help-10

[green compass]

all the math makes sense

[timid silo]

yea but how is this right but not the other thing

when i did the exact same thing

[green compass]

let me show you

the original equation is x/2 + 3 = 4

there is no 7

[timid silo]

wait

why is the + a -?

[green compass]

wdym

[timid silo]

here

[green compass]

you're subtracting 4 from both sides of the equation

[timid silo]

x/2 + 4 = 3

x/2 = 3 - 4

[green compass]

x/2 + 4 = 3
x/2 + 4 - 4 = 3 - 4
x/2 + 0 = 3 - 4

[timid silo]

that doesnt make sense

[green compass]

how so

[timid silo]

ok i have another question on my study paper

so by your method i can get it right

(x/4) + 6 = 7
(x/4) = 7 - 6
(x/4) = 1

but it cant be 1

[spiral maple]

why not?

[timid silo]

oh forgot to do something

[green compass]

but it is

then you multiply both sides by 4 and get X on its own

[timid silo]

but x = 1 isn't an answer on the sheet

[green compass]

that's because you have x/4 = 1

you're looking for x

[timid silo]

(x/4) + 6 = 7
(x/4) = 7 - 6
(x/4) = 1
x = 1 * 4
x = 4?

[spiral maple]

yes

[green compass]

yessir

[timid silo]

i just dont know how

[green compass]

wdym

[spiral maple]

what dont you get?

[timid silo]

idk

just everything

[green compass]

ask your teachers then

if you don't understadn

[spiral maple]

gonna need you to try and be more specific then, cause we cant work with nothing

[green compass]

everything is too vague for us

be specific

[timid silo]

ok next question
(x/5) + 2 = 8
(x/5) = 8 -2
(x/5) = 6
(x/5) = 6 * 5
x = 30

?

[spiral maple]

yes.

[green compass]

yep

[timid silo]

x/3 + 5 = 10
x/3 = 10 -5
x/3 = 5
x/3 = 5 * 3
x = 15

[green compass]

yes

[spiral maple]

You're just doing the same algorithm with different numbers...

[green compass]

same concept different numbers

[timid silo]

ok I think i got it

next question is
x/3 + 1 = 16
x/3 = 16 - 1
x/3 = 15
x/3 = 15 * 3
x = 45?

[green compass]

yes

[spiral maple]

again, you're just doing the exact same thing.

[timid silo]

i feel smart smart

[green compass]

you're just doing the same thing

clearly you know what to do

[spiral maple]

You're also not really demonstrating learning with these questions

[timid silo]

yea but i just wanna be extra sure

[timid silo]

what do you mean?

[green compass]

you said you don't understand what you're doing

being able to compute the answer doesn't mean you know what you did

[spiral maple]

Btw, I dont need an echo

You dont need to repeat everything I say

[spiral maple]

but anyway, you're doing the exact same problem, just with different numbers

can you solve something like $\frac{2x}{5}+3=7$?

[warm shale]

Mosh

[timid silo]

i think

ok no

no idea

only when x is alone

[spiral maple]

so like I said... you dont actually know what to do

you just mindlessly followed an algorithm

[timid silo]

yea...

can you please explain how to do it with the 2x over 5

[spiral maple]

you just try to isolate for x

first you move the 3 over to the other side, like how was shown to you

[timid silo]

ehh

what do you like exactly mean by
"first you move the 3 over to the other side"?

[spiral maple]

what dont you understand about it?

[timid silo]

like

what other side?

[spiral maple]

the other side of the equal sign

[timid silo]

$/frac{2x}{5}=7+3$?

[warm shale]

⛧Chumpa⛧

[spiral maple]

no

if you add 3 to both sides, you'd get 2x/5+6=10

[timid silo]

• 6?

[spiral maple]

no

If you have 3 pens, what do I have to do to leave you with 0 pens?

Take 3 pens from you right?

[timid silo]

-3

[spiral maple]

yes, you subtract 3 to cancel the addition of 3

so you have 2x/5+3-3=7-3 means 2x/5=4

[timid silo]

$\frac{2x}{5}+3-3=7-3$

[warm shale]

⛧Chumpa⛧

[spiral maple]

yes..

[timid silo]

$\frac{2x}{5}=4$?

[warm shale]

⛧Chumpa⛧

[timid silo]

?

[spiral maple]

yes.

I just said both of those but yeah

[timid silo]

so $\frac{2x}{5}+3=7$?

[warm shale]

⛧Chumpa⛧

[timid silo]

is just $\frac{2x}{5}=4$?

[warm shale]

⛧Chumpa⛧

[spiral maple]

those are equivalent, yes.

then what would you do, piggybacking off the fact the 1st step was to cancel the addition

[timid silo]

so the + was really just a -

[spiral maple]

no

not at all

[timid silo]

oh

[spiral maple]

• and - are completely different things

3-3=0, and adding 0 does nothing

[timid silo]

im talking about just in that equation

[spiral maple]

Still no

for the reason I just said

[timid silo]

can i ask you a question

[spiral maple]

dont ask to ask... just ask the question.

[timid silo]

how do you do the bot thingy

like how do i know what to type

@spiral maple

wait

am i allowed to ping you

[obtuse pebble]

@timid silo Has your question been resolved?

