help-0

[real cipher]

i mean what to do at last

i have no idea how to use this property

[uncut torrent]

you're not trying to find one solution, you're trying to find a range of solutions

i'd assume

[real cipher]

i know that but how do i said it

[uncut torrent]

you can certainly use this property for 93 and 94

by getting the numerators as 1

[real cipher]

for question no 96 i got value like -3,-2,-2/3 all greater then equal to what should i do?

no wait just greater then

x > -3, x > -2 & x > -2/3

[real cipher]

i didn't use that property in that question isn't x should be in the middle so i just solved by multiplying by x^2

[uncut torrent]

you can use that property though

like by dividing through the whole inequality to get 0<1/a<1/b

[real cipher]

wdym

by that

[uncut torrent]

hence, 10 > 3x > 0

so 10/3 > x > 0

[real cipher]

give me a minute to process

[real cipher]

did u take lcm of numerator that intersting

first time seeing it

[uncut torrent]

well

you can do the same as divide through everything by 2

to get the first fraction as 1 in the numerator

[real cipher]

one quick queston

[uncut torrent]

then by 3 to get the second fraction as a 1 in in numeator

[real cipher]

can we do 2 * 1/x < 3/5
2*X < 3/5 ?

[uncut torrent]

no

thats not in the form 1/a > 1/b

[real cipher]

okay

for question 92

divide by 2

and
3x+5 <= -3 ?

[uncut torrent]

close

[real cipher]

is it in form of 1/a if we divide by 2 i mean to say

[uncut torrent]

the left side is in the correct form, but the right isnt

[real cipher]

now can we do 3x+5 <= -3/2 ?

oh no

wait

i don't get it

[uncut torrent]

do u understand what i did in the pic

[real cipher]

yessir

[uncut torrent]

you can turn the right hand side into the correct form

because a = 1/(1/a)

[real cipher]

no

but we can take it to the left side and use other property

[uncut torrent]

you can do that too yeah

[real cipher]

this is correct

[uncut torrent]

doesnt really apply here

because neither side is 0

[real cipher]

if we take right hand side value to left hand side then?

[uncut torrent]

no because then there would be two fractions on one side

[uncut torrent]

you can combine the fractions but i think it'd be easier to do this way

[real cipher]

okay i give up
teach me

[pallid scarab]

I think it would be easier to use the following : if $a < b < 0$, then $\frac{1}{b} < \frac{1}{a} < 0$

[ocean seal]

rafilou2003

[real cipher]

@pallid scarab teach me

i mean in general sense not the specific question

[pallid scarab]

You can start back from this line here

The 2nd line

[uncut torrent]

thats your teachers job :)

[real cipher]

self-studying

[uncut torrent]

ah

khan academy ?

[real cipher]

no a math book

[pallid scarab]

Let's go back to $\frac{1}{3x+5} \leq -\frac{3}{2} < 0$

[ocean seal]

rafilou2003

[real cipher]

okay

[pallid scarab]

That's brave

[real cipher]

been a two months

[uncut torrent]

i mean try khan academy

[pallid scarab]

Ok, I don’t know if you've seen this rule with inequalities before, but this is an important rule :

[real cipher]

i have account but video and exercise are to passive this book have tons of question and well written

okay

[pallid scarab]

If $a \leq b < 0$, then $\frac{1}{b} \leq \frac{1}{a} < 0$

[ocean seal]

rafilou2003

[real cipher]

yeah I get that
1/2 > 1/4

[pallid scarab]

Yeah but be careful this is when both a and b are of the same sign

[real cipher]

explain

[pallid scarab]

This rule only works if either both a and b are positive or they are both negative

[real cipher]

okay

[pallid scarab]

Can you think of an example where this rule becomes false ?

[real cipher]

oh so that why < 0 and > 0 are there

[pallid scarab]

For example, a = -1, b = 1

[real cipher]

i got so that 0 at the end is it for that reason ?

[pallid scarab]

a < b, however 1/a < 1/b

[pallid scarab]

Yes

[real cipher]

okay

[real cipher]

if they are of same sign

[pallid scarab]

Let's now apply this rule to here

[pallid scarab]

I took example a = -1 and b = 1 so in this case it didn't work

[real cipher]

we are on same page then

okay moving on

[pallid scarab]

Now, can you apply the rule we just saw to this ?

[real cipher]

they are of same sign since they both are less then zero

but there is -3/2 do we have to somehow make it -1/2 ?

[pallid scarab]

Nono

let 1/(3x+5)

Be equal to a

[real cipher]

okay then b?

[pallid scarab]

And -3/2 = b

So we have $a \leq b < 0$

[ocean seal]

rafilou2003

[pallid scarab]

What can we deduce from this ?

[real cipher]

so
$$ \frac{1}{\frac{-3}{2}} \leq 3x+5 < 0 $$
something like that

never use latex before : PP

[pallid scarab]

Well that's nice enough

Let's not forget that 3x+5 < 0, might be relevant later

[ocean seal]

JXHN

[pallid scarab]

Yes

[real cipher]

$$ \frac {-2} {3} \leq 3x+5 < 0 $$

[pallid scarab]

Nice

[ocean seal]

JXHN

[real cipher]

now we just solve for x?

