help-27

[teal perch]

Augment the basis matrix of B to the left

Then the basis matrix of C to the right

Solve until the left is an identity matrix

[limpid ridge]

so the method utilized here, yes? https://www.emathhelp.net/en/calculators/linear-algebra/transition-matrix-calculator/?a=[[1%2C2%2C1]%2C[0%2C1%2C-1]%2C[1%2C0%2C4]]&b=[[1%2C1%2C1]%2C[0%2C1%2C1]%2C[0%2C0%2C1]]

[teal perch]

Yes.

If you just want computational answers

Not the theory

[limpid ridge]

i want both

but isn't it just a method

in the other one i just solve for the exact same thing but using the systme of equations

[teal perch]

There is a theory behind it. But as you said, you don't have time, nor do I.

That's the whole meaning of why we augment stuff

Why everytime when we augment stuff the left side becomes an identity matrix

[limpid ridge]

ok i'll learn it tomorrow

for now let's just solve this question

so

we have the

how do I now

Calculate the coordinates of the vector v_C in the C basis given its coordinates v_B = (5, 1, -2) in the basis B

[teal perch]

Use the equation $[v]_C=[P]_B^C[v]_B$

[woven radish]

Applejakop

[limpid ridge]

so just multiply the transition matrix by v_B = (5, 1, -2) ?

[teal perch]

Yes

[limpid ridge]

i got (2, 6, -3)

[teal perch]

Well then that's prob the answer

[limpid ridge]

could you now please explain the above

what did i do wrong first

[devout snow]

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[cobalt rover]

Hi, im wondering how to find the first quartile from a data set, like what I know is how to find it from like when the q1 number set is odd (just add both divide by 2), but what if the q1 has even amount? Is it just the middle number in there?

[molten cedar]

assuming n is the number of values in your data follow these steps:

p = n × 1/4
if p is a whole number;

take the average of pth value × (p+1)th value
else;
round p to nearest whole number and find the pth value

[cobalt rover]

Whats p?

[molten cedar]

a variable we use to calculate the first quartile

[cobalt rover]

Oh

How would I know if P is a whole number or not

[molten cedar]

all you need to do is n ×1/4

n is the number of values in the data set

[cobalt rover]

Oh ok

So what's 1/4

0.25?

[molten cedar]

here is an example:
{20, 22, 30, 45, 46, 89, 98, 99}
n = 8
p = 8 × 1/4
p = 2

p is a whole number so;

find the average of 2nd value (p) and 3rd value (p+1) = 20+22/2 = 21 is your first quartile

[cobalt rover]

Wait so how do u calculate 8 x 1/4

[molten cedar]

always divide n by 4 to find p

n is 8 here because there are 8 elements in the data set

[molten cedar]

8 × 1/4 = 8/4
?

[cobalt rover]

Ohh

Mb

Ok

So

[cobalt rover]

Can u explain last part

[molten cedar]

check this one more time

[cobalt rover]

Icl I'm finished for the test

I'ma just ball it

I don't understand a thing

[devout snow]

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[devout snow]

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[proper terrace]

for #3 whats domain and range

i forgot

[foggy vault]

Domain is the extent of which x exists

on that graph

so on what interval does the value x exist

[proper terrace]

so is it from -4 to 4

[foggy vault]

yes but remember to look at the endpoints

its non inclusive

so you can either write in interval notation as (-4,4)

or in just regular notation as -4 < x < 4

Then range is just the same but with the y values

[proper terrace]

y is -3 < y < 5 right ?

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[restive river]

find the area of | 3x | + | 4y | = 12 without draw th figure

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

[frosty cradle]

Try squaring both sides of the equation

[restive river]

i already tried it

[devout snow]

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

[restive river]

im stuck with what i do next it

because i dont get anything with that

[devout snow]

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

[restive river]

.close

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[weary shale]

I need help understanding how systems work in physics

[weary shale]

I got the acceleration right by setting the system mass and acceleration equal to the given force

but I could not for the life of me find the Ft between blocks 1 & 2

I tried reworking it, but I feel like im getting nowhere

<@&286206848099549185>

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[north hollow]

https://cdn.discordapp.com/attachments/1136709808704209150/1240500625125539850/v12044gd0000ci55do3c77ubs36unq3g.mov?ex=6646c9bc&is=6645783c&hm=0ffcc277ac90314a9673a939ecf94c39ae94640b59f1e9655adc138ce5a89461&

[devout snow]

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[inland pulsar]

with the requirements that a function f must be continuous, have the points f(0) = 0 and f(1) = 0 and the area below the curve must be one, what is function with the least arc length that fulfills these requirements?

[devout snow]

@inland pulsar Has your question been resolved?

