help-10

[timid silo]

To find perimeter

We do 2x+4y+0.5x right

[crimson wave]

we need to get the A of the 2 cubes

[ruby sequoia]

so first find area

[timid silo]

I did it

😭

[ruby sequoia]

what is it

[timid silo]

Well

It’s a prove question

You get 4y = 100/x -x/2

Just wanting to confirm for these type of questions, when it says 0.5 rad we just use 0.5 as theta

And not something in pi right?

[ruby sequoia]

180 degree is pi radian

180/pi is 1 radian

[timid silo]

Yh

[ruby sequoia]

if it says 0.5 radian

you dont take it as theta

[timid silo]

Yeah

[ruby sequoia]

it is 0.5*180/pi

[timid silo]

It worked by just making it theta

[ruby sequoia]

well it is basically an angle

you can assume it as other angle and put the value in terms of radian

[timid silo]

I didn’t use any pi

[ruby sequoia]

how did you use the area that is given?

[timid silo]

U find the perimeter

[ruby sequoia]

can you show your work?

[timid silo]

😭

It’s sort of messy

[ruby sequoia]

you already turned it into radian

what i thought that theta degree but you used theta as radian

(r^2)/2 is the area for 1 radian

[timid silo]

Yh

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wild vigil]

A boat crosses a river with a speed v, = 22km/h
Calculate the the Time it takes to cross if the flow velocity of the Water is 2m/s. The river is 70m wide.

[wild vigil]

so i did 22 :3,6 to get it into 6,1 m/s

now what?

[obtuse pebble]

@wild vigil Has your question been resolved?

[astral radish]

v=s/t so t=s/v so 70/(6.1+2)

around 8.6 seconds

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[teal turret]

,w 4pi((6.5)^(3)) * (1/3)

[warm shale]

Results provided by WolframAlpha

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

Requested by northsteve

[timid silo]

pog

[teal turret]

how are u still doing this stuff

how long have u been doing these problems

[timid silo]

I joined 2 days ago, Ive been learning everything about this I got 7 things left to learn

Now they making me learn to do that

[teal turret]

no i mean, werent u doing this all day yesterday?

have u slept lol

[timid silo]

Basically I wake up, try my best, go here, go to work, sleep then repeat

Atleast I got 7 left

[teal turret]

u have a final coming up or something?

[timid silo]

Yeah Im practicing

Atleast im getting better

lol

[teal turret]

thats good

putting in the effort

[timid silo]

rn im stuck on The similarity ratio of similar solids is 5 : 4.

I don't know what it means

They want me to find the surface areas and volume from that

I dont have the shape

Thats my issue

[shy ruin]

yeah thats right

since area is units^2

and volume is units^3

so u did it right

[timid silo]

why didnt they include a shape

[shy ruin]

well

this would work for any shape

with this ratio

[timid silo]

160?

Im confused on this one

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[obtuse pebble]

@sour elm Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sour elm]

<@&286206848099549185>

umm

[hazy leaf]

which part do you need help with

[obtuse pebble]

@sour elm Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[sour elm]

d

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dark whale]

What is the formula for finding your altitude if you know your current air pressure, tempature, and you know sea levels temp/airpressure?

[dark whale]

.close

[obtuse pebble]

Channel closed

Closed by @dark whale

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[iron nexus]

Could someone please enlighten me on question 60 and what it means for a function F of A to be linear in both colums/rows or alternating? I have been trying to get my head around it but I still don't really understand. All I "understand" is that I have to look at the rows and columns of A and do something about it.

[worn yoke]

to be linear in the first column means that if you hold the second column vector constant, the function distributes over addition and scalar multiplication of the first column (which is what it means to be linear)

in other words, for it to be linear in the first column, find out if it is true that [ F \begin{pmatrix} a_1 & b \ c_1 & d \end{pmatrix} + F \begin{pmatrix} a_2 & b \ c_2 & d \end{pmatrix} = F \begin{pmatrix} a_1 + a_2 & b \ c_1 + c2 & d \end{pmatrix} ] and is it true that [ k F \begin{pmatrix} a & b \ c & d \end{pmatrix} = F \begin{pmatrix} ka & b \ kc & d \end{pmatrix} ]
similar applies to the second column, and to the rows

[warm shale]

cloud

[iron nexus]

wow this is crystal clear

thank you so much

[worn yoke]

we would also say that it is alternating on the columns if having two identical columns would make the function 0 no matter what

[iron nexus]

I'm trying to understand how to get started with exercise though, is it a linear combination that we are effectively looking for?

[worn yoke]

if you have this or something like it, you want to verify that computing the function on the left side based on the entries of each matrix is the same as computing it on the right side

note that if it is "linear on the columns" that would mean it has to satisfy that linearity condition on every column

[iron nexus]

what does F(A) = bc mean? I understand now what F(A) means but I still can't figure out how that linearity could result in bc..

