help-27

[restive river]

right?

[misty crest]

yea

negative infinity doesn’t imply it’s the negation of the limit for positive infinity

what happens when u plug in negative infinity for the two terms with x

like the two fraction terms

because the constants aren’t affected on

obviously

[restive river]

they go to 0 from the negative?

[misty crest]

no

negative exponents flips the fraction

like 2^-2=1/2^2

and (1/2)^-1=2

so if we plug in negative infinity it flips the fractions

and they both go to infinity yes

so we get infinity/infinity

but

which one will be larger

the top or bottom

[restive river]

top

[misty crest]

why

[restive river]

because 9 > -1

[misty crest]

no

10000000009 isn’t much different than 9999999999

it’s off by 8

as u the numbers get larger and larger the quotient wouldn’t be affected

by a difference of 8

like if u has (x+8)/x

as x goes to infinity

the limit is one

it doesn’t matter

[cold bough]

knief you have just written lot of stuff!

[misty crest]

indeed

r u taking my place

[cold bough]

no i wrok out just now

work*

[restive river]

so it approaches 1?

[misty crest]

no

[restive river]

damnit

[misty crest]

because what do u notice about the exponents

for the two fraction terms

[cold bough]

and you have helper tag

[restive river]

i mean the one you provided is 1

[misty crest]

yes

but not ur question

so

what do u notice

about the exponents

for the fraction terms

[restive river]

before or after divide byt x^3?

[misty crest]

after

[restive river]

they all have x

[misty crest]

but

one is e^2x

and the other is e^x

e^2x is going to be larger than e^x yes

[restive river]

yeah

[misty crest]

so as x goes to infinity

infinity

the e^2x will outgrow the e^x

thus

the numerator grows larger than the denominator as x goes to infinity so the quotient goes to infinity

does this make sense

[restive river]

yeah

[misty crest]

ok that’s it?

[restive river]

actually I have no idea what you just said but the answer is right

[misty crest]

lol

[restive river]

i will work on this more over the weekend

[misty crest]

glad i can help

have a good one

[restive river]

thanks

.solved

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[devout snow]

@restive river Has your question been resolved?

[smoky nimbus]

I'm pretty sure it's the perimeter of the shaded part

Not the overall shape

This

Simplify sqrt(288)

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✅

[fair igloo]

...it's not? It's just (4)(4)/2

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[fair igloo]

Lol

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[raven swift]

The function m
, defined by m(h)=300⋅(3/4)^h
, represents the amount of a medicine, in milligrams, in a patient’s body. h
represents the number of hours after the medicine is administered.

[raven swift]

if this helps

well if administered on noon
thats 12 pm
4 hours and 30 min
pass?
how do i express that in minutes?
240 min +30 min
270 min
so
300*(3/4)^270/60 ?
82 milligrams?

[obsidian raptor]

yeah, just like that

[raven swift]

sure?

@obsidian raptor wait i think its worng

,calc 270/60

[woven radish]

Result:

4.5

[raven swift]

thats 4.5 of an hour

[obsidian raptor]

yea

and in the graph is like 87 or so

well, that by sight

,calc 300*(3/4)^(270/60)

wait what

[woven radish]

Result:

82.204755124851

[obsidian raptor]

yea there it is

[raven swift]

thank you

.close

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[median seal]

Can someone just explain to me how to get from the top step to the 2nd step

[cinder bobcat]

get a common denominator of t^2 in the square root, then complete the square on the numerator

[median seal]

TY

.close

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[narrow kite]

Need write this without fraction so I can find k'

[narrow kite]

how do i achieve this for the 2/x^2

[winter patrol]

negative exponent law

[narrow kite]

yes but what happens to the 2?

2x^-2?

[weak lodge]

Yes, k(x) = 2x^-2 - 3x^-1

[winter patrol]

2 stays multiplied to 1/x^2 or whatever that converted to

[weak lodge]

yeah, you can imagine 2/x^2 as 2 * (1/x^2)

[narrow kite]

ah welk thatd make sense

just split the fraction in two, to make it visible for myself

ty

[devout snow]

@narrow kite Has your question been resolved?

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[eager acorn]

hello

[eager acorn]

i dont understand

is it 2pi divided by 1/4

which is 8pi

[quasi violet]

yes

[eager acorn]

what do i do with 8pi

where are the decimals

😢

[cinder bobcat]

use your calculator

[eager acorn]

oh

LOL

WAIT

is this 2pi divided by 1/5

[cinder bobcat]

yes

[eager acorn]

31.41592654

is it rounded up like

31.42

ok wait

do i plug in 0?

or do i make it 0

[quasi violet]

x intercept so y = 0

[eager acorn]

because if i make y=0 isnt that finding the y-intercept

[strange nimbus]

You want x = 0.

[quasi violet]

no that's the x

[strange nimbus]

Oh, wait.

