help-26

[scenic flare]

from (x-9)*(x²+bx+c)

meaning -9*c = -729

therefore c=81

[orchid wyvern]

so x^2 and x values are pulled from values that multiply together to get the product of x^0?

[scenic flare]

not sure what you mean by that, you can only get a product with x⁰ if both terms in the product contain no x

from (x-9)*(x²+bx+c)
you can only multiply the 9 from the left factor with the c from the right factor

[orchid wyvern]

prob phrased it wrong but what im saying is the unknown values that u would be looking for would be factors that multiply to 729

cuz 81 * 9 is 729

[scenic flare]

exactly, for c

now, if you're for instance looking for all products of first degree, meaning they contain x¹

you can multiply x * c and -9 * bx

if this becomes clear then it'll probably feel more intuitive for you how to get these values

[orchid wyvern]

so how would for example x^3 -1 = 0 work

[scenic flare]

yes, so if we want to factor out a linear term (meaning a term with degree 1)

then we take out a common factor

-1 doesn't have any common natural factors except 1

we could also take -1, won't really matter

so let's take -1:

our linear factor which we extract is (x-1)

now we'll again look for the other remaining factor, which must have degree 2, since the initial term had degree 3

x³-1 = (x-1)*(ax²+bx+c)

x³ only has a 1 in front of it, therefore a must be 1

c must be 1, since -1 * 1 = -1

and b may seem tricky, since the initial term doesn't have anything with x² or x¹, but again just picture there being zeros

1x³+0x²+0x¹-1x⁰

which means for x¹, that 1*c + (-1)*b = 0

we already know c is 1

thereby
b = -1/-1 = 1

x³-1 = (x-1)*(x²+x+1)

[glossy acorn]

@orchid wyvern Not sure if this will help but I did it in the grouping method^

[orchid wyvern]

gotcha, ty for the help

[scenic flare]

hope either helps, for the grouping method you start by prime-factorization for all values in the initial term

and then group accordingly

a third approach would be similar to the first one, except that you just multiply both new terms

which is the method taught in germany

[topaz sinew]

@orchid wyvern Has your question been resolved?

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[low crest]

Sum of first 3n positive integers is 150 more than sum of first n positive integers. Find sum of first 4n positive integers

i already set it up with the formulas for arithmetic series (3n+1)*3n/2 etc etc and the equation i got was 8n^2+2n-300

that gives a decimal for n so im confused

[low crest]

or maybe i'm just bad at math and n is 6 💀

but how though

[prisma mesa]

n is 6

[low crest]

if i apply the formula i get -1 +- sqrt601/8

ik

but how

[drifting swift]

and the equation i got was 8n^2+2n-300
this isn't an equation.

[prisma mesa]

How did you manage to get this?

[low crest]

ok =0

[drifting swift]

show all of your work.

exactly how you wrote it.

[prisma mesa]

8n^2+2n-300 = 0 Up to this point it's correct, I checked it

[low crest]

how do i solve that quadratic

i got 4n^2+n-150=0 by dividing by two

[drifting swift]

show your work

"show your work" means "show your work" not "try guessing random shit and barking out equations with no rhyme or reason"

[low crest]

so if i apply the quadratic formula i get

i get
[-1+-sqrt(600+1)]/8

[drifting swift]

you aren't showing your work

[low crest]

..

fine

[3n(3n+1)-n(n+1)]/2=150
9n^2 + 3n - n^2 - n = 300
8n^2 + 2n - 300 = 0

and here we are

[drifting swift]

,w expand 3n(3n+1)-n(n+1)

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ann_dec

[low crest]

oh yeah that's a typo

[drifting swift]

alright

so you got that down to 4n^2 + n - 150 = 0, which we now know is correct and is not just random

now tell me:

what is the value of b^2 - 4ac for this?

[low crest]

ohhhhhhhh-

uh

1+2400=2401

[drifting swift]

yes

and what is sqrt(2401)?

[low crest]

which would turn into 49

so 48/8

[drifting swift]

yes, sqrt(2401) is 49

[low crest]

and that's 6

ty

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[neon iron]

So kinda doubtful regarding f)

First thought was that it would be $p \oplus q$

[neon iron]

But that's not the case seemingly. Why is that?