[inland pulsar]

<@&286206848099549185>

i tried some functions with some lowish arc lengths

like 6x(1-x)

but i am struggling to get well below an arc length of 3, which is my goal

i think a 2.5-2.6 would be ideal unless something lower can be achieved

[earnest kite]

A function that resembles a semi circle would be best, no?

since circles have the highest area to circumference

[devout snow]

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[wicked zodiac]

if $f\left(x\right)\ =x\cdot\ln x+x^{2}-3x+2$ and i know the monotone of the function, decreasing from (0,1) and increasing from (1,+inf), i need to solve the inequality f(x) > f(3)

[woven radish]

Striker

[wicked zodiac]

how do we go about these inequalities when f is not just increasing or just decreasing

[devout snow]

@wicked zodiac Has your question been resolved?

[wicked zodiac]

@fervent swallow

[meager hollow]

find the value of the function near the end points of the intervals

like as x becomes close to 1, the functions value is around 0

if we choose any f(x) where 0<x<1 this f(x) will be less than 0

since it is a decreasing function

so we can stop looking over there for a bit

now look at f(3)

its value is 3ln3+2

[meager hollow]

for any x>3, f(x) will be greater than this

since it is increasing in (1,+inf)

hopefully this gives you an idea

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[bright juniper]

ABCDS pyramids base is ABCD square and all the 4 faces of the pyramid is a equalavent triangle need to find angle ASC

[bright juniper]

<@&286206848099549185>

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[fresh basin]

properties of kites

[fresh basin]

1. if MU = 2x cm, and MT = 4 cm. What is the measurement of MU?
   a. 4 cm
   b. 8cm
   c. 16 cm
   d. 48 cm

[molten cedar]

what do you think the answer is?

[fresh basin]

4cm?

[molten cedar]

yeah

[fresh basin]

but how do u get the answer? i just guessed

is it just MU = MT?

[fresh basin]

then u solve for x?

[molten cedar]

lmao you see these?

[fresh basin]

ya

[molten cedar]

they indicate that these two line segments are equal to each other

[fresh basin]

got ya

[molten cedar]

that's how we know MU = MT

[fresh basin]

can u help me with # 2

[molten cedar]

sure

[fresh basin]

2. if US = 20 cm, what is TS = x+4. What is X?

[molten cedar]

what does TS equal?

[fresh basin]

US

[molten cedar]

yeah

and what does US equal

[fresh basin]

TS

US = TS

[molten cedar]

yeah

[fresh basin]

so 20 = x + 4?

[molten cedar]

yep

[fresh basin]

then u transpose?

x + 4 = 20?

[molten cedar]

solve for x

[fresh basin]

x = 20 - 4?

x = 16?

nicee

what about number 3? W
What is the perimeter of the kite MUST?

how do i measure the perimeter?
A. 4 cm
B. 8 cm
C. 16 cm
D. 48 cm

[molten cedar]

i'm assuming this kite is the kite in the question 1 and 2

first of all

what is perimeter?

[fresh basin]

the length of something?

maybe a line or angle

[molten cedar]

yeah total length of the kite
and it is MU + US + ST + TM

[fresh basin]

so how do i calculate the total length?

[fresh basin]

whats the degree

or measurement

[molten cedar]

so we already know all lengths from questiom 1 and 2

[fresh basin]

ya

[molten cedar]

what's the length of MU?

[molten cedar]

we don't need degrees for this question

[fresh basin]

4cm

[molten cedar]

yup

[fresh basin]

oh ok

[molten cedar]

what about MT?

[fresh basin]

4cm?

[molten cedar]

yup

what about US and TS?

[fresh basin]

16cm

?

oh wait no

[molten cedar]

yup

[fresh basin]

thats the x

oh rlly?

[molten cedar]

oh

[fresh basin]

okok

[molten cedar]

right

it is 20

[fresh basin]

20cm

?

[molten cedar]

so

to find the perimeter

[fresh basin]

okayy

[molten cedar]

all you need to do is add these 4 values:

4 + 4 + 20 + 20

[fresh basin]

48cm?

is the perimeter

so its D

what about # 4? @molten cedar

[compact python]

i thin k #4 might have a slightly typo

it should be 10cm not 10cm^2

[fresh basin]

oh yea prob

so how do i solve it

[compact python]

ok

so to calculate the area of a kite

[molten cedar]

@fresh basin are these questions from your teacher?

[fresh basin]

ya

[molten cedar]

there are so many mistakes smh

[compact python]

there is 2 ways

[fresh basin]

ikr

[compact python]

fr lol

[fresh basin]

show me the easier one

[fresh basin]

i think she stayed late at night to do these

[compact python]

use the formula pq/2

[fresh basin]

how

[compact python]

which is ms*ut/2

[fresh basin]

oh okok

ms = ut

[compact python]

no

[fresh basin]

so its

[compact python]

ms doesnt equal to ut

[fresh basin]

90/2

[compact python]

yes

[fresh basin]

90/2 = 45

so its 45cm

[compact python]

yus

[fresh basin]

nice

i thought it was gonna be hard lol

lets finish # 5

[compact python]

and the other method is to split the kite into 2 triangle

[compact python]

which i personally think its easier

[compact python]

typo again 💀

use the formula

pq/2

[fresh basin]

ahahahahaaa

oh sooo

76/2 = 38cm

[compact python]

yes

[fresh basin]

nicee

tysm!

