help-19

[onyx cloak]

subtract the 15k from total cost, you'll get how much it costed to travel with the load, then just divide by 8.5 because you need to find cost per kilometer

[blissful musk]

What is the point of the 85lb

The weight

[onyx cloak]

maybe just a bluff

[blissful musk]

Ok

[onyx cloak]

idk either

im just guessing

[blissful musk]

Np

So yea we get it y=mx+b form and isolate for x

[low wagon]

I would appreciate it if you give me the result

[onyx cloak]

!noans

[odd edge]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[low wagon]

Po mamenme la ñema rapa su madre to el mundo en ete maidito servidor de mierda coño 😀😁🥳😇

[sharp oak]

cool

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@mystic saffron Has your question been resolved?

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[mystic saffron]

how the ε/2 + ε/2 got added in the end

😅

[half garnet]

Triangular inequality

[mystic saffron]

yes i got that part

how that 2 came

[half garnet]

Yes

[half garnet]

Wdym

[mystic saffron]

like in the denominator

of epsilon

[half garnet]

WE choose it

[mystic saffron]

ahhh

is that so that in the later steps

it end up

[half garnet]

Yes

[mystic saffron]

making a whole epsilon

so that we can prove

it

oh yeh

kinda make sense

[half garnet]

👍

[mystic saffron]

YES whole thing

make sense now

damn

thank u

so much

gonna close it now

.close

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[graceful berry]

not sure where to start

[honest turtle]

Find a point on the line, make a vector out of it with the known point.

[graceful berry]

what should i do next?

[honest turtle]

Two vectors, a known point, what next?

[graceful berry]

i got only 1 vector by making s = 0 and and substract it with the point

which i got (0,-5,-2)

[honest turtle]

And you could get multiple points from the line, by, for instance, substituting s = 1.

An plane equation is just normal and a point on the plane.

[honest turtle]

Two vectors, and use vectorproduct to find the normal.

[graceful berry]

is there more than one way to find the equation of a plane as i use a(x-xo)+b(y-yo)+c(z-zo)=0

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[paper yarrow]

i need to convert hexadecimal to base 32 , so i started with decimal however i dont know how to continue

[somber jackal]

this is kinda hard to read
what's the table at the top? is it relevant?

[paper yarrow]

thats the conversion table from hexadecimal to decimal

top table is done

i tried to then divide the positive by 32

and multiply the negative by 32

now i have a remainer of 29 24 15 , 14

[somber jackal]

if i understand correctly those are your digits?

[paper yarrow]

yes

[somber jackal]

your method is very different from mine

but that's the question done then

should 14 be D and whatnot?

[paper yarrow]

yea that is what im not sure here

never saw how the table works of base 32

also there are many versions of it lol

doesnt really help

[somber jackal]

this is definitely the common convention so i'd put that

past that i can't really help you guess lol
use an online calculator to check if your result is right though

[paper yarrow]

thats where it gets weird

it shows something totally different

what is ur way of solving those?

https://d3i71xaburhd42.cloudfront.net/8e1a32bdb80fbec987f35bf571d9d0fff7f35b29/9-Table3-1.png

is this the correct table?

[somber jackal]

30479 / 32^1 = 29 remainder 783
783 / 32 = 24 remainder 15
15 / 32^0 = 15 remainder 0
can keep going obviously but i rounded to the ones place

[somber jackal]

oh it is that one
i assumed 0-9 would be at the start but i was wrong
this is the most commonly used one

[paper yarrow]

Thanks!

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[primal flume]

На параллельных плоскостях а и в выбрано по паре точек А1, А2 и В1, В2 соответственно так, что прямые А1В1 и А2В2 пересекаются в точке О, лежащей между плоскостями. Найдите ОА, если А1В1 = 6 см, ОВ2: OA2 = 3:1

[primal flume]

On parallel planes a and b, a pair of points A1, A2 and B1, B2 are selected, respectively, so that straight lines A1B1 and A2B2 intersect at point O lying between the planes. Find OA if A1B1 = 6 cm, OB2: OA2 = 3:1

I got to the semblance of triangles

but since I have OA1/OB1=0A2/OB2=A1A2/B1B2

but I have nothing but 3/1

So OA1/OB1=3/1=A1A2/B1B2

and I don’t understand how to get any segment

<@&286206848099549185>

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[primal flume]

.close

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[frigid vine]

I get that its no because 7 is a possibility but how do I align my thinking with the explanation

[odd edge]

@frigid vine Has your question been resolved?

