help-39

[pseudo gulch]

i see. the thing is, i did this problem as practice a month ago, and i know i did it but i dont remember how i did it anymore. maybe i did use this substitution thing, but feels foreign in my brain so idk how i did it before

[tacit mulch]

I didn't watch the video that admire posted, but I suppose you could use the squeeze theorem in a similar way to how you'd prove lim(x->0) sin(x)/x = 1

But if you've already proved that limit, then the substitution is much simpler

[pseudo gulch]

but i dont think squeeze theorem works for this question

[pearl pond]

@pseudo gulch Has your question been resolved?

Channel closed

Closed by @pseudo gulch

Use .reopen if this was a mistake.

[pseudo gulch]

i guess theres no other way?

thats weird man

[pearl pond]

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.

[calm oxide]

How do I trace backwards through the graph to find the shortest path? I've found the stage number and permanent label for each vertex

[pearl pond]

@calm oxide Has your question been resolved?

[calm oxide]

<@&286206848099549185>

[vestal tapir]

hm

i'm not sure you can

you have to remember which vertex was the best

can't figure it out in hindsight

[pearl pond]

@calm oxide Has your question been resolved?

[pearl pond]

@calm oxide Has your question been resolved?

[pearl pond]

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.

[ashen ridge]

help with this would be appreciate, im having trouble what my first step would be

[buoyant panther]

chain rule

outer funtion is ln(x) and inner is 6x^2 - 19

[ashen ridge]

this rule?

[stone robin]

idk if this helps at all

[pearl pond]

@ashen ridge Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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.

[jolly parrot]

Sakai

[pearl pond]

Channel closed

Closed due to the original message being deleted

[tawny kraken]

This equality is true or should I change de sign or the inequality at the last ?

For example when I have, X < Y we have log(X) < log(Y) or log(X) > log(Y) ?

[fiery vortex]

<

[tawny kraken]

Okay thanks

[pearl pond]

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.

[covert badge]

Hello

[covert badge]

$$\lim _{n\to +\infty }\left(\left(\sum _{p=1}^n:\frac{1}{\sqrt{:p}}\right)\frac{1}{\sqrt{2n}}\right)$$

[jolly parrot]

ANASTECH

[covert badge]

How can I prove that this limit equals 2 ?

given that the limit of $$\sum {p=1}^n:\frac{1}{\sqrt{p}}$$ is infinity

[jolly parrot]

ANASTECH
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[round temple]

By realizing this is actually a Riemann sum

[manic kite]

the limit is sqrt(2)

$$\lim_{n \to \infty} \frac{1}{\sqrt{2n}}\left( \sum_{p=1}^n \frac{1}{\sqrt{p}} \right)
= \lim_{n \to \infty} \frac{1}{\sqrt{2}n} \sum_{p=1}^n \frac{1}{\dfrac{p}{n}} = \frac{1}{\sqrt{2}}\int_0^1 \frac{1}{\sqrt{x}} ,dx$$

[jolly parrot]

秋水

[covert badge]

which is 2 right ?

[manic kite]

,w Limit[Sum[1/Sqrt[p], {p, 1, n}]/Sqrt[2 n], n -> Infinity]

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 秋水#5244

[covert badge]

I see

actually my knowledge is limited since we've just started the lesson at class

how can I prove that this limit is sqrt(2) without using any advanced knowledge like reinmann sum

[manic kite]

you can use this

[manic kite]

https://math.stackexchange.com/questions/995110/how-to-prove-the-inequality-2-sqrtn-1-−-2-le-1-frac-1-sqrt-2-frac-1

[covert badge]

I've actually proved this instead : $$2\left(\sqrt{n+1}-\sqrt{n}\right)\le \frac{1}{\sqrt{n}}\le 2\left(\sqrt{n}-\sqrt{n-1}\right)$$

[jolly parrot]

ANASTECH

[covert badge]

in the first part of the exercise

[manic kite]

so you can get this

then use sandwich theorem to do the limit

[covert badge]

idk why can this help me

[manic kite]

this is
$1 +\frac 1 {\sqrt 2}+\frac 1 {\sqrt 3}+ \dots +\frac 1 {\sqrt n}$

[jolly parrot]

秋水

[covert badge]

yes

this expression diverges

but i need to prove that when it is divided by sqrt(2n), it converges towrds sqrt(2)

[covert badge]

wait

,w Limit[Sum[1/Sqrt[p], {p, 1, n}]/Sqrt[n], n -> Infinity]

[jolly parrot]

Results provided by WolframAlpha

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

Requested by ANASTECH#3984

[manic kite]

$$\frac{2 \sqrt{n+1}-2}{\sqrt{2n}} \leq\frac{1}{\sqrt{2n}}\left( \sum_{p=1}^n \frac{1}{\sqrt{p}} \right) \leq\frac{2 \sqrt{n}-1}{\sqrt{2n}}$$

[jolly parrot]

秋水

[covert badge]

this is the expression I need

i thought it was sqrt(2n) but it's sqrt(n)

which gives the limit equals 2

[manic kite]

ok

$$\frac{2 \sqrt{n+1}-2}{\sqrt{n}} \leq\frac{1}{\sqrt{n}}\left( \sum_{p=1}^n \frac{1}{\sqrt{p}} \right) \leq\frac{2 \sqrt{n}-1}{\sqrt{n}}$$

[jolly parrot]

秋水

[covert badge]

so how to get that inequality

[covert badge]

from this

[manic kite]

let n=1,2,3, ..., n

then add them,

[manic kite]

you can read this link

$$2\left(\sqrt{2}-\sqrt{1}\right) \leq \frac{1}{\sqrt{1}}$$
$$2\left(\sqrt{3}-\sqrt{2}\right) \leq \frac{1}{\sqrt{2}}$$
$$\dots$$
$$2\left(\sqrt{n+1}-\sqrt{n}\right) \leq \frac{1}{\sqrt{n}}\$$

[jolly parrot]

秋水

[manic kite]

$$2\left(\sqrt{2}-\sqrt{1}+ \sqrt{3}-\sqrt{2}+\dots \sqrt{n+1}-\sqrt{n} \right) \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n}}$$
$$2(\sqrt{n+1}-1) \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{n}}$$

[jolly parrot]

秋水

[covert badge]

That's really nice !

thank you

.close

[pearl pond]

Channel closed

Closed by @covert badge

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.