[fossil coyote]

oplus means XOR

[neon iron]

Yes

[desert pulsar]

Logic gates

Scary

[fossil coyote]

If p is true and q is false, the thing is true

And if p is false and q is true, it’s true also

[neon iron]

Yes

[fossil coyote]

But if p is true and q is true, XOR would render it false

Oh wait shoot it’s f not d lemme reread

[neon iron]

I think you are misunderstanding my issue... The answer isn't p XOR q

[fossil coyote]

Ok just do it sequentially using ^, v, or ~ only

Btw the answer isn’t p XOR q because if you do truth tables, if p is false but q is true

Oh, actually you do need arrows for “if” also

Oh no you’re right p XOR q is equivalent

[neon iron]

[
(p \vee q) \wedge (p \implies \neg q)
]

[fossil coyote]

Ye

[neon iron]

Would this be it?

Is that equivalent to XOR I assume?

[fossil coyote]

Based on my truth tables, it should be

[neon iron]

I see yeah that's fair

I wonder why it's even necessary to have it if we can represent it using other connectives

Well regardless

[hasty trail]

I mean the last sentence is basically "both cannot be true at the same time"

[neon iron]

Yes that's why I thought XOR At first

Oh although I have a question regarding

[fossil coyote]

My guess is they probably just don’t want people using XOR because it’s not as common as the three basic ones

[neon iron]

g)

I know it is iff

[fossil coyote]

Yep

[neon iron]

But I really don't understand what "necessary and sufficient" means lol

[hasty trail]

They are both implications

In different directions

[fossil coyote]

Basically “necessary” is one way and “sufficient” is the other way

[neon iron]

Ah

Is that the terminology behind it?

[fossil coyote]

Necessary I think is <= and sufficient is =>

[hasty trail]

If a then b

Then b is necessary for a

And a is sufficient for b

or something like that

[neon iron]

Hmmmm

Oh yeah

a->b was represented as "a is sufficient for b" in my textbook

Okay dug out from the text this makes sense ty

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[ashen eagle]

What have you tried so far?

First always classify your ODE

[muted stone]

Okay

So my ODE

I would assume is linear

I haven’t tried anything I’m just scared to do something

Maybe factor something out?

Integrate first?

[ashen eagle]

Yes, it's linear

What is its order?

Can you say why it is linear?

[muted stone]

It’s linear because I don’t see another y in front of the y

Okay so it is a 1st order

[ashen eagle]

It is a linear combination of y and some of its derivatives up to some order

You must have a method for solving first-order linear ordinary differential equations

Use it. You'll get a solution function up to a constant. You can calculate the constant with the intial value. Let me know if you stuck

[ashen eagle]

I use that method too.

It's good

[muted stone]

What’s the correct initial form

[ashen eagle]

Click (1)

and read

[ashen eagle]

So, do you understand what the "correct" intial form is?

[topaz sinew]

@muted stone Has your question been resolved?

[muted stone]

No

[ashen eagle]

Essentially this means that the y' coefficient must be one

So what you have yo do?

[topaz sinew]

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[queen crown]

Need help on 11

[queen crown]

. Close

.close

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[neon iron]

alr

u have to isolate x

yk how to do that right?

ok

show me what you would get

multiply both sides by b

to remove the denominator

then you have to isolate x

get x by itself

by removing any other variables and transposing them to the other side

so you would add v to both sides

that should give you your answer

what did you get?

ok so.

it should be 2y = x - v

then adding v would give you your answer

[supple sun]

Why two it’s b

[neon iron]

oh yea

mb

its by = x - v

im stoopid

then adding v to both sides

you should have x isolated

giving you ur answer.

no

thats no right

u ADD v

not multiply

so it should be

x = by + v

not x = vby

yes

the opposite of addition is subtractiong

and vice versa

u get the jist

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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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[empty falcon]

mi is miles right?

65 ------> 1 hours
25 ------> ????

(25×1)/65

[frail reef]

Tbh it doesn’t even matter what mi means but yes, im assuming miles

[woeful drift]

!nosols

[topaz sinew]

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.

[empty falcon]

this should work

[restive inlet]

apply the relation between speed, distance and time

[empty falcon]

but we dont have distance @restive inlet

[restive inlet]

we do ...

[empty falcon]

where

[restive inlet]

in the question

[empty falcon]

oh...

[restive inlet]

that was intended for the op

you shouldn't be doing their work for them

[empty falcon]

im sorry

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[limber orbit]

hey I need help w this

<@&286206848099549185>

.close

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[sharp trail]

Need help with Kinematics, especially figuring out the formulas

v0 = intiial velocity
vf = final velocity
a = acceleration
delta x = change in distance (displacement)
delta t = change in time

<@&286206848099549185>

[frosty belfry]

@sharp trail im not good at motion questions but this is my attempt

[sharp trail]

appreciate it

[sharp trail]

You mind at having a go for number 10?

