help-10

[leaden ibex]

Knowing what I know from economics

It's a quantity on the x-axis and a price on the y-axis graph right

[meager elm]

Yes, that’s correct

[leaden ibex]

So basically what you have is y = mx + b, as you alluded to

[meager elm]

I tried Qs = Qd but I can’t do that because I can’t add like terms

I kinda forgot how y = mx+b works lol.. sorry if I ask dumb question

questions*

[leaden ibex]

I am almost 100% sure they must be equivalent.

For demand, set b = 300, m = -24

For supply, set b = 120, m = 8

[meager elm]

How do you know which number is mx and which is b

[leaden ibex]

Does the task contain any info besides that?

[leaden ibex]

well, m would be the coefficient. m is the coefficient next to the independent variable

Since price is a function of quantity, quantity must be the independent variable

[meager elm]

It just asks to plot the demand and supply curve

[leaden ibex]

On the same graph...or can you just draw two separate ones?

*plot

[meager elm]

Same graph

[leaden ibex]

Well then, there's no way around equating Q_d and Q_s

As far as I see

[meager elm]

😭

I think there’s a way to solve the equation but Idk how

[leaden ibex]

Why solve it

You just have two straight lines

Calculate two points for the supply curve

connect them

two points for the demand curve and connect those

[meager elm]

How do I calculate two points

[leaden ibex]

Take the demand curve

P = 300-24Qd

when Qd = 0, P = 300

When Qd = 10, P = 300-240 = 60

and now you have two points

(0;300) and (10;60)

[meager elm]

Thank you

I was thinking I would make P = 0

[leaden ibex]

You would also get a point like that

but it'd make calculations harder

[obtuse pebble]

@meager elm Has your question been resolved?

[obtuse pebble]

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[timid silo]

hi

[timid silo]

hi guys so look at this im doing my hw.

due in 2 hours.

my teacher gave me a video explaining this.

but the video is 4 minutes and Im stuck on the last question.

could you guys please assist me I have no idea.

I started with.

for question 2 since it goes with 4/

4(3)^3 + 6(1) + 9(ignore 9 because its a constant)

12(3)^2 + 6(1)

then I did

108+6

and with the second one my head just hurts.

IDK what to do can someone give me like a starting problem

[leaden ibex]

So wait...you have found the general function for a(t), right.

[timid silo]

I know I have to do power rule of differation.

yeah.

[leaden ibex]

so

just calculate a(3)

[silk condor]

You...... just plug in?

[timid silo]

a(t) = v'(t)

v

[leaden ibex]

no, you have a(t)

now t=3, so calculate a(3)

[timid silo]

v'(t) is 12t^2+6

right.

oooooooooooooooooooooo

ur so smart

[leaden ibex]

Don't overcomplicate your life

[silk condor]

You have your a(t) = 24t right there

[timid silo]

so would it be like this.

a(3) = 24(t)

and ignoring the 6 because its a constant yes?

so.

[leaden ibex]

no

[timid silo]

?

why would I ignore the 6.

[leaden ibex]

$a(t) = 24t \implies a(3) = 24\cdot 3$

[warm shale]

Remavas

[timid silo]

so.

24x3

24 48 72

a(3) = 72

o.

I see.

Thank you.

Im gonna try the next problem now.

Ill show u guys if i get it all right,

Thank U

I got it all right.

I did second one but im so dumb

Lol

missed a +

..

well TY guys I finally got the hang of this.

[obtuse pebble]

@timid silo Has your question been resolved?

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[river zephyr]

I only get negative values for this^

Could someone show me the steps after the last line of working

[idle thunder]

how did you go from 1 to 2 ? @river zephyr

oh you multipled nvm im so blind

[river zephyr]

Here’s the answer btw

[raven spire]

$$\sin (3B + 1) = 0.4$$

[river zephyr]

[raven spire]

$\implies 3B + 1 = \sin^{-1} 0.4$

[raven spire]

is just wrong

[river zephyr]

Oh

[raven spire]

Given $B \in [0, \pi] \implies (3B + 1) \in [1, 3\pi + 1]$

[raven spire]

,calc pi/3

[warm shale]

Result:

1.0471975511966

[raven spire]

UwU

Now, that you've figured that's the interval for 3B + 1

You should first calculate arcsin(0.4)

,w arcsin(0.4)

[raven spire]

0.41 helps

[river zephyr]

I don’t get this line

[raven spire]

what's the problem?

Given the interval of B

I just derived the interval in which 3B + 1 can lie in

Now, that you've figured that arcsin(0.4) = 0.41, which certainly does not fall into the acceptable interval

we proceed to check if π - arcsin(0.4) is acceptable

,calc pi - 0.41

[warm shale]

Result:

2.7315926535898

[raven spire]

which, indeed is acceptable

[river zephyr]

Uhuh

But there’s 2 other answers

[raven spire]

obviously

next, we proceed to check if 2π + arcsin(0.4) is in the interval

,calc 2*pi + 0.41

[warm shale]

Result:

6.6931853071796

[raven spire]

which, obviously is because our interval maxes at 3π + 1

and finally, the last candidate is:

2π + (π - arcsin(0.4))

which again, is clearly, obviously, trivially, less than
3π + 1

so there you go with your three solutions

π - arcsin(0.4), 2π + arcsin(0.4), 3π - arcsin(0.4)

[river zephyr]

Oooo wait lemme process this Hahaha

Wait but the answer says 2.67

[raven spire]

I only calced the value of 3B + 1

,calc pi - 0.41/3

[warm shale]

Result:

3.0049259869231

[raven spire]

btw, that's not the solution tbh, more like them doing some sorta black sorcery

you should focus on how they're doing it instead

if theta is a possible value, then π - theta is a possible value for 3B + 1 as well

[river zephyr]

HAHAHAAH

Okk I’ll try asking my teacher for working as well

Tysm

.close

[obtuse pebble]

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[calm pollen]

Working on transformations with polynomial graphs. I keep hearing 'reflected in the x axis' Is this the same as reflecting across the x axis?

