help-38

[wary basin]

I believe so

[stray ruin]

0,009999

Is that right

[wary basin]

0.0999

I think

[prime crystal]

like 0.0 0 1
^ones ^tens ^hundred

where is thousandth

[wary basin]

0.000 --> ones, tenths, hundredths, thousandths in order from left to right

[prime crystal]

ah

[wary basin]

think about multiplying 0.001 by 1000, it would equal 1 so that's why

because it is equivalent to 1/1000

[prime crystal]

yeah

[stray ruin]

Wait why is it not 0,009999

[wary basin]

0.01-0.0001

[stray ruin]

Won’t it go down a step ten

[wary basin]

you're right yep

I may have mistyped

[stray ruin]

Okay so how many nines do I put

[wary basin]

but you have too many nines

[stray ruin]

0,0099

Okay wait I’ll try it

ITS CORRECT

YAY

THANK YOU

Before I close this the next question is very confusing

I don’t understand what I’m supposed to do

[wary basin]

largest sum would be the biggest difference so choose the smallest and then the largest number

smallest sum would be the smallest difference so look for the smallest jump between two numbers (you can order them from smallest to largest to see this)

[stray ruin]

So 0,5 minus 0,1

Will that be the biggest

[wary basin]

it would be 0.5 and -0.6

-0.6 is the smallest

by the way

I would suggest using periods "." instead of commas

for decimal places

unless your teacher has indicated otherwise

[stray ruin]

Wait why -0,6

Won’t that make it small

The sum

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[wary basin]

oh sum nvm

take the two biggest numbers

my apologies

0.5+0.4

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[manic gyro]

No idea what the questiong is trying to ask me

[wind grail]

Sorry

[manic gyro]

its fine

[dull pilot]

LMAO-

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[manic gyro]

just a bump

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[fleet bear]

Formatting errors certainly don't make it clear. Your question though:

If f(x) is differentiable, f&-1 (x) is not necessarily differentiable. For example, if f'(x) = 0 at any point

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[sage anvil]

guys

i struggle to comprehend this completely

howcome e^c is c?

i thought the interaction between e and ln turn these e^ln something = something

[rough goblet]

oh yeah

for integrals

when we're solving diff equations

we just replace e^c with c usually

because yk both are constants

[sage anvil]

😮

[rough goblet]

it's a weird convention, i know

[sage anvil]

crazy convention

[rough goblet]

fr

[sage anvil]

but I must pass

19th

mathe xam

[rough goblet]

?

[sage anvil]

💀

[rough goblet]

yeah ur gonna have to use this sadly

[sage anvil]

ye

[rough goblet]

wha?

[sage anvil]

alot of math theory also

idk how to remember all that

[rough goblet]

your point being

[sage anvil]

very big struggle upon upcoming math exam

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[viral flint]

What I wanted to do is generate a fraction that has an repeating pattern by just first defining what the repeating pattern is and infinitely shifting it to the right until it converges, so far I've gotten a pretty good groundwork for recreating 1/3 with a summation formula, but idk the process of finding what it converges to. I've been looking at videos for summation notation and they only go into determining if one diverges or converges, and never go into finding out what it converges to

[viral flint]

I've gotten most of the pieces of the puzzles in place I'm just missing the final piece

[zinc ginkgo]

Factor the 3 out and use geometric series

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[viral flint]

Hm, how do I go about finding the common ratio?

[wraith hinge]

it is 1/10

[zinc ginkgo]

That's just one side of the formula

[viral flint]

Ah ok

[wraith hinge]

after that u put the 3 out of the sum

[viral flint]

So you multiply the sum by 3 is basically the answer

[wraith hinge]

yes

[viral flint]

Uh oh, I think I did it wrong

[wraith hinge]

dafuk

[viral flint]

I built the geometric series wrong

[wraith hinge]

this is ur serie right ?

[viral flint]

Yes

[viral flint]

I'm pretty sure I misinterpreted what they meant in this

[wraith hinge]

so u put 3 in factor and u have sigma (1/(10^n))

it is a formula to know the sum of the terms of a geometric serie

[viral flint]

There we go, now I need to figure out how to then convert this to 1/3 since I know this converges to 1/3

[wraith hinge]

0.333...=1/3 so what is the problem ?