[worn yoke]

basically that means that the function takes in a 2x2 matrix and returns the product of the top right and lower left entry (based on the letters they provide on the matrix)

[iron nexus]

wow I don't know if I'm stupid or it's actually abstract

sorry

I'm still thinking XD

[worn yoke]

so in this example, we could check for linearity in the first column by checking whether [ \underbrace{F \begin{pmatrix} a_1 & b \ c_1 & d \end{pmatrix}}{b c_1} + \underbrace{F \begin{pmatrix} a_2 & b \ c_2 & d \end{pmatrix}}{b c_2} \overset{?}{=} \underbrace{F \begin{pmatrix} a_1 + a_2 & b \ c_1 + c_2 & d \end{pmatrix}}{b(c_1+c_2)} ] as well as [ k \underbrace{F \begin{pmatrix} a & b \ c & d \end{pmatrix}}{bc} \overset{?}{=} \underbrace{F \begin{pmatrix} ka & b \ kc & d \end{pmatrix}}_{b(kc)} ]
and then do the same for the second column

[warm shale]

cloud

[iron nexus]

do I have to check both conditions are does one condition implies the other one checks out

[worn yoke]

we have to check it distributes over both addition and scalar multiplication, although you can check both at the same time by checking whether [ k_1 \underbrace{F \begin{pmatrix} a_1 & b \ c_1 & d \end{pmatrix}}{b c_1} + k_2 \underbrace{F \begin{pmatrix} a_2 & b \ c_2 & d \end{pmatrix}}{b c_2} \overset{?}{=} \underbrace{F \begin{pmatrix} k_1 a_1 + k_2a_2 & b \ k_1 c_1 + k_2 c_2 & d \end{pmatrix}}_{b(k_1 c_1+k_2 c_2)} ]
which is only one equation rather than two, but also a bit more unwieldy

[warm shale]

cloud

[iron nexus]

okay, beautiful

thanks!

.close

[obtuse pebble]

Channel closed

Closed by @iron nexus

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[robust tundra]

can someone tell me if my integrals are correctly written

I do have the answer, but without the steps.

[shy ruin]

starting with your second step

you need parenthesis around 5x - x^2

or split it into two separate integrals

[robust tundra]

Ah ye, good catch

[shy ruin]

the first one 5x can have the 5 taken out

[robust tundra]

take 5 out and then wrap the rest in ()?

[shy ruin]

but the second one doesnt have a 5 to take out

holdon

[robust tundra]

This shouldnt be in an integral anymore should it?

since its already in its primitiv form?

[shy ruin]

$5 \int_{0}^{5} x \mathrm{d}x - \int_{0}^{5} x^2 \mathrm{d}x$

[warm shale]

Potatomonke

[robust tundra]

Ah so I shoud split them up

[shy ruin]

yes

if you want to take out the 5

[robust tundra]

oki let me try, give me a min or 2

[shy ruin]

mhm

[robust tundra]

forgot the range on second one

fixed that

oh, and let me try another method and you let me know if its proper if thats okay!

nvm, its a dead end, anyway, is this better?

[shy ruin]

mhm

[robust tundra]

Approved?

[shy ruin]

yeppp

[robust tundra]

ill take it thanks for your help both

.close

[obtuse pebble]

Channel closed

Closed by @robust tundra

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[grim agate]

.close

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[molten yew]

I need help w this

theres a cone with diameter of 18 an height of 12

so r=9

[shy ruin]

any other info?

[molten yew]

and then on the right is the view from the bottom

and were supposed to find theta (the angle shown)

[shy ruin]

umm

I don't think theres enough information

[molten yew]

hmm

how do i rotate the pic

[daring rock]

,rotate

[warm shale]

Couldn't find an attached image in the last 10 messages.

May 17, 2024, 10:58 PM

[molten yew]

,rotate

[molten yew]

aight

[daring rock]

Yeah this is not enough information

Are you given the volume or anything else?

[tranquil beacon]

^

[molten yew]

you can solve for the volume of the whole thing ig

[shy ruin]

but you cant find the value of theta with the given data

u sure theres nothing else given?

[molten yew]

does it change anything if the length of the lines connecting to make the angle are 9 cm

[molten yew]

what else do you need for it?

[daring rock]

Nah we can already see that, the radius is 9

[molten yew]

yeah ik

[shy ruin]

literally anything

[daring rock]

You need the volume of the solid, or maybe the surface area

[shy ruin]

^

[molten yew]

but we dont know how much of it is missing

so we can find it

[shy ruin]

its bottom view so just the radius

[molten yew]

yeah

lets imagine we get the volume

what would be the formula

[shy ruin]

$\mathrm{Volume_{given}} = \frac{1}{3} \pi {r}^2 h \cdot \mathrm{scalefactor}$

then u solve for scale factor

and multiply either 2pi or 180 by the scale factor

depending on if you want radians or degrees

yeah

wait

[molten yew]

whats the scale factor tho

[shy ruin]

thats what we need to solve for

[molten yew]

wedont get that either

[shy ruin]

we need to find the scale factor or whatever u wanna call it

thats hwy we need more info

[molten yew]

what info is needed

[shy ruin]

volume

or surface area but then thats a diff formula

or the slant height like faiyrose was saying

we need some other information

[daring rock]

Think of it like this, if theta is 180 degrees, you have half a cone, if theta is 270 degrees, you have 3/4 of a cone, etc

The scale factor is just whatever fraction of the cone you actually have

[molten yew]

this is the formula if the volume is given?

[shy ruin]

mhm

left side is given volume

i shouldve prob stated that

[warm shale]

Potatomonke

[molten yew]

and whats the formula if surface area is given

[shy ruin]

i doubt they would make you do that

it would be pretty awkward

[molten yew]

wdym

[shy ruin]

it would be like

holdon i gotta think abt this

[molten yew]

can u send me his in dms cuz i lowkey need to go

im going out w a girl

supposed to be there in 1+

10

[shy ruin]

sure

[molten yew]

lifesaver bro

cuz this was on a test

that my friends took today

and i have it on monday

so i have the whole weekend to solve this

[shy ruin]

yeah i doubt they would make u solve this based on surface area

just know it would be a bit overcomplicated

most likely they would give a volume or something of the like

maybe even an area for the circle at the base

[molten yew]

If anyone would do it

It would be my school

Guess what grade im in based on this

[obtuse pebble]

@molten yew Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[grim agate]

Hail math gurus. I am working on an election problem that is getting into game theory. Is anyone around available to offer some discussion and possibly suggestions?

