help-49

[last slate]

In this instance 1 to -1 (displacement)

Then?

Would it be 2?

[lime matrix]

at t = 4, the displacement is -1

at t = 2, the displacement is 1

[last slate]

Yes, indeed

[lime matrix]

since it's final - initial, you get -1 - 1 = -2

[last slate]

Got it

-2/2

Thus -1

Correct?

[lime matrix]

yup

that's the velocity from t=2 to t=4

do the same for the next one, from t=4 to t=10

[last slate]

Which is 2

Over 6

1/3

Correct?

[lime matrix]

yes

[last slate]

Sketch a velocity time graph

They are weird

[lime matrix]

also, is this physics?

[last slate]

Mechanics

Maths ^

[lime matrix]

i see

[last slate]

Now for the velocity time graph

Create different sections in the graph for when the displacement changes

[lime matrix]

yes

[last slate]

Then place them in accordance to their velocity in a new graph

[lime matrix]

yeah

[last slate]

I got this

Time to make the graph

Don’t mind the bad writing

And now for d

[lime matrix]

well

[last slate]

Constant velocity is 0

[last slate]

Which is 2<t<3

PH YEAH

I forgot

Fixed

[lime matrix]

👍

constant velocity is when the slope is flat

[last slate]

For D, is that the answer?

[last slate]

0 velocity

So 2<t<3

New Q

[lime matrix]

i would say on 0<t<10

[last slate]

That wouldn’t be constant velocity

[lime matrix]

constant velocity would be when the velocity has a straight line

and isn't changing

[last slate]

Yeah

0 to 10

Changes a lot

[lime matrix]

mb, you could say that it's a constant velocity from t=0 to t=1, from t=1 to t=2, from t=2 to t=4, and from t=4 to t=10

since the lines there are flat

[last slate]

It changes

By flat, they need to be horizontal

[lime matrix]

yes

these are horizontal

[last slate]

Hmm

Maybe you’re right

[lime matrix]

👍

[last slate]

Fixed

[last slate]

What do I do for b

[lime matrix]

i'm sorry that i can't help more, but i have to go as i''ll be busy

[last slate]

Ah okay

Nw bro

[lime matrix]

displacement is the area under the curve for a velocity vs. time graph

[last slate]

Have a good one

[lime matrix]

you too!

[last slate]

Take care 🙂

How do I work out the distance?

If anyone can help

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[pliant sorrel]

i did the area of the rectangle - the area of the two small triangles = the area of the triangle of the lower half

so 15 - 1.5 - 3 = 10.5

but idk how to calculate the area of the top half of the tirangle, the part protruding the square

[midnight plank]

@pliant sorrel Has your question been resolved?

[pliant sorrel]

<@&286206848099549185>

[last slate]

Here

So, the base is equal to 5-1-2=2

[pliant sorrel]

yes

[last slate]

We can use analytic geometry to solve it. Let $y_1=\frac{-3x}{2}+\frac{15}{2}$ and $y_2=3x$.

[grand pond]

Palahoo

[last slate]

So, we can equivale the both functions

Get the distance between the E point and the x axis

[pliant sorrel]

wait where u get hose valies

values

[last slate]

Do you know calculus?

[pliant sorrel]

no

[last slate]

You can find it without calculus, but it'll help

Ok, no problem

Using the y=ax+b model

The a will be the height divided by the lenght

In this case, -3 (because it is descending) divided by 2 (the lenght)

[pliant sorrel]

um isnt that the line graphing equation

how does that relate to this

[last slate]

Analytical Geometry

[pliant sorrel]

i only know distance and midpoint formula

[last slate]

See?

So, with y1, the a will be equal to 3 divided by 1

As y1(0)=0, b=0

The a of y2 will be equal to -3 (because it's descending) divided by 2

So, we have that $y_1=3x$ and $y_2=\frac{-3x}{2}+b$

[grand pond]

Palahoo

[last slate]

To find b, we know that y2 is equal to zero at x=5

[pliant sorrel]

um isnt y1 the one decreasing

is y1 the line starting from A

[last slate]

Yes, it's starting from A

It's increasing

Getting bigger

[pliant sorrel]

ohhh

i see

wait how did u get -3/2

for y2

[last slate]

Because it decreases -3 unites by 2 units of x axis

[pliant sorrel]

nvm

yeah

carry on

[last slate]

What's nvm?

[pliant sorrel]

never mind

[last slate]

Ok, thank you!

