help-38

[fathom idol]

Chat prívate ok

Okey?

[smoky yacht]

?

Do you mean dm?

[fathom idol]

Yes

why

[smoky yacht]

Ok come

[hazy wagon]

the answer is 2

u want me to explain?

[smoky yacht]

It is askinh ratio💀

*asking

@hazy wagon

[hazy wagon]

ok

still 2

2/1 is a ratio too

[smoky yacht]

Explain

[hazy wagon]

first of all u have to write them all in one kind

[smoky yacht]

Ok

Next

[hazy wagon]

like B=4A

[smoky yacht]

Hmm next

[hazy wagon]

cuz if you multiply 1/4B with 4 you find the B

[smoky yacht]

Yea now i got it

Ty

@hazy wagon

: )

[hazy wagon]

then you have to do the same thing for c

[celest eagle]

I hope you're old enough for discord

[smoky yacht]

Hmm since 2021

.close

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[hazy wagon]

np

: )

[smoky yacht]

: )

[brittle gulch]

hi

[hazy wagon]

hi

[trim jolt]

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[latent valve]

hello

[latent valve]

2x^3 + 3x^2 - 17x - 30

how do I long divise that from x+2

I got -15x-30

but not sure if it's right

[lilac flame]

just multiply (x+2)(-15x-30) and see what's the remainder

[latent valve]

ohhh

okay

@lilac flame I got it wrong

can you do that math thing where you get the bot to show me how to do the math

<@&286206848099549185>

[undone sierra]

not an english native but

does that mean what do u get after diving that expression by x +2?

[latent valve]

yes

@undone sierra

.close

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[undone sierra]

welp

[undone sierra]

.close

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[fierce wharf]

how do you get the square root of an equation with only two terms though?

ex. x²=5x

or equations like 3x² - 5 = 4

[fierce wharf]

[sharp spindle]

divide both sides by x

For the first one

[fierce wharf]

okok

for the second one?

[sharp spindle]

Move the 5 to rhs

[fierce wharf]

continue

[sharp spindle]

Divide by 3 and square rt

[fierce wharf]

5 ÷ 3?

[wind cloak]

NOTE: By dividing by x on both sides you assumed that x cannot be zero

[sharp spindle]

If you move 5 to rhs what will you get?

[fierce wharf]

-5

[wind cloak]

but you have to take another case for x=0

[sharp spindle]

The -5 will become positive and it will be added with the 4

[sharp spindle]

Oh yh right

[fierce wharf]

ohh

okay continue

[sharp spindle]

What do you get from here

[fierce wharf]

-1

[sharp spindle]

It was -5 on the lhs, so it will become +5 on the rhs

[fierce wharf]

oh mb so

x² = 4-5

?

chat i cant understand this without the illustration

[sharp spindle]

4+5

[fierce wharf]

x² = 4+5
x² = 8

then jst find the square root of 8?

[sharp spindle]

What is 4 + 5?

[tired brook]

9

[fierce wharf]

WAIT NO

im tweeking omfg

im so sorry😭🙏

[sharp spindle]

Nah it’s ok

So 3x² = 9

[tired brook]

3x^2=9 and then divide by 3, then sqrt both sides, then it becomes plus or minus 3. x^2-5x=0 then factor x(x-5)=0. x=0 and x=5

[fierce wharf]

so i devide it by 3?

[sharp spindle]

Yep

[fierce wharf]

thanks dude

i just woke up earlier im sorry😭🙏

[tired brook]

what grade are you in?

[fierce wharf]

g9😿

[tired brook]

9th grade?

[fierce wharf]

i didnt go to class this afternoon so idk what to do

[tired brook]

i mean most kids take algebra that year so its not too bad

[fierce wharf]

😿😿

okay but how about
x² = 6x - 9
or an equation with 4 terms like
a² + b + c = d

[clear cloud]

Do you mean quadratic equation ?

[fierce wharf]

yeah

i need to find their square roots

[clear cloud]

We just say finding their roots

But i got u

well for the one you gave, you make one side 0

Let say the rhs as usual

You have x^2 -6x + 9 = 0

Hint : use indentity !

[fierce wharf]

ohh

okok i got it

thanks

[trim jolt]

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[gentle sleet]

For b this would be correct too:

(13c2)(4c2) (11c3)(4c3)?

[gentle sleet]

dam apparently not

idk why it doesnt work

oh nvm I see why did (2c1)

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[stark pebble]

For the answer to v), they say +- 0.8, but since cosθ = 0.6 (so θ = 53.1) does it not follow from the standard matrix for a reflection in the z axis that b=-sin53.1 and so b=-0.8 only?

[raw magnet]

what

cos(-53.1)

vs cos(53.1)

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[dense breach]

"One was caught by two radars on the same day with a difference of exactly 10 minutes: the first showed the time at 6:00:00 and the second showed the time at 6:10:00. It was found that the speed of the car in the first radar was 81 km/h, while, in the second radar, the speed of the car was 93 km/h (with the tolerances of 7 points already removed). Therefore, the value of the fines was R$260.32.

