help-23

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[frigid locust]

what have u tried so far?

yeah thats correct , what did u get and why are u stuck

send the variable into the denominator
$$a^{-1} = \frac{1}{a}$$

[flat frigate]

JustToPro

[frigid locust]

id first get the like terms together and add thier exponents to gether and do as u did

and then change the negatives into positive as shown

btw this answer is wrong

$$2x^{-1}y^{-1}\cdot 6u^{-4}x^{-5}\cdot 8u^4y = 2\cdot 6\cdot 8\cdot x^{-1}\cdot x^{-5}\cdot y^{-1}\cdot y\cdot u^{-4}\cdot u^4$$

[flat frigate]

JustToPro

[frigid locust]

and now we add the expoenents and we get

$$2\cdot 6\cdot 8\cdot x^{-1}\cdot x^{-5}\cdot y^{-1}\cdot y\cdot u^{-4}\cdot u^4 = 96\cdot x^{-1-5} \cdot y^{-1+1} \cdot u^{-4+4}$$

[flat frigate]

JustToPro

[frigid locust]

y and u becomes 0 in the exponenet place , so they are equal to 1

so we are left with

$$96 \cdot x^{-4} = \frac{96}{x^4}$$

[flat frigate]

JustToPro

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[high lagoon]

Hi this is for converge or diverge of sequence using limits. Could someone tell me why the part circled you have to multiply by 1/n vs plugging in your limits to 1/n+2?

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[high lagoon]

<@&286206848099549185>

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[void plaza]

I'm not sure how to work out the value for gamma

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<@&286206848099549185>

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<@&286206848099549185>

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[void plaza]

can gamma take any value?

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[white magnet]

hi I need help with this question

not sure how to approach it

[fallen thunder]

what have you tried?

[white magnet]

I tried setting the limit up but im not sure if I did it correctly

[fallen thunder]

well, first did you rewrite your thing as: $c \int_{5}^{\infty} \frac{1}{x^4} \text{ dx} = 1$

[flat frigate]

kanna

[white magnet]

yeah

[fallen thunder]

i mean then you just have $\qty[\frac{x^{-3}}{-3}]_5^{\infty}$ to evaluate

[flat frigate]

kanna

[white magnet]

oh wait im stupid lmao

I had the wrong antiderivative 😭

I got it

375?

[fallen thunder]

yeah that should be right

[white magnet]

alright thank you so much

I had one other problem I needed some help with

do you mind helping out with that one as well

[fallen thunder]

You can just ask

[solar hazel]

!status

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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

[white magnet]

I just very recently learned the concepts so its not very clear yet

and im not sure where to begin

[solar hazel]

do you know what the area under g from 1 to infinity is?

and same question for j

or actually just whether they are finite or infinite is good enough

[white magnet]

um

would it be infinite I guess

bcs x goes to infinity

[solar hazel]

which?

[white magnet]

so theres infinite area?

g

or technically wouldnt all of them

im confused lmao

[solar hazel]

well... no

something can keep growing in area without growing to "infinite area"

i don't really have the energy to help if we need to start from there tho i'm sorry

[white magnet]

I think I understand

but if ur too tired its all g nw

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[sour violet]

,rccw

[sour violet]

I have the answer but I’m having trouble seeing how the prof derived the final formula

Or in what even the final formula is

The question part A is easy it’s kinematics to find the initial velocity

Second part is asking what is the weight of m2 (x)

From what I drew up I have m_1g-ma=t and t=m_2a+m_2g

We also know that m1_g-ma=t

But idk from those three how she derived the final formula to solve for the weight

So from these

,rccw

[sour violet]

I can set the 2 and third equal to eachother

So from this do I just set Al unknowns to one side?

,rccw

[sour violet]

,rccw

[sour violet]

And I’m off by exactly that much now but my formula is way off since I would need to sqrt the 5.9 but what I’m doing rn is off the actual answer by a factor if 10

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@sour violet Has your question been resolved?

[winged flare]

sorry i can't figure out the numbers in the diagram

is the block on the right 10 kg? @sour violet

and i guess it covers the 1m in 0.9 seconds

and the mass on the left is x?

[sour violet]

Yes

The block on the right is 10kg and it covers a distance of 1 meter in .9 seconds

[winged flare]

ok so that's all the info given

and we're supposed to find x and a

[sour violet]

Ye

I used an kinematic equation to solve for a and my answer matches up

But my issuie is x my answer is off by a bit

About a factor of 10 and then 1/2

[winged flare]

ok I'll try the 1st one and see if it works

is this correct?

,rccw

[winged flare]

yeah i think so

ill try 2nd one then

[sour violet]

Ok that’s the same number I got

We got a +/- swap but that’s relative anyway

[winged flare]

yeah

[sour violet]

,rccw

[sour violet]

This is the formula I derived

[winged flare]

hm so g is 9.8?

[sour violet]

Ye

Is it@something different for you?

[winged flare]

no

pretty much the same

[sour violet]

Ik a lot of people say to round to 10 but I asked my prof she said in her class atleast she’d like us to use 9.8

[sour violet]

Any ideas?

[winged flare]

yeah i got almost 6 kg

[sour violet]

Wait

So there’s a chance the ta got it wrong

I guess 59kg does sound weird

[winged flare]

yeah

[sour violet]

Uh wel

[winged flare]

if it was 59 kg it would fall on the other side

[sour violet]

That’s what I was thinking

It would literally just slam down the other side no way 10kg pulled 59

But how did you derive that formula?

