help-38

[worthy eagle]

hai

do you remember me

!status

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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

[sudden basalt]

This is what i did so far

[worthy eagle]

write all numerators above the lastest common denominator 18x²

[rugged latch]

wait wont it be faster to just mult top and bottom by 9x^2

then long div

or whatever

[sudden basalt]

i can do that

should i?

[rugged latch]

nah just go with what @worthy eagle is saying

[sudden basalt]

thats more similar to what i have done before

[rugged latch]

ok then do that ig

[spiral socket]

$6x^{-1}$ is not $\frac{1}{6x}$

[solid kiln]

dqvidutzul

[sudden basalt]

what is it then?

[spiral socket]

what have i told you

$a\cdot x^{-y} = \frac{a}{x^y}$

[solid kiln]

dqvidutzul

[spiral socket]

a = 6
y = 1

what do you get?

[sudden basalt]

6/x^1

?

[spiral socket]

yes and x^1 is just x so 6/x

[sudden basalt]

one sec brb

[spiral socket]

now do the same for all

[rugged latch]

oh yea thats fr so just mult top and bottom by x^2, u dont need to do 9x^2

[sudden basalt]

@spiral socket

it says im right

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[earnest breach]

Honestly, I can't tell if some of the steps I did were legit. I'm not finished yet.

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[remote torrent]

i only drew a diagram idk how to prove it

[wraith hinge]

I think it is the easiest to use vectors

Hint 1:||express 2d in terms of b and c||
Hint 2:||use that x=2d+a||

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[mild flare]

For this question, I used the right corner triangle minus the small triangle except i dont know how to get the base or height of each

how do you find the base of the bigger shaded triangle, height

[spiral socket]

is abcd a square?

[mild flare]

yeah

[spiral socket]

great

the shaded region is the area of the square - area of efd - area of bac

they are both right triangles

[mild flare]

how did you find the diagonal?

nvm

you don't need it

how do you find the area of efd?

another right triangle nvm

ty

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[delicate lance]

is this right?

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@delicate lance Has your question been resolved?

[delicate lance]

<@&286206848099549185>

@bright quarry

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@delicate lance Has your question been resolved?

[delicate lance]

pls my answer is worng idk why

[bright quarry]

hi

[delicate lance]

hi

pls help

[bright quarry]

okok

what’s the 240L

ahh ok

i see

nvm

[delicate lance]

yea

the teacher didnt post the answer for this one so i dont even know what i did wrong

[bright quarry]

yea so do you know the correct answer

usually with these cost problems

you need to set up

equation for cost

and equation for other things like material/volume independently

[delicate lance]

yes

[bright quarry]

so let’s let W=width and L=length then we will do both of the width sides using brick and one of the length sides using brick the other being fencing

therefore

we know the total cost for the rectangular garden will be

384W + 240L hmm where’d you go wrong let’s see

we know LW=1000

384W+240L=C

L=1000/W

384W+240000/W =C

[delicate lance]

what does C mean ?

Is it perimeter ?

[bright quarry]

C’=384-240000/W^2

[bright quarry]

cost

[delicate lance]

oh

[bright quarry]

perimeter is just 2L+2W

[delicate lance]

oh ok

[bright quarry]

so we want to find a minimum

384-240000/W^2=0

hmm

are you sure the book is correct

[delicate lance]

maybe not

does that mean im smarter than the book yaaaaaaaaaaaaaaaay

[bright quarry]

it is 25 40

[delicate lance]

Ik it is meant for that question so

@bright quarry

[bright quarry]

book is wrong

[delicate lance]

how do I know whether to do the domain analysis or not

what does the domain analysis do for this question

[bright quarry]

domain analysis? you mean if the value you got is a valid solution

[delicate lance]

i mean

0 < x < 15 stuff

[bright quarry]

oh like yesterday

[delicate lance]

yea

[bright quarry]

not necessary for this problem

i’m curious how they got those answers

also

you can check it

like

see the price

with your values

and compare to theirs

[delicate lance]

oh ok

why they do domain analysis for this question

[bright quarry]

so they try W=29.1 and L=34.4

[delicate lance]

My brain has trouble comprehending it

[bright quarry]

yea they’re dumb

if you do 25x40 you get to total cost of 19200

if you do 29.1x34.4 you get 19430.4

[delicate lance]

never knew a textbook could be wrong

[bright quarry]

happens all the time

the first calc book i read had plenty of wrong answers

[bright quarry]

well the whole length of that side is 40 right and you’re cutting out two squares of it so if the whole length is 40 the length of the two squares can’t possibly be greater than 40 because you’re cutting them out from that side