So, the rate of "increasing" (aka a value) is equal to $\frac{-3}{2}$

[grand pond]

Palahoo

[last slate]

Because it's a "negative increasing"

Did you get it?

[pliant sorrel]

yeah

[last slate]

So, to find the b of y2, we know that y2(5)=0

[pliant sorrel]

yes

[last slate]

So we get that $y_2(5)=0=\frac{-3\cdot 5}{2}+b$, therefore $b=\frac{15}{2}$

[grand pond]

Palahoo

[last slate]

Did you get it?

Until now?

@pliant sorrel

[pliant sorrel]

um

whats -3x5/2

[last slate]

The function

[pliant sorrel]

oh

[last slate]

As we got early that a=-3x/22

[pliant sorrel]

yeah

[last slate]

Did you get it?

[pliant sorrel]

yes

[last slate]

Ok

[last slate]

So, we got these two functions

Now, we need to equivale them to get the E point

[pliant sorrel]

-3x/2 + 15/2 = 3x

[last slate]

Yes

So, what's the result of x?

[pliant sorrel]

5/3

[last slate]

Really?

Try to put this value onto the equation and verify it!

[pliant sorrel]

yes

its correct

[last slate]

Oh, sorry

I got confused here, sorry

Yeah, it's correct

So, it's just substitute 5/3 on 3x and get the height

Which is 5

5-3=2

[pliant sorrel]

(5/3,5)

[last slate]

So, we have a smaller triangle with 2 as basis and 2 as height

So, it's just get this area and sum with the bigger bottom triangle you already got

[pliant sorrel]

so its 4?

14.5?

2

12.5

[last slate]

Yes

Sorry for the "exotic" way to solve it

But was the first thing I thought

[pliant sorrel]

damn thats a complcaite method

is there a easier way to solve this?

[last slate]

I think so

[pliant sorrel]

like this question is in similar triangles section

so maybe something ot do with that

[last slate]

I already saw this problem

But I just got the answer now

[pliant sorrel]

theres another question im stuck on can u help me pls i think this ones very easy but i just cant solve it

$60$ bricklayers began a $16$-day project to build a castle. After five days, five of the bricklayers quit the job. If they work the same amount each day, how many extra day(s) (beyond the original estimate of $16$) will the remaining bricklayers need?

[grand pond]

龖

[pliant sorrel]

yeah

[last slate]

The total amount of work is equal to $60\cdot x$, where x is the amount of each worker each day

[grand pond]

Palahoo

[last slate]

Sorry

[pliant sorrel]

?

[last slate]

So, calling the total amount of work necessary for the building as Y, we get that the total days is equal to $\frac{Y}{60\cdot x}$

[grand pond]

Palahoo

[last slate]

Which is 16 with the 60 workers, according to the question

[pliant sorrel]

yeah

?

[last slate]

So, to the the x value, we can multiply $60\cdot x$ both sides, and get $Y=16\cdot 60x=960\cdot x$

[grand pond]

Palahoo

[last slate]

And, finally, x is equal to Y/960

[pliant sorrel]

yes

[last slate]

Sorry, it was not necessary, and I just realizes it now

[pliant sorrel]

huh

Solution:
After five days, 60 bricklayers could finish the job in 11 days. Since the time to finish is inversely proportional to the number of bricklayers present, 55 bricklayers would need $\frac{60}{55}\cdot 11=12$ days to finish, $\boxed1$ more than the original plan.

[grand pond]

龖

[pliant sorrel]

this was the solution on the website

i dont understand it

[last slate]

Sorry, I think my mind is not going so well right now

[pliant sorrel]

how does 60 bricklayers finish in 11 days

isnt it 55

since 5 left

[last slate]

Oh, I got it

Remember that Y was equal to 960x?

[pliant sorrel]

yes

[last slate]

So we get that 960x=60x.5+55.x.d

Where d is the remaining days

Dividing both sides by x

We get 960=300+55d

[pliant sorrel]

im confused

what is the 5 and 55

[last slate]

So d is equal to 660/55=12

The total amount of work is equal to $60\cdot x$, where x is the amount of each worker each day

[grand pond]

Palahoo

[last slate]

So, to the the x value, we can multiply $60\cdot x$ both sides, and get $Y=16\cdot 60x=960\cdot x$

[grand pond]

Palahoo

[last slate]

Then $960\cdot x = 60\cdot 5 + 55x\cdot d$, where d is the remaining days

[grand pond]

Palahoo

[last slate]

Dividing both sides by x, we get

$960= 300 + 55d$

[grand pond]

Palahoo

[last slate]

And then $d=\frac{660}{55}=12$ days

[grand pond]

Palahoo

[last slate]

About the website solution, the remaing amount of work would be done by 60 in 11

Or the $60x\cdot 11$, with the notation I used above

[grand pond]

Palahoo

[pliant sorrel]

why does 60 need to be multiplied by 5

[last slate]

Becaused we had 60 workers working by 5 days

And the remaining 55 working by d days

[pliant sorrel]

this is so confusing

[last slate]

What didn't you understand?

