#help-0

1 messages · Page 370 of 1

summer quarry
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Eh

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It equals 0?

tall topaz
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,calc 3^6 -81*3-486

ocean sealBOT
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Result:

0
tall topaz
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(this means (x-3) is a factor of the numerator)

summer quarry
#

Eh

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(x-3) still equals 0

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0/0 still doesn't make sense

tall topaz
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?

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x-3 is a factor of the numerator so factor out x-3

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do horner scheme or whatever

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in fact

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,tex \polylongdiv{x^6 -81x-486}{x-3}

ocean sealBOT
tall topaz
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,tex \polyhornerscheme[x=3]{x^6 -81x-486}

ocean sealBOT
summer quarry
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On god

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Ty ty

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,rotate

ocean sealBOT
summer quarry
#

Something like this???

tall topaz
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Yep

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Now just cancel x-3 and plug in x=3

summer quarry
#

And I can get rid of that (x-3)

white fulcrum
#

where do i get help

tall topaz
white fulcrum
#

thankss

summer quarry
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.close

lone heartBOT
#
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neat totem
#

hi

lone heartBOT
neat totem
#

ye

rancid radish
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what is quartile 3 of 10,19,21,28,34,37,40,42,43,43,45,50,54,56,61,65,70,88

neat totem
#

we struggling to know if quartile 3 is equal to 56 or 58.5

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10,19,21,28,34,37,40,42, 43,43 ,45,50,54,56,61,65,70,88
Quartile 1 quartile 2 quartile 3

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or

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10,19,21,28,34,37,40,42,43 43 43,45,50,54,56,61,65,70,88
Quartile 1 quartile 2(median) quartile 3

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we search online and results vary on each site

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ox ajas

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this is box and whisker plot

lone heartBOT
#

@neat totem Has your question been resolved?

neat totem
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no

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@rancid radish

rancid radish
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<@&286206848099549185>

neat totem
#

@compact mirage

lone heartBOT
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@neat totem Has your question been resolved?

neat totem
#

NO

lone heartBOT
#

@neat totem Has your question been resolved?

buoyant summit
#

help

lone heartBOT
#

@neat totem Has your question been resolved?

lone heartBOT
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river badge
#

how do u make an equation out of this

One of the zeros at -2, y-axis is the axis of symmetry and the value of a is 2.

barren portal
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What's the standard form of a parabola you have?

storm ridge
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What's a?

river badge
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2

river badge
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standard
Vertex Form
roots of x

lone heartBOT
#

@river badge Has your question been resolved?

storm ridge
#

(Sorry for the ping)

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twin jewel
lone heartBOT
twin jewel
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I don't know what to do

lone heartBOT
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@twin jewel Has your question been resolved?

lavish cave
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like, what's the x-value

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and what's the y-value

twin jewel
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97,2

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(7,2)

lavish cave
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7 is correct (wait, your x and y are switched)

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but try again, 2 is not correct

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you might need to click on the image to make it bigger

twin jewel
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1 i mean sorry

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lol

lavish cave
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your x and y are switched actually

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if we go to x = 1

twin jewel
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yes i just realised that too lol

lavish cave
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we draw a vertical line all the way up

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yeah ok

twin jewel
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ok

lavish cave
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now you can substitute x = 1 in

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2 * 1 + 5 and -1 + 8

twin jewel
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7 and -8

lavish cave
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shit sorry for the typo

twin jewel
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ok lol

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so that would be 7 and 7

lavish cave
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yes

twin jewel
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so thats the anwser?

lavish cave
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yeah

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they're the same

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and you should find that part a also gives 7

twin jewel
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soo sec

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how do i fit in 7

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where i mean

lavish cave
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the y value is 7

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for both parts in b)

twin jewel
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ok#

lavish cave
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and as we found before, x = 1 and y = 7

twin jewel
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v

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bruh

lavish cave
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it's the same as

twin jewel
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ok

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ty for help

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.close

lone heartBOT
#
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alpine sable
lone heartBOT
alpine sable
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I'm stuck on both of them

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I'm shit at maths so

lavish cave
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and circular sector is 60/360 * area of circle

still ibex
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but the cuboid is asking for surface area, not volume

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you could just calculate the area of each side of the cuboid and add them up

lavish cave
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shit

alpine sable
stark crater
alpine sable
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Oh thanks

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Idk the area of the sector one tho

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Idk how to do it

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I havnt done that in ages

stark crater
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If it was a full circle could u find it

alpine sable
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Probably

stark crater
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That is step 1 so start by finding that

alpine sable
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Alr

turbid sigil
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Think that a full circle has radian 2π, and area of a full circle is πr^2, so what is the area of part a circle

alpine sable
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No clue 😭

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Wait how do u get the pie symbol

turbid sigil
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You have to use the Google keyboard

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Download it

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Or latex if you know what that is

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$\pi$

ocean sealBOT
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cali5nia

stark crater
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if ur on pc u can hold alt then press 2 2 7

alpine sable
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I'm on phone

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But oh well

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Ik the area of a circle is pie x radius²

turbid sigil
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2π•1/2•r^2 is area of a full circle

alpine sable
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But idk how u get a sector of a circle