[delicate lance]

ohh

[bright quarry]

yes

[delicate lance]

so the square cant be bigger than 20

[bright quarry]

mhm

if you cut it in half

you’d get two squares of 20

there’d be no space between them you can’t fold it

[delicate lance]

ahh i see

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[golden thistle]

A circle at (0,0) has the equation x^2 + y^2 = 5.
A linear graph has the equation 2y = x + 5
Find the 2 points at which the graphs intersect.
What do i do for this question???

[golden thistle]

I got 3,4 as one of the coordinatea but im 90% sure thats wrong

[rugged latch]

if ur 90% sure ur wrong

then why dont u try and check if its wrong

@golden thistle

are u sure its x^2 + y^2 = 5 and not x^2 + y^2 = 5^2

[golden thistle]

I cant fully remember, if its 5^2 then i know i got the 3,4 part right cause its a 345 triangle

[rugged latch]

ok which version do u wanna go with

the 5 or 5^2

[golden thistle]

This i found someone who also did the paper i did

Aight yea it was 5^2 cause the radius was 5

I get the first part of getting 3,4 just not the 2nd part of 0 and 5

-5**

[rugged latch]

ok so

which part of the (-5,0) obtainment did u specifically not understand

@golden thistle

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[golden thistle]

Tbh I forgot that you could make it into a quadratic then factorise it because at first i subbed the y in wronged so i got something that woudjt factorise without a calculator

[rugged latch]

so u get how to do it now?

[golden thistle]

Yeah thankyou

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[orchid wagon]

$x^3+36a^2x-800a^3=0$

[solid kiln]

Fungus 34A05

[orchid wagon]

this isnt the original problem though, this is one of the step

my goal is to solve for x in terms of a

but im not sure how to

[cyan oxide]

that's a cubic equation in terms of x

[orchid wagon]

yeah i just realized such

cardano formula should suffice

thanks ayway

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[knotty locust]

Why is the function floating

[marble flare]

Hold on let me get my sir

@knotty locust how to solve this problem?

[knotty locust]

I suggest we differentiate

Alas that will provide us with the slope of the tangent line

[worthy basalt]

oh thats because my teacher took it from something else

[knotty locust]

You should tell her it is very confusing when you cannot tell why the function is flying

[marble flare]

Have you tried taking the derivative?

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[nimble wind]

What do i need to do to find the local min's and max's

just based on the graph above

[ripe topaz]

do u know the definition of local min/max

[nimble wind]

Yeah its when the orignal function peaks

or hits a low

but thats a graph of the derivative

the derivative is the rate of change of the function

[ripe topaz]

first condition: f'(a) = 0 or f' is undefined at (a,f(a))

you should look for that first

[nimble wind]

3

[nimble wind]

but what about the jump disconuity

and the end point at x = 0

[ripe topaz]

you mean at x = 3 ??

that's the graph of f'(x) btw not f(x)

[nimble wind]

oh wait so at 2, 4

when the line crosses the x axis

[ripe topaz]

what about undefined

[nimble wind]

at 6

[ripe topaz]

remember local min/max happens at points where the derivative is either 0 or undefined

[nimble wind]

its undefined

[nimble wind]

is it undefined at 0 also?

[ripe topaz]

right

it is defined at 0

but f'(0) is not 0

[nimble wind]

okay

[ripe topaz]

f'(0) = 3 so it can't be a local min/max

[nimble wind]

Yeah

so we now have 2,4,6

[ripe topaz]

for a point (a,f(a)) to be local max f'(x) should change signs from positive to negative at x = a

and to be a local f'(x) should change signs from negative to positive at x = a

[nimble wind]

oh my gosh your right i forgot about that

[nimble wind]

this is exactly what i was missing

thanks for the help adam!

[ripe topaz]

np

[nimble wind]

i ran into another road block

is it okay if i ask another question

[ripe topaz]

[nimble wind]

same graph as above

I need to find the concavity

[nimble wind]

what exactly do i look for?

[ripe topaz]

A function is said to be concave upward on an interval if f″(x) > 0 at each point in the interval and concave downward on an interval if f″(x) < 0 at each point in the interval.

remember f''(x) is the derivative of f'(x) which is the graphed above

[nimble wind]

I dont have f''

okay lets see

[ripe topaz]

you have the graph of f'(x)

[ripe topaz]

what does a positive/negative derivative indicate ?