[pliant sorrel]

this part

[last slate]

Did you undestand until here?

[pliant sorrel]

yeah

[last slate]

So, I just divided both sides by x

As x is different of 0

[pliant sorrel]

wdym

divide what

i dont understand why 69 x 5

60 x 5

thers 60 workers and youare multiplying 60 workers for 300 workers?

[last slate]

Calm

60 workers did the x amount of work each day for 5 days

So the total amount of work is $60x\cdot 5$

[grand pond]

Palahoo

[last slate]

60 times 5 is equal to 300

So 300x

[pliant sorrel]

so it was 60x not 60?

[last slate]

Yes

AND I got the equation 960x=300x+60xd

[pliant sorrel]

yea h iunderstand no

how dod ueven think of doing these stuff

like 3 different variables

do i just need to do more of these questions

[last slate]

Sorry, I generally use these "exotic" ways to solve questions

[pliant sorrel]

no its good i just dont know how people think of these solutions like i can never think of such a complex soltion

[last slate]

Did you undestand my solution?

[pliant sorrel]

yes

[last slate]

Do you want to undestand the website's solution (with my way as well)?

[pliant sorrel]

yeah

[last slate]

Ok

So, 60 workers did the x amount of work each day for 5 days

60 workers would do the remaining of the work working x each day for 11 days

Which is $60x\cdot 11$, using the notation above

[grand pond]

Palahoo

[pliant sorrel]

huh

isnt it 55 though

cuz 5 left

[last slate]

"would do"

[pliant sorrel]

yes

[last slate]

But, as 5 workers went out, the $60x\cdot11=660x$ of remaining work will be now done by the 55 remaining workers

[grand pond]

Palahoo

[last slate]

In one day, the total amount of work done by all the workers is 55x

So, dividing the 660x of the TOTAL remaining work by the 55x DAILY worked, we'll get 12

Did you understand?

@pliant sorrel ?

[pliant sorrel]

oh

yeah

thanks

i gotta write the solution down

[last slate]

Ok

[pliant sorrel]

60 workers did the ( x ) amount of work each day for 5 days, and 60 workers \textbf{would} do the remaining of the work working ( x ) each day for 11 days.

Because 5 workers left, the ( 60x \times 11 = 660x ) of remaining work will now be done by 55 workers. So dividing ( 660x ) of the total remaining work by the ( 55x ) daily worked, we'll get 12.

With 60 workers, it takes 11 days, but with 55 workers, it will take 12 days, giving us the answer of \textbf{1} extra day with 55 workers.

[grand pond]

龖

[pliant sorrel]

@last slate

is this good?

[last slate]

For me, that's ok

Just be aware if you need to have a formal english

Or be formal in the language you'll write it

[midnight plank]

@pliant sorrel Has your question been resolved?

[midnight plank]

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[uneven sandal]

hi I need help solving this integral for odd n

[uneven sandal]

it should just be -n! but im not sure how to show this

because for odd n if x goes to 0 then u goes to -infinity

and then its not in the form of the gamma function

.close

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[pliant sorrel]

Two rockets fly directly at each other. One travels at a speed of $8500$ miles per hour, and the other travels at a speed of $3500$ miles per hour. How many miles apart will they be one minute before they collide?

[grand pond]

龖

[fallow scarab]

Solve for when they collide

[pliant sorrel]

idk how

i know u have to set two thing equal to each other

[fallow scarab]

Yes

[pliant sorrel]

idk what to set equal to each other

[fallow scarab]

Oh no what I said is unnecessary

You just find the relative velocity

[pliant sorrel]

relative velocity?

[weak prawn]

Yea, so basically what is the relative velocity of one rocket to the other

Say I am motionless, and something is approaching me head on at 120mph

How fast is that object moving towards my relative frame of reference?