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Oh

turbid sigil
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Where the angle is 2π radians

alpine sable
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So how do u find a sector

turbid sigil
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I don't want to spoil it😭

alpine sable
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Dw

stark crater
turbid sigil
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You can Google it if you really care to

alpine sable
turbid sigil
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But it makes sense if you think about it

stark crater
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you are given a slice with 60 degrees

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How many slices do you need to form a full circle

alpine sable
alpine sable
stark crater
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yeah

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thats how many slices would form a full circle

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you only have 1 slice though, when a full circle is 6. So the area is going to be 1/6 of if it was a full circle

alpine sable
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Oh

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That's smart

stark crater
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When you have something that looks complex, see if you can relate it to a more simple shape (squares,triangles,circles) and find a relationship between them

autumn juniper
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Sorry for sneaking in , just wondering if any of you are free to help , thanks.

alpine sable
stark crater
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yeah but to be precise its not recurring if you use the true value of pi, it will be more random

alpine sable
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So should I put it in terms of pie?

stark crater
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Yeah you usually put the exact value unless they ask otherwise. In this case that means putting in terms of pi

alpine sable
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Ok thanks for all the help!

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My calculator says in terms of pi it's 27/2 pi

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Do I do 27÷2 or do I just leave it at 27/2 pi

stark crater
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both are fine. I believe most people would leave it in fraction form though so 27/2pi

lone heartBOT
#

@alpine sable Has your question been resolved?

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calm helm
lone heartBOT
calm helm
#

precalc

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no clue where to start

fervent timber
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just plug stuff into the scenario

calm helm
fervent timber
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it says in the question

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organize the info if it's confusing

calm helm
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wait no i still dont get it

echo geode
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You have h, all you need is to evalute y0 y1 y2 by inserting the values of points

calm helm
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looking at the example provided and i dont understand where they get 6 from

echo geode
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consider each point coordinates as (x,f(x)) it would help

alpine sable
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How do you get one of these bots to help you? I did the tutorial it’s not working😭

lone heartBOT
#

@calm helm Has your question been resolved?

lone heartBOT
#

@calm helm Has your question been resolved?

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sacred plume
#

I've been struggling with #5 and idk how, please someone tell me what I'm doing wrong or misunderstanding

sacred plume
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Here's my work and if you need anything else I'll be more than willing to give it

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I thought I did everything I was supposed to am I missing smth or

alpine nacelle
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the problem comes from P and P^-1 (more precisely, first column of P seems very dubious, 2nd seems ok)

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you should read your lesson about how to find P again

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idk your method but something here is wrong

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you should get
0
1 instead to have a valid column

lone heartBOT
#

@sacred plume Has your question been resolved?

sacred plume
alpine nacelle
#

the columns of P have to be a basis of eigenvectors for the space
you first column (1, 0) is not an eigenvector for the eigenvalue -1 (2nd column is eigen for 1 so it's ok)

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A*(1, 0)^t = (1, 6) not (-1, 0)

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which is why your P doesn't work

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but with (0, 1) instead it works bc A maps it to (0, -1)

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the part I screenshot is where a problem occurs

sacred plume
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Okay I got it now tysm T_T

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dusk umbra
#

ok new question

lone heartBOT
dusk umbra
#

can you have a commutator between a vector and a scalar?

mortal trellis
#

well how should that look like

lone heartBOT
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@dusk umbra Has your question been resolved?

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tranquil cipher
#

hello 🥲 im sorry if its not exactly math related (more on physics) but im really desperate rn and maybe someone here understands how this works.. does anyone know how he got the latter equation from the first equation (conservation of momentum equation)? like how he manipulated it to be like that, thank you vm!!

tranquil cipher
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KE is 1/2mv^2 and PE is mgy

outer lark
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is this part of a bigger problem?

tranquil cipher
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yes it is

outer lark
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can u post that please ? :D

tranquil cipher
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should i send the entire problem

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thank you so much!!! sure sure

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here is the problem

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and the questions are: A. Write the equation for the conservation of momentum in the inelastic collision of the bullet and the woodblock

B. Suppose that a very massive bullet (where m>>M) is fired at the wooden block pendulum and still resulted to an inelastic collision. Determine the speed of the wooden block immediately after collision.

C. At what height y will the wood block pendulum rise after the collision? Express your answer in terms of the object masses and the velocity of the bullet.

D. Suppose that we fire a bullet with mass m=10g at a speed of Vb= 400 m/s into a wooden block with mass M=5.85kg and suspended by a taut string of length L= 75.0cm. By how much height y will the wood block rise? Express your answer in cm units

E. If we carry out the ballistic pendulum test described in D. On the surface of the moon where the acceleration due to gravity is 1.63m/s^2, how many times would the measured height differ from what is obtained on the surface of the earth?

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i already have answers for A and B (though im not sure if theyre correct) and currently working on C

outer lark
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this is combined mass

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and initially we have maximum kinetic energy

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and no potential

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and we transfer all of that kinetic into potential

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giving it new height y

outer lark
tranquil cipher
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so its just kinetic energy of both masses in the left hand side of the equation?

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combined kinetic energy of the wooden block and bullet?

