help-13

[whole shuttle]

The dx and dy are switched. Otherwise the integral doesnt have sense

Since you have a sum you can apply linearity or just solve directly

[drowsy hare]

yes but is there not a rule against just changing the bounds?

[whole shuttle]

x^2 doesnt depend of y so thats why they did integral of 1 dy from 1 to sqrt(x)

[whole shuttle]

The statement of the problem is wrong, must be dy dx

[drowsy hare]

This would make the question very much easier but this will come up in an exam, would I be sure that I can just say that they have made an error?

[whole shuttle]

Yes they made an error. You cant have a variable in the bounds and integrating respect that variable. So must be dy dx. In the aolution they did that change because on your statement is wrong

[cedar kiln]

@drowsy hare Has your question been resolved?

[drowsy hare]

oh ok, thanks

[cedar kiln]

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[rain oasis]

uhh can i get help?

[rain oasis]

in a class we have 18 male student and 12 female student we want to form a group that has a president and a vice president and a secutary well lets say that the president is a male student and the secrutary is a female and the male student X doesnt not want to join a group the contains the femal student y , hhow many groups can we form in this situation

[vagrant elbow]

do (total) - (cases where x and y are together)

is there no restriction on the number of students in the group

or the composition of these groups in terms of gender

[rain oasis]

ohh what i did was count the case were y is in the group then basically x wouldent be there then the case where y is not in the group so x would join

[vagrant elbow]

what about the cases where both of them aren't there

[rain oasis]

i came to ask about that

[vagrant elbow]

you still need to answer these questions

[rain oasis]

3 student one male president , spicific vice presidnt and a femal secutay

non spicific*

[split pike]

I would make the sample space

first

[rain oasis]

that was the very first question i already did

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[wraith mauve]

So I had this request in #help-34 about this:

***A farmer (A. Thomsen) has a number of organic chickens (chickens is the common term for hens and roosters). He wants to build a chicken coop for the chickens. A Thomsen has invested in 40 m of chicken netting, and he wants to build the chicken coop next to the 30 m long chicken coop, so that part of the chicken coop forms one side of a rectangular chicken coop.

a) Create a function that indicates the area and calculate how wide the chicken coop must be in order for the area to be as large as possible***

I have done the calculations, as seen in picture 2 and the function in the first, but I didn't get an answer on what the area should be, I think the area should be 20 because that is the parabola's last point, I guess.

[wraith mauve]

found the vertex using the vertex formula, which is given by (-b/2a , d/4a) and found the discriminant (b²-4ac. )

[cedar kiln]

@wraith mauve Has your question been resolved?

[wraith mauve]

<@&286206848099549185>

[wraith mauve]

<@&286206848099549185>

[wraith mauve]

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[sharp radish]

Anyone have a video explaining this concept? Or perhaps explain to me the process. I understand the rule " Remember that when you multiply like bases, you add the exponents, and when you divide like bases, you subtract the exponents." Just don't get how they are getting to the final product

[humble karma]

They're just factoring out common factors

[sharp radish]

I was focused on the "x and y are positive integers" Thought I could put in a number and try to figure it out that way.

Where did they get the minus 1 from?

[humble karma]

Just not a particularly reliable method

[humble karma]

You have a^7-a = a(a^6-1)

Whatever a is

Like the first term is just ((xy)^y)^7

[sharp radish]

Oh man I think I figured it out

So they factored out

Dropping 7y to 6y

And because it took everything on the right side its just gonna be 1?

Why is it not just empty and just the (xy)^6y?

close.

[idle tusk]

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[misty ravine]

Please check my work

[cedar kiln]

@misty ravine Has your question been resolved?

[misty ravine]

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[frigid quartz]

Knowing that
A = M(f, B1, B2), define B = M(f, B3, B4):

A = { (2, 0), (3, 1) }

B1 = B2 = {u1, u2}
B3 = B4 = {2*u1 + u2, u2}

[frigid quartz]

How do I go on about this?

I can't figure out how the diagrams works on this when there's an "f" that I don't know about

After hours of searching, gave up and came here to ask for help.
10 minutes later, figured out the solution

.close

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[cedar kiln]

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[wicked sphinx]

so

i solved the triangle

but it's different from the answer key

so i got a = 13.035 using the law of cosines

and then since we know both A and a, we can find C using A, a, and c

right?

we can now use the law of sines

using A, a, and c

but then after i do that

i get a wrong answer

oh wai

t

there's two possible angles

nvm

guys

yeah that makes snese

now

that my angle is different from the answer key

cuz there's two possible arc angles

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[tawny drum]

I have multiple problems

[cedar kiln]

@tawny drum Has your question been resolved?

[cedar kiln]

@tawny drum Has your question been resolved?

