help-39

[pearl pond]

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[torn cipher]

how can i solve this?

[pearl pond]

@torn cipher Has your question been resolved?

[pearl pond]

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[karmic ruin]

Why is 1^∞ undefined but lim x→∞ (1^x) = 1?

[worthy lance]

That's not correct

infinite is not a number

that's why 1^infinity is undefined

because there is no definition for 1 powered to something is not a number

not sure if you know what I mean

[karmic ruin]

Yeah like infinity is a concept not a real numb

[feral sedge]

1^∞ is indeterminant is what you mean?

[karmic ruin]

A really large numb

Ya

[worthy lance]

no

[feral sedge]

because a->1 and b->∞ doesn't mean a^b -> ∞

[worthy lance]

1^∞ doesn't exist

it's like saying 1^potato

[feral sedge]

He's talking about limit forms

[worthy lance]

he made a difference in his question

between 1^∞ and limit when x->∞

[feral sedge]

∞+5 = ∞ is a shortcut for saying if a approaches infinity and b approaches 5 then a+b approaches infinity

[karmic ruin]

I typed in lim x→∞ (1^x) in the symbolab online calc and it gave me 1. That's why I'm confused

[worthy lance]

Let's make this clear. The number 1, powered to a number that grows to infinity, is 1.

[feral sedge]

^

[worthy lance]

Because it doesn't matter how big the number is, 1^a = 1

is always 1 when a >=0

so doesn't matter a -> infinity

I think you are confused with

a number that tends to 1 powered to a number that tends to infinity

[karmic ruin]

Oh so a is just approaching some really large number

[feral sedge]

The whole point of defining these operations with infinity is this property

[digital walrus]

1^x = e^(x ln(1)) thats why its indeterminant

[feral sedge]

That property is not the case for 1^infinity

[worthy lance]

??

what are u talking about deadly

[digital walrus]

do u know exponential and logarithmic?

[worthy lance]

bro

stop it

limit x-> infinity of 1^x is = 1

end of story

not indeterminant, not undefined, no nothing

it is 1

again i say

u must be confused with "SOMETHING" that aproaches to 1, powered to "SOMETHING ELSE" that approaches to infinity

that's a different story

[karmic ruin]

Ah I see now, thanks. 👍

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[pearl pond]

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[coral ruin]

im confused about this

[pearl pond]

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[wooden rain]

Here is my problem

[wooden rain]

This is what I've figured out so far

I'm not sure if summation notation is appropriate but I consider it is since it wants to find the total amount instead of a specific month within the timeframe.

Here is the original problem for clarity just in case

[rough stream]

Why do you need the last term?

You're summing from n = 1 to n = 60, no?

[wooden rain]

I don't think he'd be 60 within 60 months?

[rough stream]

Or, sorry. n = 60*12

Minus 12*17

Whatever number of months happen between those two ages

[wooden rain]

Would that be the conversion for it? Future age*months of whole year - months of whole year + initial age

ah okay I understand now, thank you!

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[lucid lark]

not sure where to start with this one

I'm thinking it's B

Or no, C

cause u have to rewrite the equation as v + (-w) which would flip the direction of w and when you bring the vectors together option C would represent the equation

I guess it could also be D though

it's definitely C or D

<@&286206848099549185>

guys is it C

<@&286206848099549185>

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[pseudo steppe]

how is this true

[brave helm]

we're taking away

n^2

from both sides

[pseudo steppe]

thanks!

i get it

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[wary junco]

hey

[wary junco]

I am doing a linear alg problem and am not sure there is a solution

here it is and here is my work

I would have assumed the anwser would be a number since it says find a and q

unless it can be a set

[steady wolf]

Nice work

[wary junco]

should I just give that equation as the solution then?

[steady wolf]

Why did you find the eigenvalues of second matrix like that

Keeping 1/5 like an outcast

[wary junco]

should I have not?

I assumed I shoudl

since that would make it 1/5x^2

and therefore it can never be equal

[steady wolf]

Well try both ways and see if you have any difference

[wary junco]

cause there is no co effiecent for it

it wouldn't work tho right?

cause then it would be 1/5x^2

and the left equation has x^2 no coeffcient

making it no solution that way

[steady wolf]

Uh no it won't be x²/5

[wary junco]

it would be x^2/25?