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[royal smelt]

anyone help im bad bad

[obtuse pebble]

@royal smelt Has your question been resolved?

[turbid marlin]

yea its a 3D drawing

solved, u can close now

.close

[turbid marlin]

thank you

split it into two rectangular prisms and add them

[royal smelt]

what?

im braindeaded

[turbid marlin]

find the volume of A

then find the volume of B

and add them together

$A=(10.1)(4.6)(5.8)$

[warm shale]

DeathWalker1

[turbid marlin]

that would be the volume of A

do the same for B

and add the two answers u get

[royal smelt]

ok ty got it

[turbid marlin]

np

[royal smelt]

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[obtuse pebble]

@zinc oxide Has your question been resolved?

[zinc oxide]

no

[dapper bloom]

What is your question?

I see the image, but I do not know what part of it you need help with is all.

[zinc oxide]

a and b'

[dapper bloom]

Yes, but what parts of these problems are giving you trouble?

Do you know the product and quotient rule for differentiating yet?

For the table they give u g'(x) for the values u need directly.

For the graph you can compute the slopes of each segment to find the values u need for f'(x)

[zinc oxide]

hmm

is a -7/16?

is b 5/2?

[obtuse pebble]

@zinc oxide Has your question been resolved?

[zinc oxide]

no

[obtuse pebble]

@zinc oxide Has your question been resolved?

[zinc oxide]

No

[keen crane]

We can solve a) with the quotient rule

let $$f(x) = u$$, therefore we get $$h'(x) = (u/x)'$$

[warm shale]

CoreByte

[keen crane]

applying the quotient rule we get $h'(x) = (u' * x - x' * u) / x^2$

[warm shale]

CoreByte

[keen crane]

we can get $f'(2)$ from the slope at $f(2)$, that being 1/2. The derivative of $x$ is 1. With this we can get $h'(2) = (0.5 * 2 - 1 * 3) / 1^2 = (1-3) / 1 = -2/1 = -2$

[warm shale]

CoreByte

[keen crane]

A similar process is used for b), where we find the slope of f(x) at x=-1 for f'(x). Since g'(-1) is already in the table, we have everything we need to use the product rule f(x) * g'(x) + f'(x) * g(x).

@zinc oxide

[zinc oxide]

ok

i did that

im not sure if it was right or now

not

[obtuse pebble]

@zinc oxide Has your question been resolved?

[obtuse pebble]

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[dark wedge]

$\root(3)(81)+2\root(3)(24)-3\root(3)(3)

[dark wedge]

Must've used the command wrong, anyways here's my current equation I'm stuck with

So I remember a bit of what we did in class and tried to split these up into their proper forms.

But I got stuck after turning 27 into 3, and 2 * 2\root8 into 2(2)?

[vapid skiff]

$\sqrt[3]{81} = \sqrt[3]{27}\sqrt[3]{3} = 3\sqrt[3]{3}$

[warm shale]

RyanDuarte56

[vapid skiff]

try to do it with the 24 too

actually, you already did

you split it into 8 and 3

that's correct

[severe marsh]

@dark wedge

[dark wedge]

well yes but, afterwards where do we go with this?

Like, we have 2 + 3sqrt3 + 4 + 2 * 3sqrt3 - 3 * 3sqrt3

It's getting a bit late so I'll show back up tomorrow some time. Thanks for the help so far though guys!

.close

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[nimble gulch]

Lets say domain of a function consists of circles and range of this function is lengths of these circles (so it is a number) . Is this function inverse function?

[obtuse pebble]

@nimble gulch Has your question been resolved?

[royal basin]

@nimble gulch when you say "is this function inverse function?" do you mean to ask "does this function have an inverse?"

[nimble gulch]

yes

[obtuse pebble]

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[royal basin]

.reopen

[obtuse pebble]

✅

[royal basin]

@nimble gulch so do i understand correctly that you are describing the following function?
f: {all circles in the plane} -> R
f(C) = length of C

if so,
then this function does not have an inverse.

you can take two circles of the same radius but with different centers, and your function will map them both to the same number.

[obtuse pebble]

@nimble gulch Has your question been resolved?

[nimble gulch]

but if radius of two circles are the same why they are different ? just because you can put them at different location on a plane ?

[royal basin]

i mean, WHERE a circle is matters just as much as how big it is, no?

[nimble gulch]

ok . in this case if coordinates of a center of circle matters then I understand. But what if we see no difference between circles if they have equal radius ?

[royal basin]

then i guess you will have your inverse...

[obtuse pebble]

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[nimble gulch]

Yes

[obtuse pebble]

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[nimble gulch]

in my book your answer is correct. so this means that circles on different location on plane are different

[royal basin]

oh, you had a book?

why is this only coming up now?

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[distant raft]

Could someone check if I'm right?
A = (-1,0) U (1, infty)
A is open on R, all points in A are interior points, and limit points are all points within the interval: [-1,0] U [1, infty)

[distant raft]

Not sure about B though, I think the only limit point is at 2

would 2 also be an interior point

[obtuse pebble]

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[obtuse pebble]

@distant raft Has your question been resolved?

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[twilit forge]

please help

[twilit forge]

how can I solve this one?

[bright scaffold]

what can you say about the chord that connects the two closest points?

[twilit forge]

i don't know?

can you give me any reference to learn about it?