[viral flint]

What manipulation do I do on this to get the fraction 1/3

Guessing it by looking at the decimal is not what I wanna do

[wraith hinge]

i see

[wraith hinge]

with this

[viral flint]

That's why I'm even touching sigma notation in the first place

[wraith hinge]

u use a1=1

r=1/10

and n=infinity

[viral flint]

Ah, I thought that a1=11/10 since I thought we were talking about the first term of the sum thing, that makes more sense what you mentioned

[wraith hinge]

yes im wrong

a1=1/10

[viral flint]

Ah, there we are

Thank you

[viral flint]

I just realized there was a dude doing it much simpler than me
https://youtu.be/Ns-dE09vzqo?si=GmYKUpCunu5irI5h

How do I close the channel?

It's resolved

[wraith hinge]

do .close

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[small glacier]

If formulas $A,B$ are satisfiable, is $A\land B$ satisfiable?

[solid kiln]

Wewwww

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[sage anvil]

Given the - is it possible to write this as this?

no

without the ^2 at the end

[burnt mulch]

$\sqrt{-\lambda^2 x^2}=i|\lambda x|$

[solid kiln]

Civil Service Pigeon

[burnt mulch]

recall that $\sqrt{x^2}=|x|$

[solid kiln]

Civil Service Pigeon

[sage anvil]

hmmm

i thouhgt like

to simplify and remove sqr root

to just ()*^(1/2) everything

but cant?

[burnt mulch]

that neglects the case where x<0

ex. $\sqrt{(-5)^2}=\sqrt{5^2}=\sqrt{25}=5$, not $-5$

[solid kiln]

Civil Service Pigeon

[burnt mulch]

(square roots have non-negative outputs if they're real)

[sage anvil]

hmm

can i show u pic

of direct thing i was doing

i send

[burnt mulch]

sure

[sage anvil]

To determine the lambda^alpha where alpha is homogenuity level of function

and i try simplify everything without sqr root

[burnt mulch]

that expression can't simplify anymore unless you count rationalising the denominator as simplification

[sage anvil]

i mean removing the square roots

[burnt mulch]

I'm aware of that, you can't cancel them out or anything of that sort

[sage anvil]

to get alpha

so this is the part i sent before

[sage anvil]

hmmm

this then wrong way to solve?

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[echo jasper]

how complex is converting b/w si prefixes?

[echo jasper]

i memorized them

how complex can it get as well

or is it just over all basic and is there like some time of formula that works

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[spice girder]

could someone explain why this is the domain of some trigonometric functions

[limpid dawn]

from where is this example?

[prime lynx]

[limpid dawn]

looks like tan(theta)

[spice girder]

t being theta

this is the full img idk why i only sent b

[limpid dawn]

pi/2, 3pi/2 ... are the root values of cos(t)

tan(t) = sin(t)/cos(t)

you need to exclude those values otherwise you cause division by 0

[spice girder]

so why is b the domain of sec(t)?

i dont see how you could get division by zero in sec(t)

[limpid dawn]

1/cos(t)

that's how secans is defined

you are dividing here by cosine again

[spice girder]

ohhhh and the equation for b would turn sec into cos therefore dividing by zero

right?

[limpid dawn]

the equation for b)?

sec(theta) = 1/cos(theta)
now you need to make sure to exclude all values that causes the denominator 0

[spice girder]

this equation

[limpid dawn]

that is the set of b)

it's a set

or call it domain

not equation

[spice girder]

i think i get it

thank you

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[pseudo fossil]

Can anyone explain to me what makes this system inconsistent?

[limpid dawn]

false statement last row

which means not every equation can be satisfied in that system

[pseudo fossil]

What makes the last row a false statement?

[limpid dawn]

Is 0 = 1 true or false

[pseudo fossil]

that 0+0+0 = 1 is false right?

[limpid dawn]

yes

[pseudo fossil]

thank you!

[limpid dawn]

thank you too

[pseudo fossil]

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[nimble stone]

checks out

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[echo jasper]

Uh so

Why does student c have both circled

😕

like its close to 10 so

ofc accurate

but do they mean by precision

[nimble stone]

the definition is right there

[echo jasper]

like

if the trial numbers are close/similar ?

[nimble stone]

if the results dont greatly vary, ie have a small range, then theyre precise

[echo jasper]

oh ok

thanks

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[wooden oriole]

What did I do wrong?

I looked over my work like 2 times, checked with symbolab and I can't find my mistake

,w partial fraction decomposition (512)/((x)(x+3)^3)

[solid kiln]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by azar314

[wooden oriole]

i think i got the constants correct

so did I make mistake on the integrating part?