[pliant sorrel]

um

idk

ive never heard of relativ velocity

[weak prawn]

That's fine, let me word it more simply

I am sitting in the middle of a highway due to poor life choices

and I see an 18 wheeler coming right at me at 120mph

if I am sitting motionlessly on the road, how fast is that truck coming at me

[pliant sorrel]

isnt it just 120mph

[weak prawn]

yea

now say I want to make it go quicker

and I begin running head on towards that truck at 5mph

and it is still moving 120mph relative to the road

how fast is the truck approaching me now that I am running 5mph towards it

[pliant sorrel]

is it 120x-5x=5xo r osmething

120-5x

[weak prawn]

in this case you just add the two velocities together

[pliant sorrel]

oh

125

[weak prawn]

yea

So the same can be done with the rockets

[pliant sorrel]

so its coming at you 12,000m/h?

[weak prawn]

Yea

so 12,000mph, we want to know how many miles away it will be at 1 minute

so simply find how much distance it will travel in 1 minute

might be easier to convert to miles per second

[pliant sorrel]

10/3 miles per second?

[weak prawn]

yea

[pliant sorrel]

so 200

[weak prawn]

then multiply the speed by time to get distance

[pliant sorrel]

per minute

[weak prawn]

yea

[pliant sorrel]

thanks

.close

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[blissful pier]

,rccw

[last slate]

i don't understand this.

[midnight plank]

@soft ocean Has your question been resolved?

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[naive flint]

hi

[errant forge]

hey

[naive flint]

Is that one okay like that

[tacit rose]

No

You're suppose to use trig ratios

Like SOH CAH TOA

[naive flint]

ohhhhhhh

Ok

[midnight plank]

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[midnight plank]

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[keen seal]

to find this derivative

uhhh

is it just me or is it super super long

[burnt flame]

my guess is there is some way to simplify this

[keen seal]

yeah

uhm

[keen seal]

cuz

tbh

im hardstuck on this

cuz

its way too long what i have

[burnt flame]

i mean technically you could do implicit differentation, and draw out a triangle to get the other trig function

i.e. write tan(y)=x/sqrt(4-x^2)

[keen seal]

yes but

[burnt flame]

ye, that's much nicer

[keen seal]

its part of a bigger equation

[burnt flame]

yea?

[keen seal]

i can stil do that?

[burnt flame]

you can split the derivative among them

d/dx (f(x)+g(x)+h(x))=d/dx f(x) + d/dx g(x) + d/dx h(x)

[keen seal]

ah

[keen seal]

wait uhm

how do i do that

i dont think ive ever done

this

likme

[burnt flame]

ok, so you know what tan(x) represents right? opposite over adjacent

[keen seal]

what after tany=x/sqrt(4-x^2)

yes

[burnt flame]

ok, then take derivative of both sides

have you done implicit differentation?

[keen seal]

yes i have

[burnt flame]

so you get d/dx (x/sqrt(4-x^2)) =...

and d/dx (tan(y))=...

you can do d/dx of tan(y) atleast, yes?

[keen seal]

yeah

so

sec^2(y)=

d/dx of x/sqrt

[burnt flame]

forgot y'

chain rule

[keen seal]

ah

so

sec^2(y) y'

=

d/dx (x/sqrt(4-x^2)) =...

[burnt flame]

yes, and that's just quotient rule, but you're wondering how to get sec^2(y)?

well, you can either use an identity (identity involving sec^2 and tan^2), or draw out the triangle

[keen seal]

ill draw a triangle

[burnt flame]

ok, so if you have tan(y)=x/sqrt(4-x^2), what does the triangle look like, with an angle of y?

[keen seal]

so opposite is x

[burnt flame]

mhm

[keen seal]

adjacent is x/sq

[burnt flame]

ye

[keen seal]

ok wait i uhm

does secy

at the end make

4-2^2/x

wait nvm]

i tought sec was cot

[burnt flame]

sec(y)=hypotenuse/adjacent

[keen seal]

i need to chemically cleanse this memory

[keen seal]

does the

hypotenuse

after everything

=2

[burnt flame]

ye, hypotenuse is 2 :)

so what should (sec(y))^2 be?

[keen seal]

wait so

sec(2/sqrt(4-x^2)^2

[burnt flame]

errr

no

what is the hypotenuse, what is the adjacent

sec(y)=hypotenuse/adjacent

square both sides

[keen seal]

2 is hypote and sqrt is adjacent

[burnt flame]

yes

sqrt(4-x^2)

[keen seal]

ytes and

when you say square both side

[burnt flame]

mhm, you get...?