[winged flare]

i just found T

[sour violet]

Mine has =m_2+m_2 which has to be a mistake

[winged flare]

yeah maybe

i found T from the net force of the 10 kg block formula

[sour violet]

I got about 6=m_2+m_2

[winged flare]

and then used the T to find x

[winged flare]

huh

how'd you get that

[sour violet]

I got two that equal T and set them equal to eachother

Then isolated two of the variables im solving for

[winged flare]

i guess this is another question?

[sour violet]

Part B of this question

Part A was solve for accl

[winged flare]

ah

[sour violet]

Part b is solve for x

[winged flare]

its much easier

just try drawing the fbd of the 10 kg block

[sour violet]

I did it’s already “solved” kinda

The issuie is

[winged flare]

T - (10)g = -(10)a

[sour violet]

In the class we got 59kg which dosent sound right

[winged flare]

oh

well what do you get for t

[sour violet]

So I’m thinking my whole formula might be wrong

[winged flare]

well ifk

[sour violet]

In the class we didn’t solve for T

Oh wait

We just did what’s equivalent to Ta mf set them equal to eachother

[winged flare]

what

[sour violet]

Shoot

Ok lemme retype that

We found the net force of T for both of the blocks and set them equal to eachother to solve for the mass of x

[winged flare]

they ARE equal to each other

[sour violet]

Yes

[winged flare]

both are the same T

[sour violet]

Yes

[winged flare]

yeah so you solve for the T from the 10 kg block

[sour violet]

,rccw

[winged flare]

T = 10(g - a)

[winged flare]

um which are m2 and m1?

[sour violet]

The way we did in class was ok these two are equal to T so set equal to eachother and solve for a

[winged flare]

ah

[sour violet]

M1 is 10kg m2 is x

[winged flare]

oh ok

[sour violet]

But the answer is like

Wonky

Mine atleast

5.9kg sounds right but I have the weird m2+m2

[winged flare]

hm lemme try

[sour violet]

So I’m assuming one of those should’ve been factored out

But yeah 59kg is batshit crazy

Cause the 10kg block is stated to move 1 meter in .9 seconds

So 5.9kg sounds more correct

[winged flare]

yeah

[sour violet]

Idk how a room full of stem students didn’t think

Hey 59kg shouldn’t be able to be dropped by 10kg

Hindsight

[winged flare]

dunno where you got the m2 + m2 from

guess you messed up the distributive property

(m2g + m2a)/(g + a) is NOT m2 + m2

just m2

[sour violet]

But why does the net vector of F= m_2(a+g)

For either

Wouldn’t it be t=ma+mg

[winged flare]

yeah that's what i wrote

[sour violet]

Oh you just made it more simple

[winged flare]

which is m(a + g)

[sour violet]

Howldup lemme write it down so I can show my ta

Ngl we had a shitfest solving this in class too

He miswrote .9 seconds at 90 seconds

[winged flare]

lmao

[sour violet]

So it must’ve been like 9.99999 kg on the left side

[winged flare]

yeah

[sour violet]

Ok ok now we have an answer that makes sense

[winged flare]

yea

[sour violet]

Thanks for the help I’ll make sure to simplify my numbers so I don’t mess myself up

[winged flare]

welcome

[sour violet]

.close

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[lone arch]

Let $u_1 = \mr{1 \ 0 \ 1 \ 2}$, $u_2 = \mr{3 \ 4 \ 2 \ 1}$, $u_3 = \mr{5 \ 8 \ 3 \ 0}$. They are linearly dependant. Let $U \coloneqq$ Span$(u_1, u_2, u_3)$ over $\mathbb Q$. Determine the dimension of $U$.

[lone arch]

How'd we do this?

This is similar to before ig @drowsy karma

We need to find out which of these u's are linearly dependant, don't we?

And then take that u out

[twilit helm]

You need only to check if two of them are LI, once they are LD 1 <= dim U < 3

[lone arch]

What if all of them were LD?

dim Span < 3

So either dim Span = 1 or dim Span = 2

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.close

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[woven quartz]

This is in Q2 so why is it not negative?

[dry imp]

3pi/8 is less than pi/2

so 1 quadran

[woven quartz]

OH! I gotcha, Was just thinking half of 3pi/4. Didnt consider it moves all the way to Q1

.close

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[spare grotto]

". (The old man in the box) A cubic box has a corner at the origin O. The three closest to O
the corners are A = (2, −6, 3), B = (6, 3, 2) and C = (−3, 2, 6). A point-shaped
ball lies at the point (5, −1, 10). Is this bullet in the drawer?
Tip: Transfer to a suitable new base. Rephrase the question of 'lying in the drawer' in ¨
algebraic terms. What finesse does the switch to a new ON base offer?"

[spare grotto]

As box dimensions are |OA| = |OB| = |OC| = 7, the marble lies inside if its new coordinates
Xf =x1, x2,x3 in the ON basis f = 1, 7{−→OA,−→OB,−→OC}are all beween 0 and 7. To find the new coordinates, one writes the position vector of the marble as a LC of the new basis vectors:
B = (5, −1, 10) = x1f1 + x2f2 + x3f3 = T Xf, T =1/7 (2 , 6, −3, −6 3 2, 3 2 6)
The transition matrix T is orthogonal, with T^−1 = T^T, so the new marble coordinates are
Xf = T^−1B = T^T B =1/7 (2 −6 3, 6 3 2, −3 2 6) (5, −1, 10) = 1/7 (46, 47 ,43)(7, 7, 7)i.e., all between 0 and 7, so, the marble lies inside."

i dont understand what he does in the solution

[steel stag]

the thing about a box is that everything inside can be gotten to by fractions of the edge vectors A,B,C (I'm just guessing 1/2)

because if any of the coefficients were not a fraction between 0 and 1, you'd be outside in that direction

the rest of the solution is them writing out (5,-1,10) in those coordiantes to get a big matrix equation applied to a vector of the new coefficients in this box basis, and solving it by taking the inverse

they also factored out a 1/7 in the beginning so that the box vectors were unit vectors, which means they have to compare to (7 7 7) in the end instead of just (1 1 1)

[spare grotto]

Oh okay thank you!