[frosty belfry]

ok

[sharp trail]

a for 3.5 should be a negative since person is decelerating i think

[frosty belfry]

yes correct

[sharp trail]

how you go from 144 - 7s to 20.57m?

[frosty belfry]

144/7

7s = 144

s = 144/7

[sharp trail]

oh ok

thanks

If you still wanna help for these I dont understand how people find the acceleration

It only tells the initial, the time, and to find the final velocity

But since I looked it up online theres a acceleration we have to find

[frosty belfry]

apply the formula

lemme send you the formulas

[sharp trail]

oh ours is different

and the fith one

so for this i do a= (vf - v0)/ delta time?? to find acceleration

[frosty belfry]

yeah change

[sharp trail]

but that would be just 16/2 which is 8

[frosty belfry]

yours just overcomplicates it imo

[sharp trail]

but acceleration is not 8

[frosty belfry]

i use v as final velocity and u as initial velocity

[sharp trail]

would it not be 8?

[frosty belfry]

u = 16m/s
t = 2
a = 9.8
v = ??

[sharp trail]

ye but how do u find 9.8

is my question

[frosty belfry]

its gravity

9.8m/s^2

[sharp trail]

so there is no formula to find a its just read question and realize its gravity?

[frosty belfry]

yes

[sharp trail]

Imma be honest I missed the first 3 weeks of this class cuz i changed from apes to this

so i was not taught that

but that makes sense

would it be a negative? cuz its on the way down after 2s or no

thanks

is this correct (srry its sideways) @frosty belfry

[sharp trail]

This right?

[neon iron]

At first glance, question 14 looks correct

[neon iron]

Question 13 looks incorrect

[sharp trail]

same question no?

or is one positive one negative

[neon iron]

what?

[sharp trail]

Like same formula to solve

just different numbers

[neon iron]

Yeah

think about

what direction the ball is going

and what direction acceleration is

[sharp trail]

ball is going up

[neon iron]

yeah

which direction is gravity

[sharp trail]

while accleration is puuling it down

[neon iron]

yeah

[sharp trail]

I asked if it was negative

[neon iron]

If what is negative?

[sharp trail]

but i dont think guy was here

the acceleration negative

which it prob is

[neon iron]

Do you think so?

[sharp trail]

ye

because ball goesup while the gravity pushes down

while the stone thrown off is both going same direction

[neon iron]

Correct

[sharp trail]

flip

,help

[thorny flame]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[neon iron]

,rotate

[sharp trail]

oh

wait lemme try and find the formulas

number 15 and 16 tricky ngl

[neon iron]

are you using the guess method?

[sharp trail]

whats that

[neon iron]

Given
Unknowns
Equations
Substitute
Solve

[sharp trail]

well

kinda

first i find given

and look for unkown

so ye

[neon iron]

What information is given to you?
What information are you being asked for?
What equations have that information?

(substitute the values into equation)
(solve)

[neon iron]

That's good!

deleted message?

[sharp trail]

I knew it was wrong so i delte it

Wouldnt the answer just be half of 19.6m

cuz 19.6 m is the total displacement

so 9.8m

and we just lf the 1s to 2s mark which is 1 sec

[neon iron]

no

You're skipping steps

use GUESS method

What information are you given

[sharp trail]

Wait

Wouldnt it be half of 9.8

[neon iron]

no

[sharp trail]

so 4.9

plus 9.8

[neon iron]

what?

[sharp trail]

14.7?

[neon iron]

I have no idea how you got that answer so idk

You

are correct

[sharp trail]

I took half of 19.6 which is 9.8 then halfed it again

[neon iron]

but i don't think the way you got the answer is the correct way to do it

[sharp trail]

oh

well im still confused on what the variables are

Im thinking they are delta t, delta x, and initial and final velocity

[neon iron]

do any of your equations have delta t?

[sharp trail]

ye

[neon iron]

which one

[sharp trail]

its delta x = vf + v0/2 (delta t)

[neon iron]

huh

[sharp trail]

lemme send pic

the 2nd one

but on my paper its delta t

[neon iron]

$$ \Delta x = \frac{v+v_0}{2} t$$ this one?

[sharp trail]

ye

but it has the triangle

[thorny flame]

᲼

[sharp trail]

obviously

next to the t

[neon iron]

I have never seen that one before

[sharp trail]

well you can show me your way

[neon iron]

Do you see the words i'm underlined

[sharp trail]

so initial is 0

[neon iron]

yeah

and it's falling

so acceleration is?