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then on the right hand side its just potential? combined potential energy ?

outer lark
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the bullets kinetic energy is responsible for moving (m+M)

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and yes it transitions all kinetic energy to potential

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at a maximum y

tranquil cipher
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but i kinda dont understand why it will be merged on the initial states since only the bullet is moving and the wooden block is stationary still the collission didnt happen yet

outer lark
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well it starts moving the instant they collide

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aka when they combine

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that is your initial moment

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final is at the peak y

tranquil cipher
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ohh wait the question is what height will be wood pendulum rise after collission i see

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so the initial is the very moment they collide

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and the final is the heighest it will go

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highest*

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Okok i seee

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for question D, it has a given value of 75 cm for the length a taught string, where will i plug that value in?

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do i need to use trigonometry for this

lone heartBOT
#

@tranquil cipher Has your question been resolved?

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gusty ocean
lone heartBOT
gusty ocean
#

I need help please

#

@

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<@&286206848099549185>

harsh crypt
#

?

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whats is the problem?

gusty ocean
#

I need help to understand and get better at doing these questions

harsh crypt
#

but is coted off on the right

gusty ocean
#

ok ill send another

harsh crypt
#

if every mile have expence of 2.25, and 20 per hour
in 1 day it haved 50 mile per hour and the gas and truck maintence is 147.50

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first we make the 2.25 per mile and add 20 per hour

gusty ocean
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so h =hour

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and g for gas/mantence

harsh crypt
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wich 2.25*5 is 10.25

harsh crypt
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now we add the 0 we removed giving 102.50

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then add 20 with become 122.50

gusty ocean
#

mhmm

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ok

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so the system of equations would be

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20 +2.25

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smth smth smth

harsh crypt
gusty ocean
#

no

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its the cost to operate

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gas plus manteniece

harsh crypt
#

oh

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22.25

gusty ocean
#

do u know wt the equations would be

harsh crypt
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i think is *50

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but im not sure

gusty ocean
#

wdym

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arent the equations like

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50h + 1457.50g

harsh crypt
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wait

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need make a system to dertime number of hours droved and the mile the truck drived

gusty ocean
#

yes

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but can u explain how to do that

harsh crypt
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here

gusty ocean
#

but 1457.50 is only for that one day he drives 50$ per hour I thought

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?

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and dont u mean g instead of m

harsh crypt
gusty ocean
#

ohhh

#

bruh

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i put g instead of m and I got it wrong

harsh crypt
#

ended?

gusty ocean
#

yeah

harsh crypt
#

use .close

gusty ocean
#

.close

lone heartBOT
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late token
#

can u put variable after a metric unit

lone heartBOT
tacit arch
lone heartBOT
# late token can u put variable after a metric unit

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

late token
#

35cmx for example

tacit arch
late token
#

35cm multiplied x

tacit arch
late token
#

a variable

ionic jewel
#

no you can't

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well you can but you shouldnt

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you shouldn't have the cm be in your problems at all tbh

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but if you do you need some indication of multiplication that's not concat

ionic jewel
#

concatenation

late token
#

.close

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granite jetty
#

How exactly do I figure out what to use for the starting guesses?

granite jetty
#

do I use a positive and negative number?

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like 2 random numbers such as -100 and 100

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I also set it to 0 first so i'm using

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x^8 - x - 3

lone heartBOT
#

@granite jetty Has your question been resolved?

granite jetty
#

nvm

#

.close

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steady tiger
#

Optimization Problem

"A cylindrical tank with a given volume V is to be made without a lid. The material for the bottom costs five times as much per unit area as that for the sides. What should be the ratio of the height to the radius for minimum cost?"

How do I solve this optimization problem? I have formulated an equation for its total cost:

C = 2πrhx + 5πr^{2}x

mighty sinew
#

Have you tried defining some variable a = h/r and somehow differentiating w.r.t it?

steady tiger
#

No i haven't tried that but am I right for getting the cost function???

#

I've tried deriving h from the volume of the cylinder:

V = πr^{2}h

h = V / πr^{2}

Then substituting h into the cost function

#

and arriving at:

C = 2Vx/r + 5πr^{2}x

#

<@&286206848099549185>

dense wigeon
#

i found a b and d but i need help with finding c and e

steady tiger
#

<@&286206848099549185>

simple grotto
#

yes

lone heartBOT
#

@steady tiger Has your question been resolved?

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wide onyx
lone heartBOT
wide onyx
#

,rotate 270

ocean sealBOT
wide onyx
#

how do i find trig ratios

#

for this

steady tiger
wide onyx
steady tiger
#

in this case u can substitute the angle in radians 2π / 9

into the trig functions like sin(x), cos(x), or tan(x)

wide onyx
#

just do sin(2pie/9)

steady tiger
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#

@wide onyx Has your question been resolved?