[pallid scarab]

Yes, I think you've done it a lot of times already

This shouldn't be anything new

[real cipher]

yeah but be around here I still have some doubt that i am unable to clear out but first let me solve this

$$ \frac {-2}{3} - 5 \leq 3x < -5 $$
$$ \frac{-2-15}{3} \leq 3x < -5 $$
$$ \frac{-17}{9} \leq x < -5/3 $$

[ocean seal]

JXHN

[real cipher]

okay i think this just about right even if it's not in standard form of inequality

[pallid scarab]

It is correct !

[real cipher]

@pallid scarab i just realized the problem we have don is 92 but there is < 0 did we just have to added it according to our intutative?

intuation(i can't spell it)

[pallid scarab]

The problem if you don’t add the < 0 is that you would lose information

When you solve inequalities like this, you have to work your way with equivalences, the symbols that look like "<=>"

This symbol means that we can freely go one way or the other without losing information

[real cipher]

hmm i don't get it what are u saying

anyway one of my friend here in server have me taught another that was not in my book but your process seems helpful since i was able to use the thing from the book that i learned

• Ann

anyway x is already in middle here

question no 93,94,95,96

[pallid scarab]

This is another way

[real cipher]

i understood that process but gave me lot of confusion since i didn't learn that way

intersection something

[real cipher]

as i was saying all x are in middle and there is another property here should we seprate the inequality and solve them separately

i did that and got many value

[pallid scarab]

Here is a better property that gives you more information :
If 0 < a < b, then 0 < 1/b < 1/a
If a < b < 0, then 1/b < 1/a < 0

[real cipher]

for question no 96 i got value like x > -3, x > -2 & x > -2/3 in this case what should i do

[real cipher]

but this is done onely to take x(unknown range) in middle right?

and then solving it

for the question i am doing all are in middle

or am i being dumb please continue explaining if u want

u are really helpful

[pallid scarab]

Well I'm sure you could apply this property I just showed you to exercices 93-96

You just have to let a = ... and b = ... the correct values

[real cipher]

correct values?

[pallid scarab]

Weird, for number 96 I find x > -1

Let's do it together to find out what went wrong

[real cipher]

hmm i maybe did algebraic mistake somewhere

okay

[pallid scarab]

So we start with $0 < \frac{1}{3x+6} < \frac{1}{3}$

[ocean seal]

rafilou2003

[real cipher]

let me do some step

[pallid scarab]

Try to use this useful tool

[real cipher]

$$ 0 < 3x+6 < 3 $$

[ocean seal]

JXHN

[pallid scarab]

Be careful

a and b are in reversed order after you take their inverse

[real cipher]

okay wait

$$ 0 < 3 < 3x+6 $$

[ocean seal]

JXHN

[pallid scarab]

Correct!

[real cipher]

do we know 3x is negative or postive?

since it < 0 it have to negative rightL

$$ 0 < 3 - 3x < 6 $$

[ocean seal]

JXHN

[real cipher]

$$ -3 < -3x < 6 -3 $$

[ocean seal]

JXHN

[real cipher]

$$ 1 > x > -3 $$

[ocean seal]

JXHN

[real cipher]

@pallid scarab i think i did some mistake somewhere i don't know where this is heading

[pallid scarab]

This was correct, why did you subtract x ?

From here you can subtract 6 and then divide by 3

[real cipher]

so,
$$ -2 < -1 < x $$

[ocean seal]

JXHN

[real cipher]

something like that ?

$$ x > -1 > -2 $$

[ocean seal]

JXHN

[real cipher]

so x > -1

i see

[real cipher]

i want to take it middle to solve it

[pallid scarab]

It isn't always the case

[real cipher]

suppose the i have two inequality and solving in one big horizontal line if i want to add +3 to any side do i have to add it on all side?

[pallid scarab]

Yes

[real cipher]

i think i got the general idea thanks gonna sleep now it's 11PM

.close

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[rigid fog]

.rotate

i need help

i dont know where ti beginning

[floral mesa]

The cosine is confusing me is it $\cos{\frac{\pi}{2}}\times(t-2)$?

[rigid fog]

yes

i believe so

[floral mesa]

based off the assumption that the stick is submerged at midnight since idk how tides work, you should be able to just solve for t by plugging in 72 for d

[stuck sinew]

yeap

so find out when d is

then solve for t

[floral mesa]

okay cosine is like that but same thing should hold true, plug in d and solve for t

I got something around 3.9 but I don't think that's quite right if the wave is traveling to the right

[lone heart]

@rigid fog Has your question been resolved?

[lone heart]

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[trail cape]

I tried solving like this, is it correct?

[lone heart]

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[last saddle]

what happens in the final line? i dont get how they get the vector eqn

[lone heart]

@last saddle Has your question been resolved?

[last saddle]

q = 2

[lone heart]

@last saddle Has your question been resolved?

[last saddle]

<@&286206848099549185>

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@last saddle Has your question been resolved?

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[visual zealot]

how do i solve this for x and y? I tried to use sin58 to get x value but it was clearly wrong any suggestions?

[lone heart]

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

[visual zealot]

it stucks at ×=10/sin58 I tried to get sin58 to get x value

[wary stream]

Because you set up the equation wrong

Do you know SOH CAH TOA?