[tawny drum]

<@&286206848099549185>

I really need help with number 14

like rn

pls

someone

nvm i can do 14 only 15

pls someone help me with 15

i have no idea

[plucky merlin]

First find the circumference of both the rollers @tawny drum

then use the fraction of (circ of 21)/(circ of 14) and multiply that fraction by 360

[tawny drum]

why though

@plucky merlin

[plucky merlin]

if you take a circle and lay it out, it will have it's kind of like travel distance

so if you find the travel distance of both circles

get a percentage of the two travel distances (with respect to the in this case (circ of 21)/(circ of 14)) that gives you how many rotations the lower of the fraction makes (circ of 14) then multiply that by 360 degrees and it changes the units to degrees

@tawny drum

[tawny drum]

um

i dont understand the part where.u put them in a fraction

@plucky merlin

what is that suppose to giv eu

[plucky merlin]

like, "how many times more"

[tawny drum]

so why do u multi buy 360

[plucky merlin]

changes the units "from times more" to degrees

[tawny drum]

u cant just change random numbers to degrees though

[plucky merlin]

these aren't random numbers

[tawny drum]

why

[plucky merlin]

they will tell you the ratio of the circumference of the circles, if you place a point on the circle and spin the circle 360degrees, the circumference is that distance, so if you compare two circumferences, you can find the answer

[tawny drum]

Thank you sir

for your kind help

Much appreciated

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[plucky merlin]

I guess you could also, use the diameters for this as well, (21/14) * 360, finding ratios in things are funky

[cedar kiln]

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[crimson sedge]

how to calculate the area of this

[marsh lake]

expand it first>

?

[crimson sedge]

$28x^{2} + 49y^{2} = 784$

[wraith dagger]

Gary

[crimson sedge]

is the radius

28

[marsh lake]

wait

why 784

[fair geyser]

you just see that it makes the thing larger

[crimson sedge]

what

[fair geyser]

28 times larger

[crimson sedge]

28*16

[crimson sedge]

yess

[marsh lake]

hold on

[fair geyser]

and square it too

[crimson sedge]

yes

dilation

[fair geyser]

hm

[marsh lake]

oh sorry mb yes its 784

[fair geyser]

i'm confused

[marsh lake]

784 is 28^2

[crimson sedge]

yeh

so is the 28

radius

of that new dilated

disc?

[fair geyser]

yeah okay, no sqauring

it's not a circle anymore

it's not the radius

[marsh lake]

Oh its not a circle

its an ellipse

[crimson sedge]

how u guys know that

[fair geyser]

it's not totally obvious that twice longer has twice the area

[crimson sedge]

hm

the proportion the x and y inflates at is not

same

i think thats what leads to

eclipse

[fair geyser]

yes

[crimson sedge]

how to calculate the

area of such

[wheat holly]

Area of ellipse = $\pi \cross ab$

[wraith dagger]

Gowtham

[crimson sedge]

can u derive it? pls

[wheat holly]

that's a standard formula

[crimson sedge]

well.. i cannot just use it

[wheat holly]

then, what about integration

[crimson sedge]

thats hmmm like

area of circle with 1 area multiplied by

bunch of rectangles

[wheat holly]

no

its the area of ellipse

[crimson sedge]

i mean

yeh

may be the dilation and the

area are related

somehow

[wheat holly]

yeah dilation is the widening

here the circle is widened to ellipse

[crimson sedge]

when the dilation is in rectangle they just get multiplied with area

right

[wheat holly]

added with the dilated area to the org. area to get the new area

just like

the ellipse

and circle

[crimson sedge]

ahhhh

so circle area + dilated part area

[wheat holly]

but it has a different way to solve this

[crimson sedge]

OHHHHHHHHHHHHHH

WAIT

the original had the area of pi

nah nvm

[wheat holly]

i dont know abt integration but using standard results just use the area of ellipse

from org. equation

a = 7 and b = 4

[crimson sedge]

hm i will have to check out how the dilation works on circle

and how that formula is

derived

[wheat holly]

k

[crimson sedge]

thanks

[wheat holly]

btw

the area is

88 units squared

[crimson sedge]

nah dw i don't need it

[wheat holly]

ok bro

time to close?

[crimson sedge]

yep

thanks

[wheat holly]

k

[crimson sedge]

.close

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[past plank]

hey so im having trouble with a question regarding rational expressions and im wondering if anyone can help me understand how to solve this: 5+sqrt3x-5 = x

[livid hound]

what's the sqrt supposed to be covering

[flint plinth]

maybe start by canceling the 5 and -5 haha

[past plank]

it covers 3x-5

[livid hound]

isolate the sqrt, then square both sides

then solve the resultant quadratic equation
and check whether any of the solutions are extraneous

[past plank]

and to isolate it i subtract it from both sides?

[livid hound]

depends what you mean by "it"

[past plank]

the sqrt

[livid hound]

then no

[past plank]

oh wait subtract the 5

[livid hound]

that won't result in the square root being isolated / by itself

yes, you should subtract 5 from both sides

[past plank]

and then combine the like terms after i square both sides? i have 3x-5 = x^2 -25 now

[livid hound]

you didn't square the right side properly

[past plank]

oh wait +25 my bad

[livid hound]

still not correct

$(a\pm b)^2 \redneq a^2 + b^2$

[wraith dagger]

ℝαμΩℕωⅤ

[crimson sedge]

$5+ \sqrt{3x}-5 = x$

[wraith dagger]

Gary

[crimson sedge]

is this the problem

[past plank]

no the -5 is under the sqrt

[crimson sedge]

u have it

here

😅

[past plank]

lol yeah

[crimson sedge]

$5+ \sqrt{3x} = x$

[wraith dagger]

Gary

[crimson sedge]

u wanna get the x inside the sqrt out

so

[past plank]

-5 under the radical sign

[crimson sedge]

square both sides

[past plank]

$5+ \sqrt{3x-5} = x$

[wraith dagger]

Nate

[past plank]

this is the equation

[crimson sedge]

ahhh

well

still

same applies here

square both sides

[past plank]

but subtract 5 from both sides?