[frigid vine]

<@&286206848099549185>

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[crisp lily]

Does anyone know how to do this one

I'm really confused

I'm trying to find a contradiction but everything i've done has failed

please

[odd edge]

@crisp lily Has your question been resolved?

[crisp lily]

nope but I figured it out :)

[frigid vine]

.close

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[outer wadi]

I am confused on this not sure where to start, I watched some vids on mod again but i’m still kind of lost

[outer wadi]

So what we are trying to prove is that

xRy is relfexive, symmetric and transitive

so theres three things we need to prove right

gonna type myself through it to see if I get anywhere

[gritty oar]

yes

show through the definition of each of the 3 properties that R satisfies them

[outer wadi]

So for proving the first thing, that it's reflexive, means we need to prove that for every element in the relation set R, there exist an ordered pair (x, x)

[gritty oar]

for every $x\in\mathbb{Z}$, $(x,x)\in R$

[clever fjord]

chebyshev's infinite pee norm

[gritty oar]

R is some subset of Z^2

[outer wadi]

but does that mean every element in Z would need to be related to itself? For example what if R was a relation on Z^2 where R is {(1,1), (2,2)}. Then R would be reflexive right

[shut flower]

No, (3,3) should be too

And (4,4)

Et cetera

Every number

[outer wadi]

oh ok

So for the reflexive one, is it reflexive because 4 divides a^2 - a^2 which is 4 divides 0?

[gritty oar]

yes

[outer wadi]

ok I'll try writing a word solution too see if i'm understanding everything right

I'm trying to figure out why a congruent b mod 4 implies a = 4q + b but I can't see it

I think I see

[odd edge]

@outer wadi Has your question been resolved?

[outer wadi]

Sorry for the messy handwriting was jotting things down

I'm not done with it but I was writing everything down to make sure I had a really clear understanding of everything

I have a question though

in my last statement I cleam that a is congruent to b mod n if there exist an integer z such that b = nz + a

but wouldn't this mean by the division algorithm that a would need to be between 0 and n -1

how can we know that a is between that

or is the division algorithm not needed here

actually maybe the final statement doesnt use the division algorithm because we only derived it from two other statements by manipulation and didn't state anything specific about b = nz + a

and this also proves that 0 is the only number that divides all integers

since q is unique

well actually

is q unique?

is a^2 = 4q + a^2 constrained to the division algorithm that says the remaineder, a^2, must be between 0 and the divisor, 4

[gritty oar]

a = b mod n by defintion means n divides (a - b) which by definition means there is some integer q with a-b = nq, adding b to both sides gives a = nq + b

for the reflexive argument

you need to show there is some integer q satisfying a^2 = 4q + a^2

subtracting by a^2 gives 0 = 4q

which integer q satisfies that?

[gritty oar]

yes the division algorithm prodivdes a unique quotient and remainder, however you're not applying division algorithm

for example $3\equiv 7\equiv 11 \mod 4$

[clever fjord]

chebyshev's infinite pee norm

[gritty oar]

since 3 = 4 * (-1) + 7

and 3 = 4 * (-2) + 11

so 3 is the dividend and the divisor is 4

but in the first example, the "quotient" is -1, and the "remainder" is 7

and in the second, q = -2, r = 11

as for your actual answer to part a), unless you haven't proved these common results about mod / divisibility in class you don't need to prove it here in the homework

the statement in a), "use the fact that..." is being provided to you, and as i said above it's just using the definitions of mod and divides

also as a shortcut, if you've shown that $\equiv$ is an equivalence relation mod n, then you can use that fact in showing R is symmetric and transitive faster

[clever fjord]

chebyshev's infinite pee norm

[gritty oar]

otherwise you're gonna need to provide integers q that satisfy the two properties

[odd edge]

@outer wadi Has your question been resolved?

[outer wadi]

Ok thank you I’ll try the other two properties

[odd edge]

@outer wadi Has your question been resolved?

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[ember shell]

A car accelerates from 14m/s to 21m/s in 6.0s. What was its accelaration? How far did it travel in this time? assuming constant acceleration

[outer wadi]

gotta use your kinematic equations

[ember shell]

ah

uhm

[outer wadi]

which equation do you have all the information of and that has acceleration where you can solve for it

[ember shell]

uhmm

[outer wadi]

what information did they give you in the question

we know we have t, 6s

what else do you have

[ember shell]