Like can you set up the proper equation?

[visual zealot]

you mean opposite/hypotenuse

[wary stream]

Yes

[visual zealot]

I tried that way

let me try again

[wary stream]

So you know that sin theta = opp/hyp, correct?

[visual zealot]

correct

[wary stream]

If you fill in what you know, what is the equation?

[visual zealot]

sin58=×/10 , I want to find x so it should be 10 times sin58 right?

[wary stream]

Yes

And what do you get?

[visual zealot]

I dont know the value of sin58

[wary stream]

Use a calculator

[visual zealot]

8.5 ish then

[wary stream]

Yes

[visual zealot]

Oh I got it

thanks!

!close

[wary stream]

.close

[visual zealot]

I did same thing as before but couldnt solve it is there something different?

[tacit arch]

,tex .sohcahtoa

[ocean seal]

riemann

[tacit arch]

identify the adjacent side to angle = 31

[visual zealot]

I tried this but I get like 0.33 for y value

[tacit arch]

!show

[lone heart]

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

[tacit arch]

,calc 5/sin(59 deg)

[ocean seal]

Result:

5.8331669860767

[tacit arch]

,calc 5/sin(59)

[ocean seal]

Result:

7.8525232418039

[tacit arch]

i have no idea how you get 0.33

but also you used the wrong trig function

[tacit arch]

read this again

[tacit arch]

and answer this question

[lone heart]

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[lone heart]

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[daring eagle]

I was doing this problem, and thought I had the right answer.

I got this

However

The judge said that the answer was 2-cbrt8y instead of just cbrt8y, and I don't know why.

picture of graph^

[placid zinc]

The shell is a cylinder with radius y, and height 2 - ³√(8y)

One side of the cylinder touches the green line, the other side touches the blue curve. So, we need the length between those two.

That's where 2 - ³√(8y) comes in

[daring eagle]

how is the height 2 - ³√(8y)?

i just dont understand where the 2 comes from.

Oh wait

it's the x=2

and it doesnt affect the bounds, since those are just 0 to 1

is this an accurate representation of why?

[placid zinc]

Perfect

[daring eagle]

great, thank you!

.close

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[astral lion]

Hello everyone, I'm seeking methodical advice, more specifically I'm reading Serge Langs calculus book (its worth mentioning that Im calculus beginner - i understand some important concepts but calculus-computation-wise im a beginner) And i stumbled upon a practice problem which says " Prove the following inequalities for all numbers x, y. 22. |x+y| ≥ |x| - |y| [Hint: Write x = x + y- y, and apply Theorem 2.3, together with the fact that |-y| = |y|. Theorem 2.3: |x+y| ≤ |x| + |y| " My question is: how should I approach this problem/how should I go about proving/disproving it?

[astral lion]

I would like to understand each step and logic behind each step so I can replicate the same logic as I continue reading

[astral lion]

I tried bunch of nonsense thats probably not even worth mentioning since I doubt it classifies as "my work" but I tried squaring both sides, i tried subtracting and even splitting into two equations

Where am i stuck? at the beginning, I have no idea how to begin solving the question. One of the things I tried and forgot to mention is that I posted the question into ChatGPT chatbox and I found out that it sucks at basic arithmetics and generates logical fallacies like " 1 ≥ -1, is not true therefore |x+y| ≥ |x| - |y| does not hold for all x,y in R"

Its worth mentioning that I understood everything clearly, prior to this practice problem in the book, even the proof of previously mentioned Theorem 2.3 and absolute value axioms

[jagged raptor]

this is equivalent to solving |x| \leq |x+y| + |y|

might make it easier to see why their hint is useful

[astral lion]

could you please rewrite the equation since i dont understand the "\leq" parameter

[jagged raptor]

$|x| \leq |x+y| + |y|$

[ocean seal]

michαel

[jagged raptor]

use the first hint on the |x| part

[astral lion]

hmm surprisingly i did not think about this possiblity, thank you very much for pointing me the correct direction i will give it a go surely.

.close

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[slender quiver]

?

[tacit arch]

don't spam

.close

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[surreal meadow]

!status

[lone heart]

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

[verbal harbor]

6

[surreal meadow]

please expand on that

[limpid turret]

[placid zinc]

Has anyone ever actually said 6 to this before? How do we go from here

[unkempt vector]

[verbal harbor]

1

[surreal meadow]

there was that mom that was getting answers for their kid lol

[verbal harbor]

first principle means chain rule correct

[surreal meadow]

no

[verbal harbor]

oh

[placid zinc]

No, means the limit definition

[verbal harbor]

\

so its not this

[placid zinc]

It is that, haha

[verbal harbor]

oh

[placid zinc]

With a lim on the beginning of it

And over an h

So it's kinda like that. A little bit.