I'm wrong.

[quasi violet]

wait what

[eager acorn]

😢

[strange nimbus]

x intercept is where it hits the x axis.

[eager acorn]

yes

[strange nimbus]

And y is always 0 there.

So, y = 0.

[eager acorn]

ohhhhhhhhhhhhhhhhhh

oh yeah

im so silly

f(x) = 8log2[0.7(x+2)]-7

0 = 8log2[0.7(x+2)]-7

7 = 8log2[0.7(x+2)]

7 = 8log2(0.7x+1.4)

what

im stuck

[strange nimbus]

I'd recommend against that.

Move the 8 to the other side.

[eager acorn]

okay

[strange nimbus]

You want to peel off everything attached to x, one by one.

[eager acorn]

7/8 = log2[0.7(x+2)]

hmmm

HELP!

[strange nimbus]

OK, what can undo a logarithm?

[eager acorn]

square root?

[strange nimbus]

No, that's an exponent.

[eager acorn]

oh

[strange nimbus]

Like x^2, then square root.

Logarithm and exponential are opposites.

If you have 2^x, that's an exponential, and you can cancel it out with a logarithm.

[eager acorn]

log the other side?

[strange nimbus]

log_2(2^x) = x.

You can also do it the other way.

If you have log_2(x), you can do 2^(log_2(x)) to get x.

Does that make sense?

[eager acorn]

i dont understand :(

[strange nimbus]

OK, so let's look at it.

Let's say we have (\log_2(55)). We want to get the 55 out. So, we do this: (2^{\log_2(55)}). That cancels out with the logarithm.

[eager acorn]

yes

do i just put it in my calculator?

[woven radish]

Chai T. Rex

[strange nimbus]

No.

[quasi violet]

that bot is cool wtf

[eager acorn]

so its just 2 now?

[strange nimbus]

No.

[eager acorn]

🥲

[strange nimbus]

(2^{\log_2(55)} = 55)

[woven radish]

Chai T. Rex

[strange nimbus]

See how that got the 55 out from inside the logarithm?

[eager acorn]

OH

7/8 = log2[0.7(x+2)]

[strange nimbus]

You can also do (2^{55} \to \log_2(2^{55}) = 55) to get the 55 out from the exponential.

[woven radish]

Chai T. Rex

[eager acorn]

2(7/8) =(0.7(x+2))

yay

🎉

[strange nimbus]

Good!

(2^{\frac78} = 0.7(x + 2))

[eager acorn]

if it was log3 the would i make it 3log3 to cancel it out

[woven radish]

Chai T. Rex

[strange nimbus]

Right.

[eager acorn]

why is it up

and not multiplying

[strange nimbus]

Because you do the same thing to both sides.

[eager acorn]

oh i thought you multiplyed it

[strange nimbus]

You did this to the right side: (\log_2(0.7(x + 2)) \to 2^{\log_2(0.7(x + 2))}).

[woven radish]

Chai T. Rex

[eager acorn]

but you put it in the exponent area

[strange nimbus]

See how you did 2 to the power of the entire thing?

Right.

[eager acorn]

okay i get it

[strange nimbus]

That's how you undo a logarithm.

[eager acorn]

2^(7/8) =(0.7(x+2))

[strange nimbus]

OK, from here, it's just older algebra stuff.

[eager acorn]

2^(7/8) =0.7(x+2)

hmmm

2^(7/8) / 0.7 = x+2

(2^(7/8) / 0.7 )-2 = x

0.620011552 = x

i feel like i did it wrong

[strange nimbus]

OK, looks good, but you want to keep it as (\frac{2^{\frac78}}{0.7} - 2) unless they tell you to round it or something.

[woven radish]

Chai T. Rex

[strange nimbus]

You can simplify it a little more with (\frac{10 \cdot 2^{\frac78}}{7} - 2).

[woven radish]

Chai T. Rex

[strange nimbus]

Oh, OK.

[eager acorn]

0.62 ~ x

[strange nimbus]

Then, yes, but you don't want 0.620011552.

Ahh, right.

That's good.

[eager acorn]

yay

this is so hard

i forgot everything

[strange nimbus]

About this part of algebra or algebra in general?

[eager acorn]

this is precalc review

like how am i supposed to do this

[strange nimbus]

Oh, OK.

Well, you have 10^x.

So, that's an exponential.

What undoes an exponential?

[eager acorn]

sqaure root

😍

[strange nimbus]

No, that's an exponent.

[eager acorn]

😢 💔

[strange nimbus]

(x^2) has the x in the base. (2^x) has the x in the exponent.

[woven radish]

Chai T. Rex

[strange nimbus]

When x is in the exponent, it's called an exponential.

[eager acorn]

ohh

[strange nimbus]

(x^2) uses square root, but (2^x) uses what?

[woven radish]

Chai T. Rex

[eager acorn]

hmmm

dividing...?