<@&286206848099549185>

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[small root]

I think it's because there are two negatives multiplied in the third term of your answer

[wooden oriole]

negatives times a negative is a positive

[small root]

And also in fourth

[small root]

Ik so write it as a positive

[wooden oriole]

okie

[small root]

I'm saying that maybe the way you wrote can't be recognised by the program as the ans

[small root]

Ok

[wooden oriole]

Yea, it didn't work

[small root]

And maybe even simplfy the terms too

Like in the fourth term

[wooden oriole]

nah, this program does take unsimplified terms

[small root]

2 cancels out

[wooden oriole]

you can leave it like that

[small root]

Oh

[wooden oriole]

yea, i am guessing I made a mistake in my constants or integration

[small root]

My test is gonna start soon, I'll be back in 3 hrs

[wooden oriole]

ok, I appreciate your assistance

[small root]

I don't know?

I'mma have to check that by doing it myself

After

[wooden oriole]

alr

.close

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[wooden oriole]

I’m going to bed now lol

I’m eastern

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[chilly meadow]

Hi, I'm stuck on these 2 problems for my math assignment. They ask me to solve the integral for the volume of the solid obtained by rotating the region enclosed by the curves and the given line.

The problems are:
y=secx, y=cscx, y=0, x=0, x=pi/2, about the x-axis. For the other problem it's about the y-axis instead.

I was thinking of using the shell method for the one about the y-axis and got radius to be x and height to be cscx-secx but I'm not sure if that's right since the integral would be hard to solve.

Any help would be appreciated!

[zinc ginkgo]

Yea just try it and show your work here

[chilly meadow]

this is what I got to so far

but how would i integrate this

[zinc ginkgo]

Which axis are you rotating around

[chilly meadow]

this one's for the y axis

[zinc ginkgo]

Yea looks like ring method will be better

This one

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

[chilly meadow]

so it's better to use washer/disk method for this?

[zinc ginkgo]

Try it

[chilly meadow]

idk its not really working

I can't integrate it

I might just ask my teacher on monday

[zinc ginkgo]

Did you try using trig identities

Shouldn't need to though if you learned the integral of sec^2 and csc^2

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[pseudo fossil]

Any <@&286206848099549185>

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[wispy sedge]

can someone help me with factoring quadratic trinomials?

[wispy sedge]

the given is x²-9x+20

[wraith hinge]

what have you tried

[wispy sedge]

uh nothing yet

[wraith hinge]

solve for roots

[wispy sedge]

i dont get it

i dont remember that being teached

im in 8th grade btw 😭

[haughty tundra]

to find roots, you want to find what value(s) of x makes the expression 0

[wraith hinge]

thats right blaze

it doesnt work that way

[wispy sedge]

huh

[wraith hinge]

just solve for roots

[cinder badge]

yea that was right

so again... do u know two numbers with product 10

and sum 9?

[wispy sedge]

wait im thinking

product is 5?

what

😭😭😭

[cinder badge]

product is 20

[wispy sedge]

OH

then what do i do

[wraith hinge]

product is 20 actually but whatever

[cinder badge]

yea mb

[wraith hinge]

its 4 and 5

[wispy sedge]

wait imma show our notes bc i dont get it

this is what they thought us

[wraith hinge]

he used the same thing

product 18 sum 9

well shouldve been -9

[wispy sedge]

i dont get it

[cinder badge]

see if roots are a and b then the eqn is x^2-(a+b)x+ab=0

[wraith hinge]

its because you dont learn math at all

you dont know formulas

[cinder badge]

so we find a and b such that product is the constant term

[wraith hinge]

go learn formulas

then do homework

[median torrent]

This is the way i approach factoring.

Assuming you have a trinomial with an x^2, not 2x^2 or 3x^2 etc

Just plain x^2

Look at our problem

x^2 - 9x + 20

There are three terms:

The x^2 (Lead Term)

-9x (middle term)

20 (constant term)

Begin by looking at the constant term, 20. Can you identify factors of 20?

[wispy sedge]

10?

or 20

[median torrent]

20

Like

2*10 = 20

1*20

Think of the factors as pairs of numbers

[wispy sedge]

OHHHH

i get it now

the answer is uh

(x - 4)(x - 5)

[median torrent]

Yea!

Nice

[wispy sedge]

thank u omg

[median torrent]

You're welcome

[prime crystal]

so the "split the middle term in such a way that their sum/difference gets u back to middle term and multiplication gives the constant term" comes from that?

[median torrent]

Well for easy ones you can just work from the constant

It's just a shortcut for the real method

[prime crystal]

ah

okay

[median torrent]

But you don't need it for the a = 1

You can do it very quickly for those

[prime crystal]

hm

[median torrent]

Not familiar enough with the way that method is derived to comment much anyway

The "ac factoring method" as i know it

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[main sigil]

how do you divide by a point? Are those angles?