[bright scaffold]

the chord will be perpendicular to both the lines.

just try to look at two skew lines as an X with one leg behind

the distance between the planes of the legs is your answer

and the points at the intersection by your perspective are the closest ones

[twilit forge]

can you show me the detailed steps?

because I don't what to do exactly

[bright scaffold]

https://brilliant.org/wiki/3d-coordinate-geometry-skew-lines/

[twilit forge]

this is for the skew lines, but what about the parallel lines?

[bright scaffold]

same thing

find any common perpendicular

[obtuse pebble]

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[obtuse pebble]

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[vocal wren]

I need to find (x,y) , x,y are real numbers

[eternal geode]

It's 3.5 or 3 , 5?

[mild ocean]

isn’t this just a system of equations

also i can barely read that

[stray cave]

[x]+{y}=3,5
x+[y]=4,5?

is it like

3.5

or (3,5)

[eternal geode]

Assuming it is 3.5 y is fractional part and and x is integer so possible value of x from 1st can be between [3,4) and from second equation it would clarify that x is 3.5 and y is 1.5

[vocal wren]

yeah it is 3.5

ok thanks

[eternal geode]

No problem

[vocal wren]

@mild ocean please stop trying to help, no one asked you

.close

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[dawn token]

how do we solve for question 1 b?

[dawn token]

isnt the answer supposed to be 7.85m/s?

[visual kelp]

This question does not make sense at all. It implies Jumbo phases through Penggy since $v_J > v_P$ after the collision, which either means further pushing or Jumbo passes through Penggy.

[warm shale]

ShatteredSunlight

[dawn token]

did it say phases through penggy?

nevermind isok i will try to figure it out

.close

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[warm phoenix]

f is function that take vector v and calculate the angle with the vector (1,0) counterclockwise. Is f continuous?

[obtuse pebble]

@warm phoenix Has your question been resolved?

[empty cypress]

think about what the values are right before and eight after (1,0)

[obtuse pebble]

@warm phoenix Has your question been resolved?

[warm phoenix]

I can't see that

[empty cypress]

so the angle is 0 directly at (1,0), right, and as you go ccw, the angle gets bigger

by the same it's almost back to (1,0) it's at almost 2pi, then it jumps back down to 0 right at that point

[warm phoenix]

Mmmmm

[warm phoenix]

so actually there is uncontinuous on the x axis?

[empty cypress]

I don't think how you said that is right

it's defined for all R^2 in the input, but the output isn't continuous

[warm phoenix]

yeah I just wonder where it isn't continous

@empty cypress

[empty cypress]

at a(1,0)

where a is a scalar

[warm phoenix]

oh so the positive x-axis?

[obtuse pebble]

@warm phoenix Has your question been resolved?

[obtuse pebble]

@warm phoenix Has your question been resolved?

[timid silo]

yeah more or less

that's the only place where we have that jump

[obtuse pebble]

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[wary lake]

I have two known vectors V and N: for some context, V represents the projected velocity of a circle and N is the normalized vector representing the contact point between that circle and a line. What operations do I need to do to acquire P, the vector component of V that is perpendicular to N and parallel to the line?

[wary lake]

I've been having some trouble looking up the operation, I forgot it a long time ago. I think it has something to do with the cross product but I'm no longer sure.

[obtuse pebble]

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[maiden mural]

Hi could someone check if this task is correctly done, the task was to minimize the given logical expression

[obtuse pebble]

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@maiden mural Has your question been resolved?

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[swift copper]

You can take out the constant for the final answer of the integral?

[mild ocean]

in this case yes

because you can write multiplication or division in a logarithmic as addition or subtraction

[swift copper]

Would it also be right if I didnt take out the 1/2?

[mild ocean]

but you can always write something like +2 as part of +C

i don’t see why not

[swift copper]

alright ty. this is my first time seeing something like this

still dont get it. like what cases would you be able to do something like that

[mild ocean]

yeah if you get two different answers when doing an integral and they have the same derivative, they’ll differ by a constant

ok let me give you an example

[swift copper]

Oh, I would have to work my way backwards. Since the derivative of either integral would still be the same?

[mild ocean]

if you’re doing a definite integral

and you have something like +5 in your normal integral

not sure how that would happen, but assume it does lol

it would get canceled when subtracting the two integrals

(F(b)+C)-(F(a)+C) = F(b)-F(a)

i probably explained that bad lol

[swift copper]

hmmm is there a video that can explain this concept?

[mild ocean]

basically the constants cancel when doing definite integrals

i can try to find something really quick

[swift copper]

ty

[mild ocean]

i found two

https://tutorial.math.lamar.edu/classes/calcII/IntegrationByParts.aspx go to example 5

well, right after example 5

https://tutorial.math.lamar.edu/Classes/CalcI/ConstantOfIntegration.aspx

and this

[swift copper]

tysm

[mild ocean]

no problem

[swift copper]

.close

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[clear urchin]

why does k = -1 here?

for this question

[nocturne minnow]

Depending on what Laplace table you use, this is one of the ways to write the transform, the other is $$te^{at} \Longleftrightarrow \frac{1}{(s - a)^2}$$

[warm shale]

dldh06

[nocturne minnow]

So that means for given expression, written in a similar form is $$\frac{1}{(s - -1)^2}$$ making a = -1

[warm shale]

dldh06

[clear urchin]

i see thank you

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[hushed raft]

@mild ocean why did u use ln here

[hushed raft]

shouldnt u use log bc theres no e?