Here's the setup. Imagine a community of nine people (A-I) that elects a leader to make decisions regarding nine issues. Each issue is only held by one person and his neighbor (hence there are issues {AB, BC, CD... HI, AI}). The current leader gives equal time to all nine issues, but loses the next election because a challenger promises to ignore three issues and devote the time to the other six. But the candidate with six issues is vulnerable to one with five, who is vulnerable to one with four. See chart.

[obtuse pebble]

@grim agate Has your question been resolved?

[obtuse pebble]

@grim agate Has your question been resolved?

[obtuse pebble]

@grim agate Has your question been resolved?

[grim agate]

Hmm

[obtuse pebble]

@grim agate Has your question been resolved?

[grim agate]

.close

[obtuse pebble]

Channel closed

Closed by @grim agate

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[winged ibex]

can you do row reduction on sub determinants?

[winged ibex]

here they dont do it

adding first row onto seconds row gives me a determant of 63

instead of -23

.close

[obtuse pebble]

Channel closed

Closed by @winged ibex

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lost swift]

can someone help me i need to find the axis of symteray, vertext, x intercepts, y intecepts, and minimum value of a parabola with an equation of y = 5x^2 +17x+6 as an extension question and idk what to do

[high lily]

would you be able to determine those above properties for something simpler like
y = x^2 + 3x + 2

[lost swift]

well i found the axis of symmetry

since its -17/10

but not sure how to get the vertext

[high lily]

technically the equation of the axis would be
**x = ** - 17/10
sub that in to get the y-coordinate of the vertex and hence the point

[lost swift]

i got (-17/10, 31/20)

[high lily]

that doesn't sound right, can you show your work

[lost swift]

oh wait i actually got (-17/10, -169/20)

i did y = 5(-17/10)^2 + 17(-17/10) + 6

and then i converted both to decimal

so (1.7,8.45)

[high lily]

**-**8.45

[lost swift]

ye

i figured the rest out i think

bc for x values

u jus use quadratic

formula

and for y u substitute the vertex value

andmin value sdhould be -8.45

as well

[obtuse pebble]

@lost swift Has your question been resolved?

Channel closed

Closed by @lost swift

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[velvet rover]

you can, but proper standard form has only one digit before the decimals

i dont usually use this format, not enirely sure. most likley not

[obtuse pebble]

Channel closed

Closed by @oblique kettle

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[austere sleet]

!help

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[austere sleet]

i dont understand

why do you jjust magically get rid of the -24

the exponent -24 why did it magically turn into a positive 24

[left vault]

$b^{-a}=\frac{1}{b^a}$

[warm shale]

The د

[left vault]

this is a rule of exponents

[austere sleet]

ok

well i never knew about that

thx

[obtuse pebble]

@austere sleet Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[burnt anvil]

I need some help

[burnt anvil]

the two lines

y = ax - 2 and y = x - 1, where a is a constant

intercept at the first quadrant

which values are possible for a?

I have sketched it but I don't know where to go from there

[swift stream]

First find the intersection point in terms of a, then see if you can make out what the possible values of "a" are

[burnt anvil]

do I set the equations equal to each other

[swift stream]

You tell me

What do hou think you should do

[burnt anvil]

well what do you mean by the intersection point in terms of a

[swift stream]

Just find the intersection point of the two lines

[burnt anvil]

alright

ax - 2 = x - 1

ax = x + 1

a = 1 + 1/x

[swift stream]

Why are you solving for a

Find the point at which the two lines intersect

[burnt anvil]

I'm not sure how you expect me to do that

we do not have the value of a

[swift stream]

Okay, so let a be unknown

[burnt anvil]

how am I supposed to find the point in which they intersect

[swift stream]

Do the algebra

[burnt anvil]

we will still have 2 variables unknown

[swift stream]

Thats fine

[burnt anvil]

I don't think I'll be able to find it

[swift stream]

Just pretend a is a number for now, and find the point of intersection

[burnt anvil]

ax - 2 = x - 1

that's where the lines intersect

[swift stream]

No, they intersect at a point

I want you to tell me what point they intersect at

[burnt anvil]

oh okay

[swift stream]

It is fine if that point has a variable in it

[burnt anvil]

ax - 2 = x - 1

ax = x + 1

ax - x = 1

x(a-1) = 1

x = 1/(a-1)

that's my x coordinate

andd

[swift stream]

Okay, now find the y coordinate

[burnt anvil]

y = x - 1

y = 1/(a-1) - 1

they intersect at

(1/(a-1) , 1/(a-1) - 1 )

[swift stream]

Yes

Now, if that intersection point is in the first quadrant

[burnt anvil]

a has to be bigger than 0

otherwise my first coordinate would be negative

[swift stream]

If a is 0.5 (bigger than 0), the first coordinate is still negative

[burnt anvil]

a has to be bigger than 1

[swift stream]

Yes

Now lets move onto the y coordinate

Solve the inequality 1/(a-1) -1 >0

[burnt anvil]

1/a-1 > 1

1 > a - 1

2 > a

[swift stream]

Right

So a must be bigger than 1 and smaller than 2

[burnt anvil]

oh alright

will I always be able to solve these types of problems this way_

my teacher recommended I solve it graphically

[swift stream]

If you have a graphing calculator at hand it may be possible to do it graphically

[burnt anvil]

yes geogebra

[swift stream]

Okay, in such a case this is quite easy

Just slide the slider for a and you will see that if 1<a<2, the intersection takes place in the first quadrant

[burnt anvil]

so im sliding it right now

but

[swift stream]

Make it 1.1, you will see that the intersection happens in the first quadrant

[burnt anvil]

oh alright

[swift stream]

This happens all the way till a=2, and beyond that the intersection point moves to the fourth quadrant

[burnt anvil]

wait

they still intersect at a = 2.1

no nvm

i zoomed in

[swift stream]

They do, but not in the fjrst quadrant

[burnt anvil]

yes

alrighty

thanks for the help

appreciate it

👍

.close

[obtuse pebble]

Channel closed

Closed by @burnt anvil

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lyric wren]

A triangle PQR is to be constructed so that the perpendicular bisector of PQ cuts the side QR at N and the line PN splits the angle QPR into two angles, not necessarily of integers degrees, in the ratio 1:22. If angle(QPR) = p degrees, where p is an integer, find the maximum value of p.