[keen seal]

now essentially sec(y)^2

oh

wait

i frogot sqrt of a fraction

[burnt flame]

you're squaring a fraction

we want sec(y)^2

we have sec(y)

[keen seal]

ah can i

^2

both side

[burnt flame]

yes.

squaring

[keen seal]

oh mb i got confused

so sec^2 is

4-x^2/x?

[burnt flame]

what is the hypotneuse

what is the adjacent

[keen seal]

mb i looked at the wrong paper

okay

so

4/4-x^2

[burnt flame]

yes!

so we have

[keen seal]

and therefore

since sec^2

is 4-x^2/x?

y' is therefore

[burnt flame]

no sec^2 is 4/(4-x^2)

[keen seal]

yes

LOL

i keep looking at the wrong one

omg

sleep deprived asf im sorry

[burnt flame]

$y'=\frac{d}{dx}\left(\frac{x}{\sqrt{4-x^{2}}}\right)\left(\frac{4-x^2}{4}\right)$

[grand pond]

Moosey

[burnt flame]

do you see why this is true?

[keen seal]

yes i do

im trying to calc this rn to see if its right

[burnt flame]

then you need to add it to your derivative of x^2e^-x^2 stuff (product rule+chain rule), and add derivative of 2p/3 (which is just 0)

[keen seal]

yes

btw the derivative of

the e

is what i wrote so far good

[burnt flame]

ye :)

wait

dropped a negative i think

-2x became 2x in next line

do you see?

so just make x^2+1 into x^2-1

and add that negative back

[keen seal]

ah thats true

but these two

like

the e and the arctan

they make such diffrent results

do i just leave them or do i try to add them

[burnt flame]

mhm

[burnt flame]

well, leave the e part as is (with corrections), and then calculate derivative of this part completely, then add them together at the end to get final result

[keen seal]

wait i could be

completely wrong but

if after implicit diff at the tany

it ends up being

sec^2(y)y'=d/dx (x/sqrt(4-x^2)

wouldnt y' be

d/dx (x/sqrt(4-x^2) divided by sec^2(y) instead

[burnt flame]

y'= d/dx(x/sqrt(4-x^2))cos^2(y)

yes

you could think of it this way as well

[burnt flame]

see how i flipped sec^2(y) here?

[keen seal]

yes

but what i mean is

isnt the sec^2

already differntiated?

[burnt flame]

ye!

so just leave it alone

only do derivative of x/sqrt(4-x^2)

[keen seal]

ohh and then i multiply them

[burnt flame]

mhm

:)

[keen seal]

okay okay ty!

just to make sure

i can just leave it as it is

and multiple right

[burnt flame]

i don't think you did the simplification right

[keen seal]

oh

wahts another way i can do it

i cant seem to find one

[burnt flame]

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

[grand pond]

Moosey

[burnt flame]

right?

i would multiply numerator and denominator by (4-x^2)^(1/2)

so

$=\frac{4-x^{2}-\frac{x}{2}}{(4-x^2)^{\frac{3}{2}}}$

[grand pond]

Moosey

[keen seal]

ah

but whats trhe uh

idk how to formulate it

like

what would be the difference simplifying the way i did and you did

[burnt flame]

how'd you go from the second line to the third line?

d/dx=line to your sqrt(4-x^2)-x line

[keen seal]

first i moved the (4-x^2)^-1/2

down

so it becomes a sqrt

[burnt flame]

you can't move it down

it's being added

[keen seal]

yes added

i added (4-x^2)^1/2

so nomi and denominator

[burnt flame]

OH WAIT

you also forgot chain rule

[keen seal]

wai tomg

[burnt flame]

what is the derivative of (4-x^2)^(-1/2)

[keen seal]

-1/2(4-x^2)^-3/2

-2x

[burnt flame]

wait, agh, i mean derivative of sqrt(4-x^2), sorry

[keen seal]

ah wait

the chain rule

[burnt flame]

yes

[keen seal]

where is it that i missed

[burnt flame]

you missed a chain rule part, so you should have an extra -2x, yes

[keen seal]

is it back in the sec?