[verbal harbor]

so like this

[placid zinc]

This should look familiar, hopefully

[placid zinc]

Yes that looks right

[verbal harbor]

ok is the answeer 1/sqrt 2x +1

[placid zinc]

I mean yes

But the work is the more important part to a question like this

[verbal harbor]

are those all the correct steps

[placid zinc]

I was under the impression you didn't send all of your work

I mean as long as you know what each step is doing, you're good

[verbal harbor]

alright thanks

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[grizzled zephyr]

could someone help me with this

[grizzled zephyr]

i got to here

[carmine reef]

what do you want to do with the derivative next

[grizzled zephyr]

idk thats why am askng

i usually have points

to substitute to find slope

[carmine reef]

What info have you been given instead of points

[grizzled zephyr]

horizontal

means x = 0

[carmine reef]

no

[grizzled zephyr]

what

[carmine reef]

that's not what horizontal means

[grizzled zephyr]

why

(0,5 ) is a horizontal line

[carmine reef]

That is a point

[median oar]

That’s a point

[carmine reef]

Not a line

[grizzled zephyr]

um

so what do i do

[carmine reef]

,w horizontal

[ocean seal]

Results provided by WolframAlpha

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

Requested by thewizardofOU#0205

[carmine reef]

ok well that's unhelpful

it just means flat

parallel to the x-axis

slope of 0

[median oar]

What can you tell me about the slope of a horizontal line

[carmine reef]

oop

[median oar]

Or not

Haha

[grizzled zephyr]

yes

so i put 0 = 2cosx(1+sinx)

now i solve for 2cosx and 1+sinx

[carmine reef]

yes probably

[grizzled zephyr]

i get (pi/2 , y) ( -pi/2,y)

[carmine reef]

if by solve for them you mean what I think

[grizzled zephyr]

how do i get the y

[carmine reef]

what did you do?

[grizzled zephyr]

2cosx= 0

arccos

of 0

[carmine reef]

ah

[grizzled zephyr]

= pi/2

[carmine reef]

I see

[median oar]

Note that cos and arccos are not exactly the same

Arccos is only defined for the principle domain

[carmine reef]

Those are indeed only some of the points

[median oar]

What do you know these points have to satisfy?

[grizzled zephyr]

can you explain this point

[grizzled zephyr]

wdym

[median oar]

That’s exactly what we’re about to do

[median oar]

You’ve found the x value by constraining “the tangent line needs a slope of 0”

What other constraint do the points have?

[grizzled zephyr]

they should be one to one function?

lol idk

[median oar]

Nah

Look at the question

[grizzled zephyr]

oh you mean they exist at each 2npi?

[median oar]

The points have to be on the graph

As in, they have to satisfy the original equation

[grizzled zephyr]

ohhh

got it

thanksss

[lone heart]

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[elfin veldt]

Hmm try using the addition formula for tan ( x + pi/4 )

[fallen verge]

you also didnt get anything wrong

you can evaluate sin(pi/4)

you can evaluate trig functions when given a value

what is tan(pi/4)?

[elfin veldt]

You can multiply both the numerator and denominator by the conjugate of 1 - tanx

1 + tan x

When you multiply you should get (tan x + 1) (1 + tan x) / (1 - tan x) (1 + tan x)

you can simplify the numerator and denominator

[elfin veldt]

You can simplify to get (tan x + 1 + tan^2 x + tan x) / (1 - tan^2 x)

[fallen verge]

did you not just evalute tan(pi/4)

[lone heart]

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[mighty berry]

How would I begin proving this? This has to do with Euler's Theorem: phi(n) is the number of relatively prime numbers between 1 and n

[mighty berry]

And phi(n) can be calculated by $\phi(n) = n(1 - \frac{1}{p_1}) \cdots (1 - \frac{1}{p_r})$, where each p is a prime factor of n.

[ocean seal]

Minighost

[mighty berry]

So, it makes sense intuitively that each p prime factor would come in a pair... But how do I prove that?

I will be working on some other problems in the meantime, please ping me if you respond!

[fallen verge]

ooh number theory

@mighty berry you around?

[mighty berry]

Yes!

[fallen verge]

ok

so a factorization of totient(n) would be as you showed above

one thing we can do is break down the factors of n and distribute them into the other factors to clear the denominators

are you following so far?

[mighty berry]

Yes, I think so...

[fallen verge]

for example, totient(12)=12(1-1/2)(1-1/3)

so we can take a factor of 2 and 3 and distribute them in

you get 2(2-1)(3-1)

which gets you 4

does that example make sense?

[mighty berry]

I think so? You make 12 into 2 * 2 * 3, then distribute 2 and 3 into the other terms?

[fallen verge]

yes

[mighty berry]

Ok!

[fallen verge]

one thing you notice is it makes the expression $(n_{new})(p_1-1)(p_2-1)...$ for any number

[ocean seal]

GarlicBredFries

[fallen verge]

with p_n being prime

you can take this even further by writing $n=p_1^{a_1}p_2^{a_2}...$

[ocean seal]

GarlicBredFries

[fallen verge]

do you know what that would make n_new?

[mighty berry]

Woah, wait, how'd you get to the $n=p_1^{a_1}p_2^{a_2}...$?

[ocean seal]

Minighost

[mighty berry]

Did you mean writing n_new like that?

[fallen verge]

thats how you represent a number as its prime factors

for example $12=2^2*3^1$

[ocean seal]

GarlicBredFries

[mighty berry]

Oh, okay. Wait...

I was going to say you could write n_new like that, but how?

[fallen verge]

if you think about it, any prime must be used once in order to clear the denominators

[mighty berry]

Once you do that, n_new will be whatever's left after taking all the prime factors out of the original n.