.close

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[still charm]

need help deriving the discriminant formula for a depressed cubic

[still charm]

D = 4p^3 + 27q^2

of the equation f(x) = x^3 + px + q

now first we find f'(x)

3x^2 + p = 0, x = +-sqrt(-p/3)

then we plug back into the original function

sqrt(-p/3)^3 + p sqrt(-p/3) + q < 0

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[near basin]

Idk where to start

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[near basin]

Can I cancel this

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@near basin Has your question been resolved?

[near basin]

No

It's been like an hour

Oh I'm dumb

Ping helpers

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[near basin]

No

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@near basin Has your question been resolved?

[lean otter]

hi

use sin, cos and tan

soh cah toa

sin = opposite over hypo

cos = adjacent over hypo

tan = opposite over adjacent

@near basin

b u use pythagoreum thoerum

c^2-a^2=b^2

thats gives u 12.6

Measure a would be sin(6/14)

measure B would be cos(6/14)

[near basin]

ok

thank you

[lean otter]

ya np

[versed wave]

@near basin

!done

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[lean otter]

.close

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[silent bloom]

bit stuck

[silent bloom]

working so far attached here

(ignore the fact i forgot a 5, just fixed that)

[thin bridge]

why expand

[silent bloom]

well, i was trying to find a simpler way to write the trig functions

[thin bridge]

works against you

you already have a product that's equal to 0

from there you can simply apply zero-product property / null factor

[silent bloom]

huh...?

[thin bridge]

huh what?

have you ever solved stuff like quadratic equations before?

[silent bloom]

yeah, but not a zero-product property nor a null factor

[thin bridge]

how would you solve something like
$$(x-42)\br{x+ \frac72} = 0$$

[flat frigate]

ℝαμΩℕωⅤ

[silent bloom]

id expand it out

$x^2-42x+\frac{7}{2}x-\frac{42\times7}{2} = 0$

[flat frigate]

Xerunox

[thin bridge]

what about something like
$$(x-123123123)(x+123125461293123) = 0$$

[flat frigate]

ℝαμΩℕωⅤ

[silent bloom]

ok wait i see what you mean now

for $(x-a)(x+b) = 0$ 2 immediate terms would be $x = a, x = -b

[flat frigate]

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

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[onyx ruin]

I am playing a game that has me roll some N number of 10-sided dice, and if I roll a certain number or higher I will get a success. I am rolling 2 such tests, one with a value of 9 and another with a value of 5. We'll denote them as Nd10v9 and Nd10v5. I can apply +1 to each result of one of these test's rolls. Is it better for me to apply that to one or the other?

My thought is that it is the same no matter what, as ultimately it represents an expected 1 additional success across 10 dice rolled. However, someone has argued that applying it to the Nd10v9 case is better because it's a greater increase in chances of success (20% -> 30%, a 50% increase).

[lost vector]

Applying the +1 bonus to the Nd10v9 test:
Originally, you have a 20% chance of success (rolling a 9 or higher) on each die.
With the +1 bonus applied, your chance of success increases to 30% (rolling an 8 or higher) on each die.

Applying the +1 bonus to the Nd10v5 test:
Originally, you have an 50% chance of success (rolling a 5 or higher) on each die.
With the +1 bonus applied, your chance of success increases to 60% (rolling a 4 or higher) on each die.

Calculate the expected number of successes for each scenario:

Applying the +1 bonus to the Nd10v9 test:

Expected number of successes = Number of dice * Probability of success per die
Expected number of successes = 10 * 0.30 = 3

Applying the +1 bonus to the Nd10v5 test:
Expected number of successes = Number of dice * Probability of success per die
Expected number of successes = 10 * 0.60 = 6

According to these calculations, applying the +1 bonus to the Nd10v5 test gives a higher expected number of successes compared to applying it to the Nd10v9 test. So itd is indeed better to apply the bonus to the Nd10v5 test in this case.

[onyx ruin]

I'm trying to maximize the number of additional successes I get. So, taking your example:

Without a modifier:
10 * .2 = 2
10 * .5 = 5

With a modifier:
10 * .3 = 3
10 * .6 = 6

So, ultimately, I am getting +1 success no matter what, and really the only thing that matters is the N (+.1 expected successes per die)

Is that right?

[lost vector]

Yea, when considering the total number of successes before and after applying the modifier, it indeed results in a net gain of +1 success regardless of which test you apply the modifier to

In both cases, you're adding +1 to the expected number of successes, but the distribution of successes across the dice changes based on which test receives the modifier. However, when you sum up the expected number of successes across all the dice after applying the modifier, it remains the same in either case.

So, in terms of maximizing the total number of additional successes, it doesn't matter whether you apply the modifier to Nd10v9 or Nd10v5. You'll always gain one additional success.

[buoyant shadow]

yeah you're right

[onyx ruin]

Okay, I have one more question... and this is what actually confused me, because there were a lot of people essentially telling me I was very wrong, and when I tried to come up with a counterexample I only managed to confuse myself more...

I am playing a game that has me roll 2 10-sided dice. On the first die, I need to roll a 9 or higher for it to be a success. For the second die, I need to roll a 5 or higher. I can apply +1 to the result of one of these test's rolls. Is it better for me to apply that to one or the other?