[fresh basin]

im gonna tell my prof to fix it ahahaha

thanks again!

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[normal lagoon]

I have tried to do this and I've been told I did it wrong/am missing a few steps.

[devout snow]

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

[normal lagoon]

Okay what do I need to do next?

[lunar harbor]

(b) is fine

[lunar harbor]

You prob just messed up your arithmetic ngl

[normal lagoon]

So its 5(mod26)?

[normal lagoon]

Okay one sec let me rewrite till the point of trying to impliment Wilson's theorem.

What do I need to do now? I can't just divide both sides by 16*15

[lunar harbor]

I meant that $$2 \cdot 14! \equiv -1 \pmod{17}$$ was fine, you just found $14! \pmod{17}$ incorrectly.

[woven radish]

Civil Service Pigeon

[normal lagoon]

Okay. What do I do from here?

[lunar harbor]

Tell me how you got 16 first

[normal lagoon]

Wilson's Theorem implies that if p is prime, then (p-1)!=-1(modp).

I'm trying to find 14! using 16!

[lunar harbor]

I know that

[lunar harbor]

I meant how you got from this to 16

[normal lagoon]

Oh... Cause 16 = -1 (mod17) and 15 = -2 (mod17)

[lunar harbor]

Not what I’m asking

[normal lagoon]

Ohhhh

[lunar harbor]

How did you get $$2 \cdot 14! \equiv -1 \pmod{17} \implies 14! \equiv 16 \pmod{17}$$

[woven radish]

Civil Service Pigeon

[normal lagoon]

My original statement? Its cause it was asking for the non-negative residue module, right? So once you find the normal module its just the inverse of the given value of b relative to mod p

modulo*

[lunar harbor]

What did you get for the inverse of 2 mod 17

[normal lagoon]

Wouldn't it be negative 2?

[lunar harbor]

No

[normal lagoon]

Okay wait. What am I missing here? I managed to get 2*14!

[lunar harbor]

The inverse isn’t simply negating it

You’re looking for what you need to multiply 2 by to get 1 mod 17

If you want the inverse

[normal lagoon]

Oh. 8

Wait let me write this down to help visualize it.

[normal lagoon]

Thank you kindly. Been stuck on this for almost 3 hours.

.close

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[opaque jetty]

HI, I am not sure how to approach this question it is true or false

[hazy yarrow]

lets start with definitions

what does it mean for a set of vectors {u,v,w} to be linearly dependent

[opaque jetty]

to be honest

i dont know

i just googled

no vector can be expressed as a linear combination of the others

[hazy yarrow]

that's linearly independent

We want to negate that

So {u,v,w} is linearly dependent == one of the vectors may be written as a linear combination of the other two

So lets just say, u may be written as a linear combination of v and w

it doesnt matter that we chose u here

we could have also chosen v or w here (try to figure out why im allowed to just pick one "without loss of generality")

What can this look like geometrically is the point of the question

it's helpful to think of a few examples in R^3

Give me any two vectors v and w in R^3

(you choose randomly and tell me)

[devout snow]

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[devout snow]

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[hazy yarrow]

😿

[devout snow]

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[opaque jetty]

Hi

[devout snow]

Please don't occupy multiple help channels.

[opaque jetty]

for vectors

{1,0,0}

{0,0,1}

2 random vectors i thoguht of

[tiny hearth]

you have 2 channels

please close one

@opaque jetty

[opaque jetty]

can i close the other one

i cant find it

[hazy yarrow]

ill do it

ok im here

[opaque jetty]

hi

me too

thank you so much

[hazy yarrow]

nw

[hazy yarrow]

Yeah these are great two vectors

lets set the first one to v and the second one to w

[opaque jetty]

yeah ok i write this down

[hazy yarrow]

Can you come up with a third vector that is linearly dependent with these two

[opaque jetty]

hmm i am not quite sure what lienar combination is

i just googled it

how about <2,0,2>

[hazy yarrow]

Linear combination of two vectors v and w is av+bw where a and b are constants

yeah

That works

or just <1,0,1> also

lets just go with what u said tho it doesnt really matter

[opaque jetty]

ok so basically i multiply both of them by a scalar then add them

[hazy yarrow]

ok so geometrically what does this look like is something we should figure out

Imagine all of the vectors that you obtain by finding av+bw for any choice of a,b

what sort of geometric shape is the set of all the vectors you can obtain by taking linear combination of v and w?

using ur choice of v and w u told me earlier

[opaque jetty]

this is quite abstract

am i supposed to draw it out

i am bad at drawing

or is there another way to tell the shape

[hazy yarrow]

Yes! this can work

Maybe use desmos or some other 3d plotter as well if it will help

do spend time doing this, it will help you

even if you think you are bad at it

the struggle is worth

[opaque jetty]

okay

thank you i will try desmos onw

[hazy yarrow]

ping me when you think you have an answer

ill be around

[opaque jetty]

thank you

@hazy yarrow triangle

i think

[hazy yarrow]

Triangle is half

Did you forget that you can take a or b to be negative?