[dapper vessel]

Really basic shell method question --Could someone help me understand why p(y)=-y instead of p(y)=y for the radius?

[azure sequoia]

Do you speak French

[dapper vessel]

No, but I can use translator

If it is easier for you

[limber ravine]

To visualize p(y) it's sometimes easier to rotate the coordinatesystem. Try to rotate the picture to the left, so the x-axis becomes the vertical axis with positive direction upwards

You'll notice p(y) looks like a straight line given by p(y) = y, but if you look at the y-axis, the negatives are to the left instead of the right where they usually are on the horizontal line

[dapper vessel]

So that will occur pretty much every time the area is located beneath the x-axis & rotated around x-axis

But on this one, for example, p(y)=y, right?

[limber ravine]

Wait, one moment. I think I've done a dumb dumb

[dapper vessel]

lol alrighty i will wait

[limber ravine]

Okay, I might not have made as big of a dumb dumb as I thought. So, it would have been p(y) = y if it had been in the first quadrant where the positive y-values are on the y-axis, but because we're in the fourth quadrant, where it's negative y-values on the y-axis, we need p(y) = -y instead

[dapper vessel]

What would happen in the event that it crosses from 1st to 4th quadrant? 2 integrals?

Like say the original problem was [0,9]

[limber ravine]

Well, I don't think we can give meaning to revolving around an axis if we cross the rotational line

[dapper vessel]

I suppose

[limber ravine]

Notice the function describing the black line is actually pretty similar to the one before. What is the equation given here?

[dapper vessel]

y=1/x

x=1/y

[limber ravine]

1/x? Is the black line in the second picture not a straight line?

[dapper vessel]

not a straight line, its just on a small scale so it looks straight

so for that one, 2 integrals are needed, according to the solution explanation

[limber ravine]

Yeah, one for the part bounded by the rectangle and one for the part bounded by the function

But the radius should be p(y) should be y - 1/8

I think

It's been a while

[dapper vessel]

Lets look at this one instead

same concept

[limber ravine]

Yes. So you see there's two parts of the shaded area

[dapper vessel]

I just actually get math explaination on it

[limber ravine]

One bounded by a rectangle and one bounded by the function

[dapper vessel]

see, i wouldve guessed the h(y)=4 - 1/y

but that is very wrong

[limber ravine]

You see the line that seperates the rectangle by the "triangle" above?

[dapper vessel]

yes yes

[limber ravine]

That is what seperates our two integral

At what y-value is this line?

[dapper vessel]

1/4

[limber ravine]

Yes

So. The radius of the "triangle" would be?

[dapper vessel]

Its whatever the y value is, no?

[pearl pond]

@dapper vessel Has your question been resolved?

Channel closed

Closed by @dapper vessel

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 flame]

out of all the tangents of this function when x > 1, find the point that has the minimal coefficient. please help./

[pearl pond]

Please don't occupy multiple help channels.

[merry python]

find the derivative

maybe even the second derivative and set that equal to 0

is it natural log in denominator btw?

[teal flame]

ive already done that but i didnt get a final answer like it should be

[teal flame]

yes

ill do it agai and ill show

idk if its worth it to keep going

[merry python]

is the function -2/e * x/ln(x)?

[teal flame]

yes

k = -2/e

[merry python]

just leave the -2/e as constant outside

only take the derivative of x / lnx

$y'=-\frac{2}{e}\cdot\frac{\ln x-x\cdot\frac{1}{x}}{\left(\ln x\right)^{2}}=-\frac{2}{e}\cdot\frac{\ln x-1}{\left(\ln x\right)^{2}}=-\frac{2}{e}\cdot\left(\frac{1}{\ln x}-\frac{1}{\left(\ln x\right)^{2}}\right)$

this is what I got

if the bot works...

[teal flame]

yea im stuck again

@merry python you got something

[merry python]

When looking at the graph of the derivative there didn't seem to be any point where the second derivative was 0

When I calculated it, e came out as the answer, but I might have made a mistake somewhere

[teal flame]

that's supposedly the answer

[pearl pond]

@teal flame Has your question been resolved?

[midnight haven]

can u help with this question?

[pearl pond]

@teal flame Has your question been resolved?

[pearl pond]

Channel closed

Closed by @teal flame

Use .reopen if this was a mistake.

[pearl pond]

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.

[sullen leaf]

How do we check for continuity in an example like this?

[honest bough]

Check if the limits at each point equals the value of the function at each point

[sullen leaf]

So plug each value into the piece of the function and check if they all equal each other?

[honest bough]

Each piece is continuous, so the limits will equal they're values, so yeah, you can do that

If I'm understanding you correctly

[sullen leaf]

So plug 1 into the 1st function?

2 into the 2nd?

[honest bough]

No

[sullen leaf]

And 2 for the last?

[sullen leaf]

Ok, then how?