After receiving the notification, the driver decided to file an appeal, looking for some argument to cancel the fine. In the justification, the driver included could be in Uniform Rectilinear Motion or Uniformly Varied Rectiliniear Motion.

A very attentive agent received the appeal to judge it and soon discovered that a first notification had occurred on a radar that was at a distance of 21 km (measured in a straight line) from the radar that issued the second notification. He recalled that the highway that contains the two radars in question does not have exits or returns between such radars and that the maximum speed allowed on this highway is 80 km/h.

4. construct, with justification, a possible function v:[0,10]→R, such that v(0) = 81, v(10) = 93 and v(t) describe the speed of the car at time t, in minutes from the moment it passes the first radar, in a manner consistent with the problem situation."

For the answer to 4., i thought the problem would give me a function or something, without it i do not know how to work, can someone help me please?

[trim jolt]

@dense breach Has your question been resolved?

[dense breach]

<@&286206848099549185>

[heavy dawn]

lets see

indeed, it seems that what they want is a function quoting "construct a possible function". it seems that the question demands the following: Given two speed camera readings 10 minutes apart with different speeds, and knowing the distance between the cameras and the speed limit, find a possible speed function for the car over those 10 minutes.

what was your answer to this question? may i take a look?

[dense breach]

first of all, thanks for your time!

[heavy dawn]

v(0) = 81 km/h, v(10) = 93 km/h,

ah no problem

[dense breach]

i didnt answer it yet

[heavy dawn]

i see

well here is a potential answer

ever heard ofa veloity function?

for the Uniformly Accelerated Rectilinear Motion or for arguments sake, UARM, we will use v(t) = at + b for instance

for v(0), we will make v(0) = 81

are you following ?

[dense breach]

yes

[heavy dawn]

we knw that v(0) = 81

[dense breach]

i was following this line of thought as well

[heavy dawn]

as such, for v(t) = at +b, b will be 81

we will then use the other v(t) which is v(10)

ok?

[dense breach]

ok!

[heavy dawn]

v(10) = 93

as such, we can conclude that v(10) = v(t) for now

therefore, v(10) = 10a + b

[dense breach]

i see

[heavy dawn]

then

we can say that since v(10) = 92, we will have 93 = 10a + 81, as b = 81

after calculting the value of a, it was determined that it turned to be 1.2

thus, a = 1.2

[dense breach]

look

[heavy dawn]

we have our final function, v(t) = 1.2t + 81, formerly (v(t) = at + b)

[dense breach]

i think i didnt say something about the problem

[heavy dawn]

did you get lost in the steps?

[heavy dawn]

yes go ahead

[dense breach]

in 2. and 3. i had to prove that the movement wasnt a URM or UVRM

which i did

[heavy dawn]

oh

[dense breach]

both

lol

[heavy dawn]

oh i see

i didnt see wasnt clearly, thought it was

so its neither URM nor UVRM correct?

[dense breach]

"2. justify that it wasnt a URM
3. justify that it wasnt a UVRM"

[dense breach]

yes

[heavy dawn]

indeed

[dense breach]

my bad

[heavy dawn]

according to the question, it was UARM correct?

oh one moment

UVRM = UARM yes?

[dense breach]

yes

[heavy dawn]

i see

however

even if they asked you to prove that it was neither URM nor UVRM in the previous question could have nothing to do with the next question yes?

look at this part

[dense breach]

i think so, i tried translating the problem

[heavy dawn]

". In the justification, the driver included could be in Uniform Rectilinear Motion or Uniformly Varied Rectiliniear Motion."

as such, there is a chance it is one of them

which we used UARM for this case

[dense breach]

i dont think so because the question asks me to justify that the driver couldnt be in Uniform Rectilinear Motion or Uniformly Varied Rectinilear Motion

[heavy dawn]

the part where he said ". In the justification, the driver included could be in Uniform Rectilinear Motion or Uniformly Varied Rectiliniear Motion." is unclear. what does the sentence "the driver included could be in URM or UVRM mean?" does it mean that he filed that he was URM or UVRM or does it mean that he himself didn't write it and the agent was giving a guess?

if this answer is solved, the question will be easy to solve

[heavy dawn]

google translate isnt accurate 😂

send the spanish question

ill try to figure it out from there

it can be an image or text

[dense breach]

it means that the driver was trying to cancel the fine by saying that he was in URM or UVRM

[dense breach]

okay!

[heavy dawn]

i see.