[opaque jetty]

right triangle

oh

wait are a and b the vectors

[hazy yarrow]

the coefficients

av+bw

[opaque jetty]

oh truee

yeah i can make a and b negative then

<-1,0,-1>

but its always some sort of triangle i think

i dont understand how i can answer the question with this though

like would same line mean same plane?

[hazy yarrow]

hmmm the shape i was looking for was a plane

specifically the xz plane

Any point on the xz plane is a linear combination of <1,0,0> and <0,0,1>

[opaque jetty]

xyz plane

?

[hazy yarrow]

nah

[opaque jetty]

oh is there a xz plane

[hazy yarrow]

because av + bw is <a,0,b>, which wont cover all of R^3

but instead just the xz plane

xz plane is the uhh

let me draw a picture (lets hope it works lol)

i suck at computer drawing

[opaque jetty]

oh that makes sense

[hazy yarrow]

Ok zoomed in that is so bad but i hope that makes sense

lolz

No y direction basically

but it's a plane

[opaque jetty]

yeah could basically remove the y i guess

[hazy yarrow]

The reason it's a plane is because v and w are linearly independent (notice now its just the two of them, i didnt mention u)

you cannot write v as a scalar multiple of w

and vice versa

[opaque jetty]

yes true

thats true

[hazy yarrow]

If perhaps you did choose v to be a scalar multiple of w, then what shape do you get when you take linear combinations of v and w?

[opaque jetty]

in order to be linearly dependent it should be a scalar multiple right

[hazy yarrow]

yeah in this case, with only two vectors in view

[opaque jetty]

ok i will change it to a scalar multiple

because that is what the question asked

[hazy yarrow]

wait

now im confused

im just spinning out hypotheticals so far

[opaque jetty]

ohh

[hazy yarrow]

i havent returned to the question yet

[opaque jetty]

okay

i thoguht it was for the question

[hazy yarrow]

To be clear this is what i am doing

i am changing your choice of v and w

since you correctly determined what the shape of all linear combinations of v and w were in your case

Now i am selecting v to be a scalar multiple of w, say v = 0,0,2

in this case, what would all the linear combinations of v and w look like?

[opaque jetty]

yeah that makes sense

let me think

well a scalar multiple of this can be 0,0,4

[hazy yarrow]

in other words, what is the geometric shape that the set {av+bw: all constants a,b} represents in space

[opaque jetty]

a z plane

[hazy yarrow]

You mean just the z axis right

a line?

[opaque jetty]

yes

just a line

literally

WAIT

[hazy yarrow]

yes...

😸

[opaque jetty]

then the question should be true

[hazy yarrow]

No no

[opaque jetty]

oh

because if i have like

[hazy yarrow]

can u post the question again actually

i lost it

[opaque jetty]

ok

wait actually

say i have like

non zero vectors

{1,2,3}

{2,4,6}

{4,8,12}?

[hazy yarrow]

In this case, what is the conclusion

[opaque jetty]

these are all linearly dependent

[hazy yarrow]

correct

and?

in view of the question you posted

[opaque jetty]

but they dont have to lie on the same line

[hazy yarrow]

in this case they do lie on the same line

The line given by a*<1,2,3> where a spans the real line

Notice though that this is just one specific case

[opaque jetty]

yes

but in order to be linearly dependent

[hazy yarrow]

Can you cook up a different example where you still have linear dependence but they dont lie on the same line?

[opaque jetty]

let me think

[hazy yarrow]

ye and sorry i interrupted u just now

[opaque jetty]

i thought it needs to be a scalar multiple

[hazy yarrow]

continue with what u were saying

[opaque jetty]

no you didnt interrupt dw

[hazy yarrow]

So scalar multiple for lineraly dependent is only between two vectors

v and w are linearly dependent if v = cw for some c

For three or more vectors you have this linear combination stuff

[opaque jetty]

oh

[hazy yarrow]

u,v,w are linearly dependent if say, u = av + bw

[opaque jetty]

wait

[hazy yarrow]

and similar expresision for more than 3 vectors

[opaque jetty]

then there must be a way to make it false

[hazy yarrow]

(really its the same thing but watever)

[opaque jetty]

like

<1,2,3>

<2,4,6>

now i need to change the last vector somehow

[hazy yarrow]

dont think too hard about it

any random vector u choose will likely do the job

[opaque jetty]

i can literally pick anything

yeah

like

<10,12,16>

[hazy yarrow]

just not anything of the form <a,2a,3a>

Yes

Bingo

so you have found that the statement in the question is ?