[nimble wind]

if the graph is moving upwards or downwards

[ripe topaz]

right, and f'' is the derivative of f'(x)

[nimble wind]

its the same for those functions?

[ripe topaz]

wdym?

[nimble wind]

just like it would be for the original function and the first derivative

the same rule applies?

[ripe topaz]

f'(x), f''(x) are just ordinary functions

[nimble wind]

Yeah

so were looking for when the graph is going up essentially?

[ripe topaz]

yeah

[nimble wind]

ohh okay

ill type those intervals in

[ripe topaz]

you can even say
h(x) = f'(x)
h'(x) = f''(x)
you need to find where h'>0

[nimble wind]

I just got when it was increasing

[nimble wind]

thanks for the help again

if i run into math problems may i DM you?

[ripe topaz]

i don't mind... but posting your question in a help channel will be way faster

[nimble wind]

Your right

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[rigid moss]

Help

[split chasm]

!status

[trim jolt]

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

[rigid moss]

I got the last option

[split chasm]

were there any steps that you did that you're concerned about

[rigid moss]

I don think so

[rigid moss]

Is It correcr?

[split chasm]

yes. though if you just want answers checked for questions like this, use a online program

[rigid moss]

.close

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[civic bobcat]

can someone please help with this

[cyan oxide]

the same way u solved (a)

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[sly hornet]

y=2x+10
y=x+7 finding poi

[swift veldt]

set the equations equal to each other

and you find the xcoord of the poi

and to find y coord it shouldnt be that bad

[sly hornet]

yeah so like 2x+10 = x+7 ?

[swift veldt]

ye

and then you will get the x coord

and it asks for a point so you need the y coord too

by just plugging whatever you get for x into one of the equations

[sly hornet]

OHH

i remember now

ty

[swift veldt]

👍

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[timber gust]

just no idea how to do

[orchid wagon]

what have you tried

[timber gust]

i just drew a diagram

i dont knwo what to do

[orchid wagon]

show your diagram

[wind folio]

<@&286206848099549185>

can I get help pwetty pwease

[timber gust]

its on paper gimme a sec

[wind folio]

ohh okay thank you

[timber gust]

looks something like this

[orchid wagon]

find the radius of the circle in terms of the distance from each chord to the center

equate that and solve

[timber gust]

how do i do that tho

<@&286206848099549185>

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[proven steppe]

confused on where to start

[raw magnet]

how is it going chef

do you know how to find a critical point usually in the case where we are given (dy/dx)

[proven steppe]

i got the criticical point but the second part i guessed

<@&286206848099549185>

[proven steppe]

help with this question instead

Don’t understand why I’m wrong

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[proven steppe]

<@&286206848099549185>

[proven steppe]

<@&286206848099549185>

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[proven steppe]

<@&286206848099549185>

[proven steppe]

i have

[simple haven]

"Once" is the key part of that quote

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[simple haven]

You probably shouldn't have 2C in your answer

The answer validator is likely only looking for +C, considering it calls out that you can't even use -C

[jade cairn]

Help me with question no 5 pls

Question no 5 @eveyone

@everyone hwlp

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[foggy wharf]

Its perpendicular chord is drawn at the midpoint of the radius of the circle. Find the angle between the rays passing through the ends of this chord

[split chasm]

!status

[trim jolt]

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

[fathom gull]

!original

[trim jolt]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@foggy wharf Has your question been resolved?

[foggy wharf]

1

or 2

[foggy wharf]

<@&286206848099549185>

Its perpendicular chord is drawn at the midpoint of the radius of the circle. Find the angle between the rays passing through the ends of this chord

<@&286206848099549185>

[fluid tundra]

hmm give me a sec

[fluid tundra]

first, do you think you can draw that

[foggy wharf]

yeah

[fluid tundra]

then do it and send it here

[foggy wharf]

here

[fluid tundra]

alright so u have two triangles

same hypotoneus and same opposites

[foggy wharf]

yes

[fluid tundra]

so that angle you're looking for

is just the double the angle of one of fhe triangles

[foggy wharf]

yeah

[fluid tundra]

pick one triangle, if the radius of your circle is r (an arbitrary number) can you label the sides of your triangle?

not all of them but two, leave the adjacent

you need only two sides

[foggy wharf]

it will be half of radius

[fluid tundra]

do you see how the hypotoneus is just a radius so it's length is r and the opposite like you said is r/2

[foggy wharf]

yes

[fluid tundra]

you can get the cos of that angle now

[foggy wharf]

yeah

[fluid tundra]

what is it?