Chance of exactly 2 successes:
1d10v9+1 = .6 * .3 = .18
1d10v5+1 = .7 * .2 = .14

Chance of exactly 1 success:
1d10v9+1 = .6 * .7 + .4 * .3 = .54
1d10v5+1 = .7 * .8 + .3 * .2 = .62

Chance of no successes:
1d10v9+1 = .4 * .7 = .28
1d10v5+1 = .3 * .8 = .24

1d10v9+1: .18 (2 successes) + .54 (1 success) + .28 (0 successes) = 1
1d10v5+1: .14 (2 successes) + .62 (1 success) + .24 (0 successes) = 1

I thought it wouldn't matter, same as the example I gave before... but my math didn't show that. 😵‍💫

[lost vector]

You are correct

You're rolling two 10-sided dice, and for the first die, you need a 9 or higher for success. For the second die, you need a 5 or higher for success. You're allowed to apply a +1 bonus to the result of one of these tests.

You've correctly calculated the probabilities of getting exactly 0, 1, or 2 successes for both scenarios: rolling 1d10v9+1 and rolling 1d10v5+1.

When comparing the two strategies:

For 1d10v9+1:
Probability of exactly 2 successes: 0.18
Probability of exactly 1 success: 0.54
Probability of no successes: 0.28
Total probability: 1

For 1d10v5+1:
Probability of exactly 2 successes: 0.14
Probability of exactly 1 success: 0.62
Probability of no successes: 0.24
Total probability: 1

Your calculations show that both strategies indeed sum up to 1, indicating that you've considered all possible outcomes. So, regardless of which strategy you choose, the total probability of success remains the same.

So, in terms of maximizing your chances of getting a success, it doesn't matter whether you apply the +1 bonus to the first die or the second die. You'll achieve the same overall probability of success....

[onyx ruin]

Well... It appears applying it to the latter is actually better, no? I have only a 24% chance at no successes. Or is there a way to show that the average number of successes across these two remains the same, due to the increased chance of getting 2 successes?

[lost vector]

damn wait a sec

[buoyant shadow]

you multiply the successes by the chance

[onyx ruin]

.14 * 2 + .62 * 1 = .9
.18 * 2 + .54 * 1 = .9

Like so?

[buoyant shadow]

,calc 0.62 + 2(0.14)
0.54 + 2(0.18)

[flat frigate]

Results:

0.9
0.9

[buoyant shadow]

yes

[onyx ruin]

Ah... thank you.

But, that's interesting!

There is a better strategy depending on what you want to chance more... 2 successes, vs at least 1

[lost vector]

ye

[onyx ruin]

So, it's not exactly the case of "do whatever you want, it doesn't matter", unless you're rolling a lot, in which case it should go towards the average anyways?

[lost vector]

If you're aiming to maximize the probability of at least one success, then applying the bonus to the die with the lower threshold (in this case, the second die, 1d10v5+1) is preferable

If you wanna maximizing the probability of achieving two successes, then applying the bonus to the die with the higher threshold (the first die, 1d10v9+1) is prolly a better choice

ye but if you're considering long-term average across many rolls, then the difference might become negligible

[onyx ruin]

Well, according the equal results of .9 above, it will average out to that expected rate.

Okay, I really appreciate y'all answering my silly question. 🙂

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[proud carbon]

i didnt understand what that A^C means

[peak estuary]

the complement of A

the elements which arent in A

[proud carbon]

oh ok thanks

cuz i thought compliment is written as A'

[peak estuary]

depends on who is writing it

[proud carbon]

never knew A^C was also a thing

thanks so much

[peak estuary]

some also write it as $\overline A$

[flat frigate]

Denascite

[proud carbon]

oh

ok

you can close this channel now btw

[peak estuary]

write .close

[proud carbon]

ill ask any other doubt later

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[proud carbon]

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[proud carbon]

how to solve this?

can someone pls tell?

[eternal carbon]

consider second equation
transpose both sides

[proud carbon]

wym transpose both sides?

[eternal carbon]

perform (A^T + B)^T operation for instance

[proud carbon]

ok

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[exotic dawn]

Hey! Can someone help me with this? No idea where to start

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[exotic dawn]

<@&286206848099549185>

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[lone arch]

Given a linear mapping $f: \bC^5 \to \bC^{2 \times 3}$, can $f$ be injective? Surjective? Bijective?

[lone arch]

So it can't be surjective, because of the dimension formula

dim C^5 = dim image(f) + dim ker(f)

5 = dim image(f) + dim ker(f)

dim image(f) = 5 - dim ker (f) <= 5

What about injective?

The kernel has to be {0_V}

And so dim ker (f) =0

Then 5 = dim image(f)

We can have that, no?

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[stiff vine]

Is this right

<@&286206848099549185>

So is it right or not

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[rancid stone]

help

[burnt wyvern]

whatdaquestion bro

[rancid stone]

i’m confused on how the integral goes from 0 to 2 to 1 to 5. like how do you get there

for number 7

[burnt wyvern]

nope sorry

cant help you i dont know integration

[rancid stone]

can someone else help

[burnt wyvern]

yeah they probably will

w a i t

[tranquil ore]

Oh you just substitute the values of x in for u

[celest ember]

You changed the variable from x to u
When x=0, u=1
When x=2, u=5
When you change the variable you have to write the limits for new variable

[tranquil ore]

What he said

[rancid stone]

ahh got it

thank you

also

is there any other way to find A1 in this

this is a partial fractions question

[runic pine]

Yep

Put x as zero

You will get eqn in terms of A1,A2,B

Since u alr know A2,B you can find A1

[rancid stone]

got it thank you

[runic pine]

!done

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[rancid stone]

.close

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[rancid stone]

.reopen

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✅

[rancid stone]

sorry i have one more question

how does long division get you that

i got something different

[severe pond]

what’d you do

[rancid stone]

ill show u let me take a pic

[junior smelt]