[opaque jetty]

false

and my explanation would be

because u,v,w are linearly dependent if say, u = av + bw

but in order for it to be on the same line

must be a scalar multiple

and a scalar multiple is not neccessary for u,v, and w

[hazy yarrow]

specifically, v must be a scalar multiple of w

But yea more or less

[opaque jetty]

ohh

yeah

by the way am i allowed to ask other questions in the same thread

or should i make new ones

I really liked how you explained it

it makes sense now

thank you so much

[hazy yarrow]

Keep thinking about this and itll make more sense i promise but for the question, a counterexample is sufficient (explanation is excellent though)

[opaque jetty]

I think I understand it tbh

[hazy yarrow]

Yeah you can keep asking in here

im glad to help

its been like four years since i have roamed the help channels so im glad i still got that dawg in me

[opaque jetty]

what motivates you to do it?

[hazy yarrow]

in the past i was just bored

now i do it because im bored AND it keeps me accountable for everything i learned

[opaque jetty]

i saw your github the questions i am asking you are clearly stuff you have done a long time ago

🤣

[hazy yarrow]

i think its good practice for me to be answering questions and stuff

[opaque jetty]

its amazing to be honest

[hazy yarrow]

idk part of it is ego thing, like, its kind of embarrassing if i cant answer questions i used to be able to answer four years ago

bc i spent so much time learning the content i might as well try to retain it

u know

[opaque jetty]

that makes a lot of sense honestly

dont you forget though

sometimes

[hazy yarrow]

I do

[opaque jetty]

or do you just know all of it

[hazy yarrow]

this is why i come to answer questions

sometimes i mess up or i end up remembering all of a sudden

I must go now though, hopefully someone else can help aaaa

sorryzz

[opaque jetty]

thank you so much 😄

I really appreciate it

have a good day

[devout snow]

@opaque jetty Has your question been resolved?

[devout snow]

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[barren sleet]

Hey all. Just looking for some help and guidance. Looking to get my GED, I only need to pass math. I'm really bad with pre algebra and algebra 1, but all I need is to figure out where to start when it comes to approaching those topics.

Any sites or material you guys recommend that would help me would be appreciated. Study tips too. I'm trying to find the best ways possible to incorporate studying in my daily life. I want to retain this information, but it has always been hard for me.

[torn vessel]

kahn academy and openstax are pretty good, free, resources that have texts and lecture series

[restive river]

i can personally vouch for khan academy

for everything up till calc 2

[restive river]

Same here

[devout snow]

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[quaint anchor]

Give me please a hint

[quaint citrus]

what have you tried

[quaint anchor]

What I've already done is: integral from 0 to 2 of ([x/(x-3) + 1]dx)

[quaint citrus]

good

so now

$\int_0^2 [\frac x{x-3} + 1 ]dx = \int_0^2 [\frac x{x-3}] dx+ \int_0^2 1 dx$

[quaint anchor]

Then I tried to substitute x - 3, but understood that won't work

[quaint citrus]

why not?

[quaint anchor]

The derivative of x - 3 = 1

So the x in the numerator lefts

[quaint anchor]

Right

[quaint citrus]

if u = x-3

du = dx

yes?

[quaint anchor]

Yes

[quaint citrus]

so now we have x in numerator

if u = x - 3

solve for x

[quaint anchor]

u + 3 = x

[quaint citrus]

good

so now we have

$\int_{u(0)}^{u(2)} \frac {u+3}u du + \int_0^2 dx$

[quaint anchor]

u + 3 *

So essentially you want to say that (u+3)/u = u/u + 3/u

Right?

[quaint citrus]

correct

[quaint anchor]

And that's the way to go

[quaint anchor]

Thank you

[quaint citrus]

np

[quaint anchor]

I didn't even know I could do such a thing

[quaint citrus]

yep, pretty useful

[stone stump]

I wanna note that at the beginning you could have either used polynomial division on (2x-3)/(x-3) or written it as (2x-6+3)/(x-3)

in general you should always try to reduce the degree of the numerator to be less than the degree of the denominator

[quaint citrus]

yea i was thinking long div but whenever i say that, someone comes and says just do the +a - a trick

[stone stump]

its essentially the same as long div obviously. just slightly faster

[devout snow]

@quaint anchor Has your question been resolved?

[quaint anchor]

What's the sense? To divide them?

[stone stump]

it gets rid of the x in the numerator

which is the annoying part

[quaint anchor]

Could you explain more?

How?