[dense breach]

put portuguese to be more accurate

[heavy dawn]

so you were correct indeed

[heavy dawn]

i see

[heavy dawn]

then give me a few minutes

[dense breach]

here

[heavy dawn]

ok thank you

[dense breach]

me and my friends were trying to use the quadratic function

[heavy dawn]

nah

its useless

here

lets start from over

we know that v(0) = 81

[dense breach]

cuz is not linear

[heavy dawn]

and v(10) = 93

[dense breach]

yes

[heavy dawn]

thats a good try, but in the wrong place

[dense breach]

hmmmm

[heavy dawn]

there are many other polynomials parabola functions that could be ebtter use other than the quadratic formula

[dense breach]

i see

[heavy dawn]

ok so lets continue

what is our question ?

can you tell me it?

it is to create a function representing the car's speed over a 10-minute period given two radar readings, one showing 81 km/h and the other showing 93 km/h 10 minutes later.

correct?

therefore

we will use a function of the form v(t) = at + b

[dense breach]

yes, and that v(t) describe the velocity of the car in t instant, in minutes, by the time that the car pass by the first radar

[heavy dawn]

yes

where a =slope and b = y-intercept

v(t) = at + b
we already know that v(t) will equal 10, from v(10) and b= 81 yes?

[dense breach]

yes

[heavy dawn]

ok

wait sorry

v(t) = v(10), not 10

yes

anyway

[dense breach]

lol

its okay

[heavy dawn]

v(10) = 10a + 81

[dense breach]

93 = 10a + 81

?

[heavy dawn]

yes!

we get 1.2

therefore

the equation would be what?

v(t) = 1.2t + 81

[dense breach]

hmmm

i see

the only thing about this

i mean

idk if it is wrong

[heavy dawn]

ah go ahead

feel free to have any questions. was my explanation not clear enough?

[dense breach]

nah, you were clear, thank you!

is just that

[heavy dawn]

yes go ahead

[dense breach]

in the third paragraph it says that the distance between the two radars were 21km

and he travelled it in 10min, right?

i calculated the average speed in the section between the two radars and it was 126km/h (if it wasnt this high he wouldnt be able to travel 21km in 10min)

so, if hes average speed was 126km/h, there is a moment indeed that he was in a instantaneous speed of 126km/h

but when he reachs the second radar, he were at 93 already removing the tolerance of 7 points

[heavy dawn]

sorry my laptop rebooted for no reason

[dense breach]

nah thats fine

[heavy dawn]

yah

[dense breach]

so what im trying to say is that

[heavy dawn]

one sec

good point

one moment

[dense breach]

1,2a + 81 is a linear formula, it only goes up

[heavy dawn]

indeed

[dense breach]

yeah i think ypu got it

and that is the problem we are facing

[heavy dawn]

thats a valid point

hmm...

[dense breach]

thats why we thought of the quadratic function

bc it is not linear

[heavy dawn]

You've calculated correctly that the average speed between the two radars is 126 km/h. This means that at some point between the radars, the car must have been traveling at a speed of at least 126 km/h.
.

However, the second radar recorded a speed of 93 km/h (after removing the tolerance). This is a clear contradiction

[dense breach]

yeah

the funcion cannot be linear

[heavy dawn]

yup

one sec

this is math + physics lol

[dense breach]

yeah

[heavy dawn]

oh my god

you were correct using the quadratic equation

after recalculating

yes

it seems you werent able to use it properly yes?

[dense breach]

a friend is having a greater time than me

[heavy dawn]

very well then. we will use v(t) = at^2 + bt +c

lol

we have an average of 126

[dense breach]

we are in a vc rn

for like

HOURS

trying to solve this

[heavy dawn]

if the speed increases non-uniformly, the equation will be v(t) = 1.8t + 81

however, this is also valid: v(t) = 0.24t^2 + 1.8t + 81

what do you think?

[dense breach]

how did you get this? did you make any assumption?

[dense breach]

we have to get a function that goes up and down

and that v(0) = 81 and v(10) = 93

ill try some values of t

[heavy dawn]

ah for fucks sake

tsk tsk

it rebooted again. ive been having these problems as of lately

ill be with you in a moment to explain in detail

[dense breach]

lol its okay

[heavy dawn]

ok back to the good ol' days

we use the quadratic formula

v(t) = at^2 + bt +c

just verifying..

lol

[dense breach]

yeah

[heavy dawn]

howd you even end up with that question

[dense breach]

the professor said that he wouldnt gonna be present today and tomorrow

and said "group up with 2 friends and try this"

thats it

lol

[heavy dawn]

quick question

does this look like the right answer to you: v(t) = 81 + 1.2t + 0.12t^2

if yes, ill explain how i got it

[dense breach]

i dont think so

[heavy dawn]

hmm

lets see

[dense breach]

i was thinking that the right answer had a - in it

[heavy dawn]

v(t) = at^2 + bt +c

[dense breach]

cuz of the break part

yknow

[heavy dawn]

v(0) = 81

v(10) = 93

v(t) = at^2 + bt +c

100c + 10b + 81 = 93

10b + 100c = 12

b = 1.2

[dense breach]

also this is about calculus I

[heavy dawn]

v(t)=v(t)=0.12t^2+ 1.2t 81

[heavy dawn]

no way in hell

here

[dense breach]

thats why we tried integrating and shi

[heavy dawn]

wait

did you say integrating

one moment

[dense breach]

yeah

[heavy dawn]

i solved it that way

look

each time i send a step, tell me if you find it correct or not kay?