14m/s to 21 m/s thats the velocity change

i think

[outer wadi]

yep

14 is the initial velocity Vo

[ember shell]

so do i use v=vknot+at

[outer wadi]

and 21 is the final velocity V

[ember shell]

mhm

[outer wadi]

yeah

and solve for a

[ember shell]

okok

so itll be

21=14+6a

[outer wadi]

yeah

[ember shell]

oki

[ember shell]

i got 7/6 :D

[outer wadi]

yeah thats right

[ember shell]

so

what is the uh like

uh

unit

for that

and also it asks me for the distance it travelled

😭

[outer wadi]

you can get your units from your equation which is good to practice and will let you know if you got the wrong answer sometimes, we generally know that acceleration standard unit is m/s^2. We can also get that from the equation because velocity units is m/s which they tell you, and time is in seconds. We can treat units almost like numbers. So we have m/s - m/s in the numerator which will be m/s, and then that divided by seconds, so (m/s)/s. Which will be m/s * (1/s) since division is same a multiplying be the recpriocal and that gives you your units of m/s^2

[ember shell]

ohhh

ok

i just forgor

:sobL

oki

and then how do i find the time :D

[outer wadi]

for distance, now that you have acceleration, velocity, and time, you want to go back and look at your kinematic equations and see which one has distance(change in x) and all the other variables you solved for or were given and then solve for the change in x

[ember shell]

ok ok

is it the like

[outer wadi]

earlier picture was being blocked heres a better one

[ember shell]

hmmm

those seem different from what i have

its just the triangle

is it the 1/2at^2 one?

[outer wadi]

that one works but using the second one might be easier since you also have all of those values

[ember shell]

ok

uhhh

thats the free falling object one

right? or

erm

oh wait nvm

uhh

what does that one do

i never actually learned it

[ember shell]

ok so how does the second one work

😓

to simplify all of htat

lol

[outer wadi]

how to simplify it?

[ember shell]

no i mean

like

how does it work

😭

im just confused on what that equation does

or what it does

pain

@outer wadi r u still there

😭

[outer wadi]

basically what this equation is doing is it's using the fact that the area under the line of the velocity vs times graph is the distance covered. We can see that the graph forms a triangle plus a rectangle and so we can use the formula for the area of a triangle and rectangle to get the area under the line. the area under this line is the area of the rectangle plus the area of the triagnle. The area of the traignle is (1/2)bh, base = time, height = V - Vo. The area of the rectangle section is b*h, Base = Vo, Height = t, we add these two areas together to get the total area under the curve which is the displacement, however they want distance which is just the absolute values of displacement in which our case our displacement is positive since the area is above the x-axis. The equation is just giving you the distance traveled

[ember shell]

ohhhh

oki oki thx

so i use the formula thing

[outer wadi]

yeah

and just plug in your V and Vo and time

[ember shell]

soo

ok

[outer wadi]

V was 21 and Vo was 14

[ember shell]

x=(21+14)/2

*6

[outer wadi]

yeah

[ember shell]

right

oki oki

[ember shell]

so 1-5 meters???

105

sorry

[outer wadi]

yeah

thats right

[ember shell]

oki

thxxx

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[thin peak]

if y=cot(5x), then $\frac{dy}{dx}+5y^2=$

[clever fjord]

Joeller

[thin peak]

$\frac{dy}{dx}+5y^2=-5csc^2(5x)+5cot^2(5x)$

[clever fjord]

Joeller

[thin peak]

I don't know where to go from here

[slow sphinx]

Csc^2x-cot^2x=1

[thin peak]

wait

fr

I never knew about this, thanks so much

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[rotund coral]

can someone tell me this take a circle with radius r and take any point on the circle and draw line connecting to the two points what is the probability that the line joinig is the diameter

[rotund coral]

<@&286206848099549185>.

[wraith heath]

hi legend

[mint mirage]

!15m

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[wraith heath]

your the same as me

ahhahahhahahahaha

[odd edge]

@rotund coral Has your question been resolved?

[steel dragon]

By any point on the circle, do you mean on the edge of the circle itself, or inside the circle

[rotund coral]

on the circumfrenc of the circle

[steel dragon]

The probability is infinitely small, essentially can be said to be zero

[rotund coral]

if zero then you meant that thier is no line like that but definetly at least one will come

[steel dragon]

If any two points are on the circumference of the circle, there are an infinite number of line segments that could connect any two points so it is close to zero

As there are less possible diameters than possible segments

[rotund coral]

so it just to small

[steel dragon]

This is probability, not the number of diameters. The answer is infinitely small

Its so small that it can be formally said to be “zero”

[rotund coral]

k

thanks

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[lapis oasis]

this is invalid right

[lapis oasis]

it should be x = 30^0.25

[swift rock]

yes

$$\log a^n = n \log a$$

[clever fjord]

JustToPro

[lapis oasis]

yeah ty

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[river ember]

The parabola y = ×^2+bx+c intersects the x-axis at points A (1,0) and B (-3,0). Find the analytical expression of the parabola.