[stone stump]

what do you get if you do the division

[quaint anchor]

2 + 3/(x-3)

[stone stump]

so therefore $\int \frac{2x-3}{x-3} dx = \int 2 + \frac{3}{x-3} dx$

[woven radish]

Denascite

[stone stump]

which you can now integrate easier

[quaint anchor]

Oh yeah, that would be even easier

[devout snow]

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[nimble mortar]

if you want to determine if vector b is in the span of matrix A? some people use gauss method and some people row evaluation, so if you wanna get an answer there are these two ways?

[stone stump]

what exactly do you mean by row evaluation

[nimble mortar]

like this

so, row evaluation, row reduction what's the difference

[stone stump]

what do you think gauss is if not this

[nimble mortar]

no

i am confused when i find answers for if you want to determine if vector b is in the span of matrix A?

there are different ways to get an answer

if i am correct

[frosty cradle]

there might be. row reduction is probably the most straightforward

[nimble mortar]

ok thank you

.close

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[worldly latch]

,rotate

[worldly latch]

Any idea how to prove these two identities?

[devout snow]

@worldly latch Has your question been resolved?

[mellow hornet]

@worldly latch for the first question try multiplying (1-cosx) in numerator and denominator

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@worldly latch Has your question been resolved?

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[static marsh]

I need help with math

[static marsh]

Homework

<@&286206848099549185>

[brazen hawk]

Maybe show what's your problem?

[static marsh]

It’s homework is that fine

@brazen hawk

?

[brazen hawk]

You don't know how to do that?

If I have 6 apple and eat 3, how much percent of my apples have I eat?

You must do number used / number total

[runic solstice]

64 percent

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[sonic smelt]

Well, how do you know that x > 0?

[jagged kernel]

If x was less than 0, it would produce imaginary numbers

[earnest kite]

,w calc 7\sqrt(x) + 1 < 9

[woven radish]

Results provided by WolframAlpha

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

Requested by hrdxpqurcxktdbanql

[earnest kite]

it is possible for x to = 0

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[restive river]

can someone please help

[restive river]

ive gotten it into a double integral using greens theorem, do I need to convert to polar coords?

[acoustic leaf]

polar coordinates would probably be the easiest to evaluate for that region, yes

[restive river]

does my working and bounds make sense?

[acoustic leaf]

don't forget the jacobian r in your dA

[restive river]

I forgot about the jacobian, I've only learnt it recently, when should it be used?

[acoustic leaf]

dA = dx dy = r dr dθ, the jacobian for polar is r

[restive river]

oh my word is that why jacobian falls under change of variables section

sorry I was dumb and didnt even realize what the jacobian was used for but now it makes sense

okay awesome I got the correct answer

thank you @acoustic leaf

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[limber wren]

Hi for part b I was able to get to this stage but I'm not sure how I find out what I should set Xn to

[devout snow]

@limber wren Has your question been resolved?

[restive bramble]

Whom needs help

[limber wren]

<@&286206848099549185>

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@limber wren Has your question been resolved?

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[tall knoll]

What have you tried?

[devout snow]

@quiet flint Has your question been resolved?

[hexed beacon]

why did you do 0.5 * 9.28^2 * tan(111.1)?

[devout snow]

@quiet flint Has your question been resolved?

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[stark ocean]

is this incorrect?

[quaint citrus]

Show work

I’d expect a plus sign in there somewhere

Hmm

@stark ocean

[stark ocean]

ugh i just closed ms paint but ok

[swift pumice]

I’ll give you a few hints: know the area of a sector formula and then low of cosines/sines

[swift pumice]

Volume of cone + cylinder

[quaint citrus]

Yes it’s incorrect btw

[stark ocean]

ok well heres what i did can u tell me where i went wrong

[quaint citrus]

U were right until the last step

$\frac {240\pi}{3} + \frac {16\pi \cdot 4\sqrt 3}{3} = \frac {240\pi + 64\pi \sqrt 3}{3}$

U can’t add those pi’s together

Cuz the 16pi is attached to the 4sqrt3

@stark ocean

See if u can simplify the second numerator

Then go from there

[stark ocean]

sorry im rusty and this might sound ignorant but

is 64pisqrt3 something that can occur

or is that a nono

[quaint citrus]

Why not

Yea that’s exactly what we gotta do

[stark ocean]

idk sorry lol

ok

[stark ocean]

ok thank you sm

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[stark ocean]

i got 1/5 correct and i asked my friend and he got all the same answers as me so now were both confused. would really appreciate the help

[final storm]

inside part is only 6m wide

so radius is 3m, volume = 99pi

[foggy vault]

uh

the inside dimension is 8ft

without the walls

[stark ocean]

ohhh shit i read it wrong

[foggy vault]

it seems fine

the first one

[stark ocean]

i thought it meant that the empty space within the cistern had a diameter of 8

[foggy vault]

because it straight up says those are the inside dimensions

[final storm]

wait yea, mb, concrete doesnt go inside

[foggy vault]

that is what it should mean

[sharp root]