[lyric wren]

I dont see how a perpendicular bisector could split an angle into a ratio of 1:22

can someone draw out a diagram that makes this possible or makes it at lealt hypothetical

[distant shadow]

Hi

[lyric wren]

hello

no matter how i draw a perpendicular bisector

i dont see a way to split an angle into 1:22

unless the answer is wanting the highest multiple of 23 within 180 which is 161

[distant shadow]

Um

Should look something like this

We need to maximize Ang QPR

[lyric wren]

yeah no i see that

but whenever i try to extend the angle

theres just no physical way to be split it nicely

[distant shadow]

QPN : NPR = 1:22

[lyric wren]

so im thinking theres a hypothetical way yk

[distant shadow]

we need to solve the question right ?

[lyric wren]

yeah

but if we just maximise qpr then its just 161 for angle p

[distant shadow]

see if we minimize QPN

[lyric wren]

mhm

[distant shadow]

QPR will be macimized

[lyric wren]

okay

[distant shadow]

I think we have to do something with that right angled triangle

QPN be k°

[lyric wren]

nothing says it has to be a right angled triangle

you can say qpn is 1/23p

[distant shadow]

PN = x

[distant shadow]

Perpendicular bisector

[lyric wren]

true

[distant shadow]

right angle so

[lyric wren]

wait no

we have different diagrams

[distant shadow]

Um why ?

[lyric wren]

my diagram is

cutting a line in the middle creating 2 right angles

and that line has to hit the opposite corner

to make an angle

[distant shadow]

there will always be a right angled triangle

PN is the hypotenuese

[distant shadow]

Can you send it

[lyric wren]

ill send one

gimme a se

c

[distant shadow]

Okk

[lyric wren]

no sorry i read the question wrong

thats on me

ok now i get the question

[distant shadow]

Okkk

We have considered that point D ok

Now see if we minimize ND then we shall have max QPR , (i think so)

[lyric wren]

agreed

lemme redraw it though

[distant shadow]

Ok

[lyric wren]

ok

so

basically

i can now draw the diagram

[distant shadow]

Yea

[lyric wren]

except i cant see why its not just 161

[distant shadow]

umm !

[distant shadow]

I cant get you

[lyric wren]

because angle p ius a multiple of 23

and that means we can just maximise it

under 180

or is that completely wrong

[distant shadow]

I am really confused

[lyric wren]

ok

so

questions says

[distant shadow]

I think we must

[lyric wren]

angle p is split into 1:22

[distant shadow]

Make some relation

Using the given angles which are restrains

[lyric wren]

which means angle p must be a multiple of 23

[distant shadow]

Yeah

[lyric wren]

and given that the angle is an int

the highest possible value

is 7x23

[distant shadow]

No i understood what you did

[lyric wren]

yeah

[distant shadow]

But im asking if its true

[lyric wren]

whats true

[distant shadow]

Ahh okk

Yeah we can extend the other two sides to match

Our ratio

[lyric wren]

yeah bro im too tired for this

thanks for your help

i needa sleep

[distant shadow]

But i

But

Im not sure

If its solved this way

[lyric wren]

its probably not

?

but like

[distant shadow]

I though we would use calculus

[lyric wren]

what else can we do

[distant shadow]

And theres a theoretical approach

[distant shadow]

If i figure out something

I will dm you ok

[lyric wren]

alright

tysm!

[distant shadow]

k bro byee !

Gn

[lyric wren]

byee

gngn

[obtuse pebble]

@lyric wren Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fallow crest]

Is it possible to find equation of tangents drawn from a point outside of parabola to it?

[fallow crest]

.close

[obtuse pebble]

Channel closed

Closed by @fallow crest

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[empty ledge]

can someone explain to me sin and cos being divergent and convergent

[upbeat gazelle]

What is the limit of pi/n when n goes to infinity?

[tame narwhal]

write out some of the terms

[empty ledge]

cos(0) = 1

so its convergent since its less than infinity gotcha

are these correct

[obtuse pebble]

@empty ledge Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scarlet forge]

While solving this problem using double integrals I have a confusion in taking the limits while drawing a strip parallel to x axis , from where do I take my y limits is it from the point of intersection or from below it ?

[tame narwhal]

what do you mean "below" it?

[velvet rover]

the y limits will be the 2 functions you have gotten in the question

[tame narwhal]

also this doesn't look like a double integral problem

[scarlet forge]

Area using double integral

I mean when I take a strip parallel to x axis y limits would be going from -6 to 6 right ?

[frosty river]

Lower y limit is x²-6x+3, upper is 2x-9

[scarlet forge]

Thats when the strip is taken parallel to y axis isnt it ?