[crimson sedge]

yeh

[past plank]

ok which then has

$\sqrt{3x-5} = x-5$

[wraith dagger]

Nate

[crimson sedge]

yeh

[past plank]

so then square both sides which should be i think

[crimson sedge]

yes

do it

[past plank]

$3x-5 = x^2 -25$

[wraith dagger]

Nate

[past plank]

right?

[sweet shore]

do you know how to expand (x-a)(x-b)

[past plank]

FOIL

right?

[crimson sedge]

bruh

[sweet shore]

yeah

[past plank]

yeah ik how to do that

[crimson sedge]

$(a\pm b)^{2} = a^{2} + b^{2} \pm 2ab$

[wraith dagger]

Gary

[sweet shore]

but when you square both sides the RHS isn't $x^2 -5^2$ but rather $(x-5)^2 = (x-5)(x-5)$

[past plank]

oh

[wraith dagger]

mikak

[past plank]

ok that makes sense

ok so it becomes then $3x-5 = x^2 - 10x +25$

[wraith dagger]

Nate

[crimson sedge]

yeh

now u can figure this

out

pretty standard

quadratic

[past plank]

awesome thanks

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[crimson sedge]

i figure it out

[cedar kiln]

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[crimson sedge]

its like u have a inner circle

[crimson sedge]

of area pi then u multiply it with the amount by which

it expands

which gives pi * ab

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[vital matrix]

how would i find the derivative of this?

[kindred storm]

Use the chain rule.

[cerulean sail]

(and convert to radians first!)

[stark aspen]

differentiate the outside function and then multiply the derivative of the inside

[vital matrix]

right so it becomes sin(pi*x/60)

yeah so would the answer be (pi/60)*cos(pi*x/60)?

sorry i have no clue how to use that math typesetting bot

[wraith dagger]

Chai T. Rex

\[\frac{\pi}{60} \cos\qty(\frac{\pi}{60} x)\]

[vital matrix]

jesus 💀

thats a lot to write just for typesetting

[kindred storm]

Though you should check if it's really degrees. The degree symbol would generally be on a number and inside the parentheses for sine.

[vital matrix]

mhm

[cerulean sail]

[they may want the angle in your final answer to be in degrees]

[cerulean sail]

[I would think that they mean the whole 3x is a quantity in degrees]

[vital matrix]

so is this correct?
and in degrees it would be pi/60 * cos (3x)?

[kindred storm]

What I put is correct when the 3x is in degrees.

If it's radians, it's 3 cos(3x).

[vital matrix]

yep

yeah i think thats it

thank you @kindred storm @cerulean sail

[kindred storm]

You're welcome.

[vital matrix]

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[dire geode]

!redir

[cedar kiln]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[dire geode]

also uhhh <@&268886789983436800> i ain't watching that video

[vital matrix]

what is bro doing 💀

[celest seal]

i aint either

[wanton sail]

I mean it wasn't inappropriate but

Not exactly a math question

It was just a kid dancing

[livid hound]

while on crack

[wanton sail]

Okay I only watched the video, I didn't read

[livid hound]

any mention of that is clearly something not appropriate here

[cedar kiln]

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[rapid salmon]

having trouble with this one: i substituted z as x+yi into each, and put them into form tan-1(b/a) and tried to solve from there

but tan(-pi/2) is undefined so

[raw gulch]

notice that these two sets of complex numbers form two half-lines in the complex plane

first sketch the graph

then you easily notice where is the point z

[rapid salmon]

ok ill try that thanks

thanks it worked 🙂

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[modest dove]

Could someone help walk me through this question.

I have done the topic in the past, just really really rusty on it right now.

[idle tusk]

vieta's formula

[modest dove]

I don't recall doing anything like that

I have a list of rules in a reference sheet here

[idle tusk]

show?

[modest dove]

$\alpha+\beta+\gamma=-\frac{b}{a}$

[wraith dagger]

Jshy

[modest dove]

ive got the one with like aB+ + ay + By =c/a

aBy=-d/a

[idle tusk]

yep that's vieta

same thing

[modest dove]

Right, so how can I start with this question

[lyric narwhal]

Vieta is the name of the mathematician who discovered these relations

[idle tusk]

one of your roots, let's say gamma, is the sum of two of the other ones

so $\alpha + \beta + (\alpha + \beta) = -\frac{b}{a}$

[wraith dagger]

artemetra

[idle tusk]

simplify from here

[modest dove]

Right, i see that in the answer sheet of the question but i do not understand why this is the case

[idle tusk]

lol

that's the point of the task

[idle tusk]

so while in general it's not true, we are given that it's the sum of two other roots

[modest dove]

so are we saying that gamma = alpha + beta?

[idle tusk]

yes

[idle tusk]

^

[modest dove]

so are alpha, beta and gamma all considred "roots"

[idle tusk]

yes

[modest dove]

Right okay

$2(\alpha+\beta)=-\frac{b}{a}$

[idle tusk]

with a minus

before b/a

[wraith dagger]

Jshy

[idle tusk]

mhm

[modest dove]

right, yes

the -b/a

[idle tusk]

what's your b?

[modest dove]

comes from the rule?

the a + b + y = -b/a?

[idle tusk]

no

b is the coefficient in front of x

look at your equation

[modest dove]

Sorry I meant in general

how did we know to form the original equation

to equal -b/a

[idle tusk]

so how did we get to this?

is that what you are asking?

[modest dove]

How did we know to create this using that rule is what I am confused with

Is it just because the question asks for a sum

[idle tusk]

yes

[modest dove]

intuition tells us to use a + B + y = -b/a?