Second one is wrong

[foggy vault]

you were right for first

[stark ocean]

guys which is it

[foggy vault]

first one is right

second is wrong

[sharp root]

Radius of whole thing should be 5

[foggy vault]

for the second one the radius becomes 5 and the height becomes 12

because its 8 wide on the inside (inner diameter)

adding 1 to each side makes the diameter 10

divide by 2 you get radius 5

[stark ocean]

so hold on the second one is just 25pi*11 tho right

[foggy vault]

no

25pi*12

the bottom also gets 1 foot added

11+1 = 12

[stark ocean]

thank you so much bro

i read that and was like "wtf does that mean" and just did it without the extra foot on the bottom

[foggy vault]

ah lol

its fine

then the third one you got wrong because you got the second wrong

[stark ocean]

yepyepyep

[foggy vault]

you just have to redo it with the new outer volume

[stark ocean]

i got it from here than you

.close

[foggy vault]

yeah

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[restive river]

@fervent swallow

Bruh they need to change that guys name

<@&286206848099549185>

[cobalt tundra]

Yea it's right

[restive river]

Thank

[devout snow]

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[restive river]

Nice

Im 2 for 2

🔥

Well

it was going well

I just hit a brick wall

Use the diagram of a regular polygon to find the area:

I forget how to do it on polygons lol

[sage burrow]

separate it to 5 triangles

[restive river]

43?

Im having a hard time

[restive river]

So for 43 is right?

ft or ft^2

I don't know the difference

[sage burrow]

one is for length, one for areas.

[restive river]

so we should use ft^ 2

[sage burrow]

yes

[restive river]

Ill put this in my notes, thank you for helpin

[devout snow]

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[restive river]

I think it may be 69

The ratio of their heights is ( h_2 / h_1 = 20/28 = 5/7 ). the ratio of their areas is ( (A_2 / A_1) = (h_2 / h_1)^2 = (5/7)^2 ).

Hard to type

[jade oak]

Yes

[restive river]

cause 90 x 4 is 360

right?

[teal perch]

I think so.

[restive river]

I do to lol

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@restive river Has your question been resolved?

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[tepid tulip]

help pls

[random temple]

what if f(x) does weird stuff between integers

[tepid tulip]

wdym

[long sundial]

Yeah

[opaque nova]

ah, u could use the classic example of f(x) = 1/n

[long sundial]

You can usually bound the area of a curve below the sum, unless it doesn't behave normally

[long sundial]

That doesn't converge for both

[tepid tulip]

oh true

[random temple]

what if f(n)=1/n^2 for integers and 1/(n+0.5) for n+0.5 and it interpolates between

[tepid tulip]

i mean i dont need to prove that it converges for both

[opaque nova]

oh yeah, my bad

[random temple]

no but it diverges for both so it doesn't work for this problem

[tepid tulip]

oh wait nvm

im trippin

[random temple]

yeah

[tepid tulip]

hm

[random temple]

I think this idea might work

[gritty juniper]

y=e^x

[tepid tulip]

im confused by the last half

[long sundial]

Its not continuous?

Also it'll be a pain to actually show it doesn't converge

[tepid tulip]

😭

[random temple]

I was thinking you could interpolate between the two

[tepid tulip]

last week of school and my teacher pulls this shit out

[random temple]

riemann sum bounding?

[long sundial]

It would converge if you do that

[random temple]

really?

[long sundial]

Should do

[opaque nova]

oh sin(pix) would probs work

[tepid tulip]

shi i ont even know what that is 😭

[random temple]

oh wait this is facts

[long sundial]

This is the idea u need to use, perhaps something to do with cos

[long sundial]

True

Goated answwr

[opaque nova]

yeah, trig comming in clutch once again

[tepid tulip]

goat

[gritty juniper]

wait why wouldn't this work? It's continuous on (1, infinity) right?

[random temple]

I was thinking weierstrass function or some shit 💀

[random temple]

but it diverges

the sum diverges

[tepid tulip]

yeah idk how sum of e^x would convege

[tepid tulip]

im so slow how would this converge

ik that it does i forgot how to prove it

[random temple]

if you sample it on the integers it's always 0

and sum of 0 converges

[tepid tulip]

i get that but by nth term test it diverges right

[random temple]

wdym

[tepid tulip]

cuz if u do lim n-> inf sin(pi x) its undefined

and by nth term test that means it diverges

[random temple]

sin(pi n)=0 always

so limit of sin(pi n)=0

[tepid tulip]

hm

[tepid tulip]

is this a identity or am i beign stupid 😭

[random temple]

unit circle

[tepid tulip]

fuck ur right

[random temple]

integer multiples of pi are on the x-axis

yeah

[tepid tulip]

i tried plugging it into photomath and it kept on telling me it was diverging

[random temple]

yeah dont use calculators for limits they suck

[tepid tulip]

oh wait

i dont think thatd work

cuz it says the function f needs to be positive

shiiiiii

[rugged sparrow]

bingbong

[tepid tulip]

yo

[random temple]

take the absolute value

it still works I think

[devout snow]

@tepid tulip Has your question been resolved?