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little turret
#

hello

lone heartBOT
little turret
#

how do we know that this limit approaches 1?

static kelp
#

just replace dt with x

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and try to make a graph

#

it comes out to be something like this

little turret
#

is there anyway to do it without just plugging in smaller and smaller numbers for dt?

static kelp
#

yeah that is what we call getting a limit of a function right

#

you plug in closer and closer values to 0

#

and see how the function behaves

little turret
#

hmmm okay

static kelp
#

but this is a standard limit right

#

so its better to remember it rather than deriving it from time to time

#

samjha bhai?

naive sorrel
static kelp
#

yes

#

as it is a 0/0 form

little turret
#

so to take the derivatave of the top function i would have to know that already

static kelp
#

oh yeah okay

hasty creek
#

$\frac{e^{dt}-1}{dt} \\$ $x=e^{dt}-1 \\$ $\ln\left(x+1\right)=dt\\$ $lim_{x\to0}\frac{x}{\ln\left(x+1\right)}\\$ $lim_{x\to0}\frac{1}{\ln\left(\left(x+1\right)^{\frac{1}{x}}\right)} =\frac{1}{\ln\left(e\right)}=1$

ocean sealBOT
#

Combustion

static kelp
#

for that you need to know the first derivative rule for limits

#

@hasty creek yo how did you do that?

hasty creek
#

the steps are there lol

wintry mango
#

I have three methods:
1.pinch theorem:1/(1-x)>e^x>1+x
2.Lhopital:0/0
3.taylor expansion

hasty creek
#

the last step depends on $lim_{x\to0}\left(\left(x+1\right)^{\frac{1}{x}}\right)=e$

ocean sealBOT
#

Combustion

little turret
hasty creek
#

x + 1 = e^dt

#

take ln of both sides

little turret
#

oh yeah

#

thanks everyone

#

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primal warren
#

how can i show this limit doesnt exist? lim x->0, y->0 of sin(xy)/(x^2+y^2)

static kelp
#

as this type of limit is in the form of f(x)/g(x) where lim of g(x) tends to zero but not the same for f(x)

#

hence we can say that limit of the given function in totality does not exist for x,y tending to 0

left isle
#

,rotate

ocean sealBOT
wintry mango
#

by calculating the limit from two directions,and the results not same,we can prove that

#

for example:from y=x and y=2x

#

and the results is
0.5 not equal with 0.4

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dire gale
#

An equipment consists of nine components, each of which will independently fail with a probability of p. If the equipment is able to function properly when at least six of the components are operational, what is the probability that it is functioning properly?

dire gale
#

is p = 0.5 and q = 0.5?

lone heartBOT
#

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dire gale
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static panther
#

hi my question is find arctan(tan(3))

lone heartBOT
static panther
#

is the answer 3 - pi?

#

3 > pi/2

#

so if we subtract pi it is between [-pi/2, pi/2]

tacit arch
#

,calc atan(tan(3))

ocean sealBOT
#

Result:

-0.14159265358979
static panther
#

oh so it is 3-pi

novel lake
#

is the 3 degrees?

static panther
#

the question doesnt specify

novel lake
#

oh okay then 🙂

static panther
#

.close

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blazing skiff
#

Hey gang, so we're doing some trig and I got this question correct except for tan, I'm not sure why you can turn "1 over radical 3" into "radical 3 over 3".

hearty quartz
#

It is just equivalient fraction. Multiplication by sqrt(3). Basically it gives an idea of the value of 1/sqrt(3). E.g. take 1/sqrt(2). Multiply top/bottom by sqrt(2) => sqrt(2)/2. We know that sqrt(2) = 1.4 approx. Therefore we can say that 1/sqrt(2) = 0.7 approx

blazing skiff
#

Ohhhhhh yeah because you can't have radicals on the bottom of a fraction, so you have to multiply the top and bottom by the radical to get rid of it.

#

I forgot lol. Thank you!

#

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red patio
#

How do I do this

lone heartBOT
red patio
#

linear approximation

#

i've been trying to figure this out for hours

gray jewel
red patio
#

yeah its L(x) = f(a) + f'(x)(x-a)

#

and I got the formula at a=3

gray jewel
#

What have you got?

red patio
#

its 1-3/4(x-3)

#

but idk what to do from there

#

I cant find any resources on this either

gray jewel
#

(3.01)^2 is given as 9.06

#

So what happens if you normally plug 3.01 into the f(x) function?

red patio
#

oh shit

gray jewel
#

👀

red patio
#

you get sqrt(4.06)

gray jewel
#

Yup

red patio
#

dudeeeee

#

wait wait

gray jewel
red patio
#

so I just plug 3.01 into x now in my linearized formula?

gray jewel
#

Yup

#

That should give you the answer

red patio
#

hell yeah

#

wait hold on one sec tho lemme check

#

I get 0.925

gray jewel
#

Alright

gray jewel
red patio
#

wait yeah I multiplied wrong

#

I got it

#

0.9925

#

youre awesome thank you

#

.close

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zinc haven
#

which is the correct notation, $x = \varnothing$ or $x \in \varnothing$

ocean sealBOT
#

FungusDesu

zinc haven
#

likewise, which is the correct notation, $x = \mathbb{R}$ or $x \in \mathbb{R}$

ocean sealBOT
#

FungusDesu

zinc haven
#

an explanation would be helpful

frail grove
#

What is x?

zinc haven
#

a variable

frail grove
#

Second notation

#

It doesn't make any sense to write a variable equal to a set

zinc haven
frail grove
#

Yeah are u trying to write solutions to an equation?

zinc haven
#

yes

#

for example $x^2 + 1 = 0$, which has no real roots

ocean sealBOT
#

FungusDesu

zinc haven
#

what is S^R

ocean sealBOT
#

Adam Chebil

frail grove
#

It means the set of solutions in R is equal to empty set

#

Or simply write: no solutions in R

ocean sealBOT
#

Adam Chebil

lone heartBOT
#

@zinc haven Has your question been resolved?