[idle tusk]

look

"One of the roots of the equation ... is the sum of the other two roots"

so let our roots be alpha, beta and gamma

and we can arbitrarily pick gamma=alpha + beta

there's no reason to pick gamma specifically

but we just want the least amount of variables possible

[modest dove]

right I understand

[modest dove]

So this would result in = -k/1?

[idle tusk]

mhm

now

[modest dove]

alpha + beta = -k/2 then

[idle tusk]

yes

and!

[idle tusk]

use this fact

alpha + beta = gamma

[modest dove]

So can we assume -k/2 = gamma?

[idle tusk]

yes

in fact it's not an assumption

it's true

[modest dove]

right

[idle tusk]

(in our problem)

[modest dove]

can we use the

[idle tusk]

and gamma is one of the roots, so one of the roots is -k/2 which is what was needed to show

[modest dove]

$\alpha\beta\gamma=-\frac{d}{a}$

[wraith dagger]

Jshy

[idle tusk]

so you are done for part i)

[modest dove]

yes

moving to part 2 sorry

didnt mention that

[modest dove]

this can find a×B right?

[idle tusk]

mm

i mean you can try

but consider factoring out (x+k/2)

[modest dove]

$\alpha\beta(-\frac{k}{2})=-1$

[wraith dagger]

Jshy

[modest dove]

so then aB = 2/k right?

[idle tusk]

yes

and alpha + beta = -k/2

[modest dove]

hmm

can we use the

$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$

[wraith dagger]

Jshy

[idle tusk]

again, try it

expand gamma

multiply everything out

[idle tusk]

and use that

[modest dove]

does c have a value in this case?

since the equation dosent feature a value with just x

[idle tusk]

yep

c=0

in general if you don't see a x^whatever term, that means that its coefficient is 0

[modest dove]

Looks right?

[cedar kiln]

@modest dove Has your question been resolved?

[modest dove]

Chcked with answers, looks right

[cedar kiln]

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[dim elbow]

How do I solve this y=7sin(π/30)(37)+11

[dim elbow]

It says the anwser is 6.32

[lyric narwhal]

,w 7sin(π/30)(37)+11

[civic coral]

calculator?

[wraith dagger]

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Requested by shikhargupta1207

[dim elbow]

THat's the anwser I got too

[lyric narwhal]

please send the original question in its entirety

[dusty hazel]

,w 7sin(37/30pi) + 11

[wraith dagger]

Results provided by WolframAlpha

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Requested by sakatayaksha

[dim elbow]

Wait so I have to multiple the 37 in?

[dusty hazel]

Where is the question?

[dim elbow]

Why though, shouldn't I do multuplication left to right?

[dusty hazel]

Shouldn't your question have that written?

[dim elbow]

Right here, Q11

y=7sin(π/30)(37)+11

[dusty hazel]

I do not see that question.

[dim elbow]

It's the ferris wheel question

I wrote the anwser down already

[dusty hazel]

Oh.

[dim elbow]

You got the anwser right here

I had y=7sin(π/30)(37)+11 and got 38

you did y=7sin(37π/30)+11 and got the right anwser

But why do you multiply the 37 in first

Shouldn't you multiply left to right>

[dusty hazel]

Because it's $y = 7\sin\pl (x-15) \f{\pi}{30}\pr + 11$

[wraith dagger]

! What the hell am I doing here?

[dusty hazel]

The x-15 is already multiplied INSIDE as the argument for the sine.

[dim elbow]

This is what the equation is and you plug in 52 for x

[dusty hazel]

You surely do.

[dim elbow]

Ohh th pie/30 is the B value

It was factored out from the x

[dusty hazel]

pi*

[dim elbow]

ye

[dusty hazel]

Precisely. It's factored.

The other factor is (x-15)

And it's still with pi.

As the argument.

[dim elbow]

I understand thanks

How do I find the rest of the values in question 1a

I found the first one in Quadrant 1

By doing
2x=30°
x=15°
first anwser is pi/12

But how is the second triangle 75° ? According to anwser key

I got the the second triangle is in Quandrant 2 and the terminal angle is 165°

@dusty hazel

[dusty hazel]

Well, sin(x) = sin(pi-x)

[dim elbow]

What? wdym

How am I supposed to find the second value?

[cedar kiln]

@dim elbow Has your question been resolved?

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[tidal helm]

Can someone help me find h

[final meteor]

you don't really need h to find the area F

[tidal helm]

Then how do I find F?

[final meteor]

do you know the formula area = 1/2*ab*sinC

[tidal helm]

We don’t have trigonometry yet

How can I solve this without trigonometry

[final meteor]

is the side c=9 given

[tidal helm]

No but I calculated it

Only 7cm and 4cm are given

[cedar kiln]

@tidal helm Has your question been resolved?

[cedar kiln]

@tidal helm Has your question been resolved?

[cedar kiln]

@tidal helm Has your question been resolved?

[raven shard]

(that is not correct)

[cedar kiln]

@tidal helm Has your question been resolved?

[tidal helm]

Why??

[raven shard]

um... i guess you could do this problem using similar triangles

[molten kiln]

Well there are a few ways you can do this

@tidal helm do u want me to give you one?

(without trig and such)

[molten kiln]

yeah pretty much!

I mean in the end, trig is all about similar triangles, and in this case you are already given the similar triangles you need in order to do the calculation so you don't need to rely on trig functions to sort of generate them for you

basically, add some more lines.

Notice that F is the same as half the area of that bigger parallelogram

So all you need to do is calculate the area of that parallelogram and divide it by two.