[violet gust]

it should be length of diagonals/sides

[main sigil]

oh, that makes sense

[wraith hinge]

yea A1A2 is the length b/w the vertices A1 & A2

[violet gust]

side of an n-gon is 2rsin(pi/n) right?

so A1A2 = 2rsin(pi/n) ; A1A3 = 2rsin(2pi/n); A1A4 = 2rsin(3pi/n)

[wraith hinge]

yeah

[violet gust]

reduce all sin to sin(pi/n) and solve for sin(pi/n)

[wraith hinge]

1/sin(alpha) = 1/sin2(alpha) + 1/sin3(alpha)

solve for alpha = pi/n

got it

.close

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[hollow citrus]

My task is to figure out how they calculating the checksum digit for these numbers.
There are some numbers and their checksum digit:
check_digit(98100304302546538700150) = 7
check_digit(98100333902581499701400) = 3
check_digit(98100304302543127300150) = 6
I tried Verhoeff, Damm, Luhn and ISO 7064 MOD 11,10 but none of these returned with a correct checksum digit. Are there other methods that I should try out? Is it possible to be a customized Verhoeff with unique tables? If yes, how could I determine the correct tables?

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[lethal bone]

I understand how to find the value of a, but what is it asking me? “Find the value of a such that HL - HJ?” Does that just mean find A or is this something else?

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[meager bloom]

Yes find value of a so that HL is perpendicular to HJ

[lethal bone]

which is 90, right?

[meager bloom]

That means angle 2 + angle 3 = 90

[lethal bone]

awesome ty :)

[meager bloom]

Np

[lethal bone]

!close

[meager bloom]

.close @lethal bone

[lethal bone]

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[strange cypress]

Can someone help with exercise number 3

[river willow]

go to another channel

[rough goblet]

@strange cypress this is directed at you

everg

occupied

[strange cypress]

Kk

[sleek canopy]

ok

[solid kiln]

ren

[river willow]

what would the graph for velocity look like?

[rough goblet]

but the velocity doesn't change--displacement changes at a constant rate

what does that tell you

[river willow]

draw the velocity graph for the last 4s

your lack of brackets displeases me

[rough goblet]

...

please never use that notation again

[river willow]

the addition of N is also weird

[rough goblet]

just assume north to be positive.

and south negative

no need for units

[river willow]

if u want to write it explicitly, write north at the end

anyway that would give u the average velocity

yes

draw the velocity graph

yes

im asking you to draw a graph to help u figure it out

do u know how the velocity graph looks like given a displacement graph

do u know what velocity represents

rate of change of displacement

so if this was the graph if displacement against time

what would the graph of velocity against time look like from 0 to 1?

yes, so how would the velocity time graph look like

no

constant motion. what does that tell u about velocity

do u know what are derivatives

do u know what are derivatives?

nvm just take it as

if u see a straight line in a displacement graph

acceleration is 0

where straight line is defined as something u can draw with a ruler

a note its only in displacement graphs

if its a straight line in a velocity graph that doesnt tell u the same thing

yes?

steepness

np

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[icy hedge]

Can someone please check this? According to the answer key vy should Va should be 8.2 and Vx should be 7.58

but i cant find where my algebra is wrong

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[zinc ginkgo]

Hard to read and follow your work with all the unnecessary numbers

Just leave them in variables

[icy hedge]

alright

other than that do you see any algebraic errors?

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[vague walrus]

how?

[zinc ginkgo]

,w int 0 to 1 exp(x)/(1+x)

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[zinc ginkgo]

Doesn't look like there's an elementary antiderivative

[vague walrus]

hmm

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[vague walrus]

must be a misprint in the book

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[clever ruin]

so i've broken down g(x)=3e^-0.9t and h(x)=cos(3(pi)t)+4 but i am completely lost on what to do

[weak hemlock]

you have to find the horizontal asymptote of the function d(t) that is given there.
That is the limit, the function goes to, for infinite

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[clever ruin]

ok, i dont really understand how, they've only had me graphing these equations instead of solving for them

[clever ruin]

wait i dont understand. its just the +4 at the end? i solved and got the graphs shown between -3e^-0.9t+4 and 3e^-0.9t+4

4 is the correct answer. because it settles on 4 as the nominal/middle on the graph?

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[weak hemlock]

4 is the correct answer, because the limit of g(x)=3e^-0.9t is zero and the cosin doens't matters here. so the +4 at the end makes the limit of the function d(t) equal to 4.
Nice that you recognized it yourself.

[clever ruin]

gotcha. thank you.