[honest bough]

You're checking if lim x->1 f(x) = f(1)

[sullen leaf]

There isn’t a limit in this though?

[honest bough]

And if lim x->2 f(x) = f(2)

That's the definition of continuity

A function is continuous at c if lim x->c f(x) = f(c)

[sullen leaf]

So what do I sub in I’m confused?

[honest bough]

What's lim x->1 f(x)?

[sullen leaf]

Doesn’t that depend on the piece of the function we’re referring to??

[honest bough]

There's only one function

Made up of pieces of others

What's lim x->1^- f(x)?

[sullen leaf]

What do I do then?

[honest bough]

.

[sullen leaf]

-1?

[honest bough]

How'd you get that

[sullen leaf]

Is this the correct way to do this?

So it’s jump discontinous?

[honest bough]

https://youtu.be/JrYmEo6hoiU

[pearl pond]

@sullen leaf Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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.

[wheat rivet]

hello i need help with geomotry

[wheat rivet]

if anyone would like to get in a call i can show you what im dealing with

<@&286206848099549185>

[pearl pond]

@wheat rivet Has your question been resolved?

[pearl pond]

@wheat rivet Has your question been resolved?

[wheat rivet]

.close

[pearl pond]

Channel closed

Closed by @wheat rivet

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.

[midnight haven]

How do I do 2b

[midnight haven]

This doesn’t work and where did I went wrong?

[pearl pond]

@midnight haven Has your question been resolved?

[tacit mulch]

That's right so far

You just have to finish simplifying the coefficient

[midnight haven]

Ah alright nvm then

.close

[pearl pond]

Channel closed

Closed by @manic turtle

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.

[tribal thunder]

I have this equation

X and y represents different amount of stock bought

Question 12

That’s the equation for it

x and y have to be whole numbers.

How can i get both x and y to have whole numbers with the equation 14x + 23y = 4800

This is the chart for it but I can’t do trial and error all the way up to a 172

[cloud hull]

either x has to be a multible of 23 or y to be a mutible of 14

[tribal thunder]

80 is not a multiple of 23

Yes it works

?

[cloud hull]

oh yeah

[tribal thunder]

<@&286206848099549185>

[cloud hull]

4800-x*14 has to be a multible of 23

also wait 15 mins before tagging helpers

[tribal thunder]

Sorry I don’t get you?

Can you help me solve that out

[tribal thunder]

How do I find this number

I can’t be solving that out mid test it’d take a lot of time

[cloud hull]

for example 4800-x*14=23

the next one will be 4800-x*14=46

and you can calculate as many as you need

[cloud hull]

oh then there is no "fast" way

to get whole numbers

4800-x14=23n will get you the n-th number

[tribal thunder]

So I’m basically gonna have to do trial and error 80+ times to find the answer

💀

[cloud hull]

-x=(23n-4800)/14

for the n-th x

but this is the closest to "fast"

x=-(23n-4800)/14 is slightly faster ig

[pearl pond]

@tribal thunder Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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.

[tight garnet]

Does anyone know how to do this

[stuck arch]

Maybe post your question

[tight garnet]

Sorry the image isn’t loading idk why

A 7.5 kg suitcase being pulled with force of 27 N at 35 degrees with the horizontal

A 7N force opposes the horizontal motion

Find the acceleration

[fast zealot]

sin(theta) = F_v / F_h

cos(theta) = F_h / F_h

27 * cos (35) = F_h

,calc 27cos(35)

[jolly parrot]

Result:

-24.399689537471

[fast zealot]

radians

sadge

27cos(35) = 22.1N

22.1 - 7 = 15.1N

[tight garnet]

Wait

[fast zealot]

15.1 / 7.5 = a

[tight garnet]

Where’s you get the 7

-7

[fast zealot]

opposing force

[tight garnet]

Ohhhh

[fast zealot]

normally you would ignore friction in these examples to make it easy, but I guess here they give an opposing force of 7N

now I dont know if im 100% correct

but this seems right to me

[tight garnet]

Do you know why the 7 isn’t given an angle?

I struggle to know when I’m supposed to find x and y components or leave the N as is

[fast zealot]

horizontal and vertical vectors are split and therefor if the opposing force is mentioned only in the horizontal we apply it to only that

[tight garnet]

Oh okay that makes a bit more sense

Calc 22.1-7

[fast zealot]

have you done projectile motion yet?

suvat

[tight garnet]

Uh yes

,calc 22.1-7

[jolly parrot]

Result:

15.1

[fast zealot]

suvat makes the idea of splitting those vectors into their components of x and y

[tight garnet]

Idk who that is

[fast zealot]

oh mb

projectile motion

[tight garnet]

,calc 15.1/7.5

[jolly parrot]

Result:

2.0133333333333

[fast zealot]

I would round to 2

dont round to an answer with more precision than the values you are given in the question

[tight garnet]

Yes I did

Thank you for helping me

[fast zealot]

nice

gj

np

[tight garnet]

.close

[pearl pond]

Channel closed

Closed by @tight garnet

Use .reopen if this was a mistake.

[pearl pond]

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.

[neon pier]

hello, can someone explain to me why this isnt equal to 1/a^14/3 b^14/3

[neon pier]

i got b^20/3 too somehow but thats wrong also

[viscid oxide]

Hmm

Lets try to solve it then

[neon pier]

im cryingt

[jolly parrot]

n11

[viscid oxide]

Do you get how this happened?