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untold cloud
#

well, you raise it to the nth power so, the value in the parenthesis is just a bit greater that 1, and since n is so large, it approaches e^a

#

don't think about this one as plugging in infinity, because you cant do that and the value is going to be just a bit greater than zero for large values of n, and when you add 1 to that and exponetiate n times you get e^a

#

shii.... thats when my knowledge ends lol

jagged sigil
#

Well, there is a flaw in your argument.

#

The difference is in your expression, you are applying the limit both at the denominator n and the exponent n at once.

vivid minnow
#

1^infinity is indeterminate

jagged sigil
#

While, in the images, you are doing it separately.

#

So, you can not use association on it.

#

It is right that $\lim_{n\to\infty}{1^{n}}=1$

ocean sealBOT
jagged sigil
#

After applying the power rule, you are getting 1^inf right?

#

It is indeterminate.

#

You can use this rule when n is finite

#

not when n is infinite

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tropic obsidian
lone heartBOT
tropic obsidian
#

pls help

#

and explain if possible idk

alpine sable
#

x + 1/4 (1/x) = 1?

limpid turret
#

Let the fraction be $x$, the reciprocal is $\frac{1}{x}$. One quarter of the reciprocal is $\frac{1}{4}\times\frac{1}{x}$. The sum must be one. So to determine the fraction, just solve for $x$ in $x+\left(\frac{1}{4}\times\frac{1}{x}\right)=1$.

ocean sealBOT
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#

@tropic obsidian Has your question been resolved?

limpid turret
#

@tropic obsidian do you still need help?

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jovial cedar
lone heartBOT
jovial cedar
#

how do i do part c?

limpid turret
#

Looks like your net forces are gravity, and drag (going in opposite direction). So $F_{\text{net}}=mg-\alpha v$

ocean sealBOT
limpid turret
#

Use Newton's second law to get $ma=mg-\alpha v$

ocean sealBOT
limpid turret
#

Then you have the differential equation $\ddot{x}=g-\alpha \dot{x}$

ocean sealBOT
limpid turret
#

Have fun with your differential equation.

storm ridge
limpid turret
#

I wish

#

I'm not that cool

storm ridge
#

Lol

limpid turret
storm ridge
#

Hi

#

You can but not here

storm ridge
#

Np

lone heartBOT
#

@jovial cedar Has your question been resolved?

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@jovial cedar Has your question been resolved?

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prime seal
#

$\int e^{-|x|} ,dx$

lone heartBOT
ocean sealBOT
#

Slowaq

prime seal
#

can somebody help be do this intergal?

#

!occupied

lone heartBOT
#

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

#

@prime seal Has your question been resolved?

prime seal
#

<@&286206848099549185>

alpine sable
#

hi

prime seal
#

hi

prime seal
#

.close

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old sinew
lone heartBOT
old sinew
#

Can you explain me the idea of branch cut in argz?

#

If i choose (0,2pi]

#

i can see that, value of argz little above positive real axis is close to 0

#

and below positive real axis is close to 2pi

#

<@&286206848099549185>

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rigid sphinx
lone heartBOT
snow cobalt
#

didn't you ask this a while back and someone answered it too?

fervent timber
#

use chain rule

rigid sphinx
#

.close

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unique pendant
#

Hi,
If i have a biased coin with 60 percent chance of head and 40 percent chance of tail and flipped 100 times, what would be the range of heads I would get?

lament wyvern
#

you could get 0 heads or all 100 heads

unique pendant
#

on averge though

last walrus
#

Youd "expect" to see 60 heads but that doesnt mean you necessarily have to get 60 heads when you flip 100 times

unique pendant
#

like you said i can get any no of heads but I want to know the range of average in which 60 percent of heads will get me.

last walrus
#

Oh so maybe you mean to build a confidence interval?

unique pendant
#

ye thats it like i could get 40 heads or maybe 70 heads, and i want to know how to find that interval

last walrus
#

ok give me a sec to think about it

#

Do you know about the binomial distribution?

#

Flipping a coin 100 times is a binomial distribution with $n=100$

ocean sealBOT
last walrus
#

and if treat flipping heads as a success, we have $p=0.6$ and $q=0.4$

ocean sealBOT
last walrus
#

Most of our possibilities should land "roughly" within two standard deviations of the expected value

#

So we could create an interval where we are most likely to see the number of heads this way

unique pendant
#

okay

last walrus
#

The expected value of binomial is $np$ and the standard deviation is defined to be $\sqrt{npq}$.

ocean sealBOT
last walrus
#

so we should be able to use the formula $np \pm 2\times \sqrt{npq}$ and sub in numbers to get a reasonable upper and lower bound

ocean sealBOT
last walrus
#

I'm not sure if that makes sense to you or not 🙂

unique pendant
#

it does im just takimg some time

#

60 plus 2X 4.9
60 minus 2 X 4.9

#

is that correct?

last walrus
#

yes

unique pendant
#

tysm i get it now

last walrus
#

I get about $50$ as a lower bound and $70$ as an upper bound

ocean sealBOT
last walrus
#

and this should be valid "roughly" 98% of the time

unique pendant
#

so 2 sd is 98 percent?

last walrus
#

Correct, I am loosely using the 2SD ~ 98% rule

unique pendant
#

okay got it👍

last walrus
#

sweet!