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[graceful kite]

Does matrice similitude stay after using linear operations on the matrice

[graceful kite]

Like line or column subtracting etc

Is the new matrice (semblable or what ever its called) to the new one

Sorry for the poor terms i dont study this in english

[atomic stone]

Is there the question on paper by chance

[ionic pendant]

row and column operations do not yield a similar matrix: https://math.stackexchange.com/questions/1025193/do-elementary-row-operations-give-a-similar-matrix-transformation#1025197

[graceful kite]

Not really im sorry

Its for an exercise

[graceful kite]

Thank you

Do u know perchance what those operations keep other than the rg of the matrix

[ionic pendant]

row operations:

• preserves rank
• preserves the span of rows (row space)
• preserves the the kernel/null space
• row operations preserves which columns are linearly independent
• preserves solutions for augmented matrices

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[graceful kite]

Thank you so much

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[graceful kite]

Do u mind if i ask you in dms a question sometimes @ionic pendant ?

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[pseudo fossil]

Can anyone tell me where I went wrong in my proof?

If v ∈ V and v + v = 0, then v = 0
From v = v + v we get that -v + (v + v) = -v + v
(-v + v) + v = 0
So, v + 0 = 0 -> v = 0

So,

c0 = c(0 + 0)
c0 = c0 + c0
Subtract c0 from both sides to get
c0 - c0 = (c0 +c0) - c0
Which finally gives us
c0 = 0

as required.

[somber ginkgo]

i'm not sure what ur doing in the first part

So,

c0 = c(0 + 0)
c0 = c0 + c0
Subtract c0 from both sides to get
c0 - c0 = (c0 +c0) - c0
Which finally gives us
c0 = 0

as required.

^ but this is all fine

[pseudo fossil]

I guess that's just proving that 0 is a neutral element with respect to vector addition.

or would this be a better proof?

Well, for any vector we have u - u = 0 (vector). So, let's construct that:
C0 = 0
C(u - u) = 0
Cu - Cu = 0
Cu = Cu + 0
As the zero vector is a neutral element with respect to addition:
Cu = Cu
And this is trivially true.

[somber ginkgo]

but that's given from the axioms?

[somber ginkgo]

not sure what ur doing here

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[sick quarry]

f(x) =32x^2-16mx+15m-22
f(a)=f(b)=0
If 1 < a< 2 and 1 < b < 2
then how which numbers can m be?

[zinc ginkgo]

Plug in x=a and b and get equations for them

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[azure oak]

I'm confused on how to go about this

[wraith hinge]

Use ohm's law

V = IR

[azure oak]

so 4-6?

it is given as a wrong answer

[ionic pendant]

why -6?

[azure oak]

the current is going the other way?

[wraith hinge]

What

[ionic pendant]

the current enters on the higher voltage side and exits on the lower voltage side

[azure oak]

i see

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[flat steppe]

I need help with number 61, idek how to approach this or do it

[flat steppe]

Nvm, I get it now

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[swift forge]

having trouble with this one

any help

<@&286206848099549185>

[frail heron]

Bro chill…

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[swift forge]

@frail heron can u help

[frail heron]

@knotty locust

[knotty locust]

Yes ping me

I will clear doubt

[swift forge]

@knotty locust

help me

[winged solar]

bro chill sometimes you gotta wait

did they give you an equation

[swift forge]

no

@winged solar no one is helping me

[winged solar]

bro no one is going to help if your impatient

start by identifying variables

[swift forge]

wdym

[winged solar]

like your knowns and unknows

[swift forge]

bro idk

[winged solar]

how much was the return in investments how much was invested etc.

[swift forge]

i did

[winged solar]

what did you do for previous problems like this

[swift forge]

this is the first one

i know i’m supposed to get and equation

idk how

[winged solar]

also write down given information

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[stiff bough]

Hi I wanted some help with 3a and 3b the question says to use the lowest sig figs is it normal to count the sig figs from the conversion factor?

[stiff bough]

Basically I'm wondering if it should be 1 or 2 sig figs for 3a and 3b

[ionic pendant]

conversion factors are exact (treat them like they have infinite sig figs) so they do not change the number of significant figures

[stiff bough]

Thanks! That's what I thought!

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[echo jasper]

help plz

i got my y0 = 0

g = 9.8

y=-25

Vo=16

Got the quadratic equation of 4.9t^2+16t+25=0

Is it so far proper steps?

i used the equation to plug in with:

y=y0 + V0t + 1/2 (-g)t^2

, 1.56(9.8)+16=V

,w V=1.56(9.8)+16

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[echo jasper]

MANnnnnnnnnnnnnnnnnnnnnnnnn

answer should be like 4

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[trim kestrel]

why is it DNE

i cant find an explanation

i just got -x < 2sinx < x

not sure what to conclude from that

is it cause sin oscillates between -1 and 1

[ionic pendant]

,w graph x/(2 sin x) from 0 to 20

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[trim kestrel]

yea im clueless then

[ionic pendant]

can sin(x) = 0? what happens when the denominator approaches 0?