[strange nimbus]

Nope, remember logarithms and exponentials undo each other.

Just like exponents and roots undo each other.

[eager acorn]

log_2?

[strange nimbus]

Right.

So, your problem has (10^x).

[woven radish]

Chai T. Rex

[eager acorn]

OH

[strange nimbus]

How would you undo that?

[eager acorn]

LOG_10

[strange nimbus]

Exactly.

Do that to both sides.

[eager acorn]

x=LOG_10(675)

[strange nimbus]

Good.

[eager acorn]

yay

[strange nimbus]

Now it wants you to round it.

What do you get?

[eager acorn]

2.829303773

2.83

[strange nimbus]

Good.

[eager acorn]

theres 2 more questions

[strange nimbus]

OK.

[eager acorn]

could you help me

[strange nimbus]

Sure.

[strange nimbus]

OK, how would you start?

[eager acorn]

the formula

[strange nimbus]

OK, good.

So, a = 4 here and b = 7.

[eager acorn]

(3(7)^2-6(7)-16) - (3(4)^2-6(4)-16)

7-4

[strange nimbus]

Good.

Oh, wait.

Should be 6(4) instead of 6(2).

[eager acorn]

oops

[strange nimbus]

OK, now just do the arithmetic there.

[eager acorn]

27

theres no decimals

[strange nimbus]

OK, let me check.

[eager acorn]

is it tricking me

[strange nimbus]

Looks good.

So, it says if needed, but it's not needed.

So, just 27 is the answer.

[eager acorn]

okay

this is similar

(4(4)^4-17)-(4(2)^4-17)

4-2

does it look right?

[strange nimbus]

Yes, that looks good.

[eager acorn]

i got 480

[strange nimbus]

Yes, that's right.

[eager acorn]

yay

oh no!

i got one wrong

[strange nimbus]

Which one?

[eager acorn]

:(

[strange nimbus]

OK, how did you get that?

[eager acorn]

i put it in my calculator

[strange nimbus]

OK, what buttons did you press?

[eager acorn]

sin(0.58)

[strange nimbus]

,calc sin(0.58)

[woven radish]

Result:

0.54802393679187

[strange nimbus]

Oh, OK.

Well, you have sin(theta) = 0.58.

You want to find theta.

So, you have theta inside of sine.

What undoes sine?

[eager acorn]

i dont know

tan?

1/sin

[strange nimbus]

No, there are arc functions.

Like if you have sine, arcsine undoes it.

If you have tangent, arctangent undoes it.

On a calculator, arcsine is probably either written as asin or sin^-1.

So, you do it like this:[\sin(\theta) = 0.58][\arcsin(\sin(\theta)) = \arcsin(0.58)][\theta = \arcsin(0.58)]

[woven radish]

Chai T. Rex

[eager acorn]

oh

sin inverse?

[strange nimbus]

Right.

[eager acorn]

why

THIS IS SO HARD

where did the inverse come from :(

does it undo it?

[strange nimbus]

Yes, the inverses that undo the trig functions are the trig function with arc in front.

Or the trig function with a -1 exponent.

It's not really an exponent.

[eager acorn]

so it "undos" it?

[strange nimbus]

Yes.

It's like how logarithms and exponentials undo each other.

The trig functions and the arc versions undo each other.

[eager acorn]

oooooo

yay i understand now

[strange nimbus]

OK 🙂

[eager acorn]

thank you for helping me :)

[strange nimbus]

You're welcome.

[eager acorn]

have a good night

[strange nimbus]

You too.

[eager acorn]

bye!

[strange nimbus]

Bye.

[eager acorn]

.close

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[loud dust]

Number of O atoms in 8.25×10−3 mol Al(NO3)3.

[topaz beacon]

!status

[devout snow]

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

[topaz beacon]

ok

its best to do unit conversions in a nice line

\left( \frac{8.25×10^{-3}}{1} \right)

$\left( \frac{8.25*10^{-3}mol Al(NO_3)_3}{1} \right)$

[loud dust]

so this represents 1 mole of Al(NO3)3?

[woven radish]

oops :)

[topaz beacon]

ok there

ok how many mol O are in one mol aluminum nitrate

[loud dust]

9

[topaz beacon]

ok

gimme a sec while i set that up

$\left( \frac{8.25*10^{-3}mol Al(NO_3)_3}{1} \right) \left( \frac{9molO}{1molAlNitrate} \right)$

[woven radish]

oops :)

[topaz beacon]

i cant be bothered tow rite out the chemical formula, just trust me

ok we want just atoms O, how do we change mol to atoms

[loud dust]

Honestly I have no idea.

[topaz beacon]

what is a mole?

[loud dust]

6.022 x 10^23

[topaz beacon]

6.022×10^23 what

[loud dust]

Wait what? I thought a mole was just a number.