[neon pier]

why r there two as

yes

i get everythng but the 1/a times a^1/3

[jolly parrot]

n11

Because $a^{\frac43}=a^{\frac33+\frac13}=a^{1+\frac13}=a+a^{\frac13}$

[neon pier]

ah

that makes sense

[viscid oxide]

Wait wtf did I just do

oh yeah at the end it wont be sum but product

[neon pier]

wait

[viscid oxide]

Alright so lets go further

[neon pier]

what

im sorry but why

oh wait

[jolly parrot]

n11

[neon pier]

yeah ok i got it

[viscid oxide]

a^m × a^n = a^(m+n)

alr so

[viscid oxide]

This now becomes $\bigg{(} \frac{1}{a^2 \cdot a^{\frac13}} \cdot b^{\frac13 + 2} \bigg{)}^2$

[jolly parrot]

n11

[neon pier]

wait

why is there a^2 at the bottom

oh

is it bc u added that to the other one

okay yeah

[viscid oxide]

multiplied the a with

yes

[jolly parrot]

n11

[neon pier]

do they become 7/3

[viscid oxide]

nopsie not yet!

[neon pier]

warum

[viscid oxide]

Next: $\bigg{(} \frac{b^{2+\frac13}}{a^{2+\frac13}} \bigg{)}^2$

[jolly parrot]

n11

[neon pier]

oh wait

what

why r they on top of each other

[viscid oxide]

we multiplied the b part with 1

Like this: $\frac{1}{a} \cdot b = \frac{b}{a}$

[jolly parrot]

n11

[neon pier]

but why are u multiplying them arent u only supposed to do the b part times the other b and the a times the other a

[viscid oxide]

other b and a?

[neon pier]

yeah like

[viscid oxide]

this will kinda simplify it a bit, because now we'll take the exponents common out, and we'll only have to do the math once

[neon pier]

what do you mean

[viscid oxide]

but we already solved them and made one out of them XD

See above ^

from the start to have a recap

[neon pier]

what do u mean by "made one out of them"

[viscid oxide]

multiplied them

[neon pier]

but did u really

wait

u did

[viscid oxide]

over here

[neon pier]

okay but why are u multiplying them again

[merry stirrup]

Simplify it

[viscid oxide]

I'm not!

We are putting them in one fraction because they have the exact same exponents

[neon pier]

then why are they on top of each other

but how is that allowed

[viscid oxide]

and the motive of this is to take the exponent out at once

Like this: $\frac{a^m}{b^m}=(\frac{a}{b})^m$

[jolly parrot]

n11

[neon pier]

that doesnt make sense though

[merry stirrup]

$(a^{-1-\frac{4}{3}}{b^{\frac{1}{3}-{2}})^2$

[neon pier]

im sorry

[viscid oxide]

its an algebraic law

[neon pier]

why are u multiplying the b and a

[viscid oxide]

eee
I've been solving it above ^^
if you wanna take over then I'll leave. I have to get ready for college anyway lol

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[neon pier]

NO PLEASE DONT LEAVE

im sorry i dont mean to be rude to the other person

[merry stirrup]

Okk

If you don't have time

Then it's ok

[viscid oxide]

to make it one fraction so that the exponents are in one single fraction

Or actually, you can skip it

[neon pier]

but why

[viscid oxide]

and do it individually

[merry stirrup]

There we use a law of exponent @viscid oxide

Which is

$a^a×a^b=a^{a+b}$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[neon pier]

i think he knows bro

but

[neon pier]

i know this

[merry stirrup]

Yes so you have to use this

[neon pier]

but i still dont get why you're putting them on top of each other

[merry stirrup]

Means bro?

[merry stirrup]

Oh

[viscid oxide]

So, without putting them on top, we can now solve them, which becomes: $\bigg{(} \frac{1}{a^{\frac73}} \cdot b^{\frac73} \bigg{)}^3$

[jolly parrot]

n11

[merry stirrup]

Well $(a×b)^n=a^n×b^n$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[neon pier]

omg

wait

wait

just one second

[merry stirrup]

Yes

Analyse it by yourself

[neon pier]

okay yes

wait

[neon pier]

why is it ^3

isnt it ^2

[viscid oxide]

oh crap yes

sorry mb

[merry stirrup]

Yes

[neon pier]

ur good

one second

okay

why is it not 14/3 then

[merry stirrup]

$\bigg{(} \frac{1}{a^{\frac73}} \cdot b^{\frac73} \bigg{)}^2$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[neon pier]

what

wait

[viscid oxide]

Now we know that $(ab)^m = a^m \cdot b^m$, therefore we can solve that expression by giving the exponent "two" to both "a part" and "b part" individually.

[jolly parrot]

n11

[neon pier]

why are u multiypling them both

[merry stirrup]

,w -1-4/3

[neon pier]

oh

u right

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[merry stirrup]

,w 2+1/3

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[merry stirrup]

Got it

[viscid oxide]

Do you get this?

[neon pier]

omg i got it

[viscid oxide]

excellent

[neon pier]

but i still dont get why ur putting them on top of each other

[merry stirrup]

You can do it like that
${a}^{-7/3×2}×b^{7/3×2}$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[viscid oxide]

because its ugly to leave them individually

[neon pier]

okay but ugly is still right

right

dont u get the same thing

[merry stirrup]

Yeah

[merry stirrup]

It's right

[viscid oxide]

you see $(\frac{a}{b})^m is better looking than $\frac{1}{b^m} \cdot a^m$ lol

[neon pier]

i know i can see the check mark habibi

okay but

[merry stirrup]

you see... $(\frac{a}{b})^m$ is better looking than $\frac{1}{b^m} \cdot a^m$ lol

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[neon pier]

why are u multiplying them

warum

kyu

[viscid oxide]

simplification

[neon pier]

ok but

[merry stirrup]

We simplify it

[neon pier]

if i did this on a quiz would i still get the point

[merry stirrup]

To make it easy

[jolly parrot]

n11

[merry stirrup]

You would have options in it or you will get marks

[neon pier]

listen bro i

[viscid oxide]

alr gotta head out, or I'll be late for my class
good day y'all, and thanks Arnab for the assist!