To calculate that area, you need to use similar triangles.

I added the angle a. As you can see, there are two triangles with this angle and those triangles are actually similar triangles, since they both have a 90deg angle.

[molten kiln]

We also know that the bigger of those triangles has one side of length 7. That means that y^2+h^2=7^2. To the right of it, there's a smaller triangle. It has one side of length 4 and one side of length h. That means that (x-y)^2+h^2=4^2. Together, those two triangles make one bigger, right-angle triangle, with area 7*4/2=14. The area can, however, also be expressed as x*h. So, we get the equation: x*h/2=14 <-> x*h=28 <-> x=28/h. Now as you calculated, x is 8.062..., so we can actually find h!

8.062...=x=(7^2+4^2)^0.5. h = 28 / (7^2+4^2)^0.5
Now that we know h, we are very close, because we can also find y. y^2+h^2=7^2 <-> y^2=7^2-h^2 <-> y = (7^2-h^2)^0.5 <-> y = (7^2 - (28 / (7^2+4^2)^0.5)^2)^0.5.
Now we know y. Here comes an important part. The triangle with sides 7, y & h can be scaled down to the other triangle with the angle a marked. The factor we scale it down by is 4/7, since the longest side in the larger triangle is 7 and the longest side in the smaller triangle is 4. This means that the bottom side of the smaller triangle is 4/7 * y. The area of the parallelogram is then 4/7 * y * x and the area, F, of the triangle we're looking for is half of that!!!

So 4/7 * y * x * 0.5 = 2/7 * x * y = 2/7 * (7^2+4^2)^0.5 * (7^2 - (28 / (7^2+4^2)^0.5)^2)^0.5 = 14. The answer is 14.

@tidal helm

[molten kiln]

In case the above is intimidating:

• The two triangles which I have written the angle a in are similar triangles. This means that if you take some number, k, then k multiplied by all the side lengths in a specific one of the triangles will give you the side lenghts in the other one of them.

• Since the longest side in the bigger triangle is 7 and the shortest side in the smaller triangle is 4, this value k can be 4/7. This means that if we multiply the length of all the sides in the bigger triangle with 4/7, we get the length of all the sides in the smaller triangle.

• We are looking for the bottom side of the smaller triangle, so first we need to find y, the bottom side of the bigger triangle using the Pythagorean theorem.

• Once we have y, we calculate the area of the parallelogram and divide it by two.

[cedar kiln]

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[digital goblet]

The answer is 5050, and I'm familiar with sign changing sequences but not when the terms are squared. How to solve?

[dire geode]

find a pattern with the first few integers before doing 100

-1^2 = -1, -1^2 + 2^2 = 3, etc.

[digital goblet]

Oh ok I see the pattern

The absolute value of the number increases by 2 then 3 then 4 and so on

.close

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[stuck hatch]

Normally the speed of sound is represented as c = 343 m/s. If the speed of sound were higher, what effect would it have on wavelength? How about period?

[warped coyote]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@stuck hatch Has your question been resolved?

[stuck hatch]

@warped coyote Sorry I was afk

@warped coyote May I dm you about this math problem? There are a couple other questions before it that might help answer it but I wanted to share it to get your opinion.

[warped coyote]

You can send them all here

[stuck hatch]

Okay

[warped coyote]

so,

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[warped coyote]

Awesome, let’s see your solution

[stuck hatch]

I have #1-3 done and wanted to get those checked. Then I need help with #4

[warped coyote]

Where does the 500 come from?

And I’m not sure about rounding 33.333 to 33, but otherwise 1&2 look good

[stuck hatch]

Ohh what would you round the 33.333... to?

[warped coyote]

For 4, notice that c=speed of sound, so if you draw different curves on top of the one you’ve already drawn how would they look compared to the first one? Ie how does lambda change if c is big or small holding f constant?

[warped coyote]

33.333 lol

[stuck hatch]

Oh cool thanks!

[warped coyote]

But you’re not necessarily wrong, but if youre supposed to use it later more digits is almost always better

[stuck hatch]

Oh okay I understand

[stuck hatch]

So now I did this: 344/33.333 = 10.3201032

[warped coyote]

Okay so that makes more sense to me for the x axis of your second curve

[stuck hatch]

Oh great! Should I round that answer?

[warped coyote]

Round it however your course recommends (but also, lambda=c/f implies lambda=ct=334*0.03=10.32 exactly)

[stuck hatch]

Ohh they didn't really give me a rounding method in this class I'm taking, so I guess I'll go with 10.32 to be safe

[warped coyote]

Okay then 1-3 look good to me, so for 4 the idea is how do T and lambda change if you change c

[stuck hatch]

Ohh hmm...

[stuck hatch]

Well it seems T and lambda could increase or decrease if c changes

[warped coyote]

Have you taken calculus?

Or would you be expected to use calculus on this assignment?

[stuck hatch]

I've not taken calculus. I'm in college for music, but I'm taking a class on Acoustics which has some mathematical stuff in it.

Basically the sound branch of physics

[warped coyote]

Okay then you can just think if you have: lambda = c/f holding f constant how does lambda change if you increase c?

[stuck hatch]

Lambda increases?

[warped coyote]

Correct

Which you can interpret how?

[stuck hatch]

Hmm, not sure

if I understand.

[warped coyote]

What is lambda? What does it mean for lambda to get bigger?

[stuck hatch]

Distance grows?

Oh wait

Wavelength increases?