[lost laurel]

$6.022 \cross 10^{23}$ particles in chemistry

[woven radish]

Why am. I here

[topaz beacon]

ok sure a mole is a number

[loud dust]

ok got it

[topaz beacon]

but what im trying to say is 1 mol atom is 6.022×10^23 of that atom

[loud dust]

so it is 6.022 x 10^23 particles of that atom?

[topaz beacon]

yeah

[loud dust]

does that mean we multiply that by 9?

[topaz beacon]

lets make it a conversion factor then multiply it on the end

so 1mol O = 6.022×10^23 atoms O

we can divide to get 1 = (6.022×10^23 atoms O)/(1mol O)

[loud dust]

uhh

Calc 2 wasnt this hard

[topaz beacon]

$\left( \frac{8.2510^{-3}mol Al(NO_3)_3}{1} \right) \left( \frac{9molO}{1molAlNitrate} \right) \left( \frac{6.02210^{23} atoms O}{1 mol O} \right)$

[woven radish]

oops :)

[topaz beacon]

ok cool

when everything cancels, you get it in atoms O

[loud dust]

oh ok. so you mean adding the numerators then dividing the words in the denominator to leave O atoms

atoms O*

[topaz beacon]

wdym adding the numerators

[loud dust]

the brackets mean multiplication

we can make the brackets just surround the interior of each term on the same division line.

[topaz beacon]

ok

that works

[loud dust]

then we are also left with the numerical values right?

so then we actually multiply

[topaz beacon]

yeah

[loud dust]

ok i got 4.471 x 10^22

[topaz beacon]

that sounds about right

[loud dust]

ill check it

thats crazy

it was right

[topaz beacon]

awesome

[loud dust]

How do you know these conversion factors?

is there a list?

.close

[devout snow]

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[topaz beacon]

oh sorry

you basically want two values that are equal

e.g. 1mol Al Nitrate = 9 mol O

in this case its a mole ratio

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[civic jolt]

nevermind

.close

[devout snow]

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[trail eagle]

Multiply by conjugate / conjugate

[devout snow]

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[manic ether]

I need with this please

[manic ether]

For interval notation would it be like this -6<x<-5?

[crisp niche]

union of three open intervals, in your case, since the restricted values of the given function are: x = -6, and x = -5

sometimes, those web applications, accept, commas, between intervals, but in maths, you use symbol of an union

[manic ether]

Oh thats how they wanted

But what do we for D and E?

[crisp niche]

in D, you have to solve the equation of the form: f ( x ) = 0

that leads you to simple form -4x + 20 = 0

and

in E you also solve the equation of the form: f ( x ) = 6 / 5

[manic ether]

Im sorry im still not understanding do I plug in 3 for the equations and the solve it?

[crisp niche]

i did not write the number 3, why you mean number 3 ?

[manic ether]

Im trying to understand how to solve and still kinda lost

[crisp niche]

in D i repeat you have f ( x ) = 0 <=> -4x + 20 = 0 <=> -4x = -20 <=> x = 5

and in E, you must solve this: f ( x ) = 6 / 5

have you ever solved equations ?

[manic ether]

I have its just this process is new to me.

[crisp niche]

i see, that is not good , let me write how to start the E

[crisp niche]

that is how you should start with E :

[manic ether]

Oh one second then

[crisp niche]

ok

[manic ether]

Like this?

[crisp niche]

yes, ok, do not forget to write symbol of equality = between them

and next, move all to the left or to the right side

as youlike

zero wil be on one of sides and the rest wil be on the opposite side

[manic ether]

Like this?

Just making sure

[crisp niche]

very well )

do not forget 66x

[manic ether]

Sorry

Better?

[crisp niche]

yes super, now you can divide by common factor like 2

[crisp niche]

yes very nice:) now, you need to remind from the school, how we solve such quadratic euqations

[crisp niche]

maybe you remember quadratic formula for solutions

[manic ether]

Not finished but like this?

Ignore the extra minus sign that was a mistake.

[crisp niche]

ok but it is better to write two differet x's, since there are two solutions, do you undersatnd me ?

you use dobule symbol plus minus, i prefer to see

plus , and later minus

two diffeertn x

[manic ether]

Like that?

[crisp niche]

yes :))

[crisp niche]

very well now

[manic ether]

So -1,-40/3?