[river ember]

that is a one daily mark question

help me

[drowsy stump]

Sir do you know where to start?

You have to crucially use the information of the two points

[river ember]

I know that the parabola intersects the x-axis at points A (1, 0) and B (-3, 0).

Idk how to do next

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[cunning kindle]

For a volume element, I'm trying to find the determinate of a function. This is multilinear algebra. What I have so far is

[cunning kindle]

The question is 1.7

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@cunning kindle Has your question been resolved?

[cunning kindle]

<@&286206848099549185>

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[rigid sonnet]

I need to find the roots of the equation, I can't figure out where to start

[elfin zodiac]

Is that a 3 in the beginning?

Also, i dont see any equation

[rigid sonnet]

yes

[elfin zodiac]

Show the original question

[rigid sonnet]

find the root of the equation. if the equation has more than one root, specify the largest root

[elfin zodiac]

You didnt give any equation

Write the equation properly

[rigid sonnet]

this is what the task looks like

[elfin zodiac]

A=B is an equation

Because it has = symbol

I dont see any = symbol in your picture

[rigid sonnet]

3-|x|=|x-1|

[elfin zodiac]

I thought that was a 2 lol

[rigid sonnet]

I'm sorry, my handwriting is terrible.

[elfin zodiac]

Did u work with absolute values before?

[rigid sonnet]

Probably not. I graduated from college 4 years ago and I'm going to study at the university , so I'm preparing for the entrance exams

[elfin zodiac]

So you dont know the definition of absolute value?

[rigid sonnet]

I remember that |x|=x

[elfin zodiac]

No

That is not true

I recommend you to check the definition of absolute value before trying to solve exercises where absolute value is involved

[rigid sonnet]

|3|=3; |-3|=3

[elfin zodiac]

That is not what you said

[rigid sonnet]

I studied it at school.

[elfin zodiac]

You said |x|=x

So if x=-3 then u said

|-3|=-3

Which is false

[elfin zodiac]

I guess is time to re-study it

[rigid sonnet]

I was wrong

|x-1|=4
x=5
x=-3

[elfin zodiac]

That is not how u solve it

What about x=5?

|5-1|=4

You forgot one solution because you dont nnow how absolute values work

Just go a step back, no problem for this. Go study the absolute value and try when you did

[rigid sonnet]

Are these modules?

x is the number after the action with which the value will be obtained after

[odd edge]

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[terse vault]

help

[half garnet]

What's the question

[terse vault]

please help

[onyx cloak]

take out x^2 common from x^3 - 3x^2

x^2(x-3) + 1(x-3)

so (x^2 +1)(x-3)

[somber jackal]

the technique you're looking for is factoring by grouping

[somber jackal]

!nosols but this is right

[odd edge]

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.

[onyx cloak]

okie

[terse vault]

ok answer done

/close

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[wise condor]

Is this correct?

[wise condor]

here is the translation:

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[wild plover]

2 people are playing the game. The first player throws 6 2-sided pieces (let's say coins). If all 6 coins match, the player gets 5 points and throws again. If 5 of 6 match, the player gets 1 point and throws again. Otherwise the player passes the coins to the second player. The first player to 40 points wins. What is the expected number of throws to get to 40 points?

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[nimble summit]

so im confused as to whether S -> a will make S a final state or not. I know the rule for an accepting state should be (non - terminal ) -> lambda

[nimble summit]

B would be a final state but idk for S

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[slim raft]

How it happened?

[slim raft]

Anybody explain concept behind this in differentiation

How it equated ?

[timid cairn]

Do you see why f(0) needs to be 1

f(a+0)=f(a)*1

[icy cove]

What

[icy cove]

Did u get it or do u want help

that sir used e^x as the function

f(x)=e^x

f(x+y)=f(e^x+y)

then d(e^x)/dx is e^x

now f'(x) = e^x

So here we put x as 5

e^5

and answer ajata

ok

bye

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[mystic saffron]

what kind of precalc trig is this? i google exact value but keep getting SOH CAH TOA based videos or the chart, dont know what to search to find this method of solving?Trig apparently anaytic trigonometry reivew unit

[open crown]

Its not precalc trig, its just trig

[mystic saffron]

somebody is lyin

[slim raft]

Thanks

[mystic saffron]

ya i know that but it doesnt answer my question

[open crown]

Wait use this

[mystic saffron]

i need a key term of what to google so i can watch tutorials

[open crown]