Wouldn't that also be a prime?

[fallen verge]

it wouldnt be a prime

it would also be written as prime factors

each prime factor is used once to clear denominators

so $n_{new}$ would look like $p_1^{a_1-1}p_2^{a_2-1}...$

[ocean seal]

GarlicBredFries

[mighty berry]

Wait, you mean like, we're now at $\phi(n) = n_{new} (p_1 - 1)(p_2 - 1)...$, but $n_{new}$ is equal to what you just typed?

[ocean seal]

Minighost

[fallen verge]

yes

now if totient(n)=3, what would that imply about its factors?

[mighty berry]

I don't know.

There has to be an odd number of factors?

Oh! All of its factors divide 3! (Or, at least, all p_i - 1 | 3, for all i = 1..k)

Man, that took me way too long to realize...

Ok, I think I got the proof done.

.close

[lone heart]

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[chilly idol]

Need to find angle measure of DE

Not sure where to begin

[quiet summit]

So you know the arc length is 8.73 in and the radius is 10 in. Is there perhaps a formula for the angle?

[chilly idol]

formula isnt provided

[quiet summit]

Mhm. But there is one.

[chilly idol]

there is

[quiet summit]

Yes.

[chilly idol]

ok

[quiet summit]

what class is this for?

[chilly idol]

geomotery

Geometry *

[quiet summit]

ok perhaps this formula is in your notes?

[chilly idol]

nope

[quiet summit]

This is a pretty common and useful formula

Ok

[chilly idol]

is it the angle sector formula

[quiet summit]

yes

[chilly idol]

ok

[quiet summit]

That gives you the area

of the shape

[chilly idol]

ok

[quiet summit]

but if you are trying to find the the angle can you manipulate the formula a little bit to give you what you want?

[chilly idol]

not sure

not sure what to manipulate

[quiet summit]

arc length formula is

s = 2pir*(theta/360)

[chilly idol]

ok

im doing it right now

just to let you know

i got 493 degreees?

for theta

[quiet summit]

so intuitively, that shouldn't make sense

[chilly idol]

ya

[quiet summit]

493 is more than a 360 degree circle

[chilly idol]

yup

[quiet summit]

so probably not

lets walk through your work

[chilly idol]

ok

[quiet summit]

how did you get 493

[chilly idol]

heres my work

360 x 1.4 =493

[quiet summit]

um 8.73/20*pi is not 1.4

[sour dove]

8.73/20pi is not 1.4

[chilly idol]

let me do that again

[quiet summit]

also is this supposed to be in radians or degrees

[chilly idol]

degrees

the calculator is set to degrees

[sour dove]

20 * pi is a bit more than 60

so 1.4/(20pi) should be a small number close to 1.4/60

aka if you're getting anything larger than 1 then something went wrong

[chilly idol]

heres the calculator

[sour dove]

so it's first doing (8.73 / 20) then multiplying by pi

you need to do 8.73 / (20 * pi)

[chilly idol]

oh

now its 0.14

im getting

the answer i got is 50.4

0.14 x 360

.close

[lone heart]

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[chilly idol]

thanks

for your help

[lone heart]

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[feral sonnet]

Hello

[feral sonnet]

I got this wrong on geometry test

I was trying to prove its a trap because only 1 set of paralell sides / same slope

[vale wigeon]

"use mathematics"

do you have any feedback on your wrong proof

[feral sonnet]

Now im forgeting how to get the slope on that

This feedback doesnt help me at all

Wouldnt the slope for QU be 3/5?

if not how would I find it

[fast lichen]

y2-y1

6-9=-3

it would be odd to have positive slope for QU when its going down right

[feral sonnet]

good point

is y2 always the one on the right?

[fast lichen]

yes

[feral sonnet]

idk why but my teacher made it sound like they are the same

thanks for the help

[fast lichen]

it technically is but i like looking at it that way

[feral sonnet]

do you mind helping me with another?

I thought I got this one right I just didnt do it the simple way

[fast lichen]

distance formula

is just pythagoras for finding hyp of this right triangle

[feral sonnet]

I know how to do distance formula but wouldnt I still be correct?

[fast lichen]

oh i thought you were asking that i havent read your thing 2 sec

yes but dont convert to decimals

[feral sonnet]

alr thanks for the help im going to see if he can change my grade

[fast lichen]

$\sqrt{40}^2+\sqrt{10}^2=40+10=50$

[ocean seal]

Køter

[fast lichen]

and then sqrt(50) ofc

sure you can do that

seems kind of odd yea to grade that wrong if they didnt explicitally ask for distance formula

[feral sonnet]

so when it says simplest radical form I can say $\sqrt{49.999}

[fast lichen]

oh nvm simplest radical form

its not sqrt(49.999) though its just because you converted to decimal midway

which you should never do btw

its sqrt(50)

[feral sonnet]

alright

big help thank you

• I forgot how to un occupy channel

[fast lichen]

.close

[lone heart]

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[fast lichen]

.reopen

[lone heart]

✅

[fast lichen]

@feral sonnet i never heard of simplest radical form before so i just looked it up

apparently it you cant have a square number as a factor in the radicand

so sqrt(50) is not simplest radical form

because 50=25 * 2 and 25 is a square number

[feral sonnet]

would it be 5sqrt2

[fast lichen]

yes

.close

[lone heart]

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[wheat isle]

How should I start this?