[merry stirrup]

If you don't have options and you wrote this ans I am sure that you will get full marks

[neon pier]

may god bless your soul n11

you too arnab

u guys are the light in the dark

[merry stirrup]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 = 1

[neon pier]

danke schön

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[neon pier]

.close

[pearl pond]

Channel closed

Closed by @neon pier

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.

[signal thorn]

how do i get the value of x in 3^2x+3 - 2 = 9^x+1

[cloud hull]

well firstly if we simlify we get 3^2x=9^x

have you learned of the log function

[barren hill]

divide both sides by 3 @signal thorn

oh wait

its 3 to the power of 2x

[cloud hull]

lol

[barren hill]

never mind i thought it was multiplication

my bad guys

[cloud hull]

the solution will be even better if you know the ln function but we can make do with log

[wet osprey]

Be careful about your brackets

$3^2x+3-2=9^x+1$

[jolly parrot]

Frosst

[wet osprey]

This is what you wrote

[merry stirrup]

Is it
$3^{2x+3}$ or $3^{2x}+3$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[wet osprey]

$3^{2x+3}-2=9^{x+1}$

[signal thorn]

the first one

[jolly parrot]

Frosst

[signal thorn]

yes

[wet osprey]

This is what I assume you’re saying

[merry stirrup]

This one?

[wet osprey]

But please be careful with your brackets

[signal thorn]

this one

[wet osprey]

They are extremely important

[signal thorn]

oh ok

it’s gonna be like this right

$3^{2x+3}-9^{x+1}$=2

[wet osprey]

Look at this, replace the {} with normal brackets and remove the dollar signs

[jolly parrot]

suzy

[signal thorn]

then this

[merry stirrup]

$3^{2x} × 3^3 -2=3^{2x} × 3^2$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[signal thorn]

oh so it won’t be like this?

[merry stirrup]

This is good Actually

[signal thorn]

$3^{2x+3}-3^{2x+2}$=2

[jolly parrot]

suzy

[merry stirrup]

It is more simplified

You can write it like this

Then

[signal thorn]

the 2 is getting me confused

[merry stirrup]

$3^{2x}×3^3-3^{2x}×3^2$=2

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[merry stirrup]

It will not create any problem your question is not a positive integer problem na?

[merry stirrup]

Take 3^2x common

[signal thorn]

wait so i’ll factor 3 out right then = 2

[merry stirrup]

$3^{2x}(3^3-3^2)$=2

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[merry stirrup]

27-9

18

So

$3^{2x}=1/9$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[merry stirrup]

Or
$3^{2x}=9^{-1}$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[merry stirrup]

It's same as
$3^{2x}=3^{-2}$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[signal thorn]

oh ok

[merry stirrup]

Yes then bases are same

$2x=-2$

[jolly parrot]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[merry stirrup]

Finished

[signal thorn]

wait question

[wet osprey]

You can check by putting x=-1 in the original and see if it works

[signal thorn]

what happened to 3 here

[merry stirrup]

It's just
$3^{2x}+1=9^x+1$

[signal thorn]

oh this one’s wrong

[merry stirrup]

No this one is right

[signal thorn]

this is the right problem

[merry stirrup]

Sorry

[merry stirrup]

Out x=-1

Put*

In that equation

[signal thorn]

i have $3^{2x} {18} =2$

[jolly parrot]

suzy

[signal thorn]

this is what i have rn

(18)*

[merry stirrup]

,w solve 3^{2x+3}-3^{2x+2}=2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[merry stirrup]

-1

,w evaluate 3^{2x+3}-3^{2x+2}=2

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[merry stirrup]

x=-1

Yes no difference

[signal thorn]

oh ok i got it now

thank you

[merry stirrup]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 = 1

[jolly parrot]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[pearl pond]

@signal thorn Has your question been resolved?

Channel closed

Closed by @signal thorn

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.

[cursive kelp]

I have no idea what the classifications are just need some help thanks

[coarse dawn]

It might say in your textbook on those pages

[cursive kelp]

I couldn't find the definitions on the textbook

[midnight haven]

did you not learn in class what those terms mean?

[cursive kelp]

no classification was not mentioned at all in the lectures

and i couldn't find it in the textbook either

[pearl pond]

@cursive kelp Has your question been resolved?

[pearl pond]

@cursive kelp Has your question been resolved?

[cursive kelp]

.close

[pearl pond]

Channel closed

Closed by @cursive kelp

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.

[vestal lodge]

How can I prove the p-adic absolute value of x satisfies the triangle inequality? I know I need a couple of cases, but the case im struggling on is when x and y have different numbers of p's in their prime factorizations

[toxic lichen]

let $x = p^m a$ and $y = p^n b$ with $m > n$, then you can show $v_p(x+y) = n$

[jolly parrot]

Ann

[toxic lichen]

from which you will in fact show that $|x+y|_p \leq \max(|x|_p, |y|_p)$, which is stronger than triangle inequality

[jolly parrot]

Ann

[pearl pond]

@vestal lodge Has your question been resolved?

[vestal lodge]

is $v_p$ the same as $ord_p$?