[frosty river]

Think of whose area you are calculating

[frosty river]

Oh, sorry. Doing that way is unnecessarily overcomplicate

You should invert the functions to have x in terms of y

Then do the same

as you always do

[scarlet forge]

Confusing 🙃

[frosty river]

For inverting the parabolla, completing the square is useful

You have two functions f(x) and g(x)

If you want to integrate first with respect to x then with respect to y, you need to calculate the expresions of f^(-1)(y) and g^(-1)(y)

[scarlet forge]

I did something like this

[frosty river]

It does not look bad

It's something like that

But if the problem doesn't indicate the contrary choose the order of integration to intelligently

Doing y then x is straightforward

[scarlet forge]

Yes that's true but I just wanted to try out the other way around but I couldn't arrive at the right answer

[velvet rover]

do you have the answer key?

[scarlet forge]

My only doubt is when I took the y limits going form -6 to 6 is it the right way or I should have started it from point of intersection ?

[scarlet forge]

I only have the final answer as 32 / 3

[velvet rover]

why -6 and 6?

[scarlet forge]

I have taken the strip parallel to x axis

[velvet rover]

if you do that you will need to divide into 2 integrals, wich is wayy overcomplicated

actually 3 integrals

[scarlet forge]

Is it because below the point of intersection the initial and final limits are defined by the same function ?

@velvet rover

[velvet rover]

yup, while if you find the strips perpendicular to the y axis the area is always under one function and over another

[scarlet forge]

Ohh alr ty

.close

[obtuse pebble]

Channel closed

Closed by @scarlet forge

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[trim portal]

<@&268886789983436800>

[obtuse pebble]

Channel closed

Closed by @cursive cove

Use .reopen if this was a mistake.

[polar fossil]

bye

[trim portal]

thank you

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bleak skiff]

can someone explain me why i always get s0 > X_i when i run my code

[brazen viper]

Not without the code

[obtuse pebble]

@bleak skiff Has your question been resolved?

[bleak skiff]

.close

[obtuse pebble]

Channel closed

Closed by @bleak skiff

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[brisk rain]

on part c of this question, how do you find the second solution
i have one solution correctly as 3.68 but im not sure how to get the second solution

the equation is t = 2(cos^-1(-3/(2sqrt5)) - 0.464)

[obtuse pebble]

@brisk rain Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[spice chasm]

can u say sqrt(x^2+1) is proportional to just x

[tame narwhal]

what do you mean by "proportional"?

[smoky cairn]

no, since proportional means you they're related by multiplication of a constant

[stoic yacht]

you can say that √(x² + 1) grows asymptotically as x

[obtuse pebble]

@spice chasm Has your question been resolved?

[spice chasm]

a

[spice chasm]

like they grow at the same rate

ohhh they grow closer to the same rate when x gets bigger

[graceful ridge]

they don't grow at the same rate, their derivatives are different

[spice chasm]

isnt the +1 negligible tho

as x gets bigger

$\dv{x}\sqrt{x^2+1}=\frac{x}{\sqrt{x^2+1}}$

[warm shale]

Jash

[graceful ridge]

sure

[spice chasm]

$\lim_{x\to\infty}\pqty{\dv{x}\sqrt{x^2+1}}=\dv{x}x$

[warm shale]

Jash

[spice chasm]

oh wait i see

ty

.clsoe

.close

[obtuse pebble]

Channel closed

Closed by @spice chasm

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean spear]

How would I solve this

[obtuse pebble]

@lean spear Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[lean spear]

.reopen

[obtuse pebble]

✅

@lean spear Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[unborn laurel]

ive got 1-1 down, and im part of the way through proving h is onto by letting y be in {1,...,mn} and expressing y = qm + r using the division algorithm. now im stuck expressing a and b in terms of these two

[viral blade]

Can you use the fact that every 1-1 function between two sets of the same size is also onto

oh boy you're proving that A×B is finite

guess not

[viral blade]

*two finite sets

[unborn laurel]

we're kinda proving that they are of the same size, so nah not rly

[viral blade]

So you're given an element in {1,...,mn}

You probably want to use the division algorithm on it but it gets weird in cases with remainder 0

So if x/m has remainder 0 it means your first coordinate becomes m and your second coordinate becomes x/m - 1

and in all other cases I think it works normally

[molten grove]

i think it's slightly more nuanced than division algorithm because you can have n > m

so the remainder a can be larger than the divisor m

[viral blade]

no the remainder is the first coordinate

It can be exactly equal to m which is our weird edge case but it can't be more than m

[unborn laurel]

i think i figured it out somehow, it's kinda weird but it works

thanks for the help ^^

.close

[obtuse pebble]

Channel closed

Closed by @unborn laurel

Use .reopen if this was a mistake.

[viral blade]

np

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mild sandal]

Okay so i learned that finding terms from the end in an expansion is (n-r+2)th term

[mild sandal]

If i had to find from beginning i just need to use the general term formula right?

[obtuse pebble]

@mild sandal Has your question been resolved?

[mild blade]

Yes

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[umbral turret]

I'm lost as to how in the first u-sub it gets to u^2 + 3, when I write out the problem I get to (u/2x^3)*(1/xu) which then becomes 1/x^4 when you move the 1/2 out of the integral and cancel the u's from the numerator and denominator

[oak gyro]

$\int\frac{1}{x\sqrt{x^4-3}}dx=\int\frac{du}{2x^4}$

[warm shale]

otheol

[oak gyro]

If we have $u=\sqrt{x^4-3}$, then $u^2=x^4-3\implies u^2+3=x^4$

[warm shale]

otheol

[umbral turret]

gotcha

[oak gyro]

So $\int\frac{du}{2x^4}=\int\frac{du}{2(u^2+3)}=\frac{1}{2}\int\frac{du}{u^2+3}$

[warm shale]

otheol

[umbral turret]

.close

[obtuse pebble]

Channel closed

Closed by @umbral turret

Use .reopen if this was a mistake.