[crisp niche]

exactly 🙂

[manic ether]

Epic

[crisp niche]

deepnds on how you system approves

🙂

[manic ether]

One second

You are awesome

[crisp niche]

cool:) congrats 🙂

smiles

[manic ether]

Thank you 😄

[crisp niche]

yvw)

[manic ether]

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[fossil moth]

so just a quick question but wouldnt we be integrating (8-x^2-z^2)-(x^2+z^2)

rather than nothing?

full question

[magic pine]

int dv gives you the volume of V

int (something) dV gives you the average value of that something (scaled up by the volume of V)

what you can notice is that if you integrate with respect to y and plug in the bounds, you'll basically get what you wanted to integrate

the difference being that now it's the double integral of the expression, instead of the triple integral

[fossil moth]

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[lament schooner]

How did n^4 turn to 4

[inland seal]

The n^4 switched places with the /2 after it

Which is actually /4 bc its squares

[lament schooner]

okay ty idk why it wasnt obvious to me

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[trim dawn]

Hello, I'm having trouble finishing the rest of this problem

[trim dawn]

#7

My work so far:

I got my complementary solution but idk how to solve for the constants for my particular solution

Did I make a mistake or am I doing something wrong?

<@&286206848099549185>

The answer is c1*cos(sqrt(3)x) + c2*sin(sqrt(3)x) + (-4x^2 + 4x -4/3)e^(3x)

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[final ridge]

Is this right at all?

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[formal wyvern]

help plsss

[formal wyvern]

hello?

i need help

HELPPP PLSSSSSSS

<@&286206848099549185>

ughhhh i knew this channel was fake

this server*

HELLLLLOOOOOOOOOOOOOOOOOOOOO?

ANYBODY??????????????

HWLLLOLEEOELE?

WTVR

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[topaz beacon]

it was two minutes

its impressive how impatient you are

didnt even read the bot message

[dark minnow]

what's going on 😂

[topaz jolt]

[shrewd thistle]

OP didn't have a new user emblem... probably more than a little frustrated

[topaz beacon]

#help-2 message history says otherwise

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[jaunty gazelle]

could someone help me with this problem?

[devout snow]

@jaunty gazelle Has your question been resolved?

[jaunty gazelle]

<@&286206848099549185>

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[restive river]

Find the exact value of $\cot(\frac{5\pi}{4})$

[woven radish]

BlazeStorm81

[restive river]

How do I find the exact value of this?

I know this is equal to $\frac{1}{tan(\frac{5\pi}{4})}$

[woven radish]

BlazeStorm81

[restive river]

oh bruh nvm im a dumbass lmao

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[tribal urchin]

Triangle ABC has side AC= 4 meters and angle BAC= 32º. Square ABDE shares a side with ABC and has side length 5 meters.
What is the total area of the triangle and square?
You construct a pyramid to serve as the tomb of your great great grandmother, Patricia III of Tallahassee. She was a modest woman who only requested for her tomb to have a volume of 40m^3. What must the height of your grandmother’s tomb be in order to honor her wish?
My work is down below, I need help to see if this is right or not.

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<@&286206848099549185>

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@tribal urchin Has your question been resolved?

[tribal urchin]

<@&286206848099549185> anyone

[zealous bloom]

hi

@tribal urchin

do u know trigonometry

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[compact shoal]

help

[compact shoal]

.reopen

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[winter torrent]

??

[compact shoal]

@winter torrent

sorry im new here, didn't know how it works 😭

[winter torrent]

ok well reading is a good place to start

here

.reopen

[devout snow]

✅

[compact shoal]

yeah my bad

[faint hearth]

!status

[devout snow]

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

[compact shoal]

1

[faint hearth]

Do you know what an exponential function looks like?

[compact shoal]

Yeah

[faint hearth]

What does it look like?

[compact shoal]

f(x) = 6^x

or something like that?

[faint hearth]

Right, but it doesn't matter, it could be a 6 ,20, 31 whatever

[compact shoal]

Yeah

You're right

[faint hearth]

Also some number could be multiplied with the 6^x

[compact shoal]

yeah

[faint hearth]

So in general, an exponential function looks like f(x)=ka^x

[compact shoal]

damn

yessir

[faint hearth]

We are given that when x=0, f(0)=0.75

[compact shoal]

okay

[faint hearth]

Well what is f(0) from here?

[compact shoal]

ka^0

1?

[faint hearth]

Right, what is a^0?

[compact shoal]

1 right?

[faint hearth]

Yeah, so ka^(0)=k

[compact shoal]

damn yes

[faint hearth]

so, k=0.75

[compact shoal]

oh shit yeah

[faint hearth]

So we know our function looks like f(x)=0.75(a^x)

[compact shoal]

yeah

[faint hearth]

To find a, use the fact that it passes through the point (1,3/2)

[compact shoal]

so a is 1?

[faint hearth]

No,

[compact shoal]

how do we find a

[faint hearth]

When they say the function passes through (1,3/2), they're trying to say when x=1, f(x)=f(1)=3/2

[compact shoal]

oh so it's in the format (value, solution)?

like value of x

[faint hearth]

More like (input, output)

[compact shoal]

yes!

[faint hearth]

Yeah

[faint hearth]

What is f(1) from here?