[idle thunder]

yes

[timid silo]

it's the same thin

thing*

just different names

or wording

[idle thunder]

different wording

or poor wording even

[calm pollen]

Alright thanks, I was getting really confused and clears things up

.close

[obtuse pebble]

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[obtuse pebble]

@proven plank Has your question been resolved?

[obtuse pebble]

@proven plank Has your question been resolved?

[tardy epoch]

what have you tried?

[obtuse pebble]

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[wanton dagger]

I don't understand how to draw the graph of each function

[royal basin]

so if you were asked to draw the graph of y=2x+3, or of y=1, or of y=x^2, you wouldn't be able to do any of those?

[wanton dagger]

Um I would

Ok I will try doing that

So this is what I drew

And this is what the answer is

[autumn adder]

Hi.

@wanton dagger

[wanton dagger]

Hello

Yes

[autumn adder]

Can you come in a VC

Or

[wanton dagger]

Um ok can you give me like 2 min

[autumn adder]

Sure, ping me once youre here

[wanton dagger]

Ok

@autumn adder

I’m ready

[autumn adder]

Okay!

Join

Do close the channel when youre done.

[wanton dagger]

Yes thanks a lot

.close

[autumn adder]

Welcome!

[obtuse pebble]

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[timid silo]

Can someoen take me theough the steps to do this

[royal basin]

@timid silo if you were asked to find the angle between two vectors, would you be able to do it?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

no

[royal basin]

so you have never done problems like that before?

[timid silo]

This is my first problem like this one

[royal basin]

how familiar are you with the dot product?

[timid silo]

not familiar at all

[royal basin]

so you have never seen it before?

[timid silo]

yes

[royal basin]

strange

[obtuse pebble]

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[royal basin]

well then i guess you will just have to draw an axis-aligned right triangle with your vector as the hypotenuse

[timid silo]

the two vectors would be

5,0 and 0, -12 ?

[tardy epoch]

.reopen

[obtuse pebble]

✅

[tardy epoch]

Bad bot

[royal basin]

what "two vectors"

[timid silo]

if you want to find the angle between 5i-12j doesnt it have to be a triangle?

[royal basin]

"the angle between 5i-12j" makes no sense grammatically

theres no such thing as "between one object"

[timid silo]

so i should draw out 5i-12j first?

[royal basin]

yes...

[timid silo]

Would i connect the two points

[royal basin]

what two points

[timid silo]

isnt 5i

the vector 5,0

[royal basin]

5i is the vector (5,0) sure but you are not talking about 5i

you are talking about 5i-12j

(and you also dodged my question but whatever...)

[timid silo]

So would i just plot (5,-12)

[royal basin]

...yes sure

[timid silo]

but then it would just be a dot

[royal basin]

when you said "just plot (5,-12)" did you mean "plot the point (5,-12), refuse to do anything else and then wonder why the problem isnt solved yet"?

[timid silo]

i thought that you couldn't get an angle from the point 5,-12

[royal basin]

question-dodging again are we

[timid silo]

?

wait what was the question

[gloomy trellis]

hint: the vector and positive x-axis

[royal basin]

when you said "just plot (5,-12)" did you mean "plot the point (5,-12), refuse to do anything else and then wonder why the problem isnt solved yet"?

this question

[timid silo]

yes

[royal basin]

which you did not answer

so you DID mean that you refuse to do anything else

[timid silo]

but now i think i see

yes

[royal basin]

you REFUSE to do anything else

yes?

[timid silo]

ues

yes

but

[royal basin]

then you cannot solve the problem

solving the problem requires doing things

[timid silo]

i see something now

[royal basin]

you refuse to do things

[timid silo]

the point 5,-12

[royal basin]

if you refuse to do things then you can't solve the problem

and you refuse to do things

[timid silo]

it takes time

[royal basin]

so you cannot solve the problem now that you've refused to do anything else

[timid silo]

to solve things

I've joined up 5,-12 to the x-axis and made a triangle

[royal basin]

you've DONE THINGS

[timid silo]

its hard to tell

[royal basin]

which you told me just moments ago that you absolutely fucking refuse to do

how about you don't LIE next time

[timid silo]

I thought i had to like say "yes"

[royal basin]

you thought you had to say "yes" regardless of the content of what i was asking you???

thats bullshit

[timid silo]

no

on the part when you repeated "you refuse"

as i thought you implied that youthought I had done nothing

[royal basin]

what a convoluted presumption

[timid silo]

I'm not trying to cause drama

Im sorry on everything that I'ved done wrong so far

[royal basin]

youre not causing drama youre causing miscommunication

whether on purpose or otherwise

[timid silo]

It wasn't on purpose and I'm sorry if it soundedd like it was

[royal basin]

plot the point (5, -12) and join it to the origin, then drop a perpendicular from (5, -12) onto the x-axis

and you will have your triangle

in which you can calculate the required angle

[obtuse pebble]

@timid silo Has your question been resolved?