Not sure

[pastel steeple]

trig identities

[mystic saffron]

more specific

like it says find exact value-but no video so far has covered this method

[pastel steeple]

are you looking for practice problems?

because i dont see the point when you literally just need to memorise and apply trig identities

[mystic saffron]

no just a demonstration on how to do it

[pastel steeple]

wdym

you just find the identity that matches

[mystic saffron]

wdym

like #13 2 pie over 12 and 3 pie over 12, why specificallt those numbers

is it because they are both on the pie chart

now im noticing they all use 12 as a denominator,and 4

[pastel steeple]

because those are special trig values with exact representations

you break it apart and then use the identity to evaluate it

[mystic saffron]

umm

so is there a thing i can google to watch tutorials on this specific thing

[pastel steeple]

khan academy prob has some

[mystic saffron]

but a name

whats it called

[pastel steeple]

trig

[mystic saffron]

no, like this specifically

[pastel steeple]

evaluating trig functions

dont know

it can be called many names

just guess

[mystic saffron]

wait

i keep seeing videos with reference angles, those arent it right? causee thisdoesnt use that

[pastel steeple]

never heard of them but sounds promising

[mystic saffron]

every video covers using this, but my paper is so weird

https://www.youtube.com/watch?v=Yu6fhhahCps&list=PLJ-ma5dJyAqq0F8u3Xkn6ha-o3rW7nIt8 maybe this?

ok thank you everyone, think i found it

[mystic saffron]

👏 👍

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[royal linden]

can someone verify my answers ?

[royal linden]

i found a = -5, c = 2 and b = 14/3

[open crown]

U can use desmos

And doesnt seem like ur solution is correct

[royal linden]

wait

im dumb

i meant a = 2 and c = -5

[open crown]

Spoiler

[desert marlin]

!nosols

[odd edge]

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.

[open crown]

Bruh

He clearly knows how to do it

Btw @royal linden

Since a b c are interger

and ur b is a fraction

Lowkey u know ur solution is incorrect

[royal linden]

true

i found c by myself

im trying to do the rest

[open crown]

Lowkey know u can do it

Good luck

[royal linden]

i found a by myself

[open crown]

b left

[royal linden]

FOUND IT

amazing

it was pretty easy

i messed everything up on my first try bc i tried using the formula a(x-b)/x-c where a = horizontal asymptote, c = vertical asymptot and b = the x value for y =0

once i understood -5 = -c/2 it was easy

ty so much for the help, have a good day

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[violet basin]

im confused as to why A(3) isnt 11.5

[wise jasper]

how are you getting 11.5?

[wise jasper]

explain your thought process?

[violet basin]

so i got both A(1) and A(2) so now when I find the area from 2 to 3 i add that value to A(1) and A(2)

the same way i got A(2)

I found A(1) and then I added the area that I found for 1 to 2 to A(1)

[wise jasper]

its just A(2) + A(3)

not A(1) + A(2) + A(3)

[violet basin]

ah i see i thought i had to add them all up

so it would be 5+3

[wise jasper]

yes

[violet basin]

and then i do the same thing for finding A(4) so A(3) + A(4)?

ty for the help

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[cyan grail]

Hi

[cyan grail]

Don’t have a clue ngl

[cyan grail]

Pls

[odd edge]

@cyan grail Has your question been resolved?

[abstract quartz]

Only thing I can assume is MN=PN because the lines dont the the mark to show that their equal

[cyan grail]

Oh yay a person

[abstract quartz]

Other wise they all prove this triange is equal to the other 1

triangle*

[cyan grail]

Thank u

[cyan grail]

How about this

Damn

<@&286206848099549185>

[odd edge]

@cyan grail Has your question been resolved?

[cyan grail]

Nah bro

<@&286206848099549185>

Fuck nvm

.closd

.close

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[spare hound]

Hi, me again. I had a question about an assignment I was working on. The left side of the columns is initial speed and the right side is braking distance. The question is as follows: "It is assumed that the stopping distance is directly proportional to the square of the initial speed. Express stopping distance as a function of initial speed. " I put in the following: r=k .(a)^2 with stopping distance r, initial velocity a and an unknown constant k. Is this correct?

[jovial pollen]

Is your name just true in Latin

[spare hound]

truth

yes

[fresh ruin]

you can figure out k from the table

[spare hound]

I did it thru my calc

But idk, this isn't a proportional relation between r and a right? this is more like an exponential one

[fresh ruin]

that a on the calculator is k yea

r is proportional to a^2

[spare hound]

Oh alright, I see

One has assumed here a braking deceleration a equal to 7.5 m/s². One can prove that the braking distance is equal to the square of the initial velocity divided by 2a. The condition is then that the initial velocity is expressed in m/s. @dusk warren Do I have to put the multiplication of 3,6 to convert it from km/h to m/s under the square as well alongside v?