[compact shoal]

It's 3/2

[faint hearth]

Yes

But we also know f(x)=0.75(a^x)

So f(1)=0.75(a^1)=0.75a

[compact shoal]

oh yes

[faint hearth]

0.75a=3/2

[compact shoal]

1/5?

[faint hearth]

Solve this and find a

[faint hearth]

No, a=2

[compact shoal]

oh yes my bad

[faint hearth]

So now you know what f(x) is

[compact shoal]

yes

[faint hearth]

f(x)=0.75(2^x)

Do the same for the other problems

[compact shoal]

I will!

doing right now

[faint hearth]

For the second one, you should take the general form k(a^(-x))

Since it's decreasing as x increases

[compact shoal]

oh why -x

oh okayy

it's wrong but this is what I came up with 😭

@faint hearth

[faint hearth]

if g was what you wrote, g(2) would not be 8/9

you know g(x)=8(a^(-x))

Put x=2, g(2)=8/9

8/9=8(a^(-2))

You get a=3

So g(x)=8(3^(-x))

[compact shoal]

and h(x) is x +4

thanks!!

You are very helpful!

@faint hearth

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[rain berry]

Completely lost with this one. I know I have to do trig substitution but have no idea how

[scarlet mauve]

integrals involves sqrt(x^2-a^2) you can let x=asec(theta)

sorry edited.

[rain berry]

so x=9sec(theta)? or would I take the sqrt(9) to make it 3sec(theta)

[scarlet mauve]

that means x would be 3sec theta (a is 3 because a^2 = 9)

[rain berry]

right right

now that I have that do I plug that in for x? and then x^2-9 would be u?

[scarlet mauve]

no u sub comes pretty late in trig sub

i just learned trig sub recently but have notes on it

[rain berry]

So now that I know x = 3sec(theta), I take the derivative to make du/dx equal 3sec(theta)tan(theta)?

[scarlet mauve]

yes

[rain berry]

How do I know what u is in this problem

cause I am at the point where I got this

[scarlet mauve]

hmm show me everything u have rn

[scarlet mauve]

it should still be in respect to theta

then we can go back to x=3sec(theta)

==> x/3 = sec(theta)

[rain berry]

So I dont have a u to substitute?

[scarlet mauve]

it doesnt seem like it unless you made some miscalculation, usually u sub occurs right before you do some intergration

[rain berry]

So my answer would be sec(theta)/3?

[scarlet mauve]

no.

you gotta find what theta is so

it should be arcsec (x/3), ima see if theres some miscalculation

[rain berry]

alright thanks

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@rain berry Has your question been resolved?

[rain berry]

<@&286206848099549185>

[scarlet mauve]

^oop dont forget constant 1/3

[rain berry]

constant 1/3?

[scarlet mauve]

yeah since when you integrate 1/3 dtheta it would be 1/3 * theta + C

[rain berry]

and where do I get the arcsec(x/3)? That is the only part im confused about

[scarlet mauve]

hmm thats how theta words.

lets take for example theta = cos^-1 (1/2)

cos theta = 1/2

[scarlet mauve]

the inverse i guess you can say? cosx = y <=> x= cos^-1(y)

[rain berry]

I see. Thank you

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[hollow jasper]

How do i simplify the L

[winter torrent]

you can't beyond that point

although I think that 59 is wrong

[hollow jasper]

18+46*root 2 i

[hollow jasper]

?

[winter torrent]

take it step by step and do the squaring first

[hollow jasper]

And plus 5

[winter torrent]

ok that looks better

[hollow jasper]

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[umbral gyro]

Ok... stupid question...

See image 1
I have 2 integrals added... one going 0->X and one 0->U
I established that U=X so...
If I am correct I have 2 integrals going 0->X
SOOO if I am furthermore not incorrect... (please tell me if my assumptions are incorrect)
that means I have essentially 1 integra going from 0-> with Fx + Fu inside... (or Fx1 + Fx2)

Correct?

Ok... under the assumption that my prior assumptions are correct now my actual question...
my prof has here in his integrals from 0->X formulas... creating the Root function (hope the english term is correct) and calculating the integral... shouldnt I essentially end up with the negative of the root function?
what image 2 is showing is my solution... which differs from that of my prof... and I am confused where I made my mistake

[umbral gyro]

did I make a mistake? surely...

my entire thought process is this... if that helps

[devout snow]

@umbral gyro Has your question been resolved?

[uneven notch]

this really looks like physics

[devout snow]

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[restive river]

How do I proceed ?

[rich summit]

How could you rewrite cot(theta)?

[restive river]

1/tan(theta)

[uneven notch]

any restrictions for theta?

[restive river]

No

[rich summit]

nothing like $0 < \theta < \frac{\pi}2$?

[woven radish]

Jelle

[restive river]

It is not given in the question

[low holly]

tan(theta) is surjective so you only need to minimize x + 1/x

[rich summit]

Okay, weird. They probably mean local minimum then

[low holly]

x + 1/x doesn't have a global min

[rich summit]

let's just assume tan(theta) is positive, that means both terms are positive

[restive river]

After that ?