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[tardy epoch]

Like x=r and y=theta?

[warm shale]

Ulo

[tardy epoch]

y=r sin theta = theta seems harder to solve

Sooooo yea don't do it

Just pick the origin and call it a night

😴

[obtuse pebble]

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✅

[tardy epoch]

2pi n = r sin(2pi n) only has one solution in n

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

@amber fiber Has your question been resolved?

[high lily]

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[gilded needle]

What do the square brackets mean? Round to nearest integer? Floor? Something else?

[obtuse pebble]

@torn cloud Has your question been resolved?

[wooden solstice]

are the tops of the square brackets bare?

$\lfloor \cdot \rfloor$

[warm shale]

the_guruji

[wooden solstice]

like this? if so, then it likely is the floor function.

if not, look at previous notation in your material as well. if this question is asking you to calculate the value of S by hand, then it probably is something like the floor.

[obtuse pebble]

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[static patio]

can i apply lhospital on this ?

[static patio]

im confused because cotx does not have limit at 0

[rugged kite]

what is 1/cot(x)?

and does that have a limit at 0

[royal basin]

are you FORCED to apply l'hop?

[static patio]

yes

[royal basin]

[rugged kite]

That's stupid as shit

[royal basin]

i mean, in this form you can't

[static patio]

but

[royal basin]

this is not a 0/0 or ∞/∞

[rugged kite]

And honestly it feels like more effort to turn it into that

Anyway, what is 1/cot(x) ?

(it's just the definition of the cotangent)

[static patio]

i applied LHR on this and i got limit above

[static patio]

0

[rugged kite]

x

not 0

I'm just asking about the function

1/cot if you will

(although, 0 is what i'm getting at, but let's take it slowly)

[static patio]

hmm

[mental solstice]

a version of l'hop says you can apply it as long as the denominator has an infinite limit, with no prior assumption on the numerator

it's kind of stupid, but

it's technically l'hop

[static patio]

so, am i allowed to use lhop for this ?

[rugged kite]

You can if you really insist on it I guess

Do you know the definition of cot?

[mental solstice]

yes, but it's kind of stupid to, is my answer

[balmy mortar]

why do u need lhopital?

tan/cos

[rugged kite]

well, you kind of spoiled it, but yes

cot = 1/tan

1/cot= tan

[balmy mortar]

they already did it

[static patio]

and how to solve it without lhop ?

[rugged kite]

1/cot(x) is tan(x)

so it's the limit of tan(x)/cos(x) as x -> 0

[balmy mortar]

Lhopital wont do anything

[rugged kite]

Which is 0/1 = 0

[raven spire]

that was the original limit I suppose?

[static patio]

yes

[mental solstice]

if that was the original limit and you've already applied l'hop, why are you insisting on it again?

[balmy mortar]

you seem to misunderstand if u think u need it

[static patio]

i didnt see other solution 😦

[balmy mortar]

There is no other solution

[raven spire]

if you've already applied it once, I think you're done with whatever's compelling you to apply L'hopital.
You should just proceed naturally if it's not even an indeterminate form anymore

[balmy mortar]

lhopital is inapplicable here according to your notes

You dont have an indeterminate form

Check what it is

[static patio]

but in denominator is abs(cotx cosx) = inf

[balmy mortar]

whats 1/inf

also, do you know what the indeterminate forms are?

[static patio]

yes

[balmy mortar]

this is none of them clearly

[raven spire]

And dude >_< lmao Idek what kinda Lhopital you applied to the original Limit

[static patio]

so it needs to be only indeterminate form when i want to apply lhop ?? because the inf denominator requirement is met

[balmy mortar]

You can only apply it to specific indeterminate forms

[rugged kite]

L'H can only be applied to 0/0 or ∞/∞

[balmy mortar]

Your notes should tell you this

[static patio]

ln(cotx)/(1/sinx)

[raven spire]

$$\lim_{x \to 0} \frac{\ln (\cot x)}{\csc x} = \lim_{x \to 0} \frac{\csc^2 x}{\cot x \cdot \csc x \cdot \cot x}$$

[rugged kite]

If you encounter something like 1/∞ then it's not indeterminate, it's 0, no need for L'H

[static patio]

okay, so the limit goes to 0 because 1/inf

[balmy mortar]

you made an algebraic error

in differentiating it seems

or?

[static patio]

i dont think so

hmm

[balmy mortar]

no its correct

[balmy mortar]

same as this

[raven spire]

halfway simplifying tho consider switching cot x to cos / sin as well

[static patio]

im not familiar with csc function... 😦

[balmy mortar]

1/sin

[static patio]

we dont use them in school

[mental solstice]

cosec

csc

same thing

[balmy mortar]

its shorthand for 1/sin

[rugged kite]

inverse trig functions are just there to make calculus with trig functions look cleaner than it actually is

[balmy mortar]

so yes its the same thing

[static patio]

ok thanks a lot for help

.close

[obtuse pebble]

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[cedar ermine]

hey guys just wondering. to find the total number of item that fit in a cube with length of 100cm. do i add the number of ites for length, width and height or multiply all them ?