[a shortcut to get that is to add and subtract x from the numerator]

[rancid stone]

ik im doing this wrong

just dont know where i went wrong

[runic pine]

Bro

[severe pond]

nono

[rancid stone]

and im fine with shortcuts but wanna learn polynomial long division first

[severe pond]

x^2 goes in x times

[runic pine]

You will multiply it with x

[severe pond]

not 1 time

[runic pine]

Not 1

[rancid stone]

oh.

how did he get that remainder

now i have this but the remainder is supposed to be -x/x^2+1

[runic pine]

Quotient+ remainder/dividend

[runic pine]

That isn't remainder, that's remainder / dividend

[rancid stone]

yes but my remainder is -1

so how did he get the x on top instead of 1

[runic pine]

You wouldn't write 4/3 as 1+1 but 1+1/3

[junior smelt]

x^3/x^2

[runic pine]

(x²+1) times x is x³+x

[rancid stone]

ahh

got it thank you

i feel like an idiot

[severe pond]

it’s -x

you multiply the x

to each term

[runic pine]

it's finee

[rancid stone]

ya i got it😭

[severe pond]

when you did long division

[runic pine]

lol

[severe pond]

you’re supposed to distribute it to both x^2 and 1

[rancid stone]

no yea

but thank you for the help

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[young walrus]

I don't know what to do at all-

[limpid light]

there you are

[magic rivet]

can someone explain this question to me the wordings a little weird and helllp me come up with a question

oops wrong channel

[limpid light]

circumference is $12\pi$ inch

@young walrus right?

[young walrus]

6 is radius, isn't the circumference formula d x pi?

[limpid light]

yep

ah its d

my

[flat frigate]

DerTheo

[limpid light]

mb

[young walrus]

nw lol, ik that formula, idk anything else regarding stuff like this

[limpid light]

and you are taking $\frac{50}{360}$ of that

[flat frigate]

DerTheo

[limpid light]

so $\frac{50}{360}\times 6\pi$

[flat frigate]

DerTheo

[young walrus]

2.61799387799 is what I got so far

[limpid light]

what is a major arc? just so I know I got this right

[young walrus]

Major arc is the larger section of the circle

[limpid light]

im not so familiar with english terminology

[young walrus]

the outside of the lines

[limpid light]

area or length

[young walrus]

length

[limpid light]

k

so it would be $\frac{310}{360}6\pi=\frac{31}{36}6\pi=\frac{31\times 6\pi}{36}=\frac{31\pi}{6}$

[flat frigate]

DerTheo

[limpid light]

ima sleep now

[young walrus]

oke tty for the help

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[steel stag]

well with 4 vectors that span R4 you can basically put any answer you want

there's only one 4-dimensional vector space in R4, the whole thing

[wanton zodiac]

Oh fr?? Well my professor couldn’t explain it to us it was hilarious

[steel stag]

amazing

[wanton zodiac]

So what do I do

Or do I just put whatever

[steel stag]

umm you could put random numbers and be right like 90% of the time

I'd just do (1 0 0 0), (0 1 0 0) ... just to be safe

[wanton zodiac]

Like the identity.

?

[steel stag]

yea

just the usual basis vectors you'd do in R4

[wanton zodiac]

It didn’t work

[steel stag]

oh sorry I missed the "subset of these vectors" part so you have to choose 4 of the 5

[wanton zodiac]

4 of those 5 vectors??

[steel stag]

the 5 vectors they give you

you'll have to pick 4 and make sure they're all linearly independent

[wanton zodiac]

Honestly to this day idek what linearly independent even means 😭

Ik it has something to do with the zero vector

[steel stag]

oh it's like in linear algebra all you do is scale and add things, and linearly independent means you can't make any of the vectors with a linear combination of the others

like (1 0 0) (0 1 0) (0 0 1) (1 2 3) isn't linearly independent because (1 0 0)+2(0 1 0)+3(0 0 1)=(1 2 3)

[wanton zodiac]

I am in linear algebra but have no clue what is going on

So if it equals the zero vector then it’s dependent??

[steel stag]

here's the formal definition with 0 on the right side

dependent means you can find some c numbers that add all the vectors to 0

[wanton zodiac]

Yep that’s the definition of linear combination

[steel stag]

independent is when the vectors point in different directions enough that you can't ever cancel them out like that

[wanton zodiac]

Ohhhhhhhhhh

Ok ok that makes sense

[steel stag]

so like for the problem, these 4 actually don't work, because that 3rd element always being 0 means they span a 3d space at best

while a true linear independent set of 4 vectors must span four dimensions

like we need the (5 3 1 -1) vector to give information about the 3rd component, scaling the others always gives 0

[wanton zodiac]

So I can’t have the 4 vectors that have 0 in it?

[steel stag]

they won't be independent because (a,b,0,c) is a 3d space

there'll be some overlap somehow with 4 vectors giving information for 3 dimensions

[wanton zodiac]

I would have to include the [5 3 1 -1]

[steel stag]

right

[wanton zodiac]

Man my professor is horrible she can’t explain this to us

She tried but failed so many times

So I can put any of the 4 vectors but I have to include that first vector

Right?

[steel stag]

umm there might be another trap with how to choose the other 3

but yea the 1st must be in

now you can just ignore the 3rd component and look at the other vectors in 3d with their 1st/2nd/4th part

[wanton zodiac]

Yah this is bad I got an exam on this in 2 weeks

[steel stag]

umm with a calculator you can do this all pretty fast, just check the determinant of a box with these vectors

[wanton zodiac]

We can’t use calculators

[steel stag]

ah okay I'll just do the intuitive way

the third vector is super simple so let's try to use that one

[wanton zodiac]

So do I put the vectors into 1 matrix?