[rich summit]

Do you know about AM-GM?

[restive river]

Yes

[rich summit]

How could you apply it to tan(theta) + 1/tan(theta)?

[restive river]

AM>GM

[rich summit]

Yeah, so $a + b \geq 2\sqrt{ab}$, how could you use that here?

[woven radish]

Jelle

[restive river]

Okay got it

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[heavy basalt]

How do I prove that ABC is isosceles?
It just seems to me like there is not enough information, but I am somehow supposed to do it

[solar goblet]

i assume AC is the diameter?

[heavy basalt]

It is not

I would say if it was

Oh right AB is tangent to the circle

And that's everything that's given to me

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[heavy basalt]

<@&286206848099549185>

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[pearl silo]

hey I was wondering why does tthe integral have tto start at a. when u take the derivative of F(a) it's a constant so it will be 0 regardless if u fill in a or any number from the domain

[pearl silo]

this is fundemental theorem of calculus part 1

[stuck field]

g(a) = 0

Is something you can tell if it starts at a

Not otherwise.

[pearl silo]

why is g(a) = 0 again?

oh then the area is 0 right?

[stuck field]

yeah

[pearl silo]

but I have another question say u take for example a + 1 instead of a on the integral

this will still hold rightt?

[stuck field]

yeah

[pearl silo]

but the theorem is defined only for when it starts at a

why

because if u take the derivative of g(a+1) it becomes 0 so I can just take any number on the domain and g'(x) = f(x)

[pearl silo]

wait is the reaason why the integral has to startt from a to x because otherwise it is not continous and differentiable from [a,b]?

[stuck field]

that's not true.

[pearl silo]

it is not?

but this picture is the theorem right?

[stuck field]

If $\int_{p(x)}^{q(x)} f(t) \dd t = g(x)$ \
Then\
$g'(x) = f(q(x)) \cdot q'(x) - f(p(x)) \cdot p'(x)$

[woven radish]

! What the hell am I doing here?

[stuck field]

(in your case, when the lower bound is a constant, its derivative is zero and you're only left with the first term)

The constant could be a, a+1 or anything at all.

It does matter

As long as its a constant.

[pearl silo]

oh I think I gett it

so the picture

which states for only a

is just part of the whole theorem?

or is this like the whole theorem but for a+ 1 ettc is just what comes from the theorem?

[stuck field]

I mean they've just used a constant for a function p(x)

[pearl silo]

ye

[stuck field]

yeah so it doesn't matter if it's a or a + 1

a is just a constant

like a+1

[pearl silo]

I think I understand this too because the derivative of F(q(x)) - F(p(x)) is the integral rightt

and then u just take the chain rule

[stuck field]

yeah

[pearl silo]

but from the second fundemental theorem of calculus the integral equals to F(q(x)) - F(p(x)) but now I am using the second theorem on the first one is there another way to view it

[stuck field]

I'm not sure what you mean another way.

[pearl silo]

like is that the correct way to view what u did here?

[stuck field]

Yes, it's correct. What I did there is essentially both parts combined.

[pearl silo]

is this the fundemental theorem firsst one but then the compelte version right?

[stuck field]

I'd say so, this is the more general case.

[pearl silo]

1 more question if u have the integral of g(x) with lower bound a + 1, and I say x = a is thatt still allowed to do?

I do get a negative answer then

[stuck field]

It need not be negative.

But it's allowed for the lower bound to be larger than the upper bound.

[pearl silo]

but if I do that and the function is positive then I will get a negative answer right?

[stuck field]

no

[pearl silo]

how actually?

[stuck field]

Do you mean F(x) postive

Or f(x) positive

[pearl silo]

I meant f(x) actually but if F(x) is positive won't it be negative too then?

[stuck field]

if f(x) then yes.

For F(x) postive or negative doesn't matter. Increasing or decreasing does.

[pearl silo]

ah Isee since u take the derivative

[stuck field]

Yes.

[pearl silo]

I am just a bit confused if u take as lower bound a + 1 like if u take the derivative everything goes fine so nothing wrong with taht but will it still be continous from [a,b] or will the domain become [a+1,b]?

[stuck field]

[a,b]

Will be the domain still.

If it's g: [a,b] -> R

Then that's all that matters.

[pearl silo]

ye so the domain stays with it is the function doesn't matter

but the thing is won't there be negative integral be possible if I do that/

if I take a + 1 to a

or are negative integrals fine?

[stuck field]

Negative integrals are fine.

[pearl silo]

oh because u defined the range as R right/

[stuck field]

yes.

No

The range is isn't R

[pearl silo]

oh wait

that s co domain

mb

[stuck field]

yeah

But yeah, because the co domain is R it can be negative as well.