#

dont forget to .close

unique pendant
#

thanks

#

.close

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civic barn
lone heartBOT
civic barn
#

@rancid radish

#

@jade sierra

jade sierra
#

Bruh

civic barn
#

bruh

jade sierra
#

Idk how to do this shit don't ping me for it 😭

civic barn
#

what are u doing then

#

??

jade sierra
#

High school algebra with a bit of trig on the side

civic barn
#

ohh

#

r u in 12th class

alpine sable
#

I can help but I gtg

civic barn
#

bruhh

civic barn
coarse marlin
# civic barn

you can simplify the expression in the () using the sum to inf GP formula

civic barn
#

what is the formula

coarse marlin
#

if a, ar, ar^2, ar^3, ...... inf are in gp.
the sum of all those terms is a/(1-r)

#

in the (cos^2 + cos^4 + cos^6 + .... inf)
a is cos^2 and r is also cos^2

civic barn
#

ohhh

#

got it

#

thannkssssssssssssssssssssss

#

what do u study

#

??

coarse marlin
#

anything the uni teaches me

civic barn
#

oh

exotic canopy
coarse marlin
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slim chasm
#

Hi. Just wanna check that these equations are equal

slim chasm
#

Ek = 1/2mv^2

#

Ek = 1/2m(v2^2-v1^2)

#

doesnt seem right

exotic canopy
slim chasm
#

should it be

exotic canopy
#

idk

#

should it?

slim chasm
#

Ek = 1/2m(v2-v1)^2 ?

exotic canopy
#

i've never seen the second form

slim chasm
exotic canopy
#

only Ek=1/2mv^2

#

would be helpful if you provided context

slim chasm
#

wait ill look

#

no sorry i cant

#

imma just assume I copied it down wrong

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kindred jacinth
#

did i do this correctly ?

lone heartBOT
kindred jacinth
#

it also wanted to find the number of joules it will be used

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lean forge
#

How is Fourier transform defined for a matrix in M_n(R) (n.n matrix)?

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#

@lean forge Has your question been resolved?

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#

@lean forge Has your question been resolved?

devout egret
#

What is the standard form of the polynomial p(x)= 4x^2+2?

vale crag
#

do you have an example where ppl are talking about that or are you just wondering ?

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#

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sharp harness
#

hi guys i should check if this is a linear mapping

lone heartBOT
#

Please don't occupy multiple help channels.

sharp harness
jagged cobalt
#

is it?

sharp harness
#

can i do it like this

jagged cobalt
#

quite drawn out

#

but yeah you can do that

prisma current
#

Just need to check for scalar multiples as well

sharp harness
#

oh ok

#

another question

#

generally

#

when i have a linear mapping

#

f(0) = 0

#

is this always true?

prisma current
#

Yes, this it a requirement from linearity

jagged cobalt
#

has to be

prisma current
#

What does linearity say about f(0+0)?

sharp harness
#

f(0) + f(0)

#

?

prisma current
#

Yep

#

So we have that f(0)=f(0+0)=f(0)+f(0)

#

And so f(0)=f(0)+f(0)

#

Can you use this to determine what f(0) must be?

sharp harness
#

hm lemme check

sharp harness
#

f(0) = f(0) + 0

#

therefore f(0) = 0?

prisma current
#

Yep AUmaruSmile

sharp harness
#

noice

#

can i send you the scalar property too?

#

Hmm

#

i really thought from first view that this would be a linear mapping

#

but it isnt

#

does someone has an intuitive explanation

#

g(x+y) = 2(x+y) + 5 /= g(x) + g(y) = 2x + 5 + 2y +5 = 2x +2y + 10

foggy pecan
#

you have to prove this:

#

$g\left( \alpha\cdot x+\beta\cdot y \right)=\alpha\cdot g\left( x \right)+\beta\cdot g\left( y \right)$

ocean sealBOT
#

Joanna Angel

foggy pecan
#

start from left side and reach right side

sharp harness
#

but it fails on the first condition

#

g(x+y) = 2(x+y) + 5

#

is not the same as

#

g(x) + g(y) = 2x + 5 + 2y +5 = 2x +2y + 10

#

therefore it cant be linear?

vale crag
#

it's just that the "linear algebra" linear mapping has a stricter meaning than the usual way ppl use linear

foggy pecan
#

yes

sharp harness
#

so it cant be a linear mapping right

#

i somehow thought it is lol

vale crag
#

y=ax vs y=ax+b

#

left one is "linear algebra" linear

#

the second one isn't

sharp harness
#

oh

vale crag
#

you need y(0) = 0 in a "lin alg" linear mapping in particular

#

@sharp harness you have other questions or no?