[burnt flame]

you should have a -2x with the 1/2, when you were doing quotient rule

[keen seal]

OH

YEAH I MISSED

[burnt flame]

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

but +

because -2x

[grand pond]

Moosey

[burnt flame]

since you have -x^2

this is right ^

[burnt flame]

this is right

and you can simplify this

by multiplying numerator and denominator by (4-x^2)^(1/2)

and it simplifies to something nice :)

[burnt flame]

uhhh

[keen seal]

wait uh

oh

[burnt flame]

it should be -2x

sorry

lmao

[keen seal]

i multiplied first

[burnt flame]

calculate derivative completely, then multiply in the (4-x^2)/4

[keen seal]

idk why i multiplied the 4 with

top

mb

yeah

[keen seal]

wait how did you

remnove the 2

[keen seal]

in here

[burnt flame]

$\frac{d}{dx}(4-x^2)^{\frac{1}{2}}=-2x \cdot \frac{1}{2}\cdot (4-x^{2})^{-\frac{1}{2}}$

[grand pond]

Moosey

[keen seal]

yes

there is a 2 on the second part

i meant ont he frist

the first*

like the part before the -

[burnt flame]

wydm

the -2x and 1/2 cancel to give -x

$\frac{d}{dx}(4-x^2)^{\frac{1}{2}}=-2x \cdot \frac{1}{2}\cdot (4-x^{2})^{-\frac{1}{2}}=-x(4-x^2)^{-\frac{1}{2}}$

[grand pond]

Moosey

[keen seal]

oh mb i

moved the

1/2

down the fraction

oh

thats not even allowed

idk why i did that

oh

i see

what you mean now

ok im actually on drugs

Does it end like how it should be

[burnt flame]

why are you just dividing by 4

in last line

[keen seal]

because if i multiply the nominator by (4-x^2) it cancels out

[burnt flame]

ohhhh

[keen seal]

i think

if its good then

i can multiply

the fraction by

(4-x^2)^1/2

right

[burnt flame]

yes

and it should simplify nicely!

[keen seal]

ok

this took wayyyyyy

wayyyyyyy

longer than it shouldve

[burnt flame]

yea lmao

[keen seal]

my brain is not braibning

T T

[burnt flame]

you could've just done chain rule as well

[keen seal]

okay but ytsm!

[keen seal]

true true

anyway ty for your patience

😅

[burnt flame]

np :)

[keen seal]

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[last slate]

<enumitem>
Consider a hanging spring of negligible mass that does not obey Hooke’s law. When the spring is pulled downward by a distance $x$, the spring exerts an upward force of magnitude $\alpha x^2$, where $\alpha$ is a positive constant. Initially the hanging spring is relaxed (not extended). We then attach a block of mass $m$ to the spring and release the block. The block stretches
the spring as it falls. 
\env{enumerate}{[(a)]
\ii How fast is the block moving when it has fallen a distance $x_1$? 
\ii At what rate does the spring do work on the block at this point? 
\ii Find the maximum distance $x_2$ that the spring stretches. 
\ii Will the block remain at the point found in part (c)?
}

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[compact plaza]

I need to double check my work

I got X = 33

6 x 33 = 198

• 18

= 180

was told that no matter what it always adds up to 180

[tacit rose]

Yes it adds to 180 but how exactly did you set up your equation?

[compact plaza]

the answer is not accepted

idk why

[tacit rose]

Because that answer is wrong

[compact plaza]

I added the variables together than the non-variables than tried to multiply and subtract the 18

i might recheck the videos

i didnt understand them the first or second time

[tacit rose]

As you said, the angles sum to 180

So the equation is 2x + 4 + 8x - 14 = 180

Do you agree?

[compact plaza]

i then tried to do the opposite, afterwards which got me 198 and x = 33

[tacit rose]

Do you agree that the equation is 2x + 4 + 8x - 14 = 180?

[compact plaza]

no

[tacit rose]

Why

[compact plaza]

I learned to combine the like terms, then solve the problem

i see the issue

give me one moment to re-check some of the videos

[tacit rose]

Yes you add the like terms

[compact plaza]

yes

[tacit rose]

So what's 2x + 8x?

[compact plaza]

10x

[tacit rose]

And 4 - 14?

[compact plaza]

i knew it

thanks

i got it

.close

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[unkempt rune]

Can you cancel out imaginary numbers like this?

,rotate

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[round mural]

how can I tell its concavity?

[burnt flame]

f''(x)>0 concave up, f''(x)<0 concave down

but as for visually, concave 'u'p means 'u' shape, and concave dow'n' means 'n' shape

[round mural]

so (0,6)U(12,18)U(24,30) would all be concave down?

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[grand pond]

Mycobacterium

[halcyon trellis]

you can try graphing the equation

= 0 means above the x-axis so you can use the quadratic formula to solve ax^2 + bx + c = 0

then look at the graph

[rose trout]

If you have your two roots and the parabola is opening up (a>0), then this tells you that it will be negative in between the roots and positive on each "tail".