[cedar ermine]

wouldnt you multiply those numbers because the volume of a cube is l^3 so

<@&286206848099549185>

[obtuse pebble]

@cedar ermine Has your question been resolved?

[cedar ermine]

<@&286206848099549185>

[obtuse pebble]

@cedar ermine Has your question been resolved?

[obtuse pebble]

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[vagrant dawn]

Could somebody tell me what I'm doing wrong here?

[vagrant dawn]

The answer is E but some of the others are not even in my list of solutions

Do I need to find the solution using sin instead of cos or what?

<@&286206848099549185>

[strong vale]

Why don't you just plug them all in?

[vagrant dawn]

I know which one apparently doesn't work which is 5pi/6 from the answer sheet but that's not even in my solution set

[compact shadow]

Sin(2x)=0 or cos(2x)=0 or cos(2x)=-1/2 that’s what I got

[vagrant dawn]

I got the one for cos2x but how did you get the one for sin?

[compact shadow]

2x=y

Siny +sin2y +sin3y=

Siny(2cosy+4(cosy)^2)

[vagrant dawn]

what did you do in the end there

Can you write it in texit?

Wait I'll try something

[compact shadow]

Sorry have to go somewhere

[vagrant dawn]

oh ok

Idk how sin(2x)=0 is found could somebody tell me?

<@&286206848099549185>

[obtuse pebble]

@vagrant dawn Has your question been resolved?

[vagrant dawn]

nope

[balmy mortar]

You should be able to sketch graphs of sin cos and tan

from -pi to pi or 0 to 2pi

[vagrant dawn]

idk if the teacher would rly accept that tho

and I'm not familiar with sketching trig graphs yet

[balmy mortar]

Theres only 3 graphs you need to learn

the ones i listed

So get familiar with them

draw each once every day maybe 🤔

[balmy mortar]

The solutions to sin2x = 0 should be able to be 'written down'

working would not be expected for it

[balmy mortar]

y = sin x
y = cos x
y = tan x

[vagrant dawn]

Like I know how they look but idk how the coefficients abcd change it in for example acos(bx+d)+c . I think I should be able to solve it without graphs since the teacher didnt teach it to us yet

[balmy mortar]

no

You just need these 3 graphs

If you can sketch these 3 graphs, then you should be able to solve sin2x = 0

for example

[balmy mortar]

If you are unfamiliar with graph transformations, practice in desmos

But there is no need to be able to sketch them in particular

[vagrant dawn]

Ik how to solve for sin2x=0 but idk how we got to it from sin2x+sin4x+sin6x=0

[balmy mortar]

but is a good thing to be able to do

Ive no idea about the question

you asked for sin2x = 0

[balmy mortar]

Look at cog's solution or ask them

[vagrant dawn]

Wait... if sin2x equals 0 then does that mean sin(2kx) is always 0?

Where k is an integer

[balmy mortar]

idk, check the graph

use desmos if u need

Yes, you need to familiarize yourself with graphical transformations it seems

[vagrant dawn]

I guess I'll graph it

[compact shadow]

?I just directly calculated. sin(y)+sin(2y)+sin(3y)=siny+2sinycosy+sin2ycosy+sinycos2y+ siny+2sinycosy+2sinx(cosx)^2+siny(2(cosy)^2-1)=siny(2cosy+4(cosy)^2)

There is a very quick way if you just want to get the answer without knowing all solutions is by considering w=cos2x+isin2x then that equation holds iff w+w^2+w^3 is a real number

[obtuse pebble]

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[obtuse pebble]

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[quaint escarp]

Need a bit help how to get started at the a)

[misty terrace]

are you supposed to solve for x?

[indigo siren]

Probably

[quaint escarp]

yup

[indigo siren]

Firstly maybe +1/2

[quaint escarp]

<@&286206848099549185>

its def easy, im just dum

[indigo siren]

Do you have the answer?

[quaint escarp]

nop

brOOuu help me man

[indigo siren]

I could but I am doubting my algebra skills

[quaint escarp]

ok i solved it

next is b)

lol

[obtuse pebble]

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[reef pagoda]

Is an orthonormal vector just an orthogonal vector with the additional criterion of being a unit vector?

[reef pagoda]

I actually have a handful of questions about this but none of them should be too hard to answer I dont think

1. If a set of vectors is orthogonal and of n dimensions, does that imply it is a basis for n dimensional real space?

[haughty coyote]

Yes for both. Assuming "a set is orthogonal" <=> the vectors are orthogonal 2 by 2

[reef pagoda]

2. Is it valid to find/prove a basis via the gram schmitt algorithm given that the criteria for the algorithm apply? I know that if the inner product space is zero between all vectors then it is an orthonormal basis

[reef pagoda]

so does the matrix representing these vectors have to have a number of entries that is a square?

[haughty coyote]

That's just a complicated way of asking whether a basis for a vector space of dimension n has n vectors. The answer to which is obviously yes

[reef pagoda]

For this, Im more asking about the strength of it I guess. Im trying to write good proofs and not just "algorithm go brrr" without any explanation

[reef pagoda]

Sorry about that, im still trying to get the formalism side of things down

[haughty coyote]

Something tells me you just need to look at the definition of the algorithm. I can't really tell you otherwise as I haven't really studied it

[reef pagoda]

Hmm alright. Do you have experience with vector rejections then?