81 = a(0)^2 + b(0) + c

using quadratic formula

for now

correct?

[dense breach]

yeah, c = 81

[heavy dawn]

c = 81

yes

93 = a(10)^2 + b(10) + 81

100a + 10b + 81 = 93

ok?

[dense breach]

ok

[heavy dawn]

now get ready

its about to get messy

[dense breach]

ok

[heavy dawn]

v = 1/10 ∫^10_0 v(t) dt = 126 km/h

ok?

the 0 is at the bottom of the sign and the 10 is at the top of it

[dense breach]

i see

[heavy dawn]

do i continue? is it clear till now?

i tried to calculate this in at least 7 ways

[dense breach]

its okay you can continue

[heavy dawn]

sure

put your seatbelt on 🚙

[dense breach]

tbh i do not know how to to integers with numbers at the top and ad the bottom but yeah i know that it exists

[heavy dawn]

you can simply search it. not that hard tbh. when you understand it

[dense breach]

yeah its fine

[heavy dawn]

∫^10_0 (at^2 + bt + 81) dt = [at^3/3 + bt^2/2 + 81t]^10_0

we simplify this

to (1000a/3 + 50b + 810) - 0

then 1000a/3 + 50b + 810 = 1260

1000a/3 + 50b = 450

1000a + 150b = 1350

oh

putting some lasagna here, rice over there, fried chicken on top of the counter, bowl of spicy biryani under the fridge, we get this: v(t) = -2.34t^2 + 24.6t + 81. this is the final function. if you need the full steps just tell me

does this look okay to your standards? like the goddamn function

@dense breach damn you look like your mind just went:

[dense breach]

nah its fine lol

[heavy dawn]

https://tenor.com/view/nuke-gif-8044239

[heavy dawn]

anyway. is the function ok?

if not then....

i will have to call backup

<@&286206848099549185>

[wraith jungle]

Is there something wrong?

[heavy dawn]

yes. i tried to answer this fellows question, however it seemed it was too complicated as it turned out there are several ways to do it

ill send the question

One was caught by two radars on the same day with a difference of exactly 10 minutes: the first showed the time at 6:00:00 and the second showed the time at 6:10:00. It was found that the speed of the car in the first radar was 81 km/h, while, in the second radar, the speed of the car was 93 km/h (with the tolerances of 7 points already removed). Therefore, the value of the fines was R$260.32.

After receiving the notification, the driver decided to file an appeal, looking for some argument to cancel the fine. In the justification, the driver included could be in Uniform Rectilinear Motion or Uniformly Varied Rectiliniear Motion.

A very attentive agent received the appeal to judge it and soon discovered that a first notification had occurred on a radar that was at a distance of 21 km (measured in a straight line) from the radar that issued the second notification. He recalled that the highway that contains the two radars in question does not have exits or returns between such radars and that the maximum speed allowed on this highway is 80 km/h.

construct, with justification, a possible function v:[0,10]→R, such that v(0) = 81, v(10) = 93 and v(t) describe the speed of the car at time t, in minutes from the moment it passes the first radar, in a manner consistent with the problem situation."

For the answer to 4., i thought the problem would give me a function or something, without it i do not know how to work, can someone help me please?

guess this is my limit as a high school student huh?

[dense breach]

looks like thats the right function, but could you please do the full steps? i got a little lost

[heavy dawn]

excellent!👍

[dense breach]

tested some values of t and it worked!

[heavy dawn]

yes

no problem amigo

[dense breach]

obrigado!

how do you know this if youre in high school bro

[dense breach]

you got it lil bro

thank you!

could you please send me the full steps? @heavy dawn

[heavy dawn]

oki doki

[heavy dawn]

yes yes

ill send it in images easier

[dense breach]

okay!

[heavy dawn]

i had an ai write down my steps in an easier way

equation 1:

now

we have another equation that we will use

[dense breach]

one question

[heavy dawn]

dont remember how i got it though. cant find the step

[heavy dawn]

ye

[dense breach]

nah nvm

[heavy dawn]

go ahead

[dense breach]

nah its okay

go ahead

[heavy dawn]

fr fr. go ahead

here

i wrote it in my notebook, took a pic, and then made an ai write what i had on my notebook since my writing isnt neat

so yea, dont mind me

ask away any questions

[dense breach]

its okay! i got a little bit lost but i alr understood everything

youre a genius

[heavy dawn]

nah

im hoping i just get a good diploma fr

[dense breach]

bro is in high school

[heavy dawn]

im in first year

[dense breach]

doing better than me in college

[heavy dawn]

i wish

math is just

like damn

anyway

does this solve your question?

[dense breach]

just a question, why did you integrate here?

[heavy dawn]

cuz...lets see.....