[spare hound]

because here above v is solely expressed in km/h

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[waxen crane]

tan2t = 2sintcostsec2t Verify the identity

[glacial inlet]

what have you done?

it just takes 1 step (ideally)

u should also start fromt he left side

[waxen crane]

I subsituted the double angle formula in for tan2t

right

[glacial inlet]

nah

a better idea would be to use the fact that tan(x) = sin(x)/cos(x)

[waxen crane]

what would that look like with tan2t?

[glacial inlet]

what do you think?

just replace x for 2t

[waxen crane]

ohh so it would be sin2t/cos2t

[glacial inlet]

yes

which you can write as

,, \frac{\sin (2t)}{\cos (2t)} = \sin (2t) \cdot \frac{1}{\cos (2t)}

[clever fjord]

nyxie9151

[glacial inlet]

right?

[waxen crane]

right

[glacial inlet]

do you know how to proceed?

[waxen crane]

give me a minute

[glacial inlet]

,, \qty(2\sin(t) \cos (t)) \cdot \sec (2t)

this is what you want

[clever fjord]

nyxie9151

[waxen crane]

ahhh

ic

ok thank you, im gonna try considering the diff possibilities at the beginning

for where to go from then

i got it

.close

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[glacial inlet]

yeah u should look at the right side for "hints"

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[lethal locust]

am i doing this correctly

i’m confused

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[keen solar]

Hiii pls help i have no idea what to do 😭

[keen solar]

<@&286206848099549185>

[odd edge]

@keen solar Has your question been resolved?

[keen solar]

<@&286206848099549185>

<@&286206848099549185>

PLS GUYS 😭

OMG NVM I FIGURED IT OUT

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[midnight flax]

A box contains 80 balls numbered from 1 to 80. If 15 balls are drawn with replacement, what is the probability that at least two of them have the same number?
I have no idea how to approach this problem

[hasty rover]

each time a ball has 1/80 chance of being same

so if you pick 1 time, 1/80 chance of it to be any number

next time

its 1/80 again for it to be same

1/80 * 1/80

1/6400

i think this is the answer

@midnight flax

[onyx cloak]

that wouldn't work

[hasty rover]

i feel it but

in any of the 15 scenarios

[midnight flax]

so

[hasty rover]

2 being same is 1/6400

[midnight flax]

you're drawing 15 numbers

[hasty rover]

look i have only done basic probabilty

[onyx cloak]

better way would be to do P(number on 2 balls same) = 1 - P(number on no 2 balls same)

[midnight flax]

i feel the approach is to start off with the chances of getting all different numbers

yeah

so

[hasty rover]

h

hmm

[midnight flax]

1 - P(80x79...x66/80^15)

wait

would that make sense

[hasty rover]

thats a small number

[midnight flax]

yea

it's in percentage

so it has to be

[static totem]

should be that yes

[onyx cloak]

yes

thats correct

[static totem]

it's 0.80

no 0.75

[onyx cloak]

0.75336900012

[midnight flax]

oh

thank you so much

was the correct answer

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[ripe rapids]

.open

Why does 0.95 x 357 equal 339.15, but 357 divided by 1.05 equal 340?

[tepid pelican]

Because 1/0.95 isn't 1.05

if it was, then they would equal to the same

take 10 * 1.5 for example, it's not the same as 10 / 0.5

[ripe rapids]

Like, say someone raked 357 leaves, 5% more than someone, and I'm trying to find out how many leaves that someone raked. Can't I multiply 357 by 0.95?

[tepid pelican]

nope, you need to divide by 1.05

[ripe rapids]

I can't make sense of that in my head, especially since the answers are so close

It just doesn't make sense. 357 is 100%, multiplying by 0.95 should give me the answer cause I'm seeing what 95% of 357 is

And since she's raked 5% more leaves that makes sense

[radiant bobcat]

someone help please wtf is this

[ripe rapids]

My channel

MINE

[tepid pelican]

357 is 105% of what someone else has

[dusk basin]

it’s pythagorean theorem, go to an unoccupied channel though

[radiant bobcat]

ok my fault

can u send the channel

[dusk basin]

#❓how-to-get-help

[ripe rapids]

How would I write this problem out algebraically

[tepid pelican]

If I have 50% of what you have, its noz the same as you having 150% of what i have

[tepid pelican]

357 = 1.05x

Or 357 = x + 0.05x

0.05 sprry

[ripe rapids]

This is the same kind of logic I'm using thinking 0.95 x 357 would get me 357 divided by 1.05?