[trim kestrel]

the function gets bigger?

1/ a small number is an infinitely big number

[ionic pendant]

yes, and how many points are there where sin(x) = 0?

[trim kestrel]

uh

idk

do i count the asymptotes

[ionic pendant]

remember that sin(x) is periodic

[trim kestrel]

nvm that wont help

i thought it would help me understand

[ionic pendant]

well does the graph look like it's approaching any finite value?

[trim kestrel]

no

so thats why its DNE

[ionic pendant]

yes

it also isn't approaching either + or - infinity

[trim kestrel]

ty

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[cosmic plinth]

Can I choose point B as Po as well?

[ionic pendant]

you would get a different equation but it would describe the same line

[cosmic plinth]

so I can use A or B for Po?

[ionic pendant]

yes

[cosmic plinth]

thank you

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[vale abyss]

Im wondering if I did 5a correctly, I got 5b but 5a is confusing me a bit

[vale abyss]

Im(z) represents the entire y axis right

basically all real numbers on the y axis

so if for the function to be analytic Im(z) just has to be some constant then wouldnt f(z) be analytic

or am I misunderstanding this

since Im(z) can be a function of y then its not certain its analytic everywhere

[cedar socket]

f(z) = Im(z) means f(x+yi) = y + 0i

So u(x,y) = y and v(x,y) = 0.

[vale abyss]

so the Im(z) takes some x + iy and remove its imaginary part and keep the y in x + iy?

[cedar socket]

Yeah, and it returns y and not iy

[vale abyss]

ok let me fix my answer

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[potent edge]

Needed help with 11A

[potent edge]

this is what I have tried

DA/Dt = k1 A---> A(t) = x_oe^(-kt)

Db/Dt = -(xo-A(t))k2

but idk if i am doing it right or not

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[potent edge]

<@&286206848099549185>

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[true flare]

OMG ITS HEISENBERG

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[distant zinc]

so far thats correct.

[potent edge]

i tried to then solve the equation for B

and sub in x and y

[distant zinc]

that seemed correct to me, but I was mistaken. It doesn't take into account: amount of B also increases depending on A.

this is what you want:

from wiki on radioactive decay.

[potent edge]

can you send the link please i cant find it

so is this right

dB/dt = -K2B + K1(x_oe^-k1t)?

[distant zinc]

https://en.wikipedia.org/wiki/Radioactive_decay#Mathematics

[distant zinc]

yes

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[unreal lark]

how do u solve c i and ii

[unreal lark]

.close

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[hallow belfry]

Hi

[hallow belfry]

So I wanted to know what todo when this occure ?

[meager bloom]

I can understand first 2 equation. What about there 2? Those 2 are same

[hallow belfry]

Oh it have infinite solution ?

[hallow belfry]

I'm trying to find the solution of x and y

[meager bloom]

Well both are same equation so no you will not have infinite solution

Insufficient condition it will be

As far as I know

[hallow belfry]

I have a vague memory of my physic teacher saying infinite somthuing

something*

• my calculator say the same thing somehow ?

[slate comet]

you are trying to find x and y?

[hallow belfry]

yea

[slate comet]

the 1st formula up there is in the original question?

[hallow belfry]

The 1st system is in the question

[meager bloom]

Yeah I mean you can take any values of x and y as you have only 1 equation

[hallow belfry]

orginal question

the 2nd is me cooking

[hallow belfry]

ooo

Alr thanks all !

[slate comet]

ye i can help you

[hallow belfry]

Thanks u all

have a good sleep

!done

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[hallow belfry]

.close found out u can have infinite solution on system !

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[hallow belfry]

thanks u :D

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[unreal lark]

how solve

[naive otter]

First we need to select the two correct ones

From 7 , how many ways can you select the two to be made correct

Next all 5 can be solved incorrectly in 3 different ways

[naive otter]

Can you understand?

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[unreal lark]

how do you do this

[lone saffron]

you can find radius from surface area

then find rate of increase of radius wrt volume

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[kind portal]

Formula is Surface Area = 4pi R^2

We know surface area is 500mm^2

Substitute in and solve for R

[spark pier]

is the answer 1/50

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[spark pier]

probably how it's done

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[quaint pivot]

im a little confused on the wording of this, what am I actually looking for?