lone heartBOT
#

@sharp harness Has your question been resolved?

lone heartBOT
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rocky juniper
lone heartBOT
rocky juniper
#

would $S: \mathbb{R} \rightarrow \mathbb{R}, S(x) = x^{2}$ be fine as it is surjective every value is mapped in the co-domain but it is not injective i.e $S(-2) = 4 \land S(2) = 4$

ocean sealBOT
rocky juniper
#

so the inverse would be

#

many to one

#

i mean

#

one to many

#

since the original function is many to one

edgy merlin
golden canyon
#

Yeah, codomain shouldn't be R

rocky juniper
#

R+ ?

golden canyon
#

$\mathbb{R}_{\geq 0}$?

rocky juniper
#

ye

golden canyon
#

not that R though

ocean sealBOT
rocky juniper
#

yes

edgy merlin
#

Or any set at least 2 elements into a set with one element

#

S: {1,2,3} $\rightarrow$ {1}

ocean sealBOT
edgy merlin
#

S(x) = 1

rocky juniper
#

oh i guess thats easier

edgy merlin
#

Its simpler

rocky juniper
#

but $\mathbb{R}_\geq 0$ is fine right

ocean sealBOT
edgy merlin
#

Yeah

rocky juniper
#

ight thanks

#

i shouldnt be making those silly mistakes

#

.close

lone heartBOT
#
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tall shuttle
#

I have f(g(x)) = h(g(x)) for all x.

f,h are bijective functions from X to X.

g is a bijective function from Y to X

tall shuttle
#

Can i say that f = h?

limpid turret
#

If $f : Y\to X$ and $h:Y\to X$ are two functions, you can say that $f=h$ when $f(y)=h(y)$ for all $y\in Y$.

ocean sealBOT
tall shuttle
#

So I just need to say that g is surjective?

limpid turret
#

seems like it, yup

lone heartBOT
#

@tall shuttle Has your question been resolved?

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limpid urchin
lone heartBOT
limpid urchin
#

heu guys,i got stuck here while solving i just don't know ehat happens with x-pi/3->0

limpid urchin
#

do i move the -pi/3 so it would be x->pi/3 and then it would be 0/0 and do lhopitals rule?

limpid urchin
pallid glen
#

ill help

limpid urchin
#

please

pallid glen
#

so basically what u do is

#

maan idk

limpid urchin
#

it's alright,someone else will help thanks anyway

#

<@&286206848099549185>

#

.close

lone heartBOT
#
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young parrot
#

Discrete math:
could someone tell me how one would solve this (with steps?)

the answer is n=3 with witnesses k=1 and C=4; but with the method i'm using i get n=4

young parrot
#

my current method:

$3x^3 + (log x)^4 <= 3x^3 + x^4 <= 3x^4 + x^4 = 4x^4$

so n = 4 with witnesses k=1 and C = 4

ocean sealBOT
young parrot
#

right, should be noted that we're using log 2,

#

im not all too familiar with logarithm (my previous highschool never taught it) so there may be some simple logarithm inequality that will allow me to show that (logx^4) <= x^3; but i dont know nor do i know if its true or provable?

young parrot
fickle heath
#

log(x)^n is polylogarithmic, which is below linear

#

For any n > 1 (n = 1 makes it just logarithmic)

young parrot
#

ah, so log(x)^n <= nx?

fickle heath
#

For a big enough x

#

$log(x)^n < x \iff nlog(log(x)) < log(x)$

ocean sealBOT
fickle heath
#

log(log(x)) grows more slowly than log(x), obviously

#

So there exists x_0 such that x > x_0 => log(x)^n < x

young parrot
#

oh im following now

#

does that apply for a function f(x) as well?
log(f(x))^n < x for large enough x?

fickle heath
#

Of course not, just take f(x) = 2^x

young parrot
#

shoot, you're right

#

what about for polynomials though?

fickle heath
#

log(f(x))^n < f(x) though, if f is increasing (I think that's a sufficient condition)

#

(for large enough x, again)

#

log(x^n) ?

young parrot
#

yea

fickle heath
#

That's just nlog(x), so O(log(x))

young parrot
#

i definitely have to go back and properly learn logarithm

#

but this makes sense

#

thanks a bunch

#

.close

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summer hemlock
#

How do I graph this?

lone heartBOT
summer hemlock
#

I'm supposed to pick some x's and replace them with x

#

I'm not sure how to find which x's though

#

The x's I have currently are - 2.5 and -3.5

hot bluff
#

f(-3.5) = 2

summer hemlock
hot bluff
#

absolute value is just a V shaped graph. you know where the function = 0, so now you just need the slope, you make the V shape pointing at the 0 and use the slope to draw the rest.

lone heartBOT
#

@summer hemlock Has your question been resolved?

summer hemlock
hot bluff
#

what x's? there are technically infinite x's to find.

lone heartBOT
#

@summer hemlock Has your question been resolved?