In order to clearly show that the driver couldn’t have been traveling under Uniform Rectilinear Motion (MRU) or Uniformly Accelerated Rectilinear Motion (MRUV)

i just felt like it matched this question the best you know? its more like something you feel

with your sixth sense

your math sense

[dense breach]

we are studying integers so lets integrate things

lol

[heavy dawn]

damn we have those things in high scool

school*

[nova ocean]

yo

im his friend

we are doing a university project

dude

im still kinda lost

i ve been doing this all day lol

[heavy dawn]

lol

ye

dont blame you

took me at least 7 tries

[nova ocean]

do you think you could join a voice call and explain some parts of what you did

i can speak english dont worry

its just that i ve been trying to do this all day lol

[heavy dawn]

i dont think i can now

[nova ocean]

damn

[heavy dawn]

but maybe later since i got like a meeting with a projec team i ahve

and my laptop is like

overcrowded with tabs

my laptop will collapse at this point

but

feel free to ask questions

[nova ocean]

i guess my main question is on how you could integrate this right here

and still know that you are in the right track

because we have to write an essay on how we solved it

[heavy dawn]

holy

who writes an essay in math

[nova ocean]

i know right

[nova ocean]

fair

[heavy dawn]

one sec

ill explain

[nova ocean]

thank you so much lol

[wraith hinge]

bro our teacher smokes scary amounts of mary

[heavy dawn]

lol

[radiant pier]

yep

[heavy dawn]

here

wait shit

wrong tab

[nova ocean]

lol

[heavy dawn]

ok

anyway

do you want to talk in a group instead?

its not practical to talk here

[nova ocean]

yeah

[heavy dawn]

k

ill create one

[nova ocean]

that would be so much better

[heavy dawn]

send me a friend request

[dense breach]

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[nova pumice]

Is it true that limx->0 (1/x^2) it doesn’t exist? Wouldn’t it just be infinity?

[west sleet]

I think this limit exists

But the limit of 1/x as x approaches 0 doesn't

[tepid prism]

we only say that a limit exists if it is equal to a real (or complex number) but infinity is not a number

so the limit does not exist

sometimes people call it the improper limit, because we know what happens for really small values of x

[nova pumice]

Thank u

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[nova pumice]

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[nova pumice]

Can I get some help on how to do 1 a?

I’m not sure how to begin

I know the point slope formula and stuff

Could I actually get help with 2a.

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[nova pumice]

Someone

[ionic pendant]

given two points, can you find the slope of the line between them?

[nova pumice]

Would I just do 7398-3438 / 40?

And then the same for each interval

It gives me

[ionic pendant]

no

that's the correct formula only for the first interval

[nova pumice]

Yea then like next I would do 5622-4559 / 20-10

[ionic pendant]

yes

[nova pumice]

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[nova pumice]

Thank

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[nova pumice]

Hello I’m confused on what number 3 is asking me to do

I’m not sure what to do with x,1/1-x or whatever

[vagrant marsh]

if you plug in x= 1.5 for example, you will get a second point on the graph (Q)

then you find the slope of the line passing through the two points P and Q

[nova pumice]

How can I do that when there’s a comma though, x,1/(1-x)

[vagrant marsh]

imagine it as Q(x) = 1/(1-x)

the comma is the ordered pair (x, Q(x))

(x, 1/(1-x))

[nova pumice]

So it would just be 1.5, Q(1.5)

[vagrant marsh]

yes

[nova pumice]

ok thank

[vagrant marsh]

youre welcome

[nova pumice]

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[spring vigil]

what are the last 6 digits of 5^131313
I figured out that:

5^131313 congruent 0 (mod 5^6)
5^131313 congruent 5 (mod 2^6)

[river willow]

write out powers of 5 until you see a pattern

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[blissful geyser]

can someone check if my (c) and (d) are correct?

[burnt mulch]

You’re supposed to minimise position here

[burnt mulch]

I have no idea what you did here but it’s wrong

[blissful geyser]

i solved (c) using (a)

since the average velocity is 2 according to (a), then lasting for 6 units should give you 12

no?

or are you not allowed to do that

?

[burnt mulch]

Acceleration isn’t uniform -> you can’t do that

[blissful geyser]

ahhh

[blissful geyser]

opps i meant s(t) instead of v(t)

i didn't mean to write v(t)

but i assume even then, my (d) is wrong?

[burnt mulch]

,w minimize (t^3)/3-3t^2+8t

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@blissful geyser Has your question been resolved?

[blissful geyser]

ah also i figured out why i got 12

it's because of the interval (2,4)

if you add instead of substracting

you would be subtracting and it makes 0 sense since we're finding the 'distance'

so we either have to do

this

or

[burnt mulch]

Distance, not displacement

[blissful geyser]

simply just subtract it!

either way works

same thing xd

[burnt mulch]

What if the particle switches directions? (Spoiler alert: it does)

[blissful geyser]

no im talking about only this interval

this is the answer for (c)

[burnt mulch]

This is fine so I assume what you did is right

[blissful geyser]

take a look at the two points on the green graph

we try to calculate the distance

between those 2 points

but it's going down

so it's negative

but for distance

we always want +

so i just substracted

[blissful geyser]

yup

oki let me take a look at (d) now

i think we're good on (c), thanks

ok

farthest to the left

means most negative

aka least value

right?