[umbral turret]

.reopen

[obtuse pebble]

✅

[umbral turret]

I thought I had it, but then I got to $v = \frac{u}{\sqrt{3}}$ and I don't know how that was determined

[warm shale]

loganctanner

[umbral turret]

the closest I can get to, which I don't think is even possible, is $v = \frac{u^2 + 3}{3} = \frac{u^2}{3} + \frac{3}{3} = \frac{u^2}{3} + 1 = \sqrt{\frac{u^2}{3}} + \sqrt{1} = \frac{u}{\sqrt{3}} + 1$

[warm shale]

loganctanner

[umbral turret]

if you're still on I'm lost as to how the next sub was determined

[oak gyro]

The substitution $v = \frac{u}{\sqrt{3}}$ is just to "prepare" for the $arctan$ identity

[warm shale]

otheol

[oak gyro]

Which is $\int\frac{dx}{1+x^2}=\arctan x+C$

[warm shale]

otheol

[umbral turret]

so how was it decided $v = \frac{u}{\sqrt{3}}$ ?

[warm shale]

loganctanner

[oak gyro]

This is to make the bottom of the $\int\frac{du}{u^2+3}$ in the form of of $x^2+1$

[warm shale]

otheol

[oak gyro]

Since $u^2+3=(v\sqrt{3})^2+3=3v^2+3=3(v^2+1)$

[warm shale]

otheol

[oak gyro]

Which gets us $\int\frac{du}{u^2+3}=\int\frac{\sqrt{3}dv}{3(v^2+1)}=\frac{1}{\sqrt{3}}\int\frac{dv}{v^2+1}$

[warm shale]

otheol

[oak gyro]

As you can see, the substitution reduces the $(variable)^2+3$ into $(variable)^2+1$

[warm shale]

otheol

[umbral turret]

right

but there's a step missing

you explained why it was determined but not how

[oak gyro]

Hmmm

Let me write this out

$\int\frac{du}{u^2+3}=\int\frac{du}{3(\frac{1}{3}u^2+1)}$

[warm shale]

otheol

[oak gyro]

To make the bottom $(something)^2+1$, we let $\frac{u^2}{3}=v^2\implies \frac{u}{\sqrt{3}}=v$

[warm shale]

otheol

[oak gyro]

Generally, if we see an integral of the form $\int\frac{C_1dx}{C_2x^2+C_3}$, we usually apply this sort of thinking to try and force it into a form of $\int\frac{C_4du}{u^2+1}$

[warm shale]

otheol

[oak gyro]

$=C_4\arctan u+C$

[warm shale]

otheol

[umbral turret]

right, we're reviewing the arcs & general integration rules so I figured it would look something like that, it just wasn't very intuitive to me for some reason why it was these substitutions were being made

[obtuse pebble]

@umbral turret Has your question been resolved?

Channel closed

Closed by @umbral turret

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[graceful ridge]

Why do they say, "differentiable in neighborhood of the point in the domain", why not just in the point?

[stoic yacht]

in the context of what?

this sounds like a definition of analyticity

[graceful ridge]

holomorphic functions

complex analysis

[stoic yacht]

great, i just finished taking that class

[graceful ridge]

yea so why

[stoic yacht]

https://math.stackexchange.com/questions/3804156/differentiability-versus-analyticity-domains-for-complex-functions

this explanation online actually matches what we were taught in class

for example, consider the function given by the taylor series $\sum_{n=0}^{\infty} n!z^n$

[warm shale]

nightshade5107

[stoic yacht]

this thing converges nowhere except at z = 0

which means its radius of convergence is 0

which means although you know its behaviour at a single point, you can't construct any epsilon neighborhood around 0 where this power series converges

just because a complex function is differentiable at a single point doesn't mean that it's necessarily analytic on some domain around that point

for it to be locally represented by a taylor series, that differentiability needs to extend to some open set around that point

[graceful ridge]

wait a sec, for it to converge, |z| < 0, because
the radius of convergence is zero, but 0 isn't less than zero

[stoic yacht]

by the ratio test, this thing converges when the limit of the n+1th term over the nth term < 1

so that becomes (n+1)!z^(n+1) / (n!z^n) = (n+1)z

then you take the limit as n tends to infinity and see which z would cause that result to be less than 1

after taking the limit, you get infinity

which means regardless of z, you have no radius of convergence

which means this function converges either nowhere or at a single point and nowhere else

notice that whenever z^n is any nonzero value, the n! term dominates and the sum always diverges

but when z = 0

you're repeatedly summing 0

and the function converges to 0

which means the series has a radius of convergence of 0 and converges at 0, meaning that there is no disk of positive radius around z = 0 where the series also converges

[graceful ridge]

Oh ok I got the difference between differentiability and analyticity, let's not get my brain ripped apart any further

lol

.close

[obtuse pebble]

Channel closed

Closed by @graceful ridge

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bright ember]

can someone help me with this part b

i suck at trig

[stoic yacht]

were you able to get part A

[bright ember]

yeah

part a was okay

[stoic yacht]

great, note that cos(2*theta) = 1 - 2sin^2(theta)

using this, how could you rewrite cos(4A)?

[bright ember]

wait which identity is this

[stoic yacht]

there are a couple different versions of the cosine double angle formula

they're all equivalent

[bright ember]

oh i see it now

[stoic yacht]

cos(2x) = cos^2(x) - sin^2(x) = 2cos^2(x) - 1 = 1 - 2sin^2(x)

all of those are the same and can be derived from each other using sin^2 + cos^2 = 1

[bright ember]

yeah i see

[stoic yacht]

but specifically use this one to rewrite cos(4A)

[bright ember]

so cos4A = 1 -2sin^2(2A)

[stoic yacht]

great, what happens when you substitute that into your right hand expression in b

[bright ember]

oh!

i see it now

[stoic yacht]

think you can finsih it from there?