[tepid pelican]

1.05 is 5% more than someone ranked

However, if u call ur rank x, then 0.95x is 5% less than you rank

And 5% of your leaves isnt same amount as 5% of someones else leaves

[ripe rapids]

Ohhh I got you

That's why the answers are so close

[tepid pelican]

Yep

One counts with 5% of ur leaves and the other one counts with 5% of someones else leaves

[ripe rapids]

Alright it makes sense

Thanks

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[frank vortex]

Can someone sketch this graph for me ?

[tawdry wagon]

Literally just plug them into desmos

[frank vortex]

i dont know how to use desmos any way you could show me ?

the other graphing websites ive tried arent any good at all i cant get a good picture of the graph

yea desmos isnt working

[upbeat osprey]

do it like this

[frank vortex]

how do i get the 3 like that ?

[upbeat osprey]

note do shift + underscore to input the base

[upbeat osprey]

^this

[steady tide]

this is literally just "just google it"

[tawdry wagon]

Nah, sorry fr, but I mean kinda

[steady tide]

first and foremost, do you know how logarithmic graph works?

[frank vortex]

does this look accurate ?

[frank vortex]

yes im just trynna get an accurate graph to sketch thats all

[steady tide]

it would be helpful to sketch out the coordinate system first

[odd edge]

@frank vortex Has your question been resolved?

[odd edge]

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[maiden crow]

Hello im having issues with multiplying and dividing and adding and subtracting with significant figures

[restive delta]

Any question in particular?

[maiden crow]

yeah so if i have something like 110.3440 - 34.30

the calculator would say 76.044

but

it would only be 76.04 right?

[restive delta]

Depends on how many significant figures the question is asking for

[maiden crow]

now what about with division?

isnt there rules about that aswell

or multiplying

so same thing

110.3440 * 34.30 =3784.7992

but i would be from my understanding

3784.80

[feral glacier]

Nope

It would be 3785

You go with the lowest amount of sig figs when multiplying and dividing

34.30 has 4 sig figs so your answer must have 4

[maiden crow]

wait so before the decimal counts too?

so would it be 3785.

with the period

[feral glacier]

No just 3785

[maiden crow]

when would you make it with the dot behind it

[feral glacier]

Saying 3785. Implies there is a zero after the decimal meaning 5 sfs but that’s not really formal

[maiden crow]

can you give me another to try?

[feral glacier]

Usually you just write the zero or convert to scientific notation

What about 559.765 / 22.347

[maiden crow]

25.050

?

[feral glacier]

Yup

[maiden crow]

nice i got that one

okay

so gotta match them

same thing with addition or subtraction?

[feral glacier]

What about 600 * 59.791

[feral glacier]

No adding and subtraction is different

With those you match the amount of decimal places that’s it

[maiden crow]

wym match?

[feral glacier]

305.1 + 50.20 would be 355.3

[maiden crow]

and for this one im thinking its 35900

[feral glacier]

Nope

[maiden crow]

this one confusing

[feral glacier]

How many sf does 600 have

[maiden crow]

3

[feral glacier]

You sure?

[maiden crow]

or it 1

?

[feral glacier]

1

[maiden crow]

why is it one

?

And not 3

[feral glacier]

Because you only know information about the 100 place

[maiden crow]

so would it be

40000

[feral glacier]

Yea

[maiden crow]

really? interesting

so thats also only one sig fig

can you give me another like that?

those ones i struggle on

[feral glacier]

All you gotta do is do the equation normally then look to see which number has the lowest amount of sig figs

[maiden crow]

okay give me another to try

[feral glacier]

6.70004 + 7.008

[maiden crow]

13.71?

[feral glacier]

Close

This is adding

So you look at the decimals

[maiden crow]

0000

ooo

13.708

[feral glacier]

Yup

[maiden crow]

nice that one makes sense

wanna give me one more

[feral glacier]

Adding or multiplying?

[maiden crow]

either

I still need to practice both

[feral glacier]

0.00704 * 5.357

[maiden crow]

0.040

[feral glacier]

Close

Why did you say that

[maiden crow]

whoops 0.380

[feral glacier]

How many sf is that

[maiden crow]

4

[feral glacier]

Nope

It’s 3

[maiden crow]

0.377

[feral glacier]

Yup

[maiden crow]

damn that one got me lol

[feral glacier]

Wait

0.0377

[maiden crow]

0.0377

isnt that 5?