[wheat isle]

rip Tex

😭

[alpine sable]

Do you know what f'(x) is

[wheat isle]

derivative

[alpine sable]

and you know how to find the derivative of a function right?

[wheat isle]

yeah

[alpine sable]

so

[wheat isle]

But finding f’(x) and f’(-1) are different things right

[alpine sable]

Bro just plug in -1 as x

???

[wheat isle]

is that it

why don’t they just say f(-1)

[alpine sable]

because $$f(x)$$ is not $$f'(x)$$

[ocean seal]

Brandon H

[wheat isle]

But I’m just subbing in -1 right?

so that would just be the same as finding f(-1)?

[tacit arch]

f(x) = sin(x)

what's f'(pi/2) ?

[alpine sable]

If they're not the same how would they be the same?

[wheat isle]

sin(pi/2)?

[tacit arch]

is f'(pi/2) = sin(pi/2)?

of course not

f'(x) = cos(x)

f'(pi/2) = cos(pi/2)

$f'(x) \neq f(x)$

[ocean seal]

riemann

[tacit arch]

maybe it's easier for you to understand that if it's on one line with a not equals sign

[wheat isle]

but here if I’m just finding f’(-1) all I do is sub in -1 for x?

[alpine sable]

what is the derivative of

$$f(x)=2x^5-7x^3+5x-4$$

[ocean seal]

Brandon H

[wheat isle]

10x^4 - 21x^2 + 5

and then I sub in -1?

[alpine sable]

Yes

[wheat isle]

oh

bruh

I thought you meant to just plug in -1 only for x lol

[tacit arch]

yes. into the derivative

[tacit arch]

that's why this was asked

[tacit arch]

then this was instructed

order matters

[wheat isle]

I see

.close

[lone heart]

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[jolly blade]

Looks like you missed solutions in the form 5π/9 + 2πk/3

those are equivalent

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[rain kraken]

I've been trying to solve this but im really stuck, I need help

[rain kraken]

It's been hours and i'm not joking

[gleaming ridge]

What is the question!

[rain kraken]

I have to prepare this for inverse laplace

I can't find a way to modify the denominator properly

[gleaming ridge]

Partial Fraction Decomposition?

You have to do that

But before that, you need to find the poles of your transfer function

[rain kraken]

there should be a way to turn it to that

[gleaming ridge]

yeah yeah, you have a third order system

[rain kraken]

I see

I'll check that out thanks

[gleaming ridge]

Do you know how to solve 0.5 s^3 + 1.5 s² + s = 0

To determine the poles

Okay try!

[rain kraken]

not really

I've realized that im extremely rusty

at basic algebra

like factorization and stuff

[gleaming ridge]

No no haha

It's third degree polynomial

No formula

Well, there is, but too complicated to be used

So, we look for trivial roots

[rain kraken]

oh I see

[gleaming ridge]

substitute by values like 0, -1, 1, 2, -2

Maybe you get lucky and find a root

Then you factorize

3rd degree polynomial = 1st degree polynomial * 2nd degree polynomial

Does this ring any bells?

Allright, I'll let you work a little

Focus on finding the poles

We'll talk about the partial fraction later

[rain kraken]

ok, thank you so much

[lone heart]

@rain kraken Has your question been resolved?

[lone heart]

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[mint glen]

Let $F$ be a field, and $a\in F$, with $a\ne0$. If $n\in\mathbb{Z}^+$, prove that $a^{-n}=(a^n)^{-1}$

[ocean seal]

Platinum_Saber

[mint glen]

Should this not be a defintion?

how do we define negative exponents?

[placid swallow]

1/a^n?

[lone heart]

@mint glen Has your question been resolved?

[pliant cedar]

But isn't division defined to be multiplication with the multiplicative inverse

[lone heart]

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[tame kindle]

.close

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@tardy quartz Has your question been resolved?

[tardy quartz]

<@&286206848099549185>

[lone heart]

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[tardy quartz]

.close

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[noble notch]

how would I start this questions

[vale wigeon]

let the equation of your double-tangent line be y = ax + b

you want the equation x^4 + 2x^3 - 7x^2 + 11 = ax + b to have two double roots

[noble notch]

yeah

i got that part

but from there i’m not sure how to solve for a or b

[vale wigeon]

since this is a 4th degree equation, and two double roots counted with multiplicity already gives 4, you can say for sure that there are no more roots

so x^4 + 2x^3 - 7x^2 - ax + 11 - b = (x-r)^2 (x-s)^2

[noble notch]

ok

so i could try divide with one of the factors?

[alpine sable]

Expand the right hand side and compare by power of x: this gives you equations for a,b,r,s. For example, the term proportional to x^3 gives you the equation 2= -2(r+s). The term proportional to x^2 gives you -7 = r^2 + 4 r * s + s^2 etc.

,w solve {2 == -2 (r + s), -7 == 4 r s + s^2 + r^2, -a == -2 r^2 s - 2 s^2 r,
11 - b == r^2 s^2}

[ocean seal]

Results provided by WolframAlpha

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

Requested by Landau08#4573

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[fathom tundra]

what's wrong with my answer?