[bright ember]

i tried to prove from LHS the entire time

yeah

thanjks so much

[stoic yacht]

happy to help 🙂

[bright ember]

🙂

.close

[obtuse pebble]

Channel closed

Closed by @bright ember

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[gritty hemlock]

I had a question, does the length, width, height change when it goes to a 2d shape and are these correct?

[oak gyro]

The numeric values all look fine

Units for area should be in², and volume in³

[gritty hemlock]

Also I had a question about this?

Would the 2d shape be a sector of a circle

?

[oak gyro]

The base of a cone is a circle

[gritty hemlock]

So for the shape I just put circle

[oak gyro]

Yes

[gritty hemlock]

Alright thanks

.close

[obtuse pebble]

Channel closed

Closed by @gritty hemlock

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[solemn creek]

yo can someone explain quadratic equations too me

[fervent nexus]

Cool, can you tell me what you have done so far

[solemn creek]

i aint done nun bro i dont got the first clue what im doin and youtube aint helpin

im on the first question

[fervent nexus]

ok, do you know square numbers

[solemn creek]

ye

[fervent nexus]

so essentially the first question is asking what should x be, such that its square is 49

[visual knoll]

^move the -49 to the other side
x^2=49

[solemn creek]

do i gotta change it to like x^2=49

is that my answer?

[oak gyro]

That is a valid way to do it, yes

[oak gyro]

No

[reef yarrow]

dont do like that

[visual knoll]

it's fine for this question

[oak gyro]

You have to solve for what just x is

[reef yarrow]

else u will not getting other possibilities if for other type of q

u need to learn how to factorise

[solemn creek]

hollon ima try question b myself ight

x=3?

[visual knoll]

once you get an answer, it's good to put it back into the equation and see if it's correct

[solemn creek]

ight thanks bro

.close

[obtuse pebble]

Channel closed

Closed by @solemn creek

Use .reopen if this was a mistake.

[visual knoll]

...3 wasn't correct for b by the way

[reef yarrow]

ye

u need to use quadratic formula

[oak gyro]

All of these problems can be done with factoring

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[static abyss]

Hi, can someone just quickly explain this step to me? Why is x-1 in the arctan ?

[tender tusk]

do u know how to solve this?

so if u were to solve this normally you would add 1 then subtract 1 from the numerator

then split it into 2 fractions

then in the second fraction when u complete the square of the denominator u will get (x-1)^2+1

[high lily]

you'd actually want -2 + 2 here

[tender tusk]

ah right

mb

[oak gyro]

It's probably something like

[static abyss]

I just don't understand this

[tender tusk]

i never knew there was a general form for this

[oak gyro]

$\int\frac{2x+1}{x^2-2x+2}dx=\int(\frac{2x-2}{x^2-2x+2}+\frac{3}{x^2-2x+2})dx$

[warm shale]

otheol

[oak gyro]

Apply u-sub for bottom on left

And then factor the right with $x^2-2x+2=x^2-2x+1+1=(x+1)^2+1$

[warm shale]

otheol

[oak gyro]

And then you can integrate $\int\frac{3dx}{(x+1)^2+1}$ to get $3\arctan(x+1) + C$

[warm shale]

otheol

[static abyss]

oh alright, I understand that, what about this?

sorry its in my language but yeah

[oak gyro]

Which part are you asking about?

[static abyss]

well

how did he get the x-1 in the top without even doing anything?

in the arctan

[oak gyro]

Oh wait oops

I misread the initial problem

[high lily]

complete the square for the denominator

[oak gyro]

$\int\frac{3dx}{x^2-2x+2}=\int\frac{3dx}{x^2-2x+1+1}=\int\frac{3dx}{(x-1)^2+1}$

[warm shale]

otheol

[oak gyro]

let $u=x-1\implies du=dx$

[warm shale]

otheol

[oak gyro]

$\implies\int\frac{3dx}{(x-1)^2+1}=\int\frac{3du}{u^2+1}=3\arctan u+C=3\arctan(x-1)+C$

[static abyss]

yes okay I understand

[warm shale]

otheol

[static abyss]

one last question

so here

the numerator is just the d/dx of the denominator right

of this

oh sorry wait

of this

then why is there that square root in the denominator with 4c - b^2 ?

that part I dont get

[oak gyro]

Ok give me a bit to try and prove this identity

Because the 2x+b I believe is related but there is a bit more manipulation first

Wait

[oak gyro]

Is there more to this equation, or is it just the $\frac{1}{x^2+bx+c}$?

[warm shale]

otheol

[static abyss]

its just this

second part

[oak gyro]

Also, I find it strange how it is exponentiated

[static abyss]

yeah same

atp everything is confusing me here

I mean ig he just put in the coefficients?