[sharp trail]

so gravity 9.8 m/s^2

[neon iron]

$$v_0 = 0$$
$$a = -g$$
$$\Delta x = 19.6$$
$$t = 2$$

what else?

[sharp trail]

well the delta x is 19.6 m

why is gravity negative>

[thorny flame]

᲼

[neon iron]

Because it's falling

[sharp trail]

but gravity pushes down

wait is it -9.8?

[neon iron]

g = 9.8

[sharp trail]

oh ok

[neon iron]

which equation has all of those variables in it?

(skip what you're being asked for part)

[sharp trail]

wait lemme look

wait lemme look'

found it i think

[neon iron]

which one

Yeah

That one

So

Back to the step we skipped

what unknown value are you being asked for?

and how can you use that equation and the information you know to get the answer

[sharp trail]

um

its looking for delta x

[neon iron]

yeah

between 1 and 2

do you know calculus?

if not

[sharp trail]

no

are we gonan beusing quadratic formula for this

?

[neon iron]

no

[sharp trail]

oh

[neon iron]

You can just

find the total distance traveled to t=2

which the problem told you was 19.6

[sharp trail]

we can

[neon iron]

and subtract the distance traveled to t =1

[sharp trail]

cross stuff out

no?

[neon iron]

yeah

you're talking about v_0 right?

v_0 = 0 so you can ignore that part

[sharp trail]

wait

lemme try problem

gimme like 4 mins

nvm

to hard

[neon iron]

it's not?

you're just changing t

and evaluating

you don't really have to even evaluate

just write the math expression

[neon iron]

so you have this equation right

$$\Delta x = v_0 t + \frac{1}{2}at^2$$

you know v_0 = 0

so you can get rid of that part

$$\Delta x =\frac{1}{2}at^2$$

oops

messed up

[thorny flame]

᲼

᲼

[neon iron]

and you know a = -g

[sharp trail]

im figuring it out

ye g = -9.8

[neon iron]

g = 9.8

[sharp trail]

why u add the -

confused me

[neon iron]

because g = 9.8

but g pulls you down

and down is negative

[sharp trail]

ye but for this equation we using 9.8

[neon iron]

yeah

so

[sharp trail]

not -9.8 like we did in 13

[neon iron]

a = -g

[sharp trail]

Oh

ISE E

bruh

i feel like ive done something wrong

i got 19.6 = 8.9

we use 2 for delta t right

or no

[neon iron]

Don't plug in 19.6 for Delta x

[sharp trail]

oh ok

hald of it?

half

[neon iron]

nothing

[sharp trail]

leave it blank?

[neon iron]

yeah

just evaluate the right side as if its just a function

it will return \Delta x

you honestly only need to do it once with t =1

because you already know what it returns at t=2

[sharp trail]

so 0 = 0(2) +1/2(9.8) (2)^2

[neon iron]

yes

no

[sharp trail]

im srry

[neon iron]

the right side returns a value which is Delta x

You don't have to set Delta x to a value

just

think of it as

$$f(t) = v_0t + \frac{1}{2}at^2$$

the 19.6 they gave you is f(2)

[sharp trail]

i didn't use delta x in 0 = 0(2) +1/2(9.8) (2)^2. I set it to 0?

[thorny flame]

᲼

[neon iron]

yeah, if you set it to zero then you set delta x to 0

just

like

use the right hand side of the equals sign

ignore left

[sharp trail]

I did

V0t + 1/2 at^2 has no delta x

plz show me on paper or something

im getting confused

should i cross out 2^2

and just do 19.6=4.9

then minus 4.9 to 19.6

[sharp trail]

to change the delta x

i gotta change the t no?

[neon iron]

yes?

[sharp trail]

Doesnt show u getting 14.7 m

we trying to find the m ye

[neon iron]

yeah meters

this was the part were it returns -14.7m btw

[sharp trail]

oh ok

i got it

imma find variables for 16

its delta t
final velocity
initial velocity

and acceleration for 16

[neon iron]

delta t?

change in time?

just say t

there is no change in time in this question ( as far as i know )

[sharp trail]

I mean it says 3 seconds

so i assumed its time

[neon iron]

Yeah you have time

I don't think final velocity is needed to solve this problem

You are correct for
$v_0$, $t$ and $a$

[sharp trail]

wait

[neon iron]

You are missing something else

[sharp trail]

is it change in velocity

[thorny flame]

᲼

[neon iron]

change in velocity would require final velocity

but you don't need final velocity to solve this problem

what other variable is there?