[opaque nova]

Or sin^2(pi x) lol

If it's still the same question as before

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[jaunty abyss]

If I have x+3 times 2 would I make that 2(x+3) or (x+3) * 2 like (x+3)(x+3)

[jaunty abyss]

Idk

[stiff kiln]

"Times 2" means "multiplied by 2"

If it were (x + 3)² it would've said "to the power of 2"

[jaunty abyss]

Oh ok

No it’s just multiply

So 2(x+3) ?

Even if the 2 is on the right side rn

Can I move it

[stiff kiln]

Yup

Because a * b = b * a

[gritty juniper]

Multiplication is commutative, so you can move the operands without altering the result.

[earnest kite]

2*anything = anything * 2 = anything + anything

[jaunty abyss]

Got it

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@jaunty abyss Has your question been resolved?

[devout snow]

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[static quarry]

Whats the simplest way to solve these? “Find the measure of the are or angle indicated.”

[static quarry]

Its highschool applied geometry but i didn pay attention when explaining these so i dont know how to do it

<@&286206848099549185>

[jade oak]

Look up inscribed angle theorem

[plush pulsar]

Wsg bro

[static quarry]

Alright ill try to look that up see if i can solve them with what they got on google

[plush pulsar]

Yo what is this

Never seen this

[static quarry]

Finding the measure of an arc or angle

Im not sure like the exact name of these type of problems but i guess an inscribed angle circle

[rose rock]

Learned about in secondary school under the name "Circle theorems"

[static quarry]

Easiest way you learned to solve them? Google gives me loops not awnsers from what i seen

[rose rock]

One moment

[rose rock]

That should cover it all really

What do the angles outside the circles refer in your questions BTW?

Oh I guess it's the angular distance

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@static quarry Has your question been resolved?

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[gusty niche]

Here if m=0 and x=1 it satisfies all the conditions as the value of the expression is non negative i.e. 0

Or do we naturally assume that m>0 because it's the coefficient of x^2

[midnight grove]

yea but it doesnt hold at m=0 and x=2

[gusty niche]

OH

I DIDN'T THINK THAT

it had to be for all positive real x

idk why sorry mb

i thought 1 value was enough

/close

how to we end

the thing

[midnight grove]

its .close

[gusty niche]

.close

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[devout snow]

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[untold pivot]

having some trouble interpreting this

[untold pivot]

The units of R[x] are exactly the units of R

what does this mean?

I thought it meant $a \in R^x$ implies $ax^0\in R[x]^x$ and $b \in R[x]^x$ implies $b=ax^0$ for some $x \in R$

but this isn't true because if we are in the real numbers for example. something like $1+x$ does have an inverse

[woven radish]

slong

slong

[untold pivot]

oh nvm

.close

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[solar goblet]

A group has 4 male students A, B, C, D and 4 female students W, X, Y, Z. There are 2 male students, each of whom is taller than exactly 2 female students. Similarly, there are 2 female students who are taller than exactly 2 male students.

Arrange B, W, A, X in a line, it can be seen that their heights are descending in that order. Additionally, Y is taller than X, but is shorter than D; D is shorter than Z, Z is not as tall as C, and C is taller than B.

If D is shorter than A, which of these propositions can be false?

• W and Z are 2 female students who are taller than exactly 2 male students
• A and D are 2 male students who are taller than exactly 2 female students
• D and X are the shortest male student and shortest female student respectively
• W is the tallest female student

[solar goblet]

my current config is CBWZADYX (descending), i have tried switching around, but i cant find a case where any of the propositions is false

[solar goblet]

solved, switching W and Z renders the fourth proposition false

.close

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[devout snow]

@blissful ingot Has your question been resolved?

[woeful burrow]

what is the first step here

seems correct

well first off we need to find the critical points of f

yep

yea

only one actually

yes for the hessian

yep

yes

well doesnt seem like it

also check if that critical point is in the region

yep

ok so let's try to express the boundaries of the triangular region

yeah that's one part of the boundary down

the same thing as before

yeah the endpoints

yep

with 0 <= y <= 6 ofc

yeah now you can eliminate some

to find max and min

well youve singled out the boundary values for f in that triangular region

now you can get the min and max out of this

since you know there is one critical point of f

yep

yea exactly

hmm im not sure

no problem

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[dense jay]

sure

[restive river]

Ill post my work but it works werid typed out

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

pls

pretty please

WIth a cherry on top

[restive river]

<@&286206848099549185> Please

.close