[feral glacier]

Nope

3

[maiden crow]

4 right?

[feral glacier]

Zeros before whole numbers are not significant

Only if they are between other whole numbers

[maiden crow]

oh okay

[feral glacier]

How many sf does 0.00000009 have

[maiden crow]

1

?

[feral glacier]

Yes

What about 0.010007

[maiden crow]

5

i believe\

[feral glacier]

Yup

[maiden crow]

so addition

i just match up after the decimal?

or different too with 0s

[feral glacier]

Yes

[maiden crow]

to the left?

[feral glacier]

With addition you only care about the right side of the decimal point

And you match your answer to the lowest amount of decimal points

15.01 plus 14.09 = 29.10

Not 29.1

[maiden crow]

that makes sense cause they the same

amount of sigfigs

[feral glacier]

What about 15.1 plus 14.907

[maiden crow]

30.01

whoops

sorry

30.0

[feral glacier]

Yup

[maiden crow]

that one i had to look at for a second lol

[feral glacier]

That’s fine as long as you understand it

[maiden crow]

yeah multiplying still hard

lemme see another multiplication

[feral glacier]

Remember just do the problem normally then convert your answer to the lowest amount of sf you can

[maiden crow]

ok

[feral glacier]

2.34 * 5

[maiden crow]

10

wait

12

[feral glacier]

12

Yea

Wait

No

[maiden crow]

shit

lol

[feral glacier]

It’s 10

[maiden crow]

1

so why 10

[feral glacier]

Only 1 sf

Because 5 has only 1

[maiden crow]

so if it was 2

it would be 12

or is 12 still 1 sigfig?

[feral glacier]

If it was 5.0 instead of 5 then yes

12 is 2 sf

[maiden crow]

300 is 1

right

3000

1

[feral glacier]

Yes

[maiden crow]

40000 1

i get it

[feral glacier]

Yes

[maiden crow]

lol

interesting okay let me see another

[feral glacier]

20.04 * 30.100

[maiden crow]

603.2

[feral glacier]

Yup

[maiden crow]

sweet!

[feral glacier]

How many sf does 30.100 have

[maiden crow]

5

[feral glacier]

Yup

[maiden crow]

definitely getting better haha

[feral glacier]

Yea good job

You taking chemistry?

[maiden crow]

yeah i am lol

im working on this other equation rn

now that you mention

looking for sugar %

in coca cola

my equation is y = 0.0958(X) +5.275

x i believe is my sugar % that i measured with my hydrometer

which would be 12%

would i multiply by 0.12

or 12 😂

well i cross referenced

im forsure 12 belongs there

[feral glacier]

Multiply what by 12%

[maiden crow]

my bad

solve for y

0.0958*(12%)

[feral glacier]

Solve for y?

Or for x?

[maiden crow]

should i be solving for y?

[feral glacier]

Do you know what x is?

[maiden crow]

ill show you all i got

yeah

i believe so

let me show you what i got

[feral glacier]

Ok

[feral glacier]

Where did you get the original 12% from

[maiden crow]

So I used a hydrometer and we calibrated heights off 4%-16% liquids

Measured their height

And then compared that

[feral glacier]

Ok

[maiden crow]

To the 2 known substances and 2 unknowns

And then classified them based off that

[feral glacier]

To use the graphed data you want to solve for x

X is sugar % and you have y which is the height

So plug and chug and rearrange equation to solve for x

[maiden crow]

okay so solve for x

[feral glacier]

Yes

[maiden crow]

then on this question

do i solve for y

[feral glacier]

Which question

[maiden crow]

one sec

[feral glacier]

Ok

[maiden crow]

cropping lol

[feral glacier]

You would still solve for x

y is the height

X is sugar concentration

[maiden crow]

okay let me try one real fast lol

like coke

Did I do this right? @feral glacier

[feral glacier]

So close

[maiden crow]

where did i mistake lol

with my sigfigs huh?

[feral glacier]

Sig figs lol

[maiden crow]

where did it fall apart?

[feral glacier]

How many sf does your answer have

[maiden crow]

4

oh should have been 12.3

[feral glacier]

Yes

12.3%

[maiden crow]

12.%

12.3%

any other units i lost in there too?

i think only cm on the height right?\

[feral glacier]

Well height would not be apart of your answer

[maiden crow]

make sense

[feral glacier]

Maybe the slope has height units I dunno

But 12.3% sugar concentration is your answer

[maiden crow]

ok so i just do that for all of them basically lol

[feral glacier]

Yup

[maiden crow]

so how would i do number 2?