[mild ocean]

any base works

it would be stupid but you could use log base 17 if you wanted to lol

would still work

[hushed raft]

ohh

[hushed raft]

its bc ur just logging both sides right?

[mild ocean]

as log as you use the same base for both sides any base works

[hushed raft]

ok thanks

.close

[obtuse pebble]

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[rough bough]

i know i can do this through graphing method

but there must be better solutions such as algebraically

wha can i do to do this algebraically

[short spruce]

$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$

[warm shale]

a disappointing son

[rough bough]

oh right left me try this

@short spruce it seems alot harder this way

[short spruce]

from what i can tell by looking at it, it should simplify nicely..

[rough bough]

I expanded it down but don't seem to reduce it down

@short spruce this is what i have

but based on looking at it, i don't quite seem to see a way

[short spruce]

evaluate cos(pi/4) and sin(pi/4), reduce, simplify

multiply the sqrt2 in

[rough bough]

so it is sin(x) + cos(x)

[short spruce]

=sin(x)

[rough bough]

yeah okay

so there is only 2 solutions

[short spruce]

yes

[rough bough]

as for this one, do i do the same?

or should i just use the graph?

[short spruce]

i'd do the same, i'm not sure how to do it using the graph

i guess if the areas under the curve add to zero? but yeah not sure how to use the graph

[rough bough]

I figured it

you use the graph and find the points when they are the same

[short spruce]

oh yeah i'm stupid lmao

[rough bough]

.close

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[obtuse pebble]

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[bold arch]

Hi could someone help me out with this? Thanks in advance

[obtuse pebble]

@bold arch Has your question been resolved?

[high lily]

what have you tried?

[bold arch]

integrating the equation

[high lily]

what did you get?

[bold arch]

2e^2x +3x+c

[high lily]

parentheses and full equation pls

dy/dx = 2e^(2x) +3x+c

what did you do after

[bold arch]

10=2e^(2x) +3x+c

[high lily]

isn't quite right

dy/dx = 10 when x=0
dy/dx isn't always 10

[bold arch]

so 10=2e^(2(0)) +3(0)+c?

[high lily]

yes

[bold arch]

and c is 8?

[high lily]

yeh

[bold arch]

.close

[obtuse pebble]

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[timid silo]

Heyy! Does anyone know how to work this question?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

no😭

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[scarlet lotus]

hello, ik it isnt completely maths but can someone teach me how to graph a parabola in excel with just the function

[scarlet lotus]

i searched it but it needs me to make a table and stuff

[timid silo]

does it have to be on excel

i dont think you can do it without table of values, not sure

[scarlet lotus]

i mean i used desmos but idk how to change how much the x and y axis go up

like the intervals

i made this graph for the profit of cupcakes

[timid silo]

are you changing the scale?

so the curve can flatten or soemthing?

[scarlet lotus]

yes i want to

[scarlet lotus]

is that not allowed?

i want it to look something like this,

this is one i found about bikes

like i want my profit to be in dollars but the cost in cents

[timid silo]

ohh

hmm

you're going to have to make another function for that, i think

[scarlet lotus]

how do i do that

[timid silo]

how did you make the first function?

[scarlet lotus]

wait a sec

i made the question, then i solved it

[obtuse pebble]

@scarlet lotus Has your question been resolved?

[scarlet lotus]

oge

.close

[obtuse pebble]

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[dense moon]

Two trains start at the same time from the same place and travel in opposite
directions. If the rate of one is 6 miles per hour more than the rate of the
other and they are 168 miles apart at the end of 4 hours, what is the rate of
each?

[dense moon]

how is the answer 24 and 18?

[astral tulip]

24 is the speed of the faster train and 18 is the speed of the slower one what's wrong with the answer?

[dense moon]

no

no sir not that

i want to know, how to get those answers

how do i solve it?

@astral tulip

u there?

[astral tulip]

oh wait

sorry i was afk

[dense moon]

np

[astral tulip]

assume the speed of one train to be x , so the speed of the other train will be x+6

[dense moon]

yes

i did like that

but my answers didn't match

😦

[astral tulip]

now that they are moving in the opposite direction and starting from the same place their relative velocity will be x+(x+6)=2x+6

[dense moon]

oh

[astral tulip]

now the relative distance travelled in 4 hours is 168 miles so, (2x+6)*4=168

[dense moon]

thanks sir 🙂

[astral tulip]

now find the value of x from this equation

[dense moon]

got it

thanks a lot

[astral tulip]

ok mate good luck to you :))

[dense moon]

thanks

.close

[astral tulip]

np :))

[obtuse pebble]

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[dense moon]

.reopen

[obtuse pebble]

✅

[astral tulip]

anything else?

[dense moon]

not now, i am actually practicing for a test that is 5 hours from now

if i face any difficulty i will ask

here

🙂

[astral tulip]

you can close the chat for now reopen it whenever you need it or just dm me :))

[dense moon]

thanks

.close

[obtuse pebble]

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[tulip agate]

I know how to solve the first part I dont know the second

[royal basin]

$f(x) \in (2.99, 3.01)$ can be restated as $|f(x) - 3| < 0.01$

[warm shale]

Ann

[royal basin]

you might find this helpful

[obtuse pebble]

@tulip agate Has your question been resolved?