[steel stag]

with that and the first you have information about the 2nd and 3rd components and can ignore them for the rest of the vectors

[wanton zodiac]

Then find the determinant?

[steel stag]

yea determinant is actually just a way of testing linear independence

dependent means it's 0, independent if it's another number

[wanton zodiac]

So if the determinant is 0 then it’s dependent?

[steel stag]

yes

[wanton zodiac]

Hmmm ok ok

[wanton zodiac]

So so what is going on on here?

[steel stag]

the green part is an intuition thing for how linear independence works

you're allowed to add vectors to others all you want and it'll still have the independent/dependent property, it just shifts the constants c1,c2,... around a bit

so with (0 1 0 0) you're allowed to cancel out the 2nd components of everything else, like (-2 3 0 1) - 3*(0 1 0 0) = (-2 0 0 1), I can zero out that spot

[wanton zodiac]

Well I am already not good at this so idk

Hmm ok so is it like row reducing?

[steel stag]

the image is just trying to say you only have to care about the 1st and 4th parts now, because the red deals with the 3rd component, the green deals with the second

yea I'm just row reducing with different words lol

[wanton zodiac]

Ohhh ok see my bad that threw me off

[steel stag]

independent means you can reduce it to a diagonal with all 0's below, dependent means you can reduce to a row of 0

[wanton zodiac]

Ohhh ok ok

That makes sense

[steel stag]

final step is that for the rest of the vectors you just want 2 that aren't multiples in their 1st and 4th parts

like (-2 1) is clearly independent from (-3 2) to me

[wanton zodiac]

Yah I see that

But dam I gotta go eat rq I will be right back

[safe radish]

@wanton zodiac Has your question been resolved?

[wanton zodiac]

@steel stag I am back

@steel stag thank you sooooo much

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[wanton zodiac]

<@&286206848099549185>

[little cosmos]

Think about what Col A is by definition, and what a basis is by definition

[wanton zodiac]

Oh shit that was it I got it

Wait I have a question do I row reduce?

[little cosmos]

What do you have so far?

[wanton zodiac]

I haven’t really done anything on it yet was just trying to figure out what to do first

[little cosmos]

Ah

Hmmmm

[wanton zodiac]

Yah yah

[little cosmos]

It would work to simply eliminate vectors until the set is linearly independent

[wanton zodiac]

So should I put them into 1 matrix and reduce?

[little cosmos]

But to do that it might help to row reduce so that you can see the dimension of the column space

Yes

[wanton zodiac]

Ok I will show you what I have

This is what I got

I’m sorry it took a while

I kept messing up my arithmetic

Actually I messed up again omg

Actually nvm I am good

[little cosmos]

So you have 4 pivot points now

[wanton zodiac]

Yes

[little cosmos]

This actually makes it even easier

What’s the dimension of col A

[wanton zodiac]

In R5?

I really haven’t learned this that much so idk I am sorry

[little cosmos]

Let me rephrase this

Do you know null spaces?

[wanton zodiac]

I just learned about those idk to much about them

[little cosmos]

Alright I’m just going to tell you that # of pivots = dim col A

[wanton zodiac]

Ohhh

[little cosmos]

It’ll make more sense once you start computing null spaces

[wanton zodiac]

Ok so it’s in the 4th dimension

[little cosmos]

Yes

[wanton zodiac]

Oh ok makes sense

[little cosmos]

But what space must the column space be a subspace of?

[wanton zodiac]

Of R^m

[little cosmos]

Which is?

[wanton zodiac]

4

Right?

[little cosmos]

Yes

[wanton zodiac]

Ok 😭

Good

[little cosmos]

So it’s a 4 dimensional subspace of R^4

What’s the only 4 dimensional subspace of R^4?

[wanton zodiac]

I have not a clue

[little cosmos]

It would be R^4 itself

[wanton zodiac]

Ok but why is that

[little cosmos]

Because any m linearly independent vectors in R^m must span R^m

For example: can you take 2 linearly independent vectors in R^2 that don’t span R^2?

[wanton zodiac]

No

Wait

[little cosmos]

You cannot

[wanton zodiac]

Oh yah whoops ok I got it

Ok I actually understand that a little better now

[little cosmos]

So now all we need is a basis for R^4

[wanton zodiac]

So it would be linearly dependent if it isn’t

And in my case it can’t be the zero vector

Which would make it linearly dependent

[little cosmos]

For this problem, any basis of R^4 suffices

[wanton zodiac]

I honestly don’t know what that means

[little cosmos]

A basis is a set of linearly independent vectors that spans a given space

A simple example would be the basis {(1,0),(0,1)} for R^2

You could also take the column vectors corresponding to the pivot positions

[wanton zodiac]

Ok ok but in my case i put the vectors with the leading pivots right?

[little cosmos]

Yes

[wanton zodiac]

Omg I am a genius

[little cosmos]

Or you could put the standard basis of R^4 or any other basis of R^4

When it asks for a basis for a space, it doesn’t care what the vectors are as long as they are a basis for the space

And there are an infinite number of distinct bases for any space

[wanton zodiac]

So it could be anything as long as it’s [1 2 3 4] ?

[little cosmos]

So long as it’s 4 linearly independent vectors in R^4

In fact, a basis for some space of dimension n is any set of n linearly independent vectors in the space

[wanton zodiac]

Oh ok ok

I got it right but now I have a question

What if it asked for NUL A then what would I do then?