[blissful geyser]

yeah minimum

isn't s(t) at its minimum at t = 0?

you get (0, -4)

isn't y = -4 the minimum?

[burnt mulch]

,w minimize (t^3)/3-3t^2+8t for 0 \leq t \leq 6

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[burnt mulch]

Yeah

[blissful geyser]

so is the answer 0 for (d)?

i mean t = 0

[burnt mulch]

[blissful geyser]

wait let me double check with the graph rq

here

we only start from 0

because in real life

we can't found negative time

xd

so....

it does look like t = 0 is the minimum!

tysm now i have a much better understanding of integration! @burnt mulch

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[burnt mulch]

[blissful geyser]

oh wait just one more thing

.reopen

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[blissful geyser]

it says it's moving along a straight line xd but in fact

it's not straight lol

so i guess um

[ionic pendant]

can you post the full context?

[vagrant marsh]

imagine a car driving along a street

[burnt mulch]

I was helping just now, I’ll take over if needed

But @vagrant marsh got the idea so we should be fine

[vagrant marsh]

it can move back and forth along the street give a certain position function

[ionic pendant]

oh this is from before

[burnt mulch]

Yeah

[ionic pendant]

the graph does not depict any physical movement because the x axis is time

[blissful geyser]

yeah sorry you can scroll up for more context

[blissful geyser]

ahhh

[vagrant marsh]

or you can imagine throwing a ball straight up

[blissful geyser]

ah yeah

guys im just dum

b

[vagrant marsh]

its position can be given by -9.81(x^2) with some shifting

[blissful geyser]

what i was thinking of was

prolly the

x-y plane

but this is a

position-time plane

right?

[vagrant marsh]

yeah its easy to get confused when theres a lot diagrams going around

i dont blame you

[blissful geyser]

[vagrant marsh]

yup, position time

[blissful geyser]

well at least now i understand integration better haha

tysm guys!

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[blazing geode]

I'm depressed

[blazing geode]

im so sad

[languid bluff]

@blazing geode bro

u gud?

[blazing geode]

the question is flawed

[languid bluff]

how so?

[blazing geode]

it merely mentions that the graph passes y=5000 at certain time. the horizontal asymptote is not determined.

Therefore both A and B are the possible solution

But the solution insists that B is the one and only solution, since the asympote of the graph should be y=5

it didn't even mark the correct unit'

I'm so sad

[limpid vortex]

it's not very clear but i think the key word approached is what's causing your issues

[blazing geode]

I'm so sad

I'm deflated

[trail ingot]

mickey

[blazing geode]

thank you. I feel hope again

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[trim jolt]

@subtle topaz Has your question been resolved?

[subtle topaz]

<@&286206848099549185>

[wind cloak]

!status

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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

[subtle topaz]

1

[wind cloak]

you know about remainder theorem?

[subtle topaz]

yes

but here two questions are mixed

f(19)=99 , f(99)=19

then?

[broken osprey]

We can judge from the options that the equation is cubic

[lusty delta]

that is fully unnecessary

[wind cloak]

hey I cam back

[broken osprey]

No

[lusty delta]

no they meant cubic

but w/e it's not important

[lusty delta]

because the divisor (x - 19)(x - 99) is a quadratic, you know the remainder will either be linear or constant

[broken osprey]

Exactly

[lusty delta]

(judging by the answer choices, you can see it's linear. but ignore that for now)

[subtle topaz]

.

[lusty delta]

now, since you know P(x) = Q(x)D(x) + R(x)

(P is the polynomial. Q is the quotient. D is the divisor. R is the remainder)

you have P(x) = Q(x)(x - 19)(x - 99) + Ax + B

from here simultaneously plug in 19 and 99 to solve for A and B

[subtle topaz]

why can these be replace for values of x and y

[lusty delta]

what y

[lusty delta]

recall that you discovered this information

[subtle topaz]

yes then?

[lusty delta]

take this equation. plug in 19. plug in 99. solve for A and B

[subtle topaz]

oh

[subtle topaz]

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[dim kettle]

For (b), in order to get 3 distinct colours we can choose 1 from 20 the first, then we choose 1 from 15 because we need to substract 5 balls that are the same color as the first ball, and finally 1 from 10 with the same reason. It follows that we get 20x15x10=3000. However, the total probability is binom(20 3)=1140. Hence there is a contradiction in my approach as the total of cases is smaller than a specific one. Where did I do wrong?

[wraith hinge]

y not choose thr color first

• ur method is wrong bc

it could be like

red green yellow

yellow green red

u counting all them

[dim kettle]

Oh so ans/3!?

I mean 3000/6=500

[wraith hinge]

ye i think

[dim kettle]

I think I am confused about when to use permutations and combinations

[wraith hinge]

i woulda done 4C3 * (5)³

[dim kettle]

Ah yes

And for (c)

[wraith hinge]

casework ig

lemme think tho

[dim kettle]

Is this correct?