[warped coyote]

Okay so put all that together

[stuck hatch]

Like write all that together?

[warped coyote]

What is the answer to the wavelength part of 4?

[stuck hatch]

The wavelength would increase

[warped coyote]

Because

The the speed of sound is increasing, holding frequency constant

[stuck hatch]

Ohhh okay

So, this?

The wavelength would increase because the speed of sound is increasing, holding frequency constant.

[warped coyote]

So for the last question: can you write an equation for t in terms of c?

[stuck hatch]

Hmm

[stuck hatch]

I don't think I understand the last question

[warped coyote]

The question is asking how does c affect t, so how do you calculate t in terms of c?

[stuck hatch]

Hmmm

[stuck hatch]

Okay so I know period is T = 1/f

[warped coyote]

Yep, keep going

[stuck hatch]

And I know c = 344 m/s

@warped coyote I think I'm still confused

[bronze crag]

@stuck hatch "write T in terms of c" means that you have some equation with T by itself on one side and c on the other side (with other stuff): T = ...c.... In this case you have lambda = c/f. Then you can use T=1/f to re-write it as : lambda = c*(1/f), lambda = c*T. Then you 'solve for T by isolating T on one side: T = lambda/c

So with that equation you can see that as c gets bigger, T gets...

[stuck hatch]

Hmm, T gets bigger too?

[bronze crag]

1/10, 1/100, 1/1000...

is that getting bigger or smaller

[stuck hatch]

Ohh smaller

[bronze crag]

Yep! I Don't know anything else about your problem, I was just chiming in to explain the "solve for T" thing

[stuck hatch]

Cool. Thank you for explaining that. Lol I'm in acoustics for a music class, but I'm trying to figure out some math stuff here and there.

[cedar kiln]

@stuck hatch Has your question been resolved?

[bronze crag]

Was there more to your question?

[cedar kiln]

@stuck hatch Has your question been resolved?

[cedar kiln]

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[royal loom]

Is the laplacian the same as the laplace operator?

[crimson delta]

https://letmegooglethat.com/?q=laplacian

[royal loom]

I did

it said something about

discrete vs finite or something

nothing helpful

[crimson sedge]

Yes, the laplacian and the laplace operator are the same thing, They are both mathematical operators that do the same thing that calculate the sum of the second partial derivatives of a function in multiple dimensions

[royal loom]

I don't know what the laplacian is, and some questions ask me to compute it. In class all we did was the laplace operator is the 2nd x, + 2nd y, etc. partials

[crimson delta]

same thing

[royal loom]

Thanks

Ty

[crimson delta]

first sentence of wikipedia

[royal loom]

I got this u hater

[crimson delta]

actually click on the article

[sacred grail]

the discrete laplacian is certainly different from the continuous laplacian

[crimson sedge]

but there's a discrete version of the laplacian, often used in image processing and numerical analysis ig

[royal loom]

yeah idk why it started talking about that

[crimson delta]

there are discrete version of all differential operators

which you get by just replacing the limit with a difference quotient

[royal loom]

isn't the difference quotient a limit

lim h->0 (f(x+h)-f(x))/h

[crimson delta]

well I mean just the second part of that

aka the quotient of the differences

f(x+h)-f(x). and x+h - x

[royal loom]

ah

[crimson sedge]

https://en.wikipedia.org/wiki/Difference_quotient

[royal loom]

.close

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[warm crescent]

I'm asked to find all matrices A in R2x2 s.t A^2 = 0

[warm crescent]

book's solution is the following

however, I got

[bold hinge]

There is a special word

"Nilpotent"

[warm crescent]

i meant bc <= 0 fkwokfo

So what I would like to know is if I'm missing some "sets" of solutions described by the book's answer

we are denoting A by A = (a b, c d)

[dire geode]

Show your work

[cedar kiln]

@warm crescent Has your question been resolved?

[warm crescent]

I just noticed after assuming c = 0 or b = 0, while a^2 = 0 and d^2 = 0 for both cases, neither c = 0 implies b = 0 or the same but swapping c and b. Then I would also have (0 b, 0 0) and (0 0, c 0) as solutions. Luckily I still got those 'solutions' inside the sqrt's solutions letting c = 0 or b = 0

I guess it is correct then. But why I didn't miss those two solutions even if I had skip that step accidentally?

[dire geode]

I don't follow the whole if this variable equals 0 then these equations stuff

You have 4 equations and 4 unknowns

Might be equivalent, but just more organized

[cedar kiln]

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[warm crescent]

oh

[warm crescent]

is it because of what is written in red ?

yeah I think so

.close

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[acoustic delta]

Hey, can someone teach me how to solve this

[peak minnow]

switch tan into sin/cos

and then simplify

[cedar kiln]

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[uneven pagoda]

K is a finite field, N are natural numbers
The task is to prove that the basis of K^N is not countable.

N is infinite and countable. K^N is not countable but why do we need more then N vectors in the basis?

[cedar kiln]

@uneven pagoda Has your question been resolved?

[cedar kiln]

@uneven pagoda Has your question been resolved?

[cedar kiln]

@uneven pagoda Has your question been resolved?

[cedar kiln]

@uneven pagoda Has your question been resolved?

[uneven pagoda]

<@&286206848099549185>

[uneven pagoda]

hard question, i agree

[uneven pagoda]

<@&286206848099549185>

[pine grove]

Can’t help, apologies.