[obtuse pebble]

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[waxen kraken]

it is showing ans is option d shouldnt it be option a

[vocal wren]

no, option d is the correct answer

area of bde is a quarter of area of abc

[astral tulip]

How?

[vocal wren]

oh right sorry

my bad

it's 2:1 yeah

[waxen kraken]

thanks

[astral tulip]

yep i too think it should be a

[waxen kraken]

ok

[astral tulip]

dude

i think d is correct not a sorry

it askes the ratio of area which is ratio of sides squared

so the ratio of area will be 2^2=4

[undone geyser]

??? but the sides of BDE isnt 1/2 of ABC

[astral tulip]

it is

[undone geyser]

how

[astral tulip]

oh yea you're right

im too dumb

[undone geyser]

isnt it just this

[astral tulip]

yep you're right

[obtuse pebble]

@waxen kraken Has your question been resolved?

[waxen kraken]

so answer is 2:1

[obtuse pebble]

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[timid silo]

,calculate 3-m

[warm shale]

The following error occured while calculating:
Error: Undefined symbol ulate

[royal basin]

.close

[obtuse pebble]

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[timid silo]

Help, someone ?

[cobalt surge]

Helllpppp

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[fast light]

Don't spam the ping

[obtuse pebble]

@cobalt surge Has your question been resolved?

[cobalt surge]

I am doing it after 15 minutes

Because no one is answering

[royal basin]

Show that the area of the triangle is given by cm^2

what does this even mean

or was that what you were going to ask?

all you did was post a problem and chose not to comment on it in any way

nor tell us what you actually wanted

[cobalt surge]

Ohh

I don’t understand where to start the question

[royal basin]

so you should've said that from the get-go instead of leaving us guessing :p

anyway, it sounds like the teacher wanted to put a number just before the "cm^2", but forgot to actually write the number.

which means we will want to find the area of the triangle.

[cobalt surge]

Ohh

Ty

.close

[obtuse pebble]

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[regal ocean]

could anyone possibly help me with this question

[regal ocean]

anyone?

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[royal basin]

don't spam the helper ping.

@regal ocean what's troubling you here?

[regal ocean]

Idk how to solve it in general

[obtuse pebble]

@regal ocean Has your question been resolved?

[regal ocean]

.close

[obtuse pebble]

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[worn forge]

attempting to find the asymptote of f(x) = -2x^2 + 5x - 4 / x + 3

[worn forge]

i know i have no horizontal asymptote and my vertical is -3

but i need to still find the slant

dont know how to do so

i figured it out now but i have a remainder in my long division

i have gotten -3 for mey vertical asymptote

i also have -2x+11 for my slant but i have a remainder

i dont know how to deal with the remainder

[mild ocean]

remainder doesn’t matter, it’s still an asymptote at x = -3 so i don’t know why it’s wrong lol

oh wait

[worn forge]

yeah

its c

[mild ocean]

there’s two answers with -3

yeah

[worn forge]

yeah

so remainder doesnt matter?

i know it may not in this sense but would it matter for other problems?

[mild ocean]

not sure, i don’t remember what slant means in this context

i just know it doesn’t matter for the vertical ones

[worn forge]

hmm

oh well i figured it out

haha

this online school is trash and im lookikng up a shit ton of tutorial videos

but ty for double checking

[mild ocean]

if you’re done with the channel, do .close so other people can use it @worn forge

[worn forge]

yeah i was just about to

.close

[obtuse pebble]

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[swift token]

It was given this graph, whose function is x³+Ax²+Bx+C. Need to find the values of A, B, C.

I'm not sure if I'm anywhere near the solution.
I first started to find the points that touch the x line, which are a(0;0) and b(2;0).

I proceeded to replace the x values with the ones of a and b. From this I know c = 0, so x³+ax²+bx =0. Then x(x²+ax+b)=0 which leads to x= 0 or x²+ax+b=0. By far am i doing it right?

[obtuse pebble]

@swift token Has your question been resolved?

[swift token]

<@&286206848099549185>

[mild ocean]

@swift token there’s two slopes of zero in that graph, so if you know derivatives you can use those

also you can find two x values where the cubic is equal to 1

and then you can maybe do a system of equations

actually you can find three x values where the cubic is equal to 1

so basically use derivatives and or do system of equations

actually no, from looking at the graph you can see that at x = 1, the cubic is equal to approximately 1

at x = 2, it’s equal to 0

[swift token]

Mhm

[mild ocean]

and at x = 3, it’s equal to approximately 3

[swift token]

Perhaps i might need a 3rd point, which i cant find an exact one

[mild ocean]

so you can use that system of equations instead, since it’s all integer coefficients

and, using derivatives, at x = 2, the cubic is equal to 0, but the derivative, aka 3x^2+2ax+b = 0

[swift token]

Yea, so 4+2a+b = 0 if x = 2

[mild ocean]

so i just gave you 4 equations, that should be enough i hope

actually it definitely should be lol

[swift token]

Alright, thanks

[mild ocean]

1+a+b = 1, 8+4a+2b = 0, 27+9a+3b = 3

no problem

and if you want, 12+4a+b = 0

that’s actually way more than enough lol

[swift token]

Hrm, another thing actually, I have given the solution, which is A = -4, B = 4, C = 0, so the procedures should be like the ones you listed or?