[wraith hinge]

why u doing 5C1 * 4C1

instrad of 5C2

idont think the order matters

[dim kettle]

But isn’t it the same? Because you are choosing one from 5 and then 1 from 4

[wraith hinge]

ye but ur now considering the order

so like

a,b n b,a are two different thing

[dim kettle]

Oh yes

[dim kettle]

Wait shouldn’t it be without order?

[wraith hinge]

ye thats why u shouldnt do 5C1 * 4C1

[dim kettle]

I mean, ab and ba are the same no?

[wraith hinge]

yes bro

[dim kettle]

Ah okok

So in 3B and 2B 1Y I think I did the same mistake

[wraith hinge]

1B also imo

whyd u do 10C1 * 5C1

[dim kettle]

Because we choose 1 from 10, because we can’t choose blue and yellow —>20-5-5=10

And then we are choosing 1 from 5

[wraith hinge]

why 1 from 5

[dim kettle]

Ahh right I see the problem

[wraith hinge]

u just choose 2 balls from 10

[dim kettle]

Yes true

[dim kettle]

Thanks for your help!!!

[wraith hinge]

nice

[dim kettle]

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[dim osprey]

the leaning tower of pisa currently leans at an angle of 4 degree and has a verticle height of 55.86m, how tall was the tower when it was orignally built

[dim osprey]

so ik how to solve this

but i m little confused, should i take cos 4 or sin 86

[violet gust]

both are same numbers

[dim osprey]

@violet gust but it wont be wrong if i take sin instead of cos right?

this is the reference diagram

[violet gust]

when you change cos, you are also changing the angle to 86

so it is still correct

[dim osprey]

alright

ty

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[dim osprey]

@violet gust ACTUALLY cuz when i m checking on the internet

[violet gust]

and just in case, ur diagram shows hypotenus as 55.86 which is wrong

[dim osprey]

@violet gust no its correct

cuz the orignal height of the leaning tower of pisa will be x

[violet gust]

vertical height is not hypotenus

hypotenus is the original height x

[dim osprey]

oh yeah fck

my bad

sorry

[violet gust]

npnp, as long as you can catch the error, its fine

[dim osprey]

yess

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[topaz shard]

I'm quite unsure if what I'm doing is right, this is multiplying functions by the way

[solid kiln]

VincentBH

[minor steppe]

Otherwise everything is correct

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@topaz shard Has your question been resolved?

[topaz shard]

Ohh

Let me try again

Uh is it normal for the final answer to be THAT long

[minor steppe]

Yes and that is also the right answer.

[topaz shard]

Ohh I see

Looking at it seemed overwhelming

Thank you for the help!!

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[sudden hornet]

how do I calculate L(p,f) and U(p,f)

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@sudden hornet Has your question been resolved?

[sudden hornet]

<@&286206848099549185>

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@sudden hornet Has your question been resolved?

[wraith hinge]

What is L(p,f) and U(p,f)?

[sudden hornet]

lower reimann integral

and upper reimann integral

[wraith hinge]

I believe you mean riemann sum, not integral

[sudden hornet]

yeah

[wraith hinge]

But how did you define them?

[sudden hornet]

the partition points are given

in the set P

[wraith hinge]

For L(p,f) you need to calculate the infimum of the function at each part of the partition, right?

[sudden hornet]

yeah

but the graph has a breakpoint in one of the sub intervals

[wraith hinge]

Is that the only part where you are having trouble to compute it?

[sudden hornet]

yes

[wraith hinge]

So

The function is almost everywhere differentiable

Do you know any relationship between derivatives and minimums/maximums of functions?

[sudden hornet]

yep the derivative is 0 at the maxima or minima

[wraith hinge]

Kind of

We can only say that if the function is defined in an open interval around that point

For example, for the function f: [0,1] -> R given by f(x)=x

The derivative is never 0

but it has a minimum at x=0

That's because 0 is an endpoint so the function is not defined in an interval around x=0

[sudden hornet]

yeah we have to check endpoints too

[wraith hinge]

We can subdivide our partition further

[sudden hornet]

so like -1/3 to 0?

and 0 to 2/3

[wraith hinge]

Yes

[sudden hornet]

and solve both individually?

[wraith hinge]

in the interior points of those intervals the function is differentiable so for those they can only be a minimum if the derivative is 0

and for the endpoints you check separately

[wraith hinge]

Yep

[sudden hornet]

for infimum in -1/3 to 0

which fucntion will I use

is 0 included in the -1/3 to 0 sub interval?

[wraith hinge]

Yes

But you can check it separately since it is an endpoint

[sudden hornet]

ok thank you

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[lavish widget]

how to do this question?