[jolly parrot]

goobybalooby

[vestal lodge]

wait, so in the process of proving the triangle inequality, I will also prove that it is non-archimedean?

and now I'm wondering if the case where x and y have the same number of prime factorizations is more involved than im making it out to be lol

[pearl pond]

@vestal lodge Has your question been resolved?

[pearl pond]

@vestal lodge Has your question been resolved?

[toxic lichen]

@vestal lodge sorry for the delay on my part

i am using v_p(n) to mean the exponent of p in the prime factorization of n

[toxic lichen]

if x and y have the same exponent then x+y has the same or potentially higher exponent

[vestal lodge]

no worries, thank you Ann!! I think i can write the proof with this, though I may need more help tomorrow with another proof for this lab lol

.close

[pearl pond]

Channel closed

Closed by @vestal lodge

Use .reopen if this was a mistake.

[pearl pond]

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.

[granite dome]

someone explain me the example 1

<@&286206848099549185>

[pearl pond]

@granite dome Has your question been resolved?

[granite dome]

<@&286206848099549185>

[pearl pond]

@granite dome Has your question been resolved?

[pearl pond]

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.

[severe quartz]

What is wrong with this solution?

[pearl pond]

@severe quartz Has your question been resolved?

[flat field]

from f(-t)=-t, i dont think you can say f(x)=x

[severe quartz]

Why not? I however actually think its the -t+t ive done in the previous step which is wrong because x should be a constant inside the integration and t is a dummy variable

[flat field]

let say g(x) = x^2

and you know that g(1)=1
and from that you are saying g(x)=x

[flat field]

yeh that part also looks sus now

[severe quartz]

You might be right, im not sure

t can be substituted as anything but x

So would t act as a constant for f(x)?

[flat field]

hmmm i am no longer sure

i will think
in the meantime someone else can look at it

[severe quartz]

f(x) is not x

and f(-t) is also not -t

i think the first step was the mistake

f(x)= x(48/23)+ 18/23

[pearl pond]

@severe quartz Has your question been resolved?

[severe quartz]

<@&286206848099549185>

[plush bramble]

t is an integration variable. it's independent of x

similarly x is independent of t, so you can't just set it equal to -t

[severe quartz]

i know that x and t dont depend on each other

[severe quartz]

But why cant i plug in -t in the function?

I agree that the cancellation to 0 is wrong

[plush bramble]

t isn't a free variable. it's a dummy integration variable

[severe quartz]

So the t inside the integration would be different from the t i plug into the function right? As in if i did plug in t the x s would change to t, but the t previously in the integration would be different from the t i just input in the function?

[plush bramble]

So the t inside the integration would be different from the t i plug into the function right?
yes. also that'd be incredibly confusing to have the same variable represent two different things

[severe quartz]

This was definitely a trap question, cuz it was under integer type so i thought 3 must be correct 😭

[plush bramble]

lol

yea you're supposed to do derivatives, and recognize that f(x) is a linear function, then do a bunch of tedious integrals to solve for the slope and intercept to get some two digit number for p+q .

for a further hint, find f(0), f'(0)

[severe quartz]

The thought process is correct?

[plush bramble]

yea i got the same answer

[severe quartz]

Thanks i get my mistake

.close

[pearl pond]

Channel closed

Closed by @severe quartz

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.

[topaz swift]

$(x-2)^2+(y-2)^2=4 ; y=2+\frac{1}{x-2}$

[jolly parrot]

Digicat195

[topaz swift]

how do I solve the system equation?

i got $x^2-4x+\frac{1}{(x-2)^2}=1$

[jolly parrot]

Digicat195

[topaz swift]

but dont know how to get the 4 solutions

[pearl pond]

@topaz swift Has your question been resolved?

[pearl pond]

@topaz swift Has your question been resolved?

[restive furnace]

after subbbing y in terms of x, bring everything to LHS(dont expand (x-2)^2 )and u'll see LHS will become perfect sq

[topaz swift]

what do u mean with LHS

[restive furnace]

left hand side

[topaz swift]

I cant get the perfect square

[pearl pond]

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.

[warm ridge]

Is answer A correct here?

[warm ridge]

<@&286206848099549185>

[pearl pond]

@warm ridge Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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 sundial]

Help

<@&286206848099549185>

[midnight haven]

HELLO

[pearl pond]

@unborn sundial Has your question been resolved?

[unborn sundial]

hi

<@&286206848099549185>

[midnight haven]

hi

[unborn sundial]

can u help

[midnight haven]

on what?

[unborn sundial]

this

[midnight haven]

oh sorry im not in that topic right now, im finding help for finding area and perimeter of shapes

[unborn sundial]

ok

[pearl pond]

@unborn sundial Has your question been resolved?

[pearl pond]

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.

[tall mantle]

To find the period of cos√x via contradiction method

[tall mantle]

Is n=0 allowed

Like are u allowed to equate n to 0

[fiery vortex]

“Wah Huat”

[tall mantle]

'Wahi Baat' means 'same thing' in hindi

[fiery vortex]

ohhh

Wah Huat means smth in my language

lmao

okok good luck

[tall mantle]

How can u know that it's the same thing for n=2, 3,etc

By just looking at the expressions

.close

[pearl pond]

Channel closed

Closed by @tall mantle

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.

[hearty roost]

Hey everyone, I'm completely unsure as to how to start a modular arithmetic problem:

[pearl pond]

@hearty roost Has your question been resolved?

[fathom turret]

you should try to find the period of the series {Fn (mod 5)}

[hearty roost]

Sorry, what’s a period?