[oak gyro]

Probably so

[static abyss]

if you put in the coefficients of this into the square root

you get x - 1

in the end

okay well thank you for taking the time @oak gyro

[oak gyro]

No problem

I have learned that that arrow usually means "leads to"

So perhaps it is saying that seeing a denominator of that form usually leads to a solution of that form

[static abyss]

ohh alright, good to know 🙂

thanks again bye

.close

[obtuse pebble]

Channel closed

Closed by @static abyss

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[raw escarp]

am stuck on this how can i solve it

i dont under stand this step

[pulsar quarry]

$e^{x-1}=\frac{e^x}{e^1}$

[warm shale]

yajat

[raw escarp]

i am slow on thos so if we take the lower e and pu it up e^x * e^-1 isnt that e^-2x

[pulsar quarry]

$\frac{\frac{e^x}{1}}{\frac{e^x}{e^1}}$

[warm shale]

yajat

[oak gyro]

$\frac{e^x}{\frac{e^x}{e^1}}=\frac{e^x\cdot e^1}{e^x}=e^1=e$

[warm shale]

otheol

[pulsar quarry]

here u go

thanks otheol

[raw escarp]

ohh ok thank you

[pulsar quarry]

np

!done

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[raw escarp]

.close

[obtuse pebble]

Channel closed

Closed by @raw escarp

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[verbal obsidian]

L hospital is not working and i am crying

[true trail]

Use expansion

[obtuse pebble]

@verbal obsidian Has your question been resolved?

Channel closed

Closed by @verbal obsidian

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[graceful ridge]

What is the difference between a minimal polynomial and an annihilating polynomial?

[rich plume]

🤨

[graceful ridge]

😭

[kind hawk]

the minimal polynomial is the monic annihilating polynomial of smallest degree

[graceful ridge]

Oh, so the difference is just between the degree?

they both annihilate but one is of the smallest degree?

ty

.close

[obtuse pebble]

Channel closed

Closed by @graceful ridge

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[half grail]

Lol is this one that hard

[wintry swift]

assume y and n are constants, then you have an quadratic equation in x, right?

[obtuse pebble]

@half grail Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sleek bear]

This question was on my exam 2 days ago on friday

I got an answer but it’s crazy, 24 900 Pa/s 😂

But the cylinder is very small though so maybe it’s reasonable

(And btw I just realized that p = 2.46/V comes from the gas law, pV = nRT, so in this case nRT = 2.46)

The volume of the cylinder sorry (not the cylinder’s volume)

[pine barn]

I think this question is just related rates

We're given the equation on how to find pressure with volume

[sleek bear]

Yeah you can use the chain rule

p depends on the volume (V) and the volume depends on time (t)

[pine barn]

The objective of the problem is to find whether p' is positive or negative when 1/2 v_f is v_0

[sleek bear]

p’ is positive

Because the pressure increases when the volume decreases

I did this

1. The height in decreasing (that’s why the volume is decreasing too). The height was 0.122 m and decreases with 0,02 m/s, so the height h = 0.122 - 0.02t

2. Plug that in for h in the formula for the volume of a cylinder, V = pir^2h. So you get V(t) = pi*r^2(0.122 - 0.02t)

3. Plug that in in the equation for pressure. Then you get p = 2.46/pi*r^2(0.122 - 0.02t). That’s the equation for pressure you wanna use. It explains how the pressure depends on the time instead of the volume

4. We’re interested in how many seconds in takes for the volume to decrease by half. Half of the volume is pir^2(0.122 -0.02t(half)) And half of the volume is also pir^20.122/2. They are equal, and if you solve the equation an cancel pir^2 you get t = 3.05 s.

5. Differentiate p(t), aka what is p’(t). Aka how fast is the pressure changing at any given moment t? Use the quotient rule (this is the hardest part)

I ended up with this

6. And then the answer is p’(3.05), how fast is the pressure changing when the volume has decreased by half (3.05 s). Plug in 3.05 s

And you’ll end up with 24 903.98 which is 24 900 Pa/s

Crazy answer but was I right? <@&286206848099549185>

[obtuse pebble]

@sleek bear Has your question been resolved?

[sleek bear]

I want to know if this is right 🙂

[pine barn]

Yeahh

It's right

Here's what I did

[obtuse pebble]

@sleek bear Has your question been resolved?

[sleek bear]

Niice, thank you so much! 👍🏻

[obtuse pebble]

Channel closed

Closed by @sleek bear

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dusk yew]

@static abyss when you look at it through an addition of ordinates when 4/X gets infinitely close to 0, the other part x-5 is pretty much the curve. You can graph it might help to visualise.

[static abyss]

that's when x goes to infinity and -infinity not 0

of the original function

of course, it gets cloer and closer to its asymptote

since there is no intersection

I understand what you are saying but even if you get a function like this on the paper you're gonna have to first calculate the asymptote and then that will tell you almost everything anyways

[obtuse pebble]

@dusk yew Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pastel wraith]

How do i convert from cartesian equation of a quadratic to parametric equation (in maple)?

[pastel wraith]

like this:

(you can ignore the part in the middle) my question is how I can calculate the parametric equations on the bottom starting from the cartesian equation on the top

[obtuse pebble]

@pastel wraith Has your question been resolved?

[pastel wraith]

<@&286206848099549185>

[obtuse pebble]

@pastel wraith Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[marsh summit]

?

[obtuse pebble]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[winged ibex]

when doing synthetic division are the algebraric multiplicty of the roots equal to the highest power of the polynomial?

[coral nest]

im not sure what this has to do with synthetic division, but in $\mathbb{C}$ you have the fundamental theorem of algebra that basically tells you that

[warm shale]

LF

[obtuse pebble]

@winged ibex Has your question been resolved?

[winged ibex]

says what bruh

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[serene mural]

It says "Solve the equation:". This is an exponent equation I stuck at, explanations are welcome.

[sage geode]

Rewrite 25^x as (5^x)^2, notice that the equation is then a quadratic equation where 5^x is the unknown

Ah wait there's more than that

Divide Both sides by 4^x first

[serene mural]

I know, but I stuck at 10^x and 4^x

[sage geode]

And the same trick applies afterwards

[serene mural]

Oh, but what about 25? (or we still convert it to (5^x)^2 ?)