[mild ocean]

probably check using those

using the first equation, you can see that a = -b

so that looks potentially correct

[swift token]

I know I can, but im tryin to stay away from those, i'll check it myself later

Thanks for the help

[mild ocean]

no problem

[swift token]

.close

[obtuse pebble]

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[meager osprey]

Hello, I am asked to prove by recurrence this expression

[meager osprey]

Previously we have established that

So in the correction, the recurrence passes by this formula but I struggle to understand why

Because technically, here the recurrence happens every 2 steps, right? My question is, how can we use 2n-1 instead of 2n to establish recurrence, I thought that was illegal

[wary junco]

What’s your concern?

The even terms of the sequence are redundant

If you know the first term, you know the third, then you know the 5th and so on

So you can use induction to prove the claim just for the odd terms

[meager osprey]

Ah, I thought we only could do one n at a time and the n itself can be odd

if.. that makes sense?

that the number we calculate for n can be odd itself

but we must pass by every step of n

Basically I dont see how a formula that only proves every 2nd term can prove the whole thing

[wary junco]

You’re not asked to prove it for the whole thing, you’re asked to prove it only for the odd terms

The statement is false for the even terms, you’re approximating square root of 2 with this sequence

The even terms squared will be bigger than 2

[meager osprey]

Oh, because It's 2n+1?

[wary junco]

Yeah that’s just the odd terms

[meager osprey]

ah, okay, that makes sense

I have another small question

In the same vein

I am asked to prove

And we find that

...because we computed xn+1 into the formula we found previously, right?

[wary junco]

That’s just using the recursion yeah

[meager osprey]

So the next term of x2n+1

will be that

Recursion is really confusing, need to do more exercises I think :S Hard to understand

[wary junco]

This is a hard example to first learn it

Do you know the easier examples

[meager osprey]

Uh, 1/n, we saw more or less how to do it in class

:S

more or less

[wary junco]

Okay goodluck

[meager osprey]

thanks

Do you have any other examples i can look up on youtube or something? to learn?

the easier ones?

[obtuse pebble]

@meager osprey Has your question been resolved?

[obtuse pebble]

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[toxic elm]

how to find the gradient of this line?

[dawn parrot]

ratio of vertical change to horizontal change (y1-y2)/(x1-x2) for 2 points on the line
parallel lines have same slope

[obtuse pebble]

@toxic elm Has your question been resolved?

[toxic elm]

okay

@dawn parrot is my calcualtion correct?

[dawn parrot]

seems good

[toxic elm]

okay thank you

.close

[obtuse pebble]

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[timid silo]

"An object moves along a straight line so that at any time t, 0 <= t <= 9, its position is given by x(t) = 7 + 6t - t^2. For what value of t is the object at rest?"

I forgot how to do this. originally thought it was when x(t) = 0 but i guess not

[dry gate]

you mix up position with velocity i guess
velocity is meant to be 0 at this time

[timid silo]

oh so i find v(t) and set it equal to 0?

[dry gate]

yep

[timid silo]

oh

so its at t = 3, damn

what a simple question lol. thanks

[obtuse pebble]

@timid silo Has your question been resolved?

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[urban summit]

how do we prove these triangles are similar

i think dab cab are congruent

but i cant find another one

[plain owl]

You need to prove CAB and DBC similar correct?

[urban summit]

im not sure

it didnt specify

shouldnt it be abd and abc

since they look simliar

[plain owl]

Those definitely aren't similar triangles though

[urban summit]

so wat should i do

im confused

[plain owl]

You know the conditions to show similarity?

[urban summit]

yes

two angles right?

[plain owl]

Yes. It works here

[urban summit]

can u show me

how do u know which triangles were using?

[plain owl]

Consider triangles CAB and DBC

[plain owl]

Well, because they are the only ones which are similar here.

[urban summit]

arent three different triangles made

how do u know its not th eother one

[plain owl]

ADB doesn't even look like it scales to become CAB

[urban summit]

oh ok

i get it

pls continue

[plain owl]

Angle ACB and angle BCD are congruent

[urban summit]

because theyre common right?

[plain owl]

Yes

[urban summit]

ok

i need one more angle

thats wat i cant figure out

[plain owl]

And, since CAB and DBC are isosceles, they have one more pair of equal angles namely BDC and ABC

[urban summit]

wat angle theorm is taht

is there like a name for it

[plain owl]

Well, you know that angles opposite to equal sides of isosceles triangle are equal?

[urban summit]

oh

i didnt think of it that way

[plain owl]

@urban summit Type .close if you are clear

[urban summit]

one more thing

if u do that last hting u did

wouldnt u be assuming the triangles are already similar?

[plain owl]

No. I am not using the fact that they are similar

[urban summit]

oh

i kinda get it

[plain owl]

We use the fact that angle ACB=ABC and, BCD=BDC

Which gives us another equality ABC=BDC

[urban summit]

oh i get it now

ty

.close

[obtuse pebble]

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[chilly ledge]

i'm trying to formulate a formula for this one

so, the first person that draws has a probability of (n-1)/n to not draw their own name

but the person after that has a probability of (n-2)/(n-1) - the probability that their name has already been drawn

and so on

am i on the right track with that train of thought and if so, can someone give me a hint how to express this part in a formula "the probability that their name has already been drawn"

[tired shell]

wrong track, that probability is too complicated

are there questions after a. and b.?