[sharp trail]

how can it be displacement

there is no m

[neon iron]

There is

The object falls from the bridge and hits the water

this is a distance between the bridge and water right?

[sharp trail]

ye

[sharp trail]

so we are looking for the displacement?

[neon iron]

this is a two part problem

[sharp trail]

do we find velocity

to find the disaplacmeent

or other way round

[neon iron]

you find the initial velocity yeah

but

first

you need to figure out the height of the bridge

( maybe there is a faster way )

[sharp trail]

so a is a= -g

[neon iron]

DO you want me to tell you how to find the height of the bridge?

yeah a = -g

it's falling

[sharp trail]

wait

4.5 seconds and 3 seconds

why is there two of them

[neon iron]

it's a two part problem

one time is for the first part

the other for the second part

[sharp trail]

whats our t

[neon iron]

do you know what the first part of the problem would be?

so

[sharp trail]

no

[neon iron]

you drop a coin right

you didn't throw it or anything

you just dropped it

what's its initial velocity?

[sharp trail]

0

[neon iron]

and it affected by gravity

and falls for 4.5 seconds

how far did it fall?

[sharp trail]

2.17 m

idk tbh

i gues

[neon iron]

how did you get 2.17? why are you guessing?

[sharp trail]

i did gravity divided by seconds

[neon iron]

does that give you $$\Delta x$$ ?

[thorny flame]

᲼

[sharp trail]

no

[neon iron]

so

okay

I typed out the

Given and unkowns for you

what equation do you use

[sharp trail]

3rd one

[neon iron]

yes

plug in your known values to find your unknown one

[sharp trail]

13.9

[neon iron]

how?

[sharp trail]

for v0t do i just cross all that out

[neon iron]

yes

because

[sharp trail]

since initial velocity is 0

[neon iron]

v_0 = 0

[sharp trail]

i did 1/2 of 9.8

which si 4.9

plus 9

[neon iron]

why plus 9

[sharp trail]

3^2

is 9

[neon iron]

yeah

but

its not addition

wait no

t = 3 is for the second part

you're still on the first

[sharp trail]

oh

so its

25.15

[neon iron]

No

[sharp trail]

is it not addition

[neon iron]

$$at^2 = a * t^2 \neq a + t^2$$

[thorny flame]

᲼

[sharp trail]

multiply

[neon iron]

yes

[sharp trail]

instead of adding them?

so 4.9 x 20.25?

[neon iron]

$$xx = x * x \neq x + x$$

[thorny flame]

᲼

[neon iron]

sure

[sharp trail]

99.23 m

[neon iron]

Nice

[neon iron]

what distance in the problem is this?

[sharp trail]

the coin

from the bridge to the river

[neon iron]

yeah

[sharp trail]

am i right?

so displacement is 99.23m?

[neon iron]

Yes

I might have lied when i said two parts

this is three part problem

So second part

Do you know what it is?

[sharp trail]

the stone throwing problem?

[neon iron]

What information do you need to solve the problem but don't currently know

[neon iron]

yes

[sharp trail]

well

we dont know the final velocity

[neon iron]

we aren't there yet

go back a sentence

what information in the second sentence is not given to you

[sharp trail]

throwing

so displacemenbt

[neon iron]

no

it's throw into the river

you found the displacement in the first part

[sharp trail]

Some negiative initial velocity

[neon iron]

yeah

[sharp trail]

the initial?

velocity

[neon iron]

mhmm

[sharp trail]

but wouldnt that just be 0

[neon iron]

you don't know what that is right?

[neon iron]

Why would the velocity be zero if you threw it

[sharp trail]

nvm

we know the initial velocity is a - though

[neon iron]

yeah you know it's negative

What other information did they give you ( or that you can assume )

[sharp trail]

the acceleration

is -g

[neon iron]

then what equation can you use to find the initial velocity

[sharp trail]

1st

[neon iron]

no

[sharp trail]

4th

[neon iron]

why are you guessing

What information do you know

What information are you looking for

which equation has that information in it?

[sharp trail]

the acceleration
the displacement
the time

is what we know

[neon iron]

what are you looking for

[sharp trail]

initital velocity

[neon iron]

which equation has all of those in it

[sharp trail]

3rd one

😢

[neon iron]

Yes

before you try to **substitute ** for initial

how would you rearrange the equation so that the initial velocity is isolated on its own side of the equation

[sharp trail]

kcf?

keep change flip?