[foggy wharf]

i can divide this half radius

by radius

rigth

[fluid tundra]

exactly

and that's the fun part

it doesn't matter which circle you choose, that angle will always be 60

the whole angle is just 120

[foggy wharf]

aaa

yes

rigth

thanks

[fluid tundra]

glad i helped 😁

[foggy wharf]

.close

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[topaz pagoda]

can anyone tell me the formula for finding wavelength?

[proven hound]

In terms of what

λ=c/ν is one of em

[tall venture]

If I want to calculateUc in volts and the saturation current, do I need to calculate Ic in volts as well? Because I experimented with Ic in kilo volts

[proven hound]

! occupied

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[tall venture]

Here is my data pack

[topaz pagoda]

what does c/v mean

[proven hound]

c is speed of wave

v is frequency

[topaz pagoda]

in my school f is the frequency

then v is for wave velocity

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[abstract shell]

can someone help? i need to solve this when lim x->0 but without using lhopitals

[split chasm]

consider the factorisation for a difference of two cubes to rationalise the denominator
(this is similar to what you'd do if instead you had a square root where you'd use conjugates which is an application of the difference of two squares)

[abstract shell]

like this?

[split chasm]

,rotate

[split chasm]

yes

[abstract shell]

this doesnt really get me anywhere, its still 0/0

[split chasm]

don't forget theose closing )

expand and simplify the denominator

[steady tulip]

Maybe we have to substitute another variable?

[split chasm]

no need

[abstract shell]

i get this, but i have no idea what do do with it

[split chasm]

simplify the denominator

[abstract shell]

im sorry how do i do that

[split chasm]

by not overthinking

[abstract shell]

so just x?

[split chasm]

yes

and then you can apply the limit identity for
sin(kx)/x

[abstract shell]

oh my God yes youre a legend

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[alpine stirrup]

can anyone help me with this math equation i dont even know where to start

[barren mountain]

you could find a,b

[barren mountain]

by using the disibility

[alpine stirrup]

do u have an example

[barren mountain]

you can change this equation into the xy=m form (with m is a constant value)

[alpine stirrup]

hmm alright

i found

we can factorize it

into

(b-1)(a+3)=-3

[steady tulip]

This is more of a general knowledge question, here you should shift -a to RHS and take b common which will leave u with
b(3+a)=a from here it's clear that a will be negative as it's lesser than b and a the option which fulfils this equation and criteria is 4 so I think it may be 4

[steady tulip]

Not sure tho I could be being a complete idiot

[alpine stirrup]

alright that is logical

since its like a local math olympiad question

and here they rly like to use cheap tricks

lemme try if it can work

[steady tulip]

Yea these types are only used in Olympiads mostly

Alr

[alpine stirrup]

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[vast tusk]

i have no idea how to approach these questions even though i kinda know the concepts

please help me out

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[vast tusk]

no

[mint rapids]

wtf is this

[vast tusk]

yea same😭

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[hearty spade]

how does one do this?

[blissful bison]

you have to apply the polar coordinates, that is a classical exercise.

[proven sluice]

Area of circle with radius 6 - area of circle with r 4 in 1st quad

[hearty spade]

does that integral calculate area? your solution seems a bit too easy compared to the other answer

[proven sluice]

Yeah it does

Int from 0 to 6(√6-x²) - int from 0 to 4(√4-x²)

I suppose this how it's done?

[hearty spade]

idk. i think the "hard" part is finding the correct boundaries.

like from the set

[proven sluice]

Idk that ans sounds complex to me too

[blissful bison]

that is not about area here lol

[proven sluice]

Idont even know what it meant haven't studied

[blissful bison]

look here:

read and learn

and write it in exercise book

[hearty spade]

this is no way near what we learned abt this stuff. idk it might be correct

[proven sluice]

Wdym? Like it did say 'area of circle ring'? Idk I'm unsure abt this stuff

[blissful bison]

your task is classical exercise for polar

open the book and read there

and if your instructor does not realize that, then let him or her learn too

[hearty spade]

there was no mention of polar stuff in the lecture for this stuff. maybe lecture used different methods

[proven sluice]

I mean how are u so sure

[digital bison]

We don't know what you're learning. You have to tell us.
There's a few ways you can do this problem, some very advanced, some very basic.
Without you telling us what you have learned to this point, we're going to be blindly throwing darts at methods.