[haughty coyote]

projections you mean ?

I saw the gram schmit algorithm once. What I mean is that I didn't really study linalg that far yet

[reef pagoda]

kind of. Rejections are the (from my understanding) the compliment of a projection such that the projection + the rejection = an orthogonal vector

[haughty coyote]

" an orthogonal vector" orthogonal to what ?

[reef pagoda]

to the vector that is being projected onto

[haughty coyote]

so (proj v onto u + rej v onto u) . v = 0 ?

[reef pagoda]

from what I understand yes. This is essentially what the gram schmitt algorithm does

a snip from wikipedia vector projection article

[haughty coyote]

so rej v onto u orthogonal to u ?

ok

[reef pagoda]

yea, that isnt how I think of it since I think of it more in terms of an algorithm than a concept since that is simpler for me but yep

aight im going to close the thread thanks

.close

[obtuse pebble]

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[slender carbon]

if left side is always bigger

and x cancels out

b has to be equal to like negative 3

but when I solve it I get b=10

[raven spire]

Did you simplify the equation a bit?

[slender carbon]

-10*

[slender carbon]

ya Ill type it out

2x+7-x=4x+2b-3x-3b

x+7 = x-b

wait

oh I did it wrong

b=-7 is supposed to be the answer then

I think

[raven spire]

⁉️

[slender carbon]

what does this mean

[raven spire]

my question remains unanswered

[raven spire]

.

[slender carbon]

isnt x+7 = x-b

it simplified

[raven spire]

Is x + 7 = x - b your final simplification of the equation?

[slender carbon]

no

it would be b=-7

[raven spire]

wrong

consider re reading the question and tell me why b = -7 and 7 = -b are actually different equations

the left side is always 3 more than the right side

there's a sense of right and left sides of the equation, so you can't just casually flip the sides

[slender carbon]

hmm

so we have to keep what is on the left to the left

and the same for the right

[raven spire]

now, let me ask you again

what is the most simplified form of the equation

[slender carbon]

-7=b

[raven spire]

wrong

please reconsider

[slender carbon]

ok

[raven spire]

and refrain from doing any alterations to either sides of the equations

simplify each side separately

[slender carbon]

x+7=x-b

[raven spire]

Yes

now, my question is

what is the left side of the equation

[slender carbon]

x+7

[raven spire]

what is

the right side of the equation

[slender carbon]

x-b

[raven spire]

Given,

the left side = right side + 3

can you solve that for me please?

[slender carbon]

x+7=x-b+3

b=-4

[raven spire]

Yes

that should be your answer

[slender carbon]

ok thanks

.close

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[mighty jackal]

how would I factorise x^3-3x+2

[mighty jackal]

It factorises to (x-1)^2 (x+2)

[high lily]

start with rational root thoerem

[mighty jackal]

But how do I spot that

ah I see

[high lily]

wdym by "how do I spot"

[mighty jackal]

I don’t know what that is

I will learn that now

I tried to use polynomial division

[high lily]

rational root theorem (in combination with factor theorem, polynomial long division or synthetic division)
is a common approach for factoring factorisable high degree polynomials

[mighty jackal]

I see

[balmy mortar]

You have to guess a root in order to factorise anything of degree 3 or above

except in some special cases

[mighty jackal]

i did guess (x-1)

[balmy mortar]

guessing usually means -2, -1, 0, 1, 2

[mighty jackal]

But I don’t understand how I’d know it was a repeated root

[balmy mortar]

You don't

You apply long division

With (x-1)

[mighty jackal]

I did

[balmy mortar]

And then factorise the resulting quadratic like normal.

[mighty jackal]

I think I might have made a calculation error

[high lily]

how'd -3x turn into 3x

[mighty jackal]

ah

I see

I was a bit stressed 🥲

okay I got it now

My bad

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[slender carbon]

is there like a cheat sheet for when to know if possible answers split into 2 like for instance
x^2 =36 can be split into +-6 for x possible values

[compact shadow]

I don’t think there is a powerful way to determine whether any given integer is a perfect square…

At least basic of algebraic number theory isn’t enough, and I haven’t got deeper yet…

But if the last digit isn’t one of those: 0,1,4,5,6,9 then it’s not a perfect square. (By considering {x^2: x from Z/10Z})

[obtuse pebble]

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[dense gulch]

Hi, I’m stuck on a linear algebra question and I’m wondering if anyone can help me? I’m not too sure on how to go about checking whether the spanning sets are equal. Can anyone help?

[serene osprey]

you can consider about their linear independence, in this case W and Z can both be generated by X and Y, since W and Z are linearly dependent we can claim that span of W is equal to span of Z, and so span of XYZ is equal to span of XYW

[warm shale]

Mat

[obtuse pebble]

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[timid silo]

how do i find the radius of these semicircles?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[sterile wyvern]

Looking the solid and the slices from above, you'd see this

Does this help you @timid silo?

[tardy epoch]

Mat with the A+ desmos skills

[timid silo]

well I eventually just decided to try 1/16x^2 - 1/2sqrt(x)

and that was it

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[timid silo]

could someone pls guide me, I did study abt taylor series in class, but im clueless how to use them in questions

[timid silo]

ping me please

[balmy mortar]

you sure thats meant to be taylor?