[quaint pivot]

is the answer just domain [-3, 4] and range [-4, 4]?

[marble wharf]

is there an (x,y) pair for every x in [-3,4] ? and for every y in [-4,4] ?

so for example x= 0.5 or y=pi ?

[quaint pivot]

what?

[marble wharf]

does 0.5 ever appear in the first position

does pi ever occur in the second position

the domain are those numbers which occur in the first position

the range are those in the second

[quaint pivot]

yeah I know each first value is an x value and each second value is a y value

what does that have to do with 0.5 and pi though

[marble wharf]

well you said the interval [-3,4]

0.5 is in that interval

[quaint pivot]

uhuh

[marble wharf]

so by that logic it should be somewhere in the first position

[quaint pivot]

okay

why does this matter

[marble wharf]

cause your answer is wrong and I want you to realize that

[quaint pivot]

okay

are you trying to say because any values other than -3, -1, 0, 2, and 4 appear in [-3, 4] the answer is wrong?

[marble wharf]

yes

the domain is just those points

in set notation {-3, -1, 0, 2, 4}

[quaint pivot]

okay so the answer would just be domain is {-3, -1, 0, 2, 4}

and range is {-4, 1, 3, 4, 4}

[marble wharf]

yes

[quaint pivot]

okay

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[subtle ether]

How do i do this

[subtle ether]

I tried to do median

45- 25

which equals 20

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[formal mountain]

I might be in bad but i think median here is 55 because is a number of the center (I'm still study math so if i'm wrong fix me).

The two center Numbers here is 50 and 60 so you add them both, After that is 110 and Because there 2 center numbers you dive by 2 and is 55.

So is 55 kg

@subtle ether

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[subtle ether]

.reopen

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✅

[subtle ether]

Thanks

It’s close

But 1/2 is the answer

[formal mountain]

That why you can't trust me with math

I'm too bad

Atleast i tried...

[subtle ether]

Hahaha you were almost there tbf

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[lethal ivy]

Help

[lethal ivy]

Plz

Do u have to write the equation in the order the instructions are given?

(The first pic is the instructions and the second is how my teacher did the question)

Plz @ me when responding

Thx

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[dry copper]

looks good yeah

[dry copper]

yeah, its a must

i give you an easy example to see

y=x, translate by 2 to the right, then scale by 1/2

y=x-2

y=2x-2

y=x, scale by 1/2, then translate by 2 to the right

y=2x

[lethal ivy]

Ok but if i think this way then if u look at the second to last line, shouldn’t y-5 be seperated from the 2 times?

[dry copper]

y=2(x-2)

y=2x-4

you see :p

[lethal ivy]

Makes sende

[lethal ivy]

@dry copper this is what confuses me

[dry copper]

no no, its correct

replace y by 2y

[lethal ivy]

@dry copper but then the expansion would be applied before vertical translation

[dry copper]

no, dont think like that about it

doesnt work like that 😛

[lethal ivy]

Wait y not

[dry copper]

its unexplainable

its not like PEMDAS

where multiplication is before +-

it just works fine, the method is legit

[lethal ivy]

Oh ok

[dry copper]

its not like PEMDAS

dont think about it that way

one more fact, interchanging translation and reflection, or interchanging translation and scaling

ruins it, not the same result

but interchanging reflection and scaling works fine, same result in the end

[lethal ivy]

What is scaling @dry copper

[dry copper]

horizontal compression/expansion

x -> ax

[lethal ivy]

Is this the same thing?

[lethal ivy]

Can u give examples. Much appreciated

@dry copper

[dry copper]

okay, interchanging translation and reflection, simple example, y=x

the translation is to the right by 2

here goes translation then reflection

y=x
y=x-2
y=-x-2

here goes reflection then translation

y=x
y=-x
y=-(x-2)=-x+2

you see, not the same

now, interchanging translation and scaling, simple example, y=x

translation is to the right by 2, and the x-scaling is by 3

here goes translation then scaling

y=x
y=x-2
y=3x-2

here goes scaling then translation

y=x
y=3x
y=3(x-2)=3x-6

not same :p

[lethal ivy]

Wait does this relate to my question? I dont see a relationship. Or is this just a useful tip

Either way much appreciated

[dry copper]

you said whether or not its okay to not follow the order

[lethal ivy]

Ooo ok

[lethal ivy]

Is this the same

[dry copper]

so im giving you when its okay to not follow the order

[lethal ivy]

Also

[lethal ivy]

But its safer to prolly just follow what is given in order

[dry copper]

either way, its always better to follow the order, dont think too much about it, just apply

[dry copper]

i dont get it, what did you do here

[lethal ivy]

Factored out the 4

So now it would be a vertical translation 7/4 up instead of 4 up

[dry copper]

what, no it isnt

[lethal ivy]

So u cant factor out whenever u want?