[neon iron]

what

[sharp trail]

or divide

[neon iron]

rearrange the equation

so that $$V_0$$ is by itself

[thorny flame]

᲼

[neon iron]

$$V_0 = ????$$

[thorny flame]

᲼

[sharp trail]

you would minus it

and put it to other side

[neon iron]

idk what it is

but i'll asumme it is what i think it is

so yes

I'm assuming you've done algebra before right?

[sharp trail]

algebra 2

[neon iron]

so you know how to solve for x

solve for V_0 like you would solve for x

[sharp trail]

lowkey im cooked

i cant think rn

forgot everything

[neon iron]

You can just not worry about rearrangeing then

[sharp trail]

by cooked i dont mean high btw

[neon iron]

Just plug in your values first

then isolate

third equation

if you were to rearrange the eqaution it would be like this btw $$\Delta x = v_0t + \frac{1}{2}at^2$$
$$\Delta x - \frac{1}{2}at^2= v_0t $$
$$\frac{\Delta x - \frac{1}{2}at^2}{t}= v_0$$

[thorny flame]

᲼

[sharp trail]

ye i got the part where u move v0 to other side

but the t underneath

i didnt get

[neon iron]

yeah it do look weird

if you were using units it would make sense tho

the units on top are meters

and t is seconds

so its

meters divided by seconds

which is velocity

[sharp trail]

so for at^2 and the t as the denominator is it 3 for both

[neon iron]

yeah

because this is the second sentence

where the person threw the stone and it hit the river in 3 seconds

[sharp trail]

and for the displacement its 99.23 correct?

[neon iron]

what do you think?

[sharp trail]

ye

but u plug it in

[neon iron]

also

it's should be negative displacement

because the river is below the bridge

[sharp trail]

k

i got -312.4

or

-47.78

[neon iron]

Yes

[sharp trail]

i did it backwards

thats why

[neon iron]

So last part

third sentence

should be easy

Given
unknowns
Equations
substitute
Solve

[sharp trail]

ok

we are looking for final velocity vf=?

we have

initial velocity
displacement
time
acceleration

wait do we need time

probably not

idk

[neon iron]

reread the problem

[sharp trail]

hm

ok

we need acceleration

time

final velocity

and displacement

to find

the last problem

[neon iron]

that's all?

which equation would that be

[sharp trail]

we can only use 4 variables no

can we use 5 variables

?

no only 4

[neon iron]

which equation

[sharp trail]

none

i dont see an equation without v0

[neon iron]

yeah

there is one that is missing that doesn't have v_0

[sharp trail]

and i didnt add that into the 4 variables'

[neon iron]

but you don't need it

[neon iron]

yeah

[sharp trail]

i see v0 for all 4

[neon iron]

anyways, i'm gonna go now. I'm sure you can solve the last step on your own.
Goodluck!

[sharp trail]

maybe number 4

[topaz sinew]

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[neon iron]

Help:
Calculating the area of the region bounded (see figure) by the curve y=cos-1(x), the straight line y=x+π2 and the x-axis we obtain:

[neon iron]

e. none of them is correct (forgot to add)

[gray ridge]

what have you tried?

[neon iron]

Using the area formula for integrals

[gray ridge]

that's good, then you're good

[neon iron]

But couldn't solve it

[topaz sinew]

Show your work, and if possible, explain where you are stuck.

[gray ridge]

?

??

???

[neon iron]

Mb, I just don't understand what to do with that green box

[gray ridge]

what have you tried?

[neon iron]

I just have written the formula, but don't know what values to write in it

[gray ridge]

please post it here, I can't understand what you don't understand without more info

like i don't know what is "the formula"

and i dont know what is "what values"

[neon iron]

this formula

[gray ridge]

okay

what did you put in f_1(x) and f_2(x)

and are you using x?

and what about a and b?

[neon iron]

f1 is y=x+pi/2 and f2=y=cos-1x

[neon iron]

yeah, why?

[gray ridge]

i see, so what you did is like

lemme type

[neon iron]

that's the issue, don't know if I should use -1 and 1 or something else

[gray ridge]

$\int y=x+pi/2 - y=cos-1x \dd x$, like that?

[thorny flame]

Biscuity

[drifting swift]

D tier notation

[gray ridge]

oh, it's because it's ∆y, you might have to use y instead of x

[neon iron]

sort of

[gray ridge]

i see

[neon iron]

oh

[gray ridge]

like, the x=cos(y) is on the right of x=y-pi/2

so, we should have
f_1(y)=cos(y)
and
f_2(y)=y-pi/2

[neon iron]

Oh

so, let me try

[gray ridge]

sure

[neon iron]

but still, what values should I use?