[little cosmos]

That is the solution set to Ax = 0

So you still want that rref

[wanton zodiac]

Well it’s not asking me to find nul a I am just curious

[little cosmos]

I do need to go now but here’s a good resource https://www.cliffsnotes.com/study-guides/algebra/linear-algebra/real-euclidean-vector-spaces/the-nullspace-of-a-matrix

Or this might be better https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk09/09_basis_for_nullspace.html

[wanton zodiac]

Thank you for the help

And this link I’ll check them out

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[stiff atlas]

In the last step should the answer be 1/13 e^(2θ) (4sin3θ/13 - 3cos3θ) + C?

I don't know how it became 2sin3θ

[hollow knot]

the entire thing was just multiplied by 4/13 from the previous step isnt it

[stiff atlas]

yes

[hollow knot]

the 1/13 is outside

[stiff atlas]

oh

so he only multiplied by 2

4

[hollow knot]

well he multiplied the inside by 4

yeah

[stiff atlas]

thank you. I didn't see it

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[cobalt tapir]

i dont get why x2 ,x4 and x6 are free

what does it mean when a varisble is free

[wise anchor]

When a variable is "free," it means that your equations do not lock down what values the variable can take. A free variable can be set to any real number

So what they do here is that they generate a parameter (s, t, u) for every free variable, to express that it can be set to anything. Notice that in the first vector, the x_2 is set to one. Whatever s is set to, that will be the value of x_2. In the second vector, x_4 is set to one. Whatever t is set to, that will be the value of x_4. Similarly for x_6.

Free variables are caused by having more variables than you have independent equations.

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[versed maple]

can someone explain where the +1 vanished to

[ocean crypt]

where

[versed maple]

last two equals

when they rewrote and simplified it just disappeared

[misty coral]

1+C is just another constant

[versed maple]

ohh okay thanks

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[severe oracle]

let A = 6. If A increases to 12, what is the percentage of change of A?

[dull sequoia]

If A is 1 and I increase it to 2 what’s the percentage increase?

[severe oracle]

It increases by 100%?

[dull sequoia]

Right

What about for your question

[severe oracle]

I thought it was 100% too but my book tells me otherwise

[junior smelt]

Do they mention A (or B or C) before that point?

[severe oracle]

here is the original question. I understand and did everything perfectly until the last part where I need to calculate the percentage of change of A

[junior smelt]

Ah I see, yea-

Why are they adding 100%

[severe oracle]

That's exactly the part that I don't understand!

[junior smelt]

Yeah I don’t think they should be, might be an error on their part?

[severe oracle]

hmm I think so. I will ask my teacher after this. But 12 is 200% of 6 right?

[junior smelt]

Yep it is

[severe oracle]

alright thank you 😊

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[spiral vapor]

,w tan(-1.1)/(3*-1.1)

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[flat frigate]

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[spiral vapor]

,w tan(-1)/(3*-1)

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[peak estuary]

there is either a #bots channel or just go directly on wolframalpha

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[spiral vapor]

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[stiff atlas]

Integration of(tanx sec^2(x))

[stiff atlas]

Can an integration have two solutions?

There were two answers I got:

tan^2(x) / 2 using tan x as u
sec^2(x)/2 using sec x tan x as u

[pine cradle]

yeah

since there is a constant factor c

when you integrate

both are correct

[stiff atlas]

oh thank you

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[west hedge]

Given a spring with a Spring Constant of 200 (N/M)

[spice grove]

is n Newton?

Use N

No, m for metre.

[west hedge]

M for Mickey

[spice grove]

😭

[magic junco]

…

[west hedge]

if some laborer being asked to extend it 10 cm longer from its original state.

How many work should he needed to pull it longer.

So I applied the Hooke's law which is known as spring force=(the spring constant) * (the extended length of the spring)

[spice grove]

That's force.

[west hedge]

it is

So F=200 * (0.1) is the force needed to be imposed on the spring, so as to extend it.

[spice grove]

(which isn't at all what you want according to the question)

[west hedge]

why?

[spice grove]

Don't you want work

[west hedge]

then I put it into the formula w=f*s where f=20 and s=10cm(0.1m)

[spice grove]

That's incorrect.

[west hedge]

why?

[spice grove]

Well,

Work = $\int_{x_1}^{x_2} \vec{F} \cdot \vec{\dd x}$

[flat frigate]

! What the hell am I doing here?

[spice grove]

For each elementary displacement, the force varies.

For example, as you're extending the spring there'll be a moment when the extension is 5cm, another when it's 6cm and so on...

The force is different each time.

[west hedge]

but the calculation is legit, 20(N) is the force needed to extend it 10cm longer

[spice grove]

20N is the force to extend it by 10cm, yes.

But it's not always 10cm.

[west hedge]

wdym, not always 10cm.

[spice grove]

You started from 0 to 1 to 2 to 10.

[west hedge]

I cannot get it. It should be correct in my opinion.

[west hedge]

what was it then, if I put the result 20(N) into the work formula. isn't it what the question was asked for.

[spice grove]

Well, springs are non impulsive. It takes time for the force in a spring to change. And the force in a spring is only dependent on the extension/compression wrt the original length.
If you're pulling it to extend it some 10 cm, it means, without any doubt you also had an extension of 2cm somewhere in between. And the force at that moment, is NOT 20N.
It's 4N.
If you break down the 10cm into infinitely many small displacements, you'll notice that for each such displacement the force is something different. It's only 20 by the time you're done.

[west hedge]

I see it now

thank you sqrt(-1)

I have no question now.

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[patent topaz]

can anyone tell me if these rules do indeed create the language in (1)?

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[cedar pendant]

I need to set the equation in terms of Y(s) so I divide the s(s-1) to the other side, but I can't figure out which side to multiply the existing fraction by. The top basically cancels out the whole equation and the bottom I'm not sure.

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@cedar pendant Has your question been resolved?

[cedar pendant]

<@&286206848099549185>

i got it elsewhere

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[pliant gorge]

What is the b and c of these two trig graphs?