[lone heart]

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[fathom tundra]

what's wrong with my answer?

idk why they expect me to do 14.5cos(2pi(t-206/365))+63.5

what's the difference if the results are the same

[alpine sable]

Did you notice that t is to be measured in years, not in days?

[fathom tundra]

well I guess the period IS defined as "1 year"

smh.. lol

.close

.close

weird

[lone heart]

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[lone heart]

@grizzled garden Has your question been resolved?

[lone heart]

@grizzled garden Has your question been resolved?

[teal ember]

clarify the question first

[lone heart]

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[safe tartan]

Hello! Need some help on an integral for sin^3x dx using u=cosx thanks!

[safe tartan]

.close

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[weary abyss]

can someone walk me through step by step on how to do this please

[lone heart]

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

[weary abyss]

1

i dont know where to begin

@mellow grail ?

[mellow grail]

i just did it so that another helper doesnt have to ask

idk how to solve ur problem, sorry

[copper delta]

what does each integral mean

[weary abyss]

idk

[copper delta]

if you were to read it out in english

[weary abyss]

i still dont know

[copper delta]

the first one is the area under the curve from -3 to 6

[weary abyss]

yes that makes sense

second one is -3 to 0

[copper delta]

yes

[weary abyss]

and so on

i dont really understand what im supposed to find

it tells me

idk how to solve actually

@copper delta?

[copper delta]

the area under the curve from -3 to 6 = area from -3 to 0 + area from 0 to 3 + area from 3 to 6

you have everything to solve now

[weary abyss]

i dont know how to solve

i know how to get the antiderivative and stuff

[serene isle]

always diagram

[weary abyss]

what do i do with the numbers

[serene isle]

5+2+x = 10

[weary abyss]

7 - 10?

oh

7 + x = 10

so x = 10 - 7?

that would be 3 but the answer is suppose to be -3

[copper delta]

oh from 3 to 0

weird q

i guess somehow that is negative

[weary abyss]

how do i get -3

[serene isle]

$\int^3_0 f(x) dx= 3$

[copper delta]

dx

not there

[ocean seal]

Poelymole

[serene isle]

lmao yeh mb

[copper delta]

lol

[serene isle]

but we have $\int^0_3 f(x) dx$

[ocean seal]

Poelymole

[serene isle]

which is kinda like the inverse

[weary abyss]

huh

[copper delta]

cus ur question sucks

[weary abyss]

not in my hands

[serene isle]

ok idk if this will help but the top one would be found as $\int f(3)-\int f(0)$ but the bottom one is $\int f(0) - \int f(3)$

obviously +c but we dont care

[ocean seal]

Poelymole

[weary abyss]

shouldnt the answer be 3

wait no

-3 because a = 3 and b = 0 right?

it would be 3 if the numbers were switched?

a = 0 b = 3

this question is so dumb

[serene isle]

if the question wasnt a trick question it would be 3 but bc they switched the top and bottom in that last integral the answer is -1x3 = -3

[weary abyss]

so because its switched, the answer is -3 but if it was normal, it would be 3

[serene isle]

yep exactly

[weary abyss]

wow

[serene isle]

hence

[weary abyss]

do you think you can help me with another question

[serene isle]

try me

[serene isle]

ok immediately i want to get rid of that root x at the bottom

[weary abyss]

would the denominator be x^1/2 + 1

x^3/2?

or do i move the denominator up top

[serene isle]

$\frac{x-1}{\sqrt{x}} = \frac{x\sqrt{x}-1\sqrt{x}}{\sqrt{x}\sqrt{x}}$

[ocean seal]

Poelymole

[weary abyss]

holy

[serene isle]

so yeah times top and bottom by root x

[weary abyss]

wait

what

should x be x^2?

im getting the antiderivative here right

[serene isle]

imo ignore that for the moment

rearrange to make it nicer

then try and integrate

bc x-1/sqrt(x) is an awful thing to integrate

if it were me i would be removing the root x from the bottom ofthe fraction by timesing top and bottom by root x

[weary abyss]

so does the bottom cancel out

[serene isle]

to make x yep

[weary abyss]

so its just x - 1?

[serene isle]

$\frac{x\sqrt{x}}{x} - \frac{1\sqrt{x}}{x}$

[ocean seal]

Poelymole

[weary abyss]

x * x^1/2 = 2x^1/2?

[serene isle]

no

[weary abyss]

x^3/2

[serene isle]

yes

x^(1+1/2)

[weary abyss]

1sqrtx is just sqrt x right

[serene isle]

yep

but sqrtx / x can be simplified too

x^1/2 \times x^-1

[weary abyss]

i have ((x^3/2) - (x^1/2)) / x

@serene isle is this right?

and i start getting the antiderivative

[serene isle]

i split these into two fractions for a reason

integration is so much simpler if you treat them separately

the first one xs cancel giving just sqrt x

the second root x / x = x^-1/2

[weary abyss]

((x^3/2)/x) - ((x^1/2)/x)

[serene isle]

so overall its just $x^\frac{1}{2} - x^{-\frac{1}{2}}$

[ocean seal]

Poelymole

[serene isle]

and then finding antiderivative is a peice of cake

[weary abyss]

so bringing the x up makes the exponent subract by 1?