[dry copper]

yeah, thats something we do everyday

even outside these function transformations

[lethal ivy]

Wait so we can factor?

@dry copper

[dry copper]

the method given by your teacher is very nice

just follow it

i didnt know it existed until today

i had my own method

[lethal ivy]

No im talking abt a different example on the notes

Why can it just be factored

How does it not change the graph @dry copper

[dry copper]

cuz translation is always (x - ...) or (x + ...)

gotta factor the 4

to have that form and actually know the amount of translation

[dry copper]

cuz you didnt actually modify the function

youre just changing how you look at it

if you distribute back, youll have it

[lethal ivy]

But like if u look at it, u can see that the vt in step 1 is different than vt in step 2

Vt is vertical translation byw

Btw

@dry copper

[dry copper]

no, its the same

dont think like that 😛

[lethal ivy]

So then

[dry copper]

always factor whats multiplied by x

[lethal ivy]

This is correct?

[dry copper]

of course, why are you doubting

[lethal ivy]

It just doesn’t make sense in my mind

How is 7 vt same as 7/4 vt

😭

[dry copper]

cuz youre still looking at it as PEMDAS

its not like it

[lethal ivy]

@dry copper wdym

By that

Oooooooooooooo

So it should be expansion first

[dry copper]

whatever, me running out of answers lol

but 7/4 is not vt, its ht

[lethal ivy]

Ht sorry

Ya

Same wit 7

[dry copper]

no, for the function 4x, +7 is a vt

[lethal ivy]

f(4x+7)

? @dry copper

[dry copper]

if 7 was a vt, it would appear outside

y= f(4x) + 7

[lethal ivy]

I meant to write it so its inside

[dry copper]

then its not a vt, its an ht

a hidden ht

[lethal ivy]

Yes

[dry copper]

you find the amount of that ht by factoring whats multiplied by x, always

[lethal ivy]

I get it now thx

Great help

.close

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[dry copper]

nice to hear that, good luck! (:

[lethal ivy]

Thx

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[shell linden]

help

[shell linden]

😦

[tidal forge]

I would suggest to separate x+7 first

into 6^7 * 6^x = 5^x

then it should be gd to do

[shell linden]

sorry about that, after that what should i do?

@tidal forge

<@&286206848099549185>

.close

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[cosmic plinth]

In a question like this where I have to find the equation of a plane, does it matter what order I choose to do the cross product?

[cosmic plinth]

Meaning can I do either v1 x v2, or v2 x v1?

[wise dagger]

note that v1×v2=-v2×v1

so if they are followed by an arbitrary constant, it doesn't matter which one's first

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[cosmic plinth]

how would I decide which one should have a minus sign?

So if I do v1 x v2, how would I know not to do -v1 x v2?

[wise dagger]

hmmm, the first thing to consider will be:

how would you use the v1×v2?

[cosmic plinth]

I want to use it to find an equation of a plane.

The cross product will give me the normal vector

and I already have a point

[wise dagger]

then the second would be

what will the equation of the plane be written as?

[cosmic plinth]

in the standard form a(x-x0)...

[wise dagger]

if you know the meaning behind these 2 considerations, you'll probably get that multiplying by a negative would not change the representation of the plane

[cosmic plinth]

ah yes since we are only dealing with direction, we could multiply the normal vector by a scalar

[wise dagger]

bingo

[cosmic plinth]

so I can say that v1 x v2 = v2 x v1?

[wise dagger]

nope

it's still negative

[cosmic plinth]

But how would I know when to place the negative

[wise dagger]

a moment please, I'm reading your example again

[cosmic plinth]

take your time

[wise dagger]

like in your example:
we have v1, v2, P(3,-4,5)
consider:
Let v1×v2 = (a,b,c)
then we have
a(x-3)+b(x+4)+c(x-5)=0
is it correct?

[cosmic plinth]

yes it is

[wise dagger]

what if we do v2×v1, we will get
v2×v1=-v1×v2,=(-a,-b,-c)

so, the new equation will be
-a(x-3)-b(x+4)-c(x-5)=0

[cosmic plinth]

ah so although not equal, I can use either v1 x v2, or v2 x v1 and still get the equation of the same plane?

[wise dagger]

since they are "equations", we can multiply both sides by an abitrary constant (other than 0)