Doesnt it look a lot like MVT statement

or IVT

[timid silo]

oh im not sure then, we got this question, from an exercise sheet just after finishing taylor

[balmy mortar]

well yes its relevant

[timid silo]

i guessed it was taylor related

[balmy mortar]

but so are MVT and IVT probably

[obtuse pebble]

@timid silo Has your question been resolved?

[warm shale]

Arkos

[timid silo]

hope im not going in the wrong direction, but I couldnt proceed

<@&286206848099549185>

ping please

[tardy epoch]

Taylor expand f around 0 to two terms, then integrate.

[timid silo]

but

im not sure how I get f''(c) from that

@tardy epoch

oh wait the error function

forgot abt that

thanks y'all

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✅

[timid silo]

.close

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[hard kelp]

got this integral cosx / 1 + sinx dx

[hard kelp]

i managed to u sub

and got it all the way down to ∫ u du

where u = 1 + sinx

integrating gives u^2 / 2

=> (1 + sinx) but i dont think this is quite right

[gloomy valve]

if you substitute $u = 1 + \sin(x)$, then your integral becomes $\int \frac{1}{u} \dd{u}$

[warm shale]

Navix

[hard kelp]

oh

oh

i see

mb mb mb

thanks thats all

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[timid silo]

on her way to her grandmothers house jackie traveled three times as many miles by train as by bus she traveled four times as many miles by bus as by foot if jackie traveledd 34 miles in all how many miles did she travel by train
can someone help me
its simple
but im dumb

[hard kelp]

t = 3b
b = 4f
f = 34

hope that helps

[obtuse pebble]

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[wind fern]

help

[wind fern]

how do i do this

[royal solar]

try draw euler/ venn diagrams. it’s pretty neat. if p implies q. then draw p as a subset of q

p if q would mean p and q are the same set in the diagram

this is just for visualising btw

@wind fern

p and t is the intersection

p or t is the union of sets

[royal solar]

oh wait. they’ve solved it. line 2 is redundancy imo

[royal solar]

seems right to me, then again

discrete maths

[obtuse pebble]

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[crystal gulch]

So i have been tasked with figuring out the new price for tomatoes. The initial price was $1.04 and the percent decrease is 22.8%. I subtracted 100% by 22.8% and I got 77.2%. I then multiplied 1.04 by 77.2% and I got 0.802. Am i doing it right?

[balmy mortar]

and then?

yes

,calc 1.04*(1-.228)

[warm shale]

Result:

0.80288

[obtuse pebble]

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[wintry stream]

can someone pls help me solve this

[obtuse pebble]

@wintry stream Has your question been resolved?

[tardy epoch]

did you draw a picture

[obtuse pebble]

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[primal panther]

I know the formula(s) for this, but Im a bit at a loss as to how to simplify them? I feel like I must be doing something wrong because of how busy the equation gets. Ive been trying |r'(t) x r''(t)| / |r'(t)|^3

[tardy epoch]

looks right?
https://tutorial.math.lamar.edu/classes/calciii/curvature.aspx

show your step by step work

[primal panther]

I did the derivatives first, so r'(t) = -3cos(t), -3cos(t), -4sin(t) and r''(t) = 3sin(t), 3sin(t), -4cos(t), then moved on to the cross product which ended up giving me 12, -(12cos^2(t)-12sin^2(t)), 0 but what stumps me is when I get to the magnitude

I guess I could reduce -(12cos^2(t)-12sin^2(t)) with trig to -12cos(2t) but that still leaves me with sqrt(144+144cos^2(2t)) but then Im not sure where to go from there

[tardy epoch]

show your work on the cross product

[primal panther]

i j k
-3cos(t) -3cos(t) -4sin(t)
3sin(t) 3sin(t) -4cos(t) then the math:

(12cos^2(t)+12sin^2(t)), -(12cos^2(t)+12sin^2(t)), (9sincos-9cossin)

oh wait, it should be -12sint^2(t) in the first block

does it all negate to 0 in the magnitude? one moment

hm, yeah im still not sure, i just did a quick look via calculator, and the simplification is confusing

[tardy epoch]

your i component has a sign error

and your j component

[primal panther]

where from? I dont see it

[tardy epoch]

https://tutorial.math.lamar.edu/classes/calcii/crossproduct.aspx

[primal panther]

oh

im still not sure, I included the block 2's added negative, i accounted for the cos negative when derived, where did i go wrong?

oh i see it, idk why i thought i had that wrong to begin with

god what a stupid mistake

lemme try again

so then its sqrt(288)

the bottom half still has me confused though

|r'(t)|^3, on a calculator that gives me 46656cos^6(t)sin^3(t)

[tardy epoch]

and is that the right answer?

[primal panther]

I just thought itd be something simpler, ill check

no it isnt

i feel like there has to be a way to simplify the bottom part but i dont know it

[tardy epoch]

even if you can simplify, which yes you can, it doesn't change the value of it

[primal panther]

i tried inputting sqrt(288)/46656cos^6(t)sin^3(t) and its apparently wrong, im not sure how that would be if theres nothing else to do

ah wait

i figured it out

[tardy epoch]

$|(-3\cos(t), -3\cos(t), -4\sin(t))| = |2\cdot 9\cos^2(t) + 16\sin^2(t)| = |2\cos^2(t) + 16|$

[primal panther]

i was just too dead tired to realize I didnt put the + sign in the calculator

im an idiot

[warm shale]

riemann

[tardy epoch]

i can't help you with your calculator skills

[primal panther]

lol

anyway, its correct, thank you so much for pointing out those mistakes

.close

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[obtuse pebble]

@copper latch Has your question been resolved?