[proven sluice]

I mean I don't understand how the other helper was so sure that they have to do it in only polar method tho? That's weird actually

[blissful bison]

do you want me to show you pages from y textbook ? or what you need? my diploma Aditi ?

[proven sluice]

I m not here for any sort of arguments

[hearty spade]

lecture for this stuff was parametrization, plane ints, and volume ints with determinants of the jacobian matrix

[proven sluice]

Like it just was little off tbh

[blissful bison]

this exercise is typical , if you use cartesian, it takes a lot of more time

[proven sluice]

Idk man it maybe correct, you can't just force them to do this way

This might be actually really time efficient your method but if they don't know what are we supposed to do

[blissful bison]

if problem would refer to the area, then i agreed with you, since geometrical aspect wud be better, but the presenc4 of the function noot equal to 1, then it is not the area

[proven sluice]

Alright they can follow your method, since I was not sure of the question from beginning itself, just said what I thought of as ans. Mb

[blissful bison]

smiles , I do not insist either 🙂

[hearty spade]

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[vast tusk]

.reopen

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[wraith hinge]

can you guys help with like negative exponents

[wraith hinge]

like explain them

<@&286206848099549185>

[dusty flax]

This video gives the algebra of expoenents, including negative exponents
While this one specifically covers negative exponents

[wraith hinge]

thanks

[dusty flax]

One important property about negative exponents is that, provided your base is not equal to 0, you can always convert a negative exponent to a positive one by taking its reciprocal
That is to say, $a^{-1} = \frac{1}{a}$ provided $a \neq 0$

[solid kiln]

curry

[dusty flax]

So if you feel comfortable dealing with positive exponents, you should convert them all to positive and work that way

[wraith hinge]

ok

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[sullen ore]

first find the second side length of the 45cm rectangle

the area is 45 right

what are it's side lengths

it does

no

no

wrong rectangle

the 45 one

what are it's sides

ok just

look at the picture

theres a line separating the 45 one and the 40 one

ok, lets assume that

so it's sides are

...

come on

what are it's sides

45 one yea

you said 6 for the long sides

so it's sides are

what is the area of a rectangle

in general

whats the formula

for rectangle area

alright

in the 45 triangle

it has two sides

which ones do we know

you could see that the entire time??

I was trying to make you notice it

anyways, so 45=5*(the other side)

do you understand why that's true

so area=length*width

area is 45

length is 5

45=5*width

yes

alright, we're getting somewhere

so the width is 9

so the part of the 13 side that is in the 45 rectangle area is 9

so what is one side of the 40 rectangle

YES

so whats the other side

fantastic

alright, so the width of the 40 cm rectangle is 10

now we can find one side of the gray rectangle

just find one side of the gray one first

we're on a roll

no no

side of the gray one

we know that the 40cm rectangle has a side with 10

for finding the side of the gray one

we're focused on the 5cm one

the 5 side of teh 45

yes

alright, lets find the other gray side

first find the sides of the 48 one

the SIDES of the 48 one

uh

not exactly

alright yes

now find the other side of the gray box

[knotty locust]

!noans

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[quiet remnant]

In the figure, ABCE is a square, BCF and CDE are equilateral triangles, and EB = 2, how long is the FD segment?

[sullen ore]

“ABCDE is a square” excuse me what is a square

[quiet remnant]

Sorry lol

I made a little mistake there hehehe

ABCE*

[orchid lagoon]

so what have you tried?

[quiet remnant]

I did the graphical representation but I can not seem to think of anything

[orchid lagoon]

so trying to solve it in a coordinatesystem?

[quiet remnant]

Well I didnt mean that, I drew it

[orchid lagoon]

ahh okay ^^' well so i gotta ask do you know of the distance formula? (basically pythagoras theorem)

[quiet remnant]

Well yeah

[orchid lagoon]

okay then my tip would be imagine the figure in a coordinatesystem, where B is on origio (0,0)

If we could figure out the coordinates of d and f we could use the distance formula

[quiet remnant]

How, I mean I am not the best at drawing tbh

[tepid hamlet]

try to take a screenshot of the image with Windows+Shift+S instead of copy and pasting it since it contains transparency and looks messed up in discord

[tepid hamlet]

ok phone photo...well i guess it's a step up anyway...

