alpine matrix]

I don't got time

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

Pls help

<@&286206848099549185>

[storm gorge]

Hi marcus

still struggling?

[midnight plank]

@alpine matrix Has your question been resolved?

[midnight plank]

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[plucky dust]

for the maclaurin series of the top function, i plugged in x^4 to the cosine series, but what would you do with the squared part?

[main current]

Cry

[plucky dust]

[main current]

Sadly there's no easy algebraic way to get products of series into one series

[burnt flame]

you could apply power reduction formula to cos()^2 originally

[main current]

Other than multiplying the members out one by one

[plucky dust]

what is that again

i suck at the trig rules

[burnt flame]

you can derive it from cos(2x)=cos^2(x)-sin^2(x)

[burnt flame]

i believe its 1/2(cos(2x)+1)=cos^2(x)

[plucky dust]

so using this rule?

yeah

[burnt flame]

yea

so cos^2(x^4)=1/2(cos(2x^4)+1)

[plucky dust]

oh ok thanks i should be able to work that out

so in general you cant really multiply series together?

[burnt flame]

there is a way

but i think it involves convolutions and whatever

[plucky dust]

yeah thats probably way too advanced for me

so i gotta stick to angle identities

ok that makes sense thanks

.close

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[last slate]

idk how they got 2.11

.solved

[midnight plank]

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[drowsy briar]

I know the answer is A but why?

[drowsy briar]

I see the equations in regular y=mx+b form but I given these options instead

[stable owl]

What is the equation in y=mx+b?

[drowsy briar]

y=2/3x+(-2)

[stable owl]

Yes. y=2/3x-2.

We want it in the form of the question

What if we multiplied both sides by 3?

What would happen?

@drowsy briar

[drowsy briar]

What do you mean by both sides?

and why thre?

[stable owl]

Like the side with y, and the side with 2/3x.

So that it cancels out the 2/3.

(2/3)(3)=2

Which is an integer.

Notice that in the question there is no fractions.

As choices.

[drowsy briar]

yuh but the choice is solely base on the answer choices?

or am i multiplying the three cuz of the 2/3

[stable owl]

Yeah.

[drowsy briar]

on

k

[stable owl]

We want to make it such that we have something similar to the answer choices.

rearrange the equation y=2/3x-2

You would agree that it is the same as 3y=2x-6 right?

I multiplied both sides by 3.

[drowsy briar]

ok

so we multiply 3 with all of the numbers

including the y

[stable owl]

Yes, both sides of the equality.

What happens when I move the 6 to the left hand side?

[drowsy briar]

you would have to add 6 to rid of it?

but u cannot add a variable without being the same?

[stable owl]

I didn't add anything.

Look.

y=(2x-6)/3

y=2/3x-2

Which is the same thing as we started with.

[drowsy briar]

i am so lost but continue, if we were to put the -6 on the other side it would be -6+3y=2x

?

[stable owl]

Why

It was -6

If we move it it would be +6

6+3y=2x

[drowsy briar]

ok

i remember now that if u flip numbers u gotta flip signs

[stable owl]

What if we move the 3y to the right hand side?

It would be 6=2x-3y

Which is the answer choice A

[drowsy briar]

ok

how does the other options not complete the answer

like could i not just do same shit we did with the other options

and get the same result or no?

[stable owl]

You could put every choice into y=mx+c

Btw, you sure you are in undergrad level math?

Just curious

[drowsy briar]

like people in college/uni that take math classes no?

i thought that what was the role was meant for

[stable owl]

Yes.

[drowsy briar]

yuh than i am, i am a psych major

and havent done any sort of algebra,arth, or mathj

in years

[stable owl]

Oh. Its more suited as people that take like calculus 3, abstract math, linear algebra, topology, complex analysis...

[drowsy briar]

havin to learn basic shit to get into a science program

[stable owl]

Right

[drowsy briar]

ahh

i got u

[stable owl]

But yeah no one is gonna judge u just because you have this on

Its good to brush on the basics.

Right you got everything I told you right?

You understood right

[drowsy briar]

i didnt know that, i just thought it was for college mens yk; ill remove it if its botherin u

but back to the quesiton

[stable owl]

No ofc not

I am just curious lol

Its not bothering me in anyway

[drowsy briar]

u said any of options will work in the y=mx+b formula

[stable owl]

I'll do one for you

Let's do option B.

3x-2y=6.

Look at the general formula of y=mx+b. x is on the right hand side. What is a logical thing to do?

[drowsy briar]

AHHHHHHHH

fuck

some of this shit is coming back to me