[tardy epoch]

ohhh use the arctan sum formula repeatedly

https://proofwiki.org/wiki/Sum_of_Arctangents

[copper latch]

Oh thanks

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[shell hull]

$$2a^2+2b^2+2c^2=2ab+2ac+2bc$$
$$Proof a=b=c$$

[warm shale]

Tanjiha

[compact shadow]

(a-b)^2+(b-c)^2+(c-a)^2=0

[shell hull]

.close

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[grave kestrel]

How to find area of base of right angle pyramid with isosceles triangle base?

[obtuse pebble]

@grave kestrel Has your question been resolved?

[high lily]

depends on what you're given

[hybrid gull]

@grave kestrel ^^

[flat anvil]

yeah can you send ur whole question

[hybrid gull]

[high lily]

e.g if you're given all/sufficient info about the triangular base, its presence in the pyramid is completely irrelevant

[obtuse pebble]

@grave kestrel Has your question been resolved?

[grave kestrel]

It is actually an example as you can see in the pic
I just can't understand about how we would solve this problem

[high lily]

,rotate

[high lily]

so yeh

e.g if you're given all/sufficient info about the triangular base, its presence in the pyramid is completely irrelevant

the diagram could be clearer but based on the work shown, 2 sides of that triangle are 15cm

[grave kestrel]

How would we determine that 15 is hypotenuse?

[high lily]

bisection theorem

a perpendicular bisector can be constructed to split the isosceles triangle into 2 congruent right triangles

[grave kestrel]

Can you tell me about where that 9 came from in the pythagoras theorem in the bottom left corner

[high lily]

a perpendicular bisector can be constructed to split the isosceles triangle into 2 congruent right triangles

what happens when you split the 18cm base into 2 parts evenly?

[grave kestrel]

9+9=18
Okay I got it
So, we will get 2 right angle triangles with 15cm hypotenuse and 9cm base and unknown perpendicular
Right ?

[high lily]

yes

[grave kestrel]

Thanks

[obtuse pebble]

@grave kestrel Has your question been resolved?

[obtuse pebble]

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[sleek sinew]

Can someone please help me with this? Number 6

I dont know if either of these are correct, or if one is, why

The second one seems weird to me because i would expect the plane to have a trajectory that looks like the first diagram

So the plane's path would be shifted slightly righr slightly up

But drawing diagram one, is the wind blowing at 320 degress anymore??

It seems like 350 degrees now or so

[obtuse pebble]

@sleek sinew Has your question been resolved?

[sleek sinew]

<@&286206848099549185>

[obtuse pebble]

@sleek sinew Has your question been resolved?

[timid silo]

sorry i haven't learnt about vectors yet

[royal solar]

write eqns in terms of is and js

[royal solar]

you know the normal stuff right? anything that goes right or left has an i coefficient

up or down would have a j coefficient

is don’t affect js and vice versa

[sleek sinew]

i dont think ive learned it that way

we are just using law of sines/cosines

and drawing the triangles/vectors and finding the angles

[timid silo]

uh yeah idk what u mean

[royal solar]

you can do it that way

[royal solar]

if you have 2x + 3. 2 is the coefficient of x

right?

[sleek sinew]

okay, but is any of my diagrams correct?

the first diagram seems more right to me, but i dont know if i can just move the vector like that

[royal solar]

is air speed the resultant of lift and ground speed?

[sleek sinew]

or if it would change the angles

[timid silo]

uh yeah

[royal solar]

nvm, it’s 2d like that

[royal solar]

your top one looks right i think

just check the angles one last time

ok, you need to add the angle

draw all the angles you know

[royal solar]

yeah soo if i threw a ball straight up in the air with a velocity y which i don’t know. the vector form would be v = yj

if i throw it at straight ahead, zero vertical velocity and if i throw it at a velocity x, then v = xi

[sleek sinew]

I dont know how to find the inside angles of the triangle

Also i feel uncomfortable with moving the vector like that and how it is still blowing 320 degrees

Even though it looks right

[royal solar]

now if i throw the same ball with the resultant of both velocities. so at an angle. v = xi+ yj

[royal solar]

head wind?

that’s coming from 320

soo

[sleek sinew]

Im just confused about the location of the headwind

It is 320 degrees but is it blowing down, diagonally, right..?

[royal solar]

yep

it’s from 320

soo it’s towards 320-180

140

and 140 is the bearing from north

[royal solar]

soo what would the angle be

[sleek sinew]

not really sure what you mean by this. i would guess 340 - 320 = 20

but dont know how that makes sense

[obtuse pebble]

@sleek sinew Has your question been resolved?

[royal solar]

i’ll draw and show ya

@sleek sinew

does that make sense

the angle made by the dotted line and 400 can be calculated by sine rule

ratio of angles = ratio of sides

and if i call that angle x

340+x is your bearing from north

[obtuse pebble]

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[static patio]

any ideas how to solve this limit ?