[fathom turret]

suppose we have a series like 1 2 3 4 5 1 2 3 4 5

the period of this is 5 since it repeats after every 5th term

[hearty roost]

Ah ok

So how would you go about finding the period for this sequence? For some reason I can’t manage to wrap around the way it defines the Fibonacci sequence

[fathom turret]

try writing down the sequence for a few terms:
it's: 1 1 2 3 0 3 3 1 4 . . .
if at some point the sequence has a "1 1" again then you know the next terms are 2 3 0 3 ... and it repeats

[hearty roost]

I see, thanks!

.close

[pearl pond]

Channel closed

Closed by @hearty roost

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.

[sturdy rose]

Help

[pearl pond]

@sturdy rose Has your question been resolved?

[pearl pond]

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.

[empty sigil]

in the year 1849 bob was 43 years old. he squared his age and found it to be the same number as the year (43^2=1849). john was able to do the same thing during the years 2000-2099. john was born in what year

[ivory swallow]

That's not possible, not all those numbers are perfect squares

[hot stone]

It's true for some year in that range

Not for all of them

[ivory swallow]

That is phrased very poorly then

[hot stone]

Agreed but thats the only interpretation that makes sense

[ivory swallow]

,w lb(2048

[jolly parrot]

Results provided by WolframAlpha

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

Requested by Mr. Gamer#6658

[ivory swallow]

Darn

[empty sigil]

i think its for the only ones possible in that range of years? im not sure

[ivory swallow]

Just guess at a perfect square in that range ig

[hot stone]

I would just square numbers greater than 43 until you find one

It's less than 50 for sure

So not a big task

[empty sigil]

oo

yea tysm! i got it

[hot stone]

Yw

To get the birth year remember to subtract the age from the current year

[empty sigil]

.close

[pearl pond]

Channel closed

Closed by @empty sigil

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.

[tame smelt]

I really need help with this

[tame smelt]

Consider a set E=......

May (x,y) belong to E
Show that x and y have the same sign

[ivory swallow]

Is that even true tho? Consider (x,y) = (1,0)

[tame smelt]

Try x,y= (1/2,1/2)

🗿

Mate

🗿🗿

Lmao

Go get your own channel

[still dawn]

don't be a jerk

don't take someone else's channel

read the how to get help

https://discord.com/channels/268882317391429632/488120190538743810

[tame smelt]

No but we will show later that not every number works in it

So for now try 1/2 for x and 1/2 for y

<@&286206848099549185>

[still dawn]

let me help you

[tame smelt]

Ok thank you

[still dawn]

that shit was pretty bs that you dealt with earlier

[tame smelt]

Yeah lmao

[still dawn]

i'm sorry you had to go through that

[tame smelt]

It's fine no problem

[still dawn]

but you know, internet stuf..................

anyways

[tame smelt]

Yeah it's hard managing the internet

[still dawn]

it seems you have some set E

and its got some rules that define it

[tame smelt]

Yup

[still dawn]

what is it again that you're looking for?

[tame smelt]

Show that x and y have the same sign

Positive positive

Negative negative

Thought about doing a contradiction approach but didn't work out

[still dawn]

but what information do we have to use? and how can we use it?

[tame smelt]

We have logic

Like that's all we have

[still dawn]

but we have the set's properties?

[tame smelt]

That's the whole excercice

That I sent

[jolly parrot]

drviolence

[still dawn]

an inequality right?

[tame smelt]

Yes

[still dawn]

can't we do inequality operations?

[tame smelt]

We can if you want

[still dawn]

and structure the logic maybe secondary?

[tame smelt]

I think so

Any ideas?

[still dawn]

so when you have an inequality of a<b<c, what does that break into?

[tame smelt]

A<b b<c a<c

[still dawn]

nice

[tame smelt]

[still dawn]

mostly we will only need a<b and b<c rather than a<c

[tame smelt]

I see

[still dawn]

but there is a logical operator between them

what logical connective lies between a<b and b<c?

[tame smelt]

Yeah x+y>=1

Lemme see

[still dawn]

here are a list of logical connectives: and, or, not. implies, biconditional

[tame smelt]

x^2+y^2<= x+y

Hmm

[still dawn]

which one is used between a<b and b<c when you break it up from a<b<c?

[tame smelt]

And

[still dawn]

exactly

[tame smelt]

I see

[still dawn]

and also corresponds to what set operators? Union intersection complement

[tame smelt]

Uhhh intersection I believe

[still dawn]

believe?

[tame smelt]

That one is and

[still dawn]

be certain

[tame smelt]

It is

🗿

[still dawn]

perfect

[tame smelt]

Lmao

Oh it's starting to come together

[still dawn]

ok

so there's a lot of things you can do to consider the conditions for E

[tame smelt]

Yeah

[jolly parrot]

drviolence

[tame smelt]

Yeah

Oh wait sorry

Ok carry on I had a mistake on theee

[jolly parrot]

drviolence

[tame smelt]

Hmm

[still dawn]

at least in geometry?

what shape?

[tame smelt]

Oh God geometry my worst enemy

[still dawn]

you know any trig identities?

[tame smelt]

Trigonometry I did study but it's on the tip of my tongue

I studied like 2 years ago so I don't remember it that much

[still dawn]

do you know about parametric equations?

[tame smelt]

What's that

I know about the absolute value if that's of use

I faintly remember it being linked with trig

[still dawn]

i find that graphing helps visualize the idea

[tame smelt]

I don't think we're allowed to add graphs to the mix

Cuz we didn't study them yet

[jolly parrot]

drviolence

[still dawn]

is important

[tame smelt]

I know that x^2+ y^2 SHOULD be positive

[still dawn]

why?

what makes it positive?