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[twin storm]

dammm this shi hard

[wraith hinge]

i found it i think

add 4 to the first eq

so a+b+c+d(1/(b+c+d) + 1/(a+c+d) + 1/(a+b+d) + 1/(a+b+c)) = 5

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[wraith hinge]

im setting s = a+b+c+d
A = b+c+d B = a+c+d C = a+b+d D = a+b+c

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[wraith hinge]

lemme see if it works tho

[wraith hinge]

s(1/A + 1/B + 1/C + 1/D) = 5

the second eq is (s-A)²/A + (s-B)²/B + (s-C)²/C + (s-D)²/D

= s²(1/A + 1/B + 1/C + 1/D) - 4*(2s) + A + B + C + D
= -3s + 3s = 0

so the answers 0 ig

https://tenor.com/view/brandon-brandon-curington-curington-gif-16362657958830388259

@lavish widget

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[rotund grove]

find number of positive integers satisfying:

8 = [n/7] + [n/49] + [n/343] + [n/7^4] + [n/7^5].........

[rotund grove]

[.] represents the greatest integer function for you kind information

this was very easy to be honest

no-thanks

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[rotund grove]

let p(A) = 15
a1, a2, a3, ..... a15 be its elements
find number of combinations like a1 * a2, a2 * a3, a2 * a7, ...... in this set

[rotund grove]

the product combinations

,w 15C2

[solid kiln]

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[rotund grove]

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[rotund grove]

,w nC2

[solid kiln]

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Requested by anshulyolo121

[rotund grove]

,w simplify N_C_2

[wraith hinge]

lol

[solid kiln]

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Requested by anshulyolo121

[rotund grove]

,w (n k)

[pearl lotus]

Wouldn't that just be equal to the no of subsets except the null set and singleton sets

[rotund grove]

I have goteen it

Each set must contain 2 elemetns only

[pearl lotus]

like take a subset and multiply it's components

[pearl lotus]

Why?

[rotund grove]

I basically had to find 15C2

[pearl lotus]

Ohh nvm it's the question

[rotund grove]

pairs of 2-2-2-2-2-

Yes

[pearl lotus]

Yup

[pearl lotus]

[trim jolt]

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[devout ledge]

Each of the digits 3, 4, 5, 6, 7, 8 are to be used once to create 2 distinct 3-digit numbers. Find the smallest possible difference between these 2 numbers.
I got it through brute force but wanted to know if there's a much easier way (which there should be). The answer is 47 by the way.

[spiral kettle]

your instincts should tell you that the hundreds place should be close together, and leave room at the extremes for other digits to be used to make them close together

like the numbers are gonna be 634 and 587

because you reserve small numbers for the 6 and big numbers for the 5

and then the tens spot for the 6 should be as low as possible and for the 5 should be as high as possible, because the 10’s place will have more impact than the ones place

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@devout ledge Has your question been resolved?

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[wraith hinge]

,rccw

[loud bobcat]

Does anyone what I’m doing wrong here?

I’m trying to find angle A

It shouldn’t be more than 30 degree since that’s where the adjacent in respect to axis x is 30 degrees

[simple haven]

I don't quite understand the diagram

[trim jolt]

@loud bobcat Has your question been resolved?

[loud bobcat]

I’m trying to find the angle A

[simple haven]

So why not use the law of cosines?

a^2 = b^2 + c^2 - 2bc cos A

thus

A = arccos((b^2 + c^2 - a^2)/2bc)

crunching the numbers I get 35.29 degrees. @loud bobcat

you said it shouldn't be more than 30 degrees.

Are you certain that your lengths correct?

[loud bobcat]

same

yea the lengths are correct

[simple haven]

then are you certain your diagram is correct?

because if the lengths are correct, there is only one possible triangle.

and that triangle has a 35.29 degree angle right there.

[simple haven]

and is this part of 1/2 or 1/3?

[loud bobcat]

1/3

[simple haven]

let me work through the problem really quickly

[loud bobcat]

Maybe my diagram is wrong

Not sure though

[simple haven]

that's the most probable result, but I'm working it out

so I have the y component of the resultant vector being positive.

and my magnitude of the resultant agrees with your magnitude.

so the angle is 35 or so degrees

@loud bobcat

[loud bobcat]

yea i got it now, thanks

.close

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[shrewd pewter]

Hello!! I need help solving this question

Solve through factoring
2x² - 9 = 7x

[shrewd pewter]

Ive gotten half way through the problem however the latter half is confusing and I need help

2x² - 9 = 7x
2x² - 7x - 9 = 0
(2x )(2 ) = 0

[shrewd obsidian]

try to write 7 = a+b where $ab=18=2 \times 9$

[shrewd pewter]

I'm so dumb, I'm sorry what?

[solid kiln]

atif

[shrewd obsidian]

youre not dumb i made a mistake earlier

[shrewd pewter]

From what I think I need to do next, I have to factor and distribute the factors of nine and do the outer inner method? And then if it equates to 7 I got the right factors?

Although this part is really confusing for me

[shrewd obsidian]

just a second

[shrewd pewter]

Ahhhhh

[shrewd pewter]

Ofc take your time. Thank you for helping me

[shrewd obsidian]

$2x^2 -9x + 2x -9 = 0$