[gray ridge]

since, we will integrate from 0 to pi/2

along y-axis

[gray ridge]

so , we will have bounds 0 to pi/2

[neon iron]

So like this?

[gray ridge]

nah

it's
cos(y)-(y-pi/2)

[gray ridge]

you wrote it in reverse

[neon iron]

mb

so the result gave me d.

let me check

yep, that's the answer

TYSM

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[neon iron]

how to draw this

[neon iron]

$750x\ +\ 1500y\le\ 18000$

[thorny flame]

Chocolate

[drifting swift]

why the extra spaces

but uh

draw the line 750x+1500y=1800 then shade in the half plane below that

[neon iron]

how can I draw the line 750x+1500y=1800

[drifting swift]

do you know how to graph straight lines from their equations in general

[neon iron]

like ?

it's better to assume I do not know, because idk

if they are y=mx+b I may be able to draw them

if they are like 1500y + 750x = 18,000 I may not be able to draw them correctly

[void vapor]

Try converting this to the form y= mx+b

[neon iron]

%30y<=15x+360%

[craggy haven]

idk why the percent signs

[void vapor]

Im not sure this is correct

There has to be a negative sign on x

[neon iron]

$30y=-15x+360$

[thorny flame]

Chocolate

[void vapor]

Yes

Can you plot this?

[neon iron]

a mistake, in my keyboard$ and % are close to each other

[void vapor]

If you cant, how do you think you can bring it to the form y=mx+b

[neon iron]

is it the same as y = mx+b? because here it's 30y

[void vapor]

It isn't the same as y=mx+b

But you can bring it to that form

[neon iron]

yes, that's my issue, I cannot solve it

CAN ANYONE ANSWER THUS OMG

A student of mass 50kg walks up a slope inclined at 30° horizontal.find the work done by the student if he walks 25m along the incline

[void vapor]

Whats stopping you from bringing it to that form?

You have 30y right

So how do you make it just y

[neon iron]

#❓how-to-get-help , you need to open a new hlep channel, because I take this one

[neon iron]

let me try

[void vapor]

You know that if you have an equality or inequality, you can multiply by a positive constant or add a number to both sides and preserve the equality or inequality right?

[neon iron]

$y = \frac{-1}{2}x+12$

[thorny flame]

Chocolate

[neon iron]

I divide all of them by 30

so I can make y = 1y

[void vapor]

My bad

Now you can plot this right?

[neon iron]

yes

Thank you.

[void vapor]

👍

[neon iron]

you are so good at explaining things

.close

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hoary creek]

all these are vectors:
AF=AB+AC; CE=2AB; BD=1/3BC
There are two questions. no1 is answered, and am stuck in no2 :
express AE as a function of AB and AC
I Answered till here:
AE=AC+CE
=2AB+AC
Is it finished or do i have to calculate it

[hoary creek]

$AF→=AB→+AC→; CE→=2AB→; BD→=1/3BC→
Express AE→ as a function of AB→ and AC→$

[thorny flame]

Minds&Moves

[hoary creek]

(Vectors above the letters)

<@&286206848099549185>

[grim void]

its finished

[hoary creek]

huh

[grim void]

also correct btw

[hoary creek]

but its 2ab, and we want ab

[grim void]

well you have 2 of them

[hoary creek]

so?

we can't divide

[grim void]

you can do ab+ab+ac

but 2ab+ac is definitely the answer that they want

[hoary creek]

alr then, i'll take your word

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[topaz sinew]

@little mulch Has your question been resolved?

[little mulch]

<@&286206848099549185>

[supple sun]

!show

[topaz sinew]

Show your work, and if possible, explain where you are stuck.

[coarse pewter]

What have you tried? / Show your working. (Discord crashed, didn't see joseph already did the !show)

[little mulch]

<@&286206848099549185>

can someone help me

<@&286206848099549185>

[pastel juniper]

dont ping individual helpers!

[potent ravine]

why not divide equation 1 by 2

[little mulch]

why would i do this

[potent ravine]

to cancel (5a-2b)

wait lemme try

[supple sun]

You have $x+y=4xb-2yb$

[thorny flame]

Joseph.P

[potent ravine]

yeah nothing from here. gotta do something else

[little mulch]

what can i do with this equation

[little mulch]

yep its kinda confusing

[potent ravine]

indeed 😭

[little mulch]

i found that (20a-5)x-(10a+14)y=18

[little mulch]

how did you fpund that

[potent ravine]

on dividing first equation by the second one

[little mulch]

haa okay

[potent ravine]

if someone answers this, do lemme know. this is hard af

[little mulch]

but i cant ping you its forbidden

[potent ravine]

i have no problem tho but yeah the rules. i'll keep lurking here ig 😂

[little mulch]

okeokeoke

<@&286206848099549185>

[potent ravine]

@little mulch u gotta ask your teacher mate

[little mulch]

got it

thanks btw

.closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[main knot]

.open

[dreamy drum]

need help with part a first

[shell seal]

Heya, @dreamy drum ! Hope all's well.
For the first part, let's consider the specific events which satisfy this condition.