[quiet remnant]

Its the best I can get lol

[orchid lagoon]

sorry i wasnt clear we arn't gonna draw to figure it out, we need to calculate the coordinates

[tepid hamlet]

can't you take a screenshot of it?

[quiet remnant]

I cant since this is training homework from past years, I dont know if I explained myself lol

[tepid hamlet]

what is the screen you are taking a photo of?

you have the image on something with a screen clearly

[quiet remnant]

Of a friend of mine, I mean I could ask him for a cleaner screenshot

But idk

[tepid hamlet]

ok, anyway

all the segments are the same lengths right?

[quiet remnant]

Yes

[tepid hamlet]

what is the red angle there?

[quiet remnant]

30

[tepid hamlet]

how did you get that?

yes, 30

[quiet remnant]

30*

[tepid hamlet]

ok, do you notice where im going with this?

what can you use that for?

[quiet remnant]

30 60 90 right triangle? idk

[tepid hamlet]

whats the measure of angle FCD?

[quiet remnant]

90?

[tepid hamlet]

yes

so you have a right triangle

you know 2 of its sides

[quiet remnant]

45 45 90 right triangle

[tepid hamlet]

guess so since its isosceles

[quiet remnant]

sqrt(2) * CD or FC

So sqrt(2)?

[tepid hamlet]

is CD or FC length of 1*?

[quiet remnant]

No

[tepid hamlet]

how did you get this

[wraith hinge]

isnt FD = 2?

[tepid hamlet]

think about this triangle

[tepid hamlet]

!nosols

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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.

[quiet remnant]

Oh oh

[tepid hamlet]

so what is FD?

[quiet remnant]

2

[tepid hamlet]

yes, basically the blue triangle is the same as the red triangle but rotated

[quiet remnant]

Yeah I saw that lol

[tepid hamlet]

or if you dont notice that you can calculate the length of the sides of the square from the diagonal and then calculate the length of the hypotenuse of the blue

[quiet remnant]

Thanks!

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[wraith hinge]

sry mb I forgot about that rule

I just did quick calculation in my head 💀

[tepid hamlet]

ya goal is to help people work through the problem themselves as much as possible, not just hand them the answer, try to meet them at their current level of understanding

[wraith hinge]

@tepid hamlet Gotcha

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[rugged latch]

x(t1) = x(t2) and y(t1) = y(t2) for t1 != t2

make a system of equations from that

$x(t_1) = x(t_2)$ and $y(t_1) = y(t_2)$ for $t_1 \neq t_2$

[solid kiln]

Stephen

[rugged latch]

at some time t1, our coordinate is (x(t1), y(t1)) and at some time t2, our coordinate is (x(t2),y(t2))

we wanna find t1 and t2 such that (x(t1), y(t1)) is the same point as (x(t2),y(t2))

so rn we have

$x(t) = t^2 - 4$

[solid kiln]

Stephen

[rugged latch]

whats x(t_1) and x(t_2)

i mean its not wrong

we just have to find the t's now

its just another step in the process

[rugged latch]

.

no

if i asked u to find x(4)

what would u do

yes

now find x(t_2)

and y(t_1)

and y(t_2)

[rugged latch]

then use these equations to solve for t1 and t2

t_2^2 - 4 **

bruh u shda said if u got the point

idt u needa solve for t if u have the pt of intersection

cuz u alrdy got dy/dx

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[wraith hinge]

$x(t_1) = x(t_2)$ and $y(t_1) = y(t_2)$ for $t_1 \neq t_2$

[solid kiln]

John Mornt

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[half bobcat]

now down

i made it x-3 and x+3

but up

how do i split it without using quadratic formula

[split chasm]

AC method

[half bobcat]

ok wait the ac method

to split this

[nimble stone]

that wouldnt be particularly useful here

[half bobcat]

i can divide the thing beside x^2

by 2

no wait

well whenever i have such thing i divide the thing beside x by 2

[nimble stone]

do you know the ac method?