[midnight plank]

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[last slate]

I would use completing the square to solve this right?

[last slate]

$\int \sqrt{4x^4 + 4x^2 + 1 } dx$

[grand pond]

Several people

[last slate]

$\int \sqrt{(2x^2+1)^2} dx$

[grand pond]

Several people

[last slate]

big brian

https://tenor.com/view/family-guy-brian-drunk-drinking-gif-27067851

.close

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[grand pond]

Mycobacterium

[burnt flame]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[burnt flame]

so you're trying to find the range of a quadratic function f(x)...or...?

and ye

ok, I see what you're doing, domain of inverse function=range of function

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@last slate Has your question been resolved?

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[meager ice]

Hey can u help me

[midnight plank]

Please don't occupy multiple help channels.

[meager ice]

Ok

Sorry

Can u help me with a few questions I don’t understand any

[lilac sage]

post your question and close your other channel

[midnight plank]

@meager ice Has your question been resolved?

[meager ice]

Yes but can I ask more questions?

[midnight plank]

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[meager ice]

.reopen

[midnight plank]

✅

[midnight plank]

@meager ice Has your question been resolved?

[meager ice]

I need help

[last slate]

What're your questions?

@meager ice

[meager ice]

I don’t understand

I’ve counted it but

I got 20 but it’s says it’s incorrect

[last slate]

What's the frequency density

[meager ice]

1

[last slate]

Between 30 and 40

[meager ice]

It#s it 0.4

[last slate]

So, as the difference between 40 and 30 is equal to 10

We know that, for we get the frequency density of this histogram, we take the frequency and divide by the class width

So, for get the frequency again, we need to MULTIPLY by the class width to get the frequency

So it's just to multiply 0.4 by 10

@meager ice

[meager ice]

Ohhh thank u so it’s 4

What about this one I’ve got the median which is 62 and 82 I think but the interquartile range not sure

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[sacred helm]

What test would I use here

[sharp coral]

limit comparison?

[midnight plank]

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[last slate]

hi

[midnight plank]

Please don't occupy multiple help channels.

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[fallow scarab]

.close

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[hexed talon]

The question is to prove cube root is continuous at x not equal to 0

I have 2 quesitons. The first one is, how did the numerator simplify to |x-c| to the second line (in red)

The second one is, how to make sense of this denominaotr in blue?

actually nvm

.close

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[inner glacier]

I'm pretty sure by "graphical utility" it means graphing calculator

how can i use my calculator to find the tangent of f(x) = -5x?

[sharp coral]

what calculator do you have?

[inner glacier]

TI-84 Plus CE

[sharp coral]

2nd-> prgm (draw), it's the 5th option

[inner glacier]

i tried Tangent(-5x) and got an error

what am i messing up?

[sharp coral]

use it when you're on the graph of your function, then it'll prompt you to select a point on the graph

[inner glacier]

ohhh got it

is there an easy way to select the x? by moving side to side with the arrows closest i can get is 1.969697

[sharp coral]

if you enter a number when it's prompting you to select a point that will give you the exact x-value you want

[inner glacier]

lol exactly what i needed

thnx

nothing much really happens when i select it

[sharp coral]

that's because the tangent line to any line is the same line

[inner glacier]

ah...

[sharp coral]

it gives you the equation at the bottom though

[inner glacier]

y = -5x + 0 i assume

[sharp coral]

yes

[inner glacier]

cool cool

tysm for the help

.close

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[west hearth]

what does anything of this problem mean

[sharp coral]

it's saying you should be able to simplify it to just sin(x), cos(x), tan(x), sec(x) or something like that

[west hearth]

i have no idea what sin cos tan or sec mean

i havent taken trigonometry

[abstract iron]

without any knowledge of trig basics i dont think solving this problem is the most imp[ortant thing to do at this time

[west hearth]

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[latent seal]

how?

practice exam problem.

is this inseperable?

how do i even approach this i forget lol

[midnight plank]

@latent seal Has your question been resolved?

[honest kelp]

that is about differentatin of Implicit function

have you ever read about it ?