[pearl silo]

I think I understand it but somehow it feels like I don't fully understand it lol

[stuck field]

Just try problems to see whether or not you do.

[pearl silo]

if I take an excercise for example how can I see the domain and range of the integral

he didn't defined it

I thinkl I know how to solve it btw

but can I just assume the domain is a = 1/2

[stuck field]

You don't have yo know the range and stuff.

[pearl silo]

oh

[stuck field]

No you may not.

[pearl silo]

I know if I take the derivative of this

it becomes just f(x) right?

[stuck field]

Yes. That 1/2 should be used in place of x to make the integral zero and solve for C.

But otherwise it's just f(x)

[pearl silo]

but since in the question it asked for x>= 1/2 the lower bound had to be bigger then 1/2 right?

otherwise I couldn't apply the theorem?

[stuck field]

I suppose so. Then I guess it is valid to say a is infact 1/2

But it's not true in general the domain to start from the lower bound of the integral.

[pearl silo]

say I call this g(x)

can I say that the domain is starting 1/2 of g(x) because of the lower bound

so not of f(x)

and from the question here I can say that the domain of f(x) starts from 1/2 right?

[devout snow]

@pearl silo Has your question been resolved?

[pearl silo]

hey I solved the excercise but, am I not assuming things which aren't explicitly being told in the excercise. Because in the theorem it says that the integral has to be cont. on [a,b] and diff. on (a,b)

[stuck field]

And?

[pearl silo]

well is it necessarily true that it is cont and diff on the given intervals?

[stuck field]

Yes

It has to.

[pearl silo]

but how could I know that?

[pearl silo]

do u also know this part?

[stuck field]

It's not visible.

[stuck field]

Yes what about it

[pearl silo]

idk if I asked this before but can I assume a = 1/2 here?

as in the domain [a,b]

domain of g(x) btw

[stuck field]

Yes

You may.

[pearl silo]

is it because of the lower bound or because of given in the question where it asks for x >= 1/2

or both?

[stuck field]

yeah

Latter

Not both.

[pearl silo]

ah right I thought so, and I have 1 last question

so how can I know the integral is cont and diff from [a,b] and (a,b) respecitively

do I check it afterwards?

it is only given that f(x) is continous and this does not imply differentiable from (a,b) right?

the only way I can think of is when u got a formula for f(x) is checking if the integral is differentiable?

or can I just assume it tbh I feel like that's the easiest

[pearl silo]

I tried to think what u mean with this, u mean it has to as in I can assume it has to
or it has to so for a different reason?

[devout snow]

@pearl silo Has your question been resolved?

[devout snow]

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[devout snow]

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[restive river]

Show that every ideal in $\Z_n$ is principal

[restive river]

how should i begin with this?

[supple knot]

Is that really the full question

[restive river]

woops no i typo'd because discord was dying

for some reason

but there ya go

maybe we can start by letting I be some non-zero ideal in Z_n

hmmm

[supple knot]

Try proving it for n = prime first

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[restive river]

Guys, can yo help me with this?
The answer is D but I want to see how it's done (step by step)

[tight spire]

substitute x=0, u will then find the value of b then substitute x=1 and also the value of b

[restive river]

Dude, why didn't I come up with that earlier

[restive river]

But I get 2 values for b
b² = 4 therefore b = ±2

[tight spire]

umm was that sarcasm

[restive river]

Ok First let's consider b = +2

[tight spire]

yeah try both the values you will much get the same answer maybe

[restive river]

Yes, +2 is the true value for b
and I will get a = 2 and b = 2 and
a⁴ + b⁴ = 32

That's right

Thanks

[tight spire]

wlcmm

[restive river]

Let's close this channel, shall I?

.close

[tight spire]

yea go on

[devout snow]

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[devout snow]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[restive river]

If I want to find the exact value of $\cos(-\frac{7\pi}{12})$ is it legal to get rid of the negative and then do $\cos(\frac{\pi}{4}+\frac{\pi}{3})$ cause if so it leads to a very different answer than what my professor got..

[woven radish]

BlazeStorm81

[lavish sigil]

shouldnt be different

all that is fine

[restive river]

mine is

[lavish sigil]

divide by the sqare root of 2

[near trout]

divide top and bottom by sqrt(2)

[restive river]

oops

wrong image one sec

its the same in the calculator

[restive river]

oh alr

i thought we wanna keep square roots in the numerator tho no?

[lavish sigil]

doesnt really matter

second result looks better

[restive river]

how do you know it looks better? i gotta know for my exam haha

im studying for it rn

[lavish sigil]

also should be

sqrt(3) - 1

I think

[lavish sigil]

whatever suits you :)

[restive river]

so would mine be considered an acceptable answer as well? haha

even tho its different

[lavish sigil]

yes it would

but its not meant to be sqrt(2) - sqrt(6)