[fading sluice]

yeah just realized

[half bobcat]

but since i have 2 beside the x^2

im kinda confused

[nimble stone]

seems you dont, if you have ax^2+bx+c, a method of factorisation is to find the factors of ac that sum to b, then split the x term into two and factorise by grouping

[half bobcat]

ye ik but isnt there a faster way to do it

like if i have x^2 + 6x + 25

lets say

i can do 6/2

[nimble stone]

whats the benefit of that

[half bobcat]

and get (x+3)^2 + 16

[nimble stone]

thats completing the square sure, but it doesnt help you here

[half bobcat]

oh

[nimble stone]

you need to factorise into products

like a(x+b)(x+c)

[half bobcat]

oh

and then i do (ax+ab)(x+c) ?

[nimble stone]

i dont see the point in that

the next step would to be cancel whatever factors are in the denom and numerators

[half bobcat]

ye

but wait

ax^2+bx+c

i shld get 2 numbers that the mul for them is b and the sum of them is c

or is it the opposite

[nimble stone]

product ac, sum b

[half bobcat]

ye

[half bobcat]

so the answer shld be

the sum of b

and the mul for ac

i learnt the long way but long ago

the thing where u go from ax^2+bx+c

to

ax^2+hx+px+c

and i learnt the one that u get the sum of the numbers and the mul of them

and u get the (x+smth)(x+smth)

oh

the long

[cyan oxide]

maybe u can start guessing, 21 is 3 * 7, so u will have something like
(ax - 3)(bx + 7) since b is positive and c is negative,
and after some thinking u would say b should be 2 and a is 1 in order to get 2x^2 and +1 for b
(x-3)(2x+7)

[half bobcat]

ye that way i learnt

[long trench]

factoring?

[cyan oxide]

and also u have (x-3) down so yeah, it has to be in top

[long trench]

can't you just use synthetic division here or no?

[half bobcat]

now ik that way but my problem is what do i do with the thing beside x^2

like in general

i usually divide the question by the thing beside it

[nimble stone]

i have said twice what to do

[half bobcat]

ye the a(x+b)(x+c)

but do i just keep it that way

[nimble stone]

no thats the result, not the method

[nimble stone]

this

youre used to having a be 1, its no real difference

you cant just divide by a

expressions dont work like that, equality has to be maintained

[long trench]

wouldnt that make it more complicated as you'll have fractions?

[half bobcat]

oh ok wait

[cyan oxide]

there is no shortcut, u either guess or the long road.
gussing for smalle equations and long road for complix ones

[nimble stone]

its not something they can do anyway, but yes

its not like you have to go through every factor pair, use some sense to guess

[half bobcat]

ahh

ok wait

[half bobcat]

im sorry i didnt understand much what u said i dont learn math in english

ok i got how u got this

so it shld be a*c

and i shld get the sum of b to be the result of a*c

then i have 2x^2 - 6x + 7x - 21

now

hm

now i take the x out of 2x^2 and 6x

so thats

and ill take the 2 too

so thats

2x(x-3)+(7x-21)

wait

2x(x-3)+7(x-3)

2x+7(x-3)(x-3)

hm

(2x+7)(x-3)(x-3)

now i have x-3 alone

i can cancel

actually

[nimble stone]

why have you created a cubic

[half bobcat]

whats cubic

oh the ()?

[nimble stone]

a polynomial with a x^3, anyway you need to get rid of one of your x-3 s

its just (2x+7)(x-3)

[half bobcat]

ye

i canceled one of them with the stuff down

now wait

i can take the - out

[nimble stone]

no, there shouldnt have been 2 at any point

[half bobcat]

wait

[nimble stone]

the numerator is (2x+7)(x-3)

you can cancel after that

[half bobcat]

i did this

[half bobcat]

and got to this

i took the 7 out

[nimble stone]

ik, its fine up to here

[half bobcat]

now its x-3

wait where was the issue then

[nimble stone]

where did any of this come from

[half bobcat]

cuz

actually ye true wait

[nimble stone]

youre saying ax+bx= (a+b)xx

doesnt make much sense

[half bobcat]

ye true

oh wait

i take the x-3 out

[nimble stone]

mhm

[half bobcat]

ah ok got it

then i cancel

oh ok damn ty

ok solved

ok tysm appreciate the help guys

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[twin shell]

How do i rotate equations useing sin and cos to find the values of cos tan etc

[fading sluice]

what do you mean?

the relationship between sin cos tan?

[high current]

What do you mean rotate

[twin shell]

like rotation

like a clock

45,90 degrees

Use the equation 8x^2 - 4xy + 5y^2 = 36 to find tan 2

@fading sluice

[trail ingot]

rotate equations

[twin shell]

hows

<@&286206848099549185>

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@twin shell Has your question been resolved?