[tame smelt]

The square

^2

[still dawn]

yes

[tame smelt]

Hmm

[still dawn]

holdon

[tame smelt]

Okay

[quick bramble]

What went wrong with chasing a c

ontradiction?

[still dawn]

my friend is hoping in

[tame smelt]

Hi

[still dawn]

show him what you were doing for a contradiction

[tame smelt]

Uhh

[still dawn]

we want to see how far you got

[tame smelt]

So we have x²+y²

No I didn't go far at all

I just found a contradiction somewhere else

Which was weird

[quick bramble]

What did you get?

[tame smelt]

Quick question

If we have x²+y²

If we were to add a square root

Would that make it |x+y|?

@quick bramble

[quick bramble]

Over x^2 + y^2?

It wouldn't.

[tame smelt]

Yeah the whole thing

Ah

Then no no contradiction found

[quick bramble]

Well you know those terms sum to less than one.

[tame smelt]

Yeah

[quick bramble]

Solve for one of them.

And then you can take a square root.

And get an absolute value like you wanted.

[tame smelt]

Oh so if both are pos and under 1 then x² is under 1 and y² is under one

Ah.

[jolly parrot]

drviolence

[quick bramble]

Even better, it's true whether or not both are positive.

[tame smelt]

Yeah

[quick bramble]

|x| ≤ 1 and |y| ≤ 1

[tame smelt]

Woah woah

Shouldnt it become |x|

Oh ok

Lmao you scared me for a sec

Okay

[quick bramble]

So now you've used one condition.

You can use this fact with you other condition and search for your contradiction.

Assuming x and y are not the same sign.

1 ≤ x+y.

[tame smelt]

Ah yes

But how could that end up showing that they both have the same sign?

@quick bramble

[quick bramble]

I'm saying assume they have different signs, and look at the expression 1 ≤ x+y.

[tame smelt]

But what if x = 6 and y= -3 they have different signs yet it is true

That is what caused me to not understand

[quick bramble]

x^2 + y^2 ≤ 1 prevents that.

[tame smelt]

Ah.

[quick bramble]

Do you know what that region looks like?

[tame smelt]

Region?

[quick bramble]

The set of (x,y) pairs that satisfy the inequality.

[tame smelt]

No but they did say to show it in a later question

For now 1/2,1/2 works

So does 0,1?

[quick bramble]

Test it.

[tame smelt]

It works yes

1/4<=1<=1

Because or =

[quick bramble]

0^2 + 1^2 = 1 ≤ 1, so 0,1 works too.

[tame smelt]

Yup

[quick bramble]

Neither x or y can be larger than 1.

Agree?

[tame smelt]

Yup

So I need to write that up in writing form

[quick bramble]

Now look at 1 ≤ x + y.

[tame smelt]

One thing is x<=1 and y<=1
So x+y<= 2 or not that confuses me

[quick bramble]

Assume one of these numbers is negative.

[tame smelt]

Oh okay

Yes

[quick bramble]

What does that say?

[tame smelt]

It's gonna be false

[quick bramble]

Because...

[tame smelt]

It would be lower than 1

[quick bramble]

Yeah, and now you have a contradiction.

[tame smelt]

Indeed

[tame smelt]

So I'm writing it up and does this work or no

[quick bramble]

That isn't relevant to the argument we've laid out.

[tame smelt]

Yeah yeah I understand the argument

I'm writing

I'm here

Trying to write x+1

X+y

[quick bramble]

Rewrite the first condition: x^2 ≤ 1 - y^2.

[tame smelt]

Okay

[quick bramble]

Taking roots: |x| ≤ sqrt(1-y^2).

[tame smelt]

Okay

[quick bramble]

Since y^2 is positive, then that right hand side is less than 1.

[tame smelt]

Hmm

[quick bramble]

That's how you get the |x| ≤ 1 result.

[tame smelt]

Ah so |x|<=1

Ah ok

We do the same for y

Right?

[quick bramble]

Yeah, it's true for y by symmetry.

[tame smelt]

Ok so now x+y

[quick bramble]

If they are opposite signs, you want to show that x+y < 1.

That contradicts 1 ≤ x + y.

[tame smelt]

Aaah

Ok so

I'm here now showed that -1<x<1

And so for y

"We have x+y>= 1
If they have different signs then x+y<=1 which is a contradiction "

[quick bramble]

You want to word that with an explicit inequality.

[tame smelt]

Hmmm

How so

[quick bramble]

So we can assume one is positive and the other negative.

Say x is positive.

[tame smelt]

Oh yeah

[quick bramble]

There are two cases to check, |x| ≤ |y|, and |y| ≤ |x|.

[tame smelt]

Something along these ligns

[quick bramble]

You can write x + y as x + (- |y|).

Or even |x| - |y|.

[tame smelt]

Why tho

[quick bramble]

Definition of absolute value.

If y is negative, then |y| = -y.

[tame smelt]

Ah

[quick bramble]

So basically, you are trying to write everything in symbols.

[tame smelt]

Hmmm

[quick bramble]

And you're trying to construct x+y < 1.

[tame smelt]

Yes

[quick bramble]

So you make whatever substitutions you need along the way.

[tame smelt]

So it would be.... x-y<1? Since |y|=-y

[quick bramble]

Since x is positive, you can say x = |x|.

[tame smelt]

Yes

[quick bramble]

And put that in.

Check the two cases of |x|-|y|.

When |x| ≤ |y| and |y| ≤ |x|.

[tame smelt]

[quick bramble]

Also using |x| ≤ 1 and |y| ≤ 1 from that first condition.

[tame smelt]

Ok lemme see

I am utterly confused and need to get my crap together lmfao