[serene isle]

yes

because

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

[ocean seal]

Poelymole

[weary abyss]

and then now i get the antiderivative right

[serene isle]

yes

at long last

[weary abyss]

i got 40/3

at long last

thank god my 6 hour study session is over

6 am

thank you

so much

goodnight

.close

[lone heart]

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[surreal oracle]

i don't know where i did wrong

the answer is -6 btw

the solution just canceled the $\frac{81}{4} x^4$ out of no where

[ocean seal]

yomiko

[upbeat hornet]

looks like it’s because they are using small angle approximations, so some things are negligible

[surreal oracle]

that makes no sense

by that logic i can also cancel out -9x^2?

and 32/x^2?

[upbeat hornet]

in the working out you showed, they keep using ≈

[alpine sable]

My teacher would have bitten me for writing sin(x) = theta

May I be your teacher?

[surreal oracle]

sure

[alpine sable]

LoL

[surreal oracle]

i get that x is really small that 81/4 x^4 would be negligible, but that means that -9x^2 is also negligible? no?

[torn elk]

In the small angle approximation, higher order terms (like x^3, x^4, etc.) are usually ignored because their contribution becomes negligible for very small x

[alpine sable]

Yeah, looks like they assumed the x⁴ term to be extremely smol so they neglected it.

[upbeat hornet]

in the step where they cross out the fourth power, you can expand it out yourself and see that in the end it is -6 plus some x terms

[alpine sable]

Yep. What they showed is a bad practice

But what is really happening is

That

They were neglecting all the x² and higher terms

In the end

So like

[(1-9x²+81/4 x⁴)-1]/(3/2 x²)

= (-9x²)/(3/2x²) + (81/4 x⁴)/(3/2 x²)

= -6 + umm 27/2 x²¿

But as x is small, x² is super small ≈ 0

So in the end

≈ -6

This is what I assume your answer provider did

[upbeat hornet]

$$ \frac{(1 - \frac92 x^2)^2 - 1}{\frac32 x^2} \newline = \frac{-9x^2 + \frac{81}4 x^4}{\frac32 x^2} $$

[ocean seal]

l33t_syncopations

[upbeat hornet]

After this step, you can divide both numerator and denominator by x^2

And then you get $$-6 + \frac{27}2 x^2$$

[ocean seal]

l33t_syncopations

[surreal oracle]

oh so $\frac{27x}{2} -6$ x approx = 0 so its just -6

[ocean seal]

yomiko

[surreal oracle]

alr

thanks

.close

[lone heart]

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[hidden heart]

i dont quite get why this is the result

[hidden heart]

i thought it would be

is it meant to be dw/dr

what is even the difference between the partial derivative of w and the derivative of w wrt r

[lone heart]

@hidden heart Has your question been resolved?

[brave stump]

huh

[upbeat hornet]

1. #❓how-to-get-help
2. you can talk about it in the relevant channel, like #get-advanced-access

[ocean hawk]

try in #discussion or #math-discussion , or others

[lone heart]

@hidden heart Has your question been resolved?

[lone heart]

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[turbid wolf]

hi

[normal holly]

what do i do next

bro

[wet lark]

!status

[lone heart]

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

[turbid wolf]

Do anyof you know what the INS and PNS number system it

is*

[vale wigeon]

googling does not seem to give any results

did you have a math problem which used those terms?

[turbid wolf]

no i am simply trying to figure out what that is as its mentioned in my admission course test

[vale wigeon]

can you show the test...?

[turbid wolf]

alll google tells me is abt nercous testem

system*

[turbid wolf]

i havnt given it yet, its part of the course

[wet lark]

also wrong channel

[vale wigeon]

ok but theres some kind of written document which mentions these "INS and PNS number systems"

yes?

[torn elk]

Provide the image of the text where you saw those terms.

[turbid wolf]

yes one sec

[torn elk]

Show the whole context.

[stuck cape]

are you from pakisthan

[torn elk]

I don't see any reason to hide information. Show the whole image.

[stuck cape]

INS seems to be International Number system , and PNS Pakisthani Number System

[vale wigeon]

last i heard, there is no letter h in the name Pakistan

but also yeah this tells us nothing

[gaunt gust]

Anyone know what the tilted < on m<C= means ?

What does the symbol mean

The help thing said to reply to math 0 which is this tab sry it looks like it’s busy

[ocean hawk]

channel taken, read #❓how-to-get-help

[stuck cape]

@gaunt gust #❓how-to-get-help

[stuck cape]

i'm from bordering state of pakistan and we write it rajasthan insted of rajastan

[turbid wolf]

ye

[gaunt gust]

Not letting me

Make a new thread

[stuck cape]

its the symbol of angel

[turbid wolf]

.close

[lone heart]

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[robust lintel]

hi, does 2a=2nd term
for the equation: tn=an^2+bn+c

[ocean hawk]

2nd term? is tn a sequence or something?

is it supposed to be a subscript n, too?

[torn elk]

Show the original question.

[robust lintel]

nvm

i think i understand

tn is the nth term correct?

[ocean hawk]

you didn't answer any of my questions...

[robust lintel]

sorry