If we use ordered pairs to represent outcomes, like $(1,2)$ for getting first a $1$ and then a $2$, then we can find the valid events to be

[thorny flame]

Drenitor

[shell seal]

$(1,1),\ (1,2),\ (2,1),\ (2,2)$

[thorny flame]

Drenitor

[shell seal]

Do you see why that is?

[dreamy drum]

yup

[shell seal]

Nice! Then we notice that these are the only 4 valid events out of the whole $6\times 6=36$ possible pairs.

[thorny flame]

Drenitor

[dreamy drum]

yup so the probability in this case would be 4/36 right?

[shell seal]

Very good

[dreamy drum]

is there a way to do it by applying a formula instead of just listing it out?

[thorny flame]

keto11

[dreamy drum]

because the next question expands upon it for n elements

[shell seal]

That's a good question! But I think in this case we should first do a and b first to see if we can pick up the pattern.

[dreamy drum]

okay

[shell seal]

The formula you provided justifies the 4/36 part, but that's only after you know how may valid options there are.
Would you like to try listing the valid pairs in b?

[dreamy drum]

for b you could end up with (1,1) (1,2) (1,3) (2,1) (2,2) (2,3) (3,1) (3,2) (3,3)

so that's 9 in total

[shell seal]

You got it!

Ooh, I think I see a pattern there

[dreamy drum]

I guess that's just 3^2

[shell seal]

It seems like it

But why?

[dreamy drum]

no clue lol

[shell seal]

That's okay haha. I'm thinking too

[dreamy drum]

Actually I guess it makes sense

[shell seal]

Let's see. You first listed the pairs in which 1 is the maximum. Only $(1,1)$ satisfies that.

Then for those where 2 is the maximum, we have
$(2,2), (2,1)$ which is like having 2 with itself and everyone before it.
Ah, and the inverted pair $(1,2)$ also.

[thorny flame]

Drenitor

[dreamy drum]

if you arrange it like this:

1 2 3 4 5 6
1
2
3
4
5
6

[shell seal]

So far, that gives us
1 + 3 pairs

[dreamy drum]

then the number of pairs expands like a square

[shell seal]

Dang, that sounds interesting. Let me see how that would look

My goodness, you're right

[hot tulip]

a

[shell seal]

This gives the pattern, actually.

The probability of having the maximum be less than or equal to $n$ would be
$\frac{n^2}{36}$

[thorny flame]

Drenitor

[dreamy drum]

yup that makes sense for $n\leq 6$

[thorny flame]

keto11

[dreamy drum]

do you know if there's any way to prove it?

at least more formally using the basic theorems

[shell seal]

Yes, and the observations made up there are the guide.
We have to use a bit of combinatorial logic but it will be a proof that holds in general.
\
Let $A$ represent the set of outcomes in which the maximum number is less than or equal to $n$. Our The probability of having the maximum be $n$ is equal to $$\frac{\left\vert A\right\vert}{\left\vert \Omega \right\vert}=\frac{\left\vert A\right\vert}{36}$$

[thorny flame]

Drenitor

[shell seal]

so we need only compute $\left\vert A \right\vert$.

[thorny flame]

Drenitor

[shell seal]

We first count the pairs which begin with $n$ and are in $A$. These are exactly $(n,n),\ (n,n-1),\ldots,(n,1)$, that is, $n$ pairs, since $\text{max}{n,i}=n\ \forall i\in{0,1,\cdots,n}$

[thorny flame]

Drenitor

[shell seal]

Then we count those pairs having $n-1$ in the first die. These must similarly be $(n-1,n),\ (n-1,n-1),\ldots,(n-1,1)$, that is, another $n$ pairs, since the maximum value in any of the pairs is still $n$.

[thorny flame]

Drenitor

[shell seal]

Continuing in this fashion we get $n$ rows for the different possible starting values, with $n$ columns, with all the possible second values. That is, $n\times n$ valid pairs.

[thorny flame]

Drenitor

[shell seal]

Hence $\left\vert A\right\vert = n^2$, which completes the proof.