[chilly ledge]

yes but we don;t have to do them

c.) Consecutively, find a general formula for arbitrary values of n.

that's the last one

[dawn parrot]

you might wanna take a look at derangement

[tired shell]

it's a nasty formula with the principle of inclusion/exclusion if you feel like learning that

[chilly ledge]

not really

this is part of the set of problems for the class where they teach us about conditional probability

[tired shell]

for b. just list all the ways it can happen and divide by the total

you can't really get the answer by multiplying things

[chilly ledge]

i've thought about this one for a while and i just can't get beyond the first probability before things get complicated

[tired shell]

what do you mean?

[chilly ledge]

ok the probability that the first person to draw DOESN'T draw their own name is n-1/n

but then, i have no clue how to move on

because obviously the first person could draw the 2nd persons name, so the probability of them drawing their own name goes to 0

so when you say that i should list all the ways it can happen... i don't know what to do really

[dawn parrot]

i just dont think you should use conditional probability at each stage this will grow very complicated

[chilly ledge]

sure, but what DO i use?

[dawn parrot]

derangements

[chilly ledge]

which i haven't been taught, so that's not how they want us to approach this

[tired shell]

I'm talking about for the n=3 and n=4 case

like 2341 is a possibility because there is no 1 in the first position, 2 in second position, etc

[obtuse pebble]

@chilly ledge Has your question been resolved?

[chilly ledge]

ok i still don't understand, but i've been doing this for hours, so i'm just going to leave it

[obtuse pebble]

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[desert warren]

Could someone help me with how to show that

[desert warren]

$$ \left \lfloor{\frac{2b}{b^2-1}}\right \rfloor = 0 $$ where b \in \mathbb{N}, \quad b \geq 3

[warm shale]

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

[desert warren]

I can see that it is so, but I don't know if I could/should do something to make it more obvious that it is so

(this is the end result of some earlier algebraic manipulation, not sure if I could continue to show it more clearly that the floor function has to be zero or I just say it is because b \geq 3 and leave it at that)

[tired shell]

there's a nice inequality you can do to show it's <1

b^2 = (b+1)(b-1), and try to adjust one of those factors so that the fraction gets a little bigger, and that lets you cancel the b in the numerator

[desert warren]

I don't understand what do you mean by adjusting factors so that the fraction gets a little bigger

I can do $$ \left \lfloor{\frac{2}{b - \frac{1}{b}}}\right \rfloor = 0 $$

[tired shell]

like change one of the (b+1) or (b-1) to a b

[warm shale]

Nurkas

[tired shell]

that's fine too sure

so 1/b < 1 so b-1/b >2 (b>=3)

so the whole thing is less than 1

[desert warren]

yeah yeah, ok that makes sense

but if I take (b+1)(b-1) and get the b as factor isn't it the same thing

I didn't quite understand how I can change one of them to b

[tired shell]

$$ \left \lfloor{\frac{2b}{(b+1)(b-1)}}\right \rfloor < \left \lfloor{\frac{2b}{b(b-1)}}\right \rfloor $$

[warm shale]

Plurmorant

[desert warren]

oh yeah

but if b = 3 that floor function will be 1?

[tired shell]

yeah it almost doesn't work

but since the inequality is strict it's fine

the previous one is <1

[desert warren]

oh yeah

that's right

I like it

thank you very much

👍

.close

[obtuse pebble]

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[ruby minnow]

How is
-2/3 (cos10t) +1/2 (sin 10 t)
Equal to

5/6 sin(10t -0.927)

[ruby minnow]

Which formula is used can someone pls help

[obtuse pebble]

@ruby minnow Has your question been resolved?

[spiral maple]

auxillary angle

[ruby minnow]

Thank you so much

.close

[obtuse pebble]

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[meager osprey]

Could someone please explain how we get the last expression?

[spiral maple]

it's a geometric series

[meager osprey]

Yes

🤔

[spiral maple]

so... they used the formula for finite geometric series

[meager osprey]

Im sorry if this is a really stupid question but because the numbers are inversed here, where r^n is meant to be applied, we would have 1 instead and the signs seem to be wrong if we apply this one?

[spiral maple]

they're equivalent

[meager osprey]

I dont see how

[spiral maple]

factor a -1 from numerator and denominator

[meager osprey]

oh

OH

dang im dumb, thanks a lot

.close

[obtuse pebble]

Channel closed

Closed by @meager osprey

Use .reopen if this was a mistake.

[meager osprey]

wait no

.reopen

[obtuse pebble]

✅

[meager osprey]

But now he dissapears

where did he go

[spiral maple]

"he"

[meager osprey]

Sorry, I am not an english speaker and nouns are gendered

I dont mean to offend if I did

[spiral maple]

no, I just have no clue what he is suppose to refer to

[meager osprey]

oh, sorry

the term at the top

[spiral maple]

$\abs{r}<1\implies\lim_{n\to\infty}r^n=0$

[warm shale]

Mosh

[meager osprey]

I see. Thank you

.close

[obtuse pebble]

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[chilly ledge]

so, the probability of the first draw is 1/5

the probability of the 2nd draw - you can pull out the first one again, or 20 that aren't broken

so that's 21/25

and then the 3rd draw has to be one of the 4 that hasn't been pulled

so 4/25

multiply those... and i get an answer that isn't correct (they give us answers but not methods)

what is wrong here?