[tulip briar]

i can

[uneven pagoda]

help pls

[eager quest]

WHaT

im in 8th

but kinda dumb

[haughty violet]

Hey im in 6th

[uneven pagoda]

It is enough to prove that the amount of decimals of an irrational number is not countable

[uneven pagoda]

Which isn't true, i guess

idk

[hot isle]

im not sure about your original question (i only have a very approximate thought) but ur right in that the amount of decimals of n irrational number is countable

cuz theres a bijection from that to the natural numbers, which is countably infinite

[harsh plume]

hello?

[cedar kiln]

@uneven pagoda Has your question been resolved?

[worldly chasm]

@uneven pagoda the key is that a basis can represent any vector in the field with a finite linear combination of basis vectors.

So we can consider a potential countable basis for the vector space: {e1, e2, ...} And then construct a vector in the space that requires an infinite number of the basis vectors to represent. This vector is linearly independent of the e_i but part of the space, but that contradicts what it means to be a basis, so our assumption must be faulty, so there is no countable basis of this vector space.

[cedar kiln]

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[sinful bear]

m_1 not possible because of 0.1 term

m_2 not possible because variance would be negative

m_3 possible

is that correct?

[raw gulch]

M(0) always exists and is equal to 1

so that explains which one you have to select

[sinful bear]

yes im saying m_1 not possible because first term is 0.1

and m_2 not possible because E(X) = -3, E(X^2) = 8 which is also not possible because then variance is -1

is that correct logic>

[raw gulch]

we have such form of this series:

so that confirms what you have written

[sinful bear]

Ok thanks

[raw gulch]

yw

[sinful bear]

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[fallen moat]

e^(cos(x)+ln(x))
just checking

[slim quiver]

ln(x) is part of the exponent?

[tropic oxide]

differentiate*

but show work for both methods

[brave aspen]

IMO easiest way to view e^f(x) form derivative is:

d/dx[ e^f(x) ] = f'(x) * e^f(x)

In other words take derivative of the exponent and multiply it back through

[tropic oxide]

and the second?

[cedar kiln]

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[bleak axle]

How would I solve this one

[tropic oxide]

do you know what a recursive definition is

@bleak axle

[bleak axle]

no

[tropic oxide]

$$a_n = \begin{cases} 3, & n=1 \ a_{n-1} + 2, & n > 1 \end{cases}$$

[wraith dagger]

Ann

[tropic oxide]

this says, in words:

"the first term of our sequence is 3, and each term after that is equal to the previous term plus 2"

do you understand?

[bleak axle]

no

like, almost but not fully

[tropic oxide]

what do you not understand about the line i said there

[bleak axle]

like, what would happen if you change the first 1 to a 2

would that mean it becomes and each term after that is one less to the previous term plus 2

[tropic oxide]

you would have $a_2 = 3$ (the second term would be 3, instead of the first), \textbf{the recursive rule would be unchanged}, and the first term $a_1$ would be left without a definition.

[wraith dagger]

Ann

[bleak axle]

ok

[tropic oxide]

did i clear up your confusion?

[bleak axle]

yes

[tropic oxide]

does any more confusion remain?

[bleak axle]

so then what happens next

[tropic oxide]

you want to find the first three terms

that is you want to find a_1, a_2 and a_3

the first term of our sequence is 3, and each term after that is equal to the previous term plus 2

do it

[bleak axle]

so like, a2 - 1 + 2 = 3 + 2 = 5

and then a3 - 1 + 2 = 5 + 2 = 7

[tropic oxide]

bad notation

when you write \verb|a2 - 1| it reads as $a_2 - 1$ and not as $a_{2-1}$ which you intended. you need to write \verb|a_(2-1)|

[wraith dagger]

Ann

[tropic oxide]

but yes.

[bleak axle]

got it

[cedar kiln]

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[crimson sedge]

can someone help me on where i shoud start

dunnowhere to begin

[hollow trail]

start by thinking about quantities that are conserved (energy, momentum). Which quantities are conserved and when?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[open mauve]

What does it mean, how does it come from, I mean lines 2

[open mauve]

If you are curious about the whole Q

<@&286206848099549185>

[idle tusk]

what exactly, $1-\frac{1}{2n+2}$?

[wraith dagger]

artemetra

[open mauve]

Ratio test

Ratio test of series an

[idle tusk]

no i get that

[idle tusk]

"what does it mean" what exactly?

can you underline or circle?

[open mauve]

I mean how did it turn into 1-(p/2n+2) + 0(1/2n+2) as n->inf

[cedar kiln]

@open mauve Has your question been resolved?

[cedar kiln]

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[hollow pier]

im stuck here and idk how to combine like terms

[hollow pier]

idk how to find the center and radius

[clear umbra]

complete the square for x and y

[hollow pier]

the formula is (x-h)^2 + (y-k)^2 = r^2

my only problem is how to combine like terms

[clear umbra]

from those equations above?

[hollow pier]

yes

[clear umbra]

theres no like terms to combine there

[hollow pier]

there is

i arranged the #5 this is the result "x^2 +4x + y^2 -2y = 20"

[hollow pier]

how to combine this to make it like the formula here

[clear umbra]

you need to complete the square for x and y

[hollow pier]

how

[clear umbra]

do you know how to complete the square

[hollow pier]

give hint

[clear umbra]

x^2 + 4x = (x+something)^2 - something

[hollow pier]

simplfying?

is it equal to (x+2)^2? am i right?