[twin shell]

heh?

[tulip cloud]

@twin shell whots the question?

[twin shell]

i emojied it

[tulip cloud]

Okie dokie

[twin shell]

this one

[tulip cloud]

Wait

Do we know what x and y is

Oh

36

Is tan 2 tanx or something?

[twin shell]

maybe not quite sure

but for these equations you have to eliminate the middle value

useing rotation

[tulip cloud]

Do you have the original question in picture or-

Rotation?

[twin shell]

uh lets say we have to rotate the equation 45 degrees

[tulip cloud]

[twin shell]

we can use that to find different values for x and y

that are equal

know what i was talking about earlier in circles

sin150 and sin 30 are the same values

you can use the same principal in the equation itself

its a bit advanced ish though

[tulip cloud]

Uh

I

[twin shell]

its fine if you dont know

[tulip cloud]

Still dont really understand how we are expected to solve the question though but lemme take a look

So

We use the ratio of y/x

To solve that equation thingy

And find tan 2 with it

What

[twin shell]

?

[tulip cloud]

So like

Uh

[twin shell]

in rotations x equals xcos plus ysin

and y equals -xsin plus ycos

[tulip cloud]

Huh

[twin shell]

its complicated

[tulip cloud]

Awh

[twin shell]

ill try again tmrw

have a good day though

[tulip cloud]

Ty

You too

I hope we both have good luck with our questions

[twin shell]

yep

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[upbeat pike]

Can someone please help me with question c

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[upbeat pike]

<@&286206848099549185>

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[zinc dove]

If you know how long it will take him, and you know the rate

Just multiply

@upbeat pike

[upbeat pike]

well I think the question wants me to use the number of days

but the equation is in years

so idk how to like change it

[zinc dove]

Convert 🤷‍♂️

Days/365 is years

years*365 is days

[upbeat pike]

look i got this

every month he pays 700

so every day he pays (700 * 12) / 365

if you multiply that by 2505 you get 57649.31

[zinc dove]

Seems right

[upbeat pike]

oh wait it is right

i was looking at the wrong answer sheet

[zinc dove]

Nice

[upbeat pike]

ggs

thanks for the help regardless

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[slate barn]

for what alpha does this series diverge?

[honest comet]

hi can anyone help me to solve this problem, the answer is 125 but i dont know how to prove it tysm

<@&286206848099549185>

[oblique valley]

!occupied

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[oblique valley]

!15min

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[honest comet]

oh sry for pinged

[oblique valley]

What did you try?

[earnest granite]

I am going to assume that at least for 0<=a<1

btw I don't now the method nor do I have any knowledge regarding series so I may be wrong but just by looking at it I suppose that is a solution

[oblique valley]

This is wrong, sorry

[earnest granite]

yea

I expected that

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[earnest granite]

😐

is it a <= 0?

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[severe oasis]

i would like to know what that symbol is supposed to mean

i see the (-) has also been removed

i cant find anything in my textbook what its function is

just the name of it would help me

this is from begining, but i think only the top picture is relevant

nvm, i think i found it

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[severe oasis]

.reopen

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✅

[severe oasis]

nvm, idk

[severe oasis]

.

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[brazen gazelle]

how do i put thsi in my graphics calculator

[delicate jasper]

plot 150 and the derivative of left side

[dapper swift]

Or do 800 * 0.0786 e^0.0786t - 150 = 0

Chain rule

And then subtract 150 on both sides so you can find the zeroes

Instead of trying to move your GDC window up to y = 150

Ofc this is solvable by hand as well

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[brazen gazelle]

is there like a way i can just put that equation straight in

cuz i did find a d/dx but its got an x after it which i have to type in the value of x for

[dapper swift]

Depends on your calculator: TIs have NSolve and Casio has SolveN

I don't think you can put it straight in though

Like IIRC it should return an error if you try to do that, idk

Test it out yourself

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[fossil cobalt]

can some1 please explain why we took the second derivatve?

[rugged latch]

look at the question

"is the rate of change increasing or decreasing"

rate of change is first derivative

[fossil cobalt]

ok we already have first derivative

and its the rate of change

why did we take second derivative

[rugged latch]

to see if its increasing or decreasing at t = 1

[fossil cobalt]

i just pluged in t=1 and y =2.5