[obtuse pebble]

@chilly ledge Has your question been resolved?

[chilly ledge]

<@&286206848099549185>

i have like 5 more that i'd like help with 😦

[sand saddle]

do you think that the defective light bulbs are labeled? "second defective light bulb is ..."

[chilly ledge]

that's what i'm assuming what they are implying

[sand saddle]

so why the probability of the first draw is 1/5 ?

[chilly ledge]

5 defective lightbulbs, 25 total

so getting a defective one on the first draw: 1/5

[sand saddle]

werent the bulbs enumerated ?

"second defective light bulb"

[chilly ledge]

honestly, i have no idea how to interpret it

[sand saddle]

ok sorry i'm not an english speaker i think i understand this question now and why your answer is wrong.

a )the product rule (prob a * prob b * prob c) is sensible to the order, so 1/5 (probability of pulling of a defective one )

[chilly ledge]

ok yes

[sand saddle]

4/5 (probability of pulling off a nice one)

so i'm interpreting the question like this, in your first 3 extractions you have to retrieve 2 defective, and 1 in good shape

[chilly ledge]

yeah

ok i have it

it's 1/5 * 20/25 * 1/5

plus

20/25 * 1/5 * 1/5

[sand saddle]

with the restrain that in your third extraction you have to retrieve the third one

yes

[chilly ledge]

the third draw needs to be 1/5

so the only change in order is in the first ones

[sand saddle]

exaclty

is that correct?

[chilly ledge]

ok so these questions are written by a non-native speaker and his english is NOT good.

yeah it is

[sand saddle]

fine, the part b) is similar but in these cases you are not replacing the bulbs

[chilly ledge]

yeah

[sand saddle]

so 1/520/24 * 4/23 + 4/5 5/24*4/23

[chilly ledge]

yeah that's what i got here too

awesome, thank you - i knew i was missing something and that product rule thing was it

[sand saddle]

nice dude i'm glad to hear that

🙂

[chilly ledge]

yeah see, i just checked that answer and it's correct

ok i'm gonna stop now, i've been at this all day and i've done more than enough i feel like

thank you for your help

.close

[obtuse pebble]

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[near veldt]

Can someone correct me if this is wrong? This is a physics formula by the way

[sand saddle]

and the acceleration $g$ ?

[warm shale]

BillyElKid

[near veldt]

Ohhh umm

Let me try again

[sand saddle]

it seems to dissapear

between 2) and 3) line

[near veldt]

The 2y comes from the multiples two

[sand saddle]

the minus sign between the 5th and 6th line?

of (2y)

[near veldt]

The minus is gone because the -g and -2y both has same negative sign so it’s positive ?

[sand saddle]

and where is $-g$? i only see -2y and g, not -g in the 5th and 6th line

[warm shale]

BillyElKid

[near veldt]

I put the negative on the top with 2y

Should I only put it to g and not whole fraction ?

[sand saddle]

that isn't really important because you have this.

what i'm saying is this

[near veldt]

the negative sign moving to the top or bottom is the same thing is that what it means ?

[sand saddle]

also, when you apply square root you have to enclose all of the remaining expression

[sand saddle]

yes

[near veldt]

which means i have to square root on everything including the g ?

[timid silo]

hi does this server help me with my math hw?

[sand saddle]

yes

[sand saddle]

use an unoccupied channel

[near veldt]

what does this means?

for the red color

[sand saddle]

$\frac{(2y+2y_{0})}{g}$

[warm shale]

BillyElKid

[near veldt]

why is it not equal ? isnt it already not equal in the first place?

[sand saddle]

because you write that in the 5th and 6th line

[near veldt]

ohhh i thought because i want to eliminate the negative sign for the t^2 on the step above it

because -t^2 and -2y both have negative in front, so to remove its positive thats why the negative is gone?

[sand saddle]

that negative sign was already eliminated

[near veldt]

ohhh

i think i made a mistake there, one thing to clarify, the -g is seperated from the t^2, yes?

[sand saddle]

yeeeep

[near veldt]

okay I'm fixing it again

im not sure if i should put the negative outside or inside the radical

[sand saddle]

inside , (remember that g is also inside the radical)

[sand saddle]

this

[sand saddle]

almost there!, remember that $2y+2y_{0}$ is not equal to $2y-2y_{0}$

[warm shale]

BillyElKid

[near veldt]

Ohhhh I think because the - mistake earlier on the t squared I put it into the +2yo xD

Here!!

[sand saddle]

yeeeeeeeeep 👍

[near veldt]

I have another question, instead of multiplying the 2 to both sides, is it okay if I multiply -2 instead and the right side would be = gt^2?

Will that be more complex since it involves a negative sign?

[sand saddle]

no, you are right, will be exactly the same, in that case you would have this

that is exacly the same because of this

[sand saddle]

this

[near veldt]

Would it be like this ?

[sand saddle]

nono, because $$-(a-b)=-a-(-b)=-a+b$$ and not $$-(a-b) \neq (a+b)$$