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[merry stratus]

How do I find the global max? I find that the derivative is -D^2 + CD which is 0 at 0 and -C. both are incorrect answers.

[obtuse plover]

dT/dD = 0

@merry stratus

[merry stratus]

how

also
if the rate of change was 0 there would be no global max
but none is not an option

[snow robin]

what dosage maximises the temperature change

means at what D is dT/dD the biggest

or smallest

the absolute value

[merry stratus]

T is the temperate change, not dT/dD

wouldn't you want the crit points to help find the global max

[obtuse plover]

we're doing question (1) right?

[merry stratus]

yea

[obtuse plover]

ok so find dT/dD

[merry stratus]

i got -D^2 + CD

[obtuse plover]

ok now set equal to 0 and solve

[merry stratus]

yea

i did

D = 0 and -C

[obtuse plover]

show how u got that

[merry stratus]

-0^2 + C0 = 0

[obtuse plover]

no

show how u got the solutions algebraically

im not asking u to prove it

from -D^2 + CD = 0

what did u use to get D=0 and D = -C

[merry stratus]

well I can clearly see plugging in D = 0 solves it

for -C I did
-D^2 + CD = 0
divide by D
-D + C = 0
D = -C

[obtuse plover]

ok lets start with the first statement "well I can clearly see plugging in D = 0 solves it"

just for future reference

if u have an equation like this

there are 2 main methods of solving for D

one is quadratic formula

and the other is factoring

if we wanted to factor, we'd get

nonono

kookiemon nooo

i seeeee it

[pure agate]

-D + C = 0

C = D

[obtuse plover]

waiitttt

[merry stratus]

bruh
this is what not sleeping does to a mf i forgor quadratic

[obtuse plover]

nah its fine just do factoring

so we get

D(-D + C) = 0

then using zero product property

we set each factor equal to 0

that means

D = 0 and -D+C = 0

thats how we do it. dont divide both sides by D, that ends up losing our solution of D = 0

here, it was pretty clear that D = 0 worked, but just for future reference

[obtuse plover]

anyways going back to this @merry stratus

if -D + C = 0

how do we solve for D

[merry stratus]

yes yes I see it

[pure agate]

His mistake was going from -D+C=0 to D=-C. An algebraic mistake.

[obtuse plover]

yea

[merry stratus]

I swear every time I come to this server it is never for a good reason

thx

[obtuse plover]

thats why i said this lol

[merry stratus]

.close

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[urban vale]

when I have something like tan^2(x) what does that ^2 mean?

[urban vale]

if I wanted to restructure tan^2(x) how could I do that

like tan(x)^2?

[umbral swan]

tan(x)² usually

[pliant egret]

it means tan(x)*tan(x)

[urban vale]

thanks a lot

[pliant egret]

yeah usually

[umbral swan]

rarely some people use it for the stacked case: tan(tan(x)) but then using tan²(x) is kind of misleading

[pliant egret]

it can sometimes mean tan(tan(x))

[urban vale]

ok

well thanks for the help, I totally get it now. bye

.close

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[sharp cedar]

Could someone help me out with this question?

[proud belfry]

what does the bar represent

[umbral swan]

^ @sharp cedar

[proud belfry]

normally it means the conjugate

but this is a real function

[stray socket]

The helll is s

[proud belfry]

shit i didnt even see the s before

[stray socket]

Oh square root?

Maybe it's square root

And like the bar is to show that they're the radicand??

[umbral swan]

that's like writing i(...) for integral

[stray socket]

that's hilarious

I should start doing that

[sharp cedar]

Yes

[sharp cedar]

Yes it is

[proud belfry]

lmfao

[umbral swan]

wut ok

[stray socket]

https://tenor.com/view/family-guy-dancing-bird-is-the-word-trashmen-surfin-bird-gif-27644963

[proud belfry]

alright just chain rule this

[sharp cedar]

Dumbfucks

[proud belfry]

LMFAO

[sharp cedar]

The book

Not yall

[proud belfry]

oh

[sharp cedar]

Mb for not adding context

[proud belfry]

yeah

what kinda of book does fuckin s

[umbral swan]

what's the bar?

[proud belfry]

what kind of book has enough formatting for top bar and not enough formatting for a fuckin square root

[sharp cedar]

Fucking james stewart

[sharp cedar]

They just tryna make it more hell for university students

[sharp cedar]

Square root

[proud belfry]

alright so you just chain rule this while being careful to treat y as a constant

[sharp cedar]

Yk whats even more confusing even after adding it

I dont know

How to solve it

[proud belfry]

so you know chain rule right

[sharp cedar]

The square root is on x and y only right

So it will be s sqrt(x^2+y^2)

Making sure

[proud belfry]

yes

[sharp cedar]

Alright

Another thing

[proud belfry]

be careful to treat y as a constant since it's with respect to x

[sharp cedar]

Yes

Alright

[sharp cedar]

what on gods earth is that

[umbral swan]

perfect math notation from the perspective of your math books' authors

[sharp cedar]

fuck james

Is that correct so far? I feel like somethings wrong

[proud belfry]

s is the square root

i think

[sharp cedar]

huh

[proud belfry]

so its not a variable

idfk

[sharp cedar]

what

😂

[proud belfry]

i still dont understand ur author's notation

is x bar sqrt(x)

or is s xbar sqrt(x)

[sharp cedar]

fuck that bastard james

the answer should be 1/5

im getting

1+3s/ 5(3+5s)

how the fuck should i cancel them

[sharp cedar]

<@&286206848099549185>

guys help me out

[proud belfry]

i dont think s is a variable

[sharp cedar]

product rule?

[proud belfry]

what is your textbook