[clear umbra]

ok good

but theres a +4

how do you get rid of it

[hollow pier]

i dont know

[clear umbra]

(x+2)^2 = x^2 + 4x + 4

you want x^2 + 4x only

[hollow pier]

ah

can it be x^2 +4x + 2^2?

[clear umbra]

thats just (x+2)^2

you still have that 4 there

[hollow pier]

using foil method?

[clear umbra]

ok see here

(x+2)^2 = x^2 + 4x + 4

we only want x^2 + 4x

what do you do

ever considered moving the 4?

[hollow pier]

im confused

[clear umbra]

you have a +4 term on the right

[opaque root]

If you have A + B but you one want A

[clear umbra]

to remove it what do you do

[opaque root]

What do you do

[clear umbra]

or here i'll ask you this

a + b = c

i only want a on the left

[hollow pier]

move the 4 to where?

[clear umbra]

you have a left and a right

the 4 is on the right

where else can you move it

[hollow pier]

left

[clear umbra]

and what happens to the 4 if you move it left

[hollow pier]

nega

tive

[clear umbra]

so what do you get overall

[hollow pier]

-4x + x^2 ?

[clear umbra]

(x+2)^2 = x^2 + 4x + 4

move the 4

what do you get

[hollow pier]

bro i just want to get how to combine terms

[clear umbra]

theres no terms to combine there

you cant combine x^2 and x together

the best you can do is to complete the square

to get to the equation you want

[hollow pier]

ok

i get it thanks

got

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

Consider a triangle ABC whereAC =12 , CB=7 , and the measure of angle BAC =25 degrees.Find the perimeter

[crimson sedge]

oh and angle BAC = 25 degs

[tropic oxide]

no need to say it twice

!status ?

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[crimson sedge]

right sorry

So I have an answer and I would like to get my work checked

[tropic oxide]

show

[crimson sedge]

so if I use the cos rule

$$AC^2 + AB^2 - 2(\cos{25^{\circ}})(AB)(AC) = BC^2$$

[tropic oxide]

BC**^2**

[wraith dagger]

Al-Mardhikwar

[tropic oxide]

yeah

[crimson sedge]

Yeah fixed it

I should get the third side

[tropic oxide]

mhm

[crimson sedge]

ANd if I add all three up, I should get the perimeter

[tropic oxide]

yes

[crimson sedge]

so my eqution is :

$$7^2 = 12^2 + m^2 -2(m)(12)(\cos{25^{\circ}})$$

[wraith dagger]

Al-Mardhikwar

[crimson sedge]

Which when simplified becomes

[tropic oxide]

hold on

oh

oh hold on we are in trouble i think

[crimson sedge]

Why what happened?

[tropic oxide]

this is your triangle right?

[crimson sedge]

yup

that would be righ

[tropic oxide]

yeah AB will have two possible values for its length...

[crimson sedge]

ohhhhh

$$m^2 - 24\cos{25^\circ}m + 106 = 0$$

[wraith dagger]

Al-Mardhikwar

[crimson sedge]

here m is the length of AB

,w plot m^2 - 24\cos{25^\circ}m + 106 = y

[wraith dagger]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by wasif_al_turki

[minor crystal]

,w solve m^2 - 24*cos(25 degrees)m + 106 = 0

[crimson sedge]

right thanks

[crimson sedge]

right

So i take the smaller value in this case

Because I need to find the smallest perimeter according to the question

so that should be

,w 7.37 + 7 +12

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by wasif_al_turki

[crimson sedge]

26.37

Correct?

[minor crystal]

where did it say smallest perimeter?

[crimson sedge]

oh the question says that in fine print

at the very bottom

😓

I just saw it

so yeah would my answer be correct?

@minor crystal

@tropic oxide

Thanks for your help btw

.close

[cedar kiln]

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[minor crystal]

yes

[minor crystal]

.close

[cedar kiln]

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[frail violet]

Is there a pattern to primes?

[tropic oxide]

what do these colored arrows signify

[frail violet]

5+7=12
13+17=30
73+97=170
173+197=370
inbetween e.g 29<->31
i wonder whether theres a pattern to the col/rows and the extra spaces/omitted number

maybe there's an already existing method that i'm not aware of

[tropic oxide]

this feels kind of coincidental?

[frail violet]

could be, that's why i'm asking whether there's any pattern that exists

[tropic oxide]

not that i personally know of.

[cedar kiln]

@frail violet Has your question been resolved?

[cedar kiln]

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[fervent jackal]

I hear that to make sure W is a vector space 1st we must check if for all scalar lambda, the product of the scalar with all vectors in W must be in W too. And for any two vectors in W, the result must be in W too. But I noticed in my notes they said that if W doesn't pass the origin then it is not a vector space. I'm kind of confused and would like a proof. Thanks

[fervent jackal]

Oh hi Ann

[tropic oxide]

if W is not empty, and is closed under addition and scaling, then it contains the zero vector.

[fervent jackal]

is it some sort of sigma algebra property?

ik it might not be relevant but yk

[tropic oxide]

it is not relevant in the SLIGHTEST!!!

for any vector v in W you have -v in W, and thus v + (-v) = 0 in W.

so if your set does not contain the origin a.k.a. the zero vector, then at least one of those closure properties must fail.

[fervent jackal]

wait let me try to think about/experiment it

oh that makes sense

for any vector v since it is closed under scalar multiplication, then -1 x v must exist hence by the closed under addition property v + (-v) must exist too

which is the O vector

okay okay I get it

thanks

.close

[cedar kiln]

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[sweet ridge]

Ummm

[sweet ridge]

Im laughing rn