[latent seal]

yes i have

[honest kelp]

then great

[latent seal]

hmmm

[honest kelp]

look

if you could find z here

then it wu dbe easier for oyu

but as you can see

it is not possible

but

impliciit fucntiosn theory helps

to find partial derivatives

withotu knwoing function

that si great concept, admit

my hint is:

take the equation of the surface and

differentiate it , once in respect to x, and once in respect to y

but you msjut also treat

z = z ( x,y )

means

z is soem fucntion of x, y

so when you differetnaite z, it will not disappear

[latent seal]

ahh

tysm

[honest kelp]

yvw 🙂

[latent seal]

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[floral apex]

If we have some nice, analytic, harmonic function of two variables, call it f(u,v)

[floral apex]

will all level curves approach a level curve of 0 at infinity?

I mean are we always able to join level curves f=C and f=D using a curve f=0 by letting f=0 join f=C and f=D at infinity

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[tall copper]

Whats the difference between arccos(-sqrt.3/2) and Arccos(-sqrt.3/2)?

[main current]

None in most cases

[tall copper]

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[tall copper]

How to find exact value of arctan(-1/2)?

[main current]

,calc atan(-1/2)

[grand pond]

Result:

-0.46364760900081

[tall copper]

Is there no radian form for this?

[meager basin]

That is the radian form

[tall copper]

Ok ty

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[last slate]

why is it $8T(\frac{n}{2})$ instead of $4T(\frac{n}{2})$

[grand pond]

Several people

[last slate]

if you divide a square into multiple squares with side lengths of n/2 you get four squares right?

[frail carbon]

Yes but you need to perform eight multiplications

Multiplying two 2x2 block matrices, you need 2 multiplications for each of the 4 entries in the product

[last slate]

You're a genius

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[sterile sail]

Am i right

[sharp coral]

how did you find the area of the triangle?

[sterile sail]

multiplying the base and height by -1/2

[sharp coral]

what is the base and height?

[sterile sail]

10 ig

[sharp coral]

both of them?

[sterile sail]

ye

[sharp coral]

but the base and height must be perpendicular

[sterile sail]

Ohh yeah

Totally forgot that

What formula do I use then?

So A= πr²?

[viral dragon]

Well yes, but it's only part of the circle. Can you see which proportion?

Or, cosine rule?

Or just sin, maybe?

[midnight plank]

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[last slate]

This is so confusing to me .

[last slate]

I cant' seem to figure out how to rearrange $ac-bd + i(ad+bc)$ into an equation that only has three multiplications

[grand pond]

Several people

[last slate]

I'm so lost

[midnight plank]

@last slate Has your question been resolved?

[shell wigeon]

It's an algorithm, not an equation

You can reuse results that you previously computed

So for example if you compute ac, you can reuse it later without it counting as multiple multiplications

@last slate

[last slate]

yeah

but how would i reuse ac

ac = ???

[shell wigeon]

That's for you to find out

[last slate]

nah

[shell wigeon]

?

[last slate]

it's not happening

can't figure it out

[shell wigeon]

The obvious thing to try is to take only two of {ac, bd, ad, bc} and have one more complex product such that you can combine them to get both ac-bd and ad+bc

The obvious first choices would be to pick ac and bd, so you can already make ac-bd

Then with ac, bd, and one more complex product, you need to make ad+bc

Hint: there is one product that uses all of a,b,c,d that expands into a sum with all of ac, bd, ad, bc

[midnight plank]

@last slate Has your question been resolved?

[last slate]

ty

[midnight plank]

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[leaden matrix]

is the radius not the limit as n tends to infinity of n^2 / n^3 ?

[honest kelp]

no, lol

[leaden matrix]

why not?

[honest kelp]

becasue R is nor 0 neither 1

[leaden matrix]

isn't it of the form
an * (x-c) ^ n
so the constant an is n^2 so R is the limit as n tends to infinity of n^2 / n^3

[honest kelp]

you forgot ab the denominator

[leaden matrix]

oh I thought anything to the power of n wasn't included in the constant

[honest kelp]

65536^n is very important here

i give you formula for Radius

$\sum_{n=0}^{\infty }c_{n}x^{n}\Rightarrow R=\lim_{n \to \infty } \frac{1}{\sqrt[n]{\left| c_{n} \right|}}$

[grand pond]

Joanna Angel

[honest kelp]

but here, also it is important

that you have :x^8n

so i suggest you to consider series like:

with t = x^8

adn when you fidn radius of this ne w series

you will know what si vlue of x

consider this series:

$\sum_{n=1}^{\infty }\frac{n^{2}t^{n}}{65536^{n}}$

[grand pond]

Joanna Angel

[honest kelp]

find its R

with my formula

and enxt

find R for main series with x

[leaden matrix]

I did it with the formula we learnt in class but im a bit stuck now

[honest kelp]

ok it is dalamebert