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spare fern
#

Could someone help me out with this?

lone heartBOT
spare fern
ocean whale
#

If the bases are the same, you can just equate the exponents and solve for x from there

spare fern
#

Here is my working:

#

$$\log \left(7^{6x}\right)=\log \left(7^{20+4x}\right)$$

ocean sealBOT
#

Lex1729

spare fern
#

$$6x\log \left(7\right)=20\log \left(7\right)+4x\log \left(7\right)$$

ocean sealBOT
#

Lex1729

spare fern
#

$$x\left(6\log \left(7\right)-4\log \left(7\right)\right)=20\log \left(7\right)$$

ocean sealBOT
#

Lex1729

untold cloud
#

what

#

take the log base 7

#

$$log_7$$

ocean sealBOT
#

alphaGengar

spare fern
untold cloud
#

because you want to cancel the 7

#

so you have just the exponent stuff

#

recall that

#

log_x(x) = 1

#

so log_7(7) = 1

spare fern
#

Oh I see

#

So x = 10?

ocean whale
#

Yes

lone heartBOT
#

@spare fern Has your question been resolved?

lone heartBOT
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wicked tree
lone heartBOT
wicked tree
#

can someone help?

#

I found Fx = 11 and Fy = -5 but idk what to do after that

placid zinc
#

Check your endpoints. That is, what is the function doing along the lines?

#

The first line is y = x - 4

wicked tree
#

mhm

#

so if I plug that into 11x-5y

#

then I get 6x+20

#

if I plug in -x-4 then I get 16x+20

#

if I just plug in 4 then I have 11x-20

#

but then what do I do with that? @placid zinc

lone heartBOT
#

@wicked tree Has your question been resolved?

wicked tree
#

<@&286206848099549185>

hasty umbra
#

@lone heart

#

int (sin(cos(x))) dx

#

simplify

wicked tree
#

.close

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drifting seal
#

is the question correct?

lone heartBOT
drifting seal
#

cuz when k 1 or even 2, it’s not divisible by 19

lone heartBOT
#

@drifting seal Has your question been resolved?

drifting seal
#

<@&286206848099549185>

silver flare
#

hey

#

how can i help @drifting seal

silver flare
lone heartBOT
#

@drifting seal Has your question been resolved?

lone heartBOT
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velvet nacelle
#

is that true

lone heartBOT
#

@velvet nacelle Has your question been resolved?

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@velvet nacelle Has your question been resolved?

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acoustic wave
#

Hi there, I am having a hard time wrapping my head around applying Richardson extrapolation in my homework assignment. I don't even know where to begin....

acoustic wave
#

Okay well, maybe I do know where to start: I need to compute Q0, however I need a small n and a large N for that, and I'm not sure what they are in this case

#

I’ve gotten to this point

#

However, I'm not 100% sure that I even need it in sum form...

#

<@&286206848099549185>

#

Perchance 0 is an unlucky number....

deep basin
#

Yeahs

acoustic wave
#

Yeahs?

deep basin
#

Wqcuzbi?

acoustic wave
#

Uh ohhh

deep basin
#

Uh ohhh

acoustic wave
#

Well, at least I have someone to talk to

deep basin
#

Wefechu

acoustic wave
#

However, I imagine taking up a help channel is not ideal, so Ima close it

#

Good luck, random person

deep basin
#

Ffgdh go Christmas no. Vuch if v

#

Big guy

acoustic wave
#

.close

lone heartBOT
#
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deep basin
#

Bbh I am not sure if I will be in a relationship or not but if you can

lone heartBOT
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inner haven
#

find cos(x) and tan(x) if csc(x) = -sqrt(6)/2 and cot(x) > 0

inner haven
#

would cos = -sqrt(2)/sqrt(6)

#

and tan = 2/sqrt2?

echo socket
#

csc(x) isn't the same as 1/cos(x) though

inner haven
echo socket
#

Can you recall what csc(x) is?

inner haven
#

1/sin

echo socket
#

Right, so you have sin(x) = -2/sqrt(6)

inner haven
#

yes

echo socket
#

Are you solving for x in [0, 2pi) or are you looking for the general solutions?

inner haven
#

general

echo socket
#

Just use (-1)^k arcsin(-2/sqrt(6)) + pik where k is an integer then

inner haven
#

oh wait actually

#

what is the difference between general and solving for x

#

This was all the problem said

#

Oh wait would cos be sqrt2/sqrt6

echo socket
#

Yes

lone heartBOT
#

@inner haven Has your question been resolved?

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scenic wing
lone heartBOT
scenic wing
#

so

#

$-cos(x)-(-cos(-x))$

ocean sealBOT
#

putridplanet

scenic wing
#

$-cos(x)+cos(-x)$

ocean sealBOT
#

putridplanet

scenic wing
#

now what

echo socket
#

Recall that cosine is an even function

scenic wing
#

wdym

echo socket
#

cos(-x) = cos(x)

scenic wing
#

why

echo socket
#

Changing the sign of the angle on the unit circle is equivalent to reflecting your point around the x-axis

#

That, however, does not change its x-coordinate

scenic wing
#

ok

echo socket
#

Hence cos(-alpha) = cos(alpha) for all angles alpha (I named it alpha here not to confuse it with the x coordinate)