[static patio]

probably lhospital should be used

[compact shadow]

Yeah

That

[static patio]

?

[compact shadow]

I mean yeah but maybe you need to adjust a little

Like y=1/x

[static patio]

hmm

i think in does not help

[compact shadow]

It helps

[static patio]

it still woud be inf/0

[compact shadow]

Otherwise the index can’t get smaller

[brittle swan]

try finding ln (L)

$L = e^{\ln L}$

[compact shadow]

No it was 0/0, after this adjustment, it becomes positive infinity/ positive infinity

[warm shale]

Chromium

[static patio]

hmm

i ended up with this

[static patio]

what is big L ?

[brittle swan]

i don't think t = 1/x is needed

[compact shadow]

No, t^70/e^(t^2)

[brittle swan]

original limit

[compact shadow]

I need the index getting smaller and smaller dude, without this adjustment 70–>72–>74–>…

[brittle swan]

try investigating $\lim_{x \to 0} \ln \qty(\frac{e^{-\frac{1}{x^2}}}{x^{70}})$

[static patio]

i dont get it, how did you come up with this

[compact shadow]

Direct calculation…

[static patio]

hmm

[warm shale]

Chromium

[compact shadow]

It’s adjusted so I get 70–>68–>66–>…

[static patio]

got it

[compact shadow]

Try the original one if you don’t believe me, the index will only get bigger

[static patio]

i will try

thanks guys, i will try both ways

[static patio]

what to do with it ?

inside ln is inf/0

[spiral maple]

no

[static patio]

?

[spiral maple]

e^-(1/x^2) goes to 0 as x goes to 0

[static patio]

-1/x^2 goes to - inf ?

[spiral maple]

as x goes to 0, yes.

[static patio]

and e ^ -inf is not -inf ?

[spiral maple]

no.

,w plot y=e^x

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Omegabet_#9771

[spiral maple]

$\lim_{x\to-\infty}e^x=0$

[warm shale]

Mosh

[static patio]

uhmmm

[spiral maple]

recall that exponentials have asymptotes

[static patio]

so i can apply lhospital directly

[spiral maple]

sure if you want to be lazy

[static patio]

what is the best way ?

[spiral maple]

t=1/x^2 probably

gets $\lim_{t\to\infty}\frac{e^{-t}}{1/t^{35}}$

[warm shale]

Mosh

[spiral maple]

then growth comparison

[static patio]

uhmm

if I chose the first way it would be necessary to apply LHR 70 times ?

[spiral maple]

yes, I believe so

Though I'm not really giving L'H a 2nd thought

[static patio]

ok thanks a lot

.close

[obtuse pebble]

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[last crow]

Hello!

[last crow]

im not sure if thats how you should calculate the pmf and then calculate the expected value and variance

or for the variance would i have to add a sigma

for the x = n part

[obtuse pebble]

@last crow Has your question been resolved?

[obtuse pebble]

@last crow Has your question been resolved?

[last crow]

.close

[obtuse pebble]

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[tender nest]

in integration by substitution,

[tender nest]

in integration by substitution,, why do they multiply dx to du

like here... why is du = 2x dx and not just "du = 2x"

where does that dx come from?

[daring rock]

If $u=x^2+5$, then $\frac{du}{dx}=2x$, so $du=2x dx$

[warm shale]

tatpoj

[daring rock]

The third line in your picture is wrong though, a 2x randomly turned into 2

[high lily]

implicit differentiation/chain rule

[obtuse pebble]

@tender nest Has your question been resolved?

[tender nest]

oh

so we can just multiply dx to the right?

[obtuse pebble]

@tender nest Has your question been resolved?

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[delicate orbit]

a question from elementary logic

[delicate orbit]

Is getting a grade of higher than 50 (out of 100) on the final examination a necessary condition for passing this course? Is it a sufficient condition?

I am just a beginner in studying logic, but I think both are appliable.

it is both a necessary and a sufficient condition

because getting a grade higher than 50 implies that I've passed and I've not passed unless I've got a grade higher than 50.

[obtuse pebble]

@delicate orbit Has your question been resolved?

[leaden ibex]

Well, each grading system is different. But is getting 50/100 the only condition to pass the course?

[delicate orbit]

if we look at a standard one, it could be the only one

[leaden ibex]

Then yes, it would be both a necessary and sufficient condition

You pass the course if and only if you get higher than 50/100

[delicate orbit]

and, since you've raised the question of the system having a possibility of not being a standard one, what would be our response

[leaden ibex]

Well, let's say the course also required you to finish all labwork (since I study physics, that is one example that comes to my mind)

Then getting a passing grade (> 50/100) is a necessary condition

but not a sufficient condition

[delicate orbit]

makes sense

[leaden ibex]

If I passed the course, it follows that I got a grade higher than 50/100.

If I got a grade higher than 50/100, it doesn't necessarily follow that I passed the course.

[delicate orbit]

yes, got it!

thanks

🙏

.close

[obtuse pebble]

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[gleaming locust]

how do i work out the area of trapexiums

[leaden ibex]

Do you know the formula for the area of a trapezium?

[gleaming locust]

nvm i just googled it

[flat anvil]

Use ‘.close’ if no more questions

[gleaming locust]

nah i acc got it

[flat anvil]

So u hav no more questions?

[gleaming locust]

i do