[steady wolf]

First multiply the matrix by 1/5 then subtract by lambda I

[wary junco]

ohh

[steady wolf]

Then find determinant

Nick don't jump on conclusions too fast

[wary junco]

lol - let me try it

[jolly parrot]

fäf

[wary junco]

I get in a similar situation but I was able to find two solutions to each

the second one is just 1/5 of the original one

so would the second one be the correct one?

(the one divided by 1/5)

[steady wolf]

No no just show me the polynomial you get

[wary junco]

same on left side and right side is x^2-2x+2/5

wait

thats wrong i made an error

its x^2-2x+19/25

[steady wolf]

Show me steps okay?

,w (x-1)(x-1) - 6/5

[jolly parrot]

Results provided by WolframAlpha

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

Requested by yes sam#5028

[steady wolf]

x²-2x-(1/5)

Idk how you got 19

[steady wolf]

Oh okay

[wary junco]

i think I got it

[steady wolf]

Good

[wary junco]

thanks for the 1/5 thing!

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[pearl pond]

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[long valley]

I dont get why gaH is an element of gH. Could someone explain?

[midnight haven]

do you mean why ga is in H?

it is by assumption, x=ga

[sharp quest]

given g in G, the coset gH = {gh | for all h in H}
Since a is in H, ga is in H by definition.

[pearl pond]

@long valley Has your question been resolved?

[tropic saddle]

dont just answer no to the bot after you got help. say what is still bugging you

[long valley]

Oh sorry i didnt mean to give the impression that i was ignoring anyone. I was just thinking over what they told me and rereading the chapter on cosets

Im still struggling with the disjointness of cosets but im not sure what it is exactly thats not clicking

I'll answer in a few tho, once i understand what i dont understand a bit better ahaha

Aha okey so long story short i completely missunderstood cosets, i thought one element of g with a subgroup gives rise to a partition.

Thanks for all the help! ❤️

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[wary junco]

I’m not sure if I am doing this right

[wary junco]

It’s a markov matrix

would this be right

or should AB and BA be 0.25

cause 0.5*0.5

[pearl pond]

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[modest silo]

my question is the following

how do i solve b(ii)

i dont understand how to apply condtional probability

[pearl pond]

@modest silo Has your question been resolved?

[modest silo]

<@&286206848099549185>

guys plss help

[pearl pond]

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[frigid burrow]

Please ping me when you're here to help, I'll be trying it myself in another app.

[blissful cloak]

When the beetle begins, he has two corners to choice from, so if the beetle moves for 1 minute, there are 2 possibilities. Now, say Mr. Beetle gets a second minute, for either of the cases there are two more cases, then there are two more for another minute for each of those cases and we're just doubling each time. Do you begin to see a pattern? If so, what about that pattern can you use to figure out how many combinations there are for 10 minutes?

[frigid burrow]

does it increase exponentially?

[blissful cloak]

<@&268886789983436800>

[blissful cloak]

Yes

[frigid burrow]

ohhh ok

[blissful cloak]

so if for the first case there are 2

for the second minute there are 4

[frigid burrow]

then 8

[blissful cloak]

for the third there are 8

yes

[frigid burrow]

yeah ok i get it

thank you!

[blissful cloak]

now just 2^10 is the number of possible combos

[frigid burrow]

yep

thank you!

[blissful cloak]

yw!

[frigid burrow]

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[compact wing]

Which Color is correct

Green or Black* or Both

[limber oasis]

well neither

[compact wing]

Wait then the Vern diagram will be empty then?

Veen*

[limber oasis]

it is equal to the empty set yes

the first is a part of C
the second is a part of C's complementary
Therefore their intersection is empty

[compact wing]

But if it was Union instead of intersection then the green Color and the universe(universal) be correct?

If the question was to shade the union and not the intersection

[limber oasis]

then it would be everything except A U (B n C)

[compact wing]

Oh

[pearl pond]

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[toxic lichen]

show work?

[glass jewel]

If I do this using the derivative way. yes one sec

6x - 6 = slope

(3,10) is point

point slope

[toxic lichen]

dy/dx = 6x - 6, yes.

the slope of your tangent line is the value of dy/dx at x=3.

and that's not 6.

[glass jewel]

y-10 = not sure how to set up this part

y - 10 = 6x - 6 (x - 3) this is wrong

[toxic lichen]

yes it is

the slope of your tangent line is the value of dy/dx at x=3.

do you understand this or not

[glass jewel]

ok I need to plug in 3 for x into 6x - 6 = 12

[toxic lichen]

yes, so m=12.

[glass jewel]

y - 10 = 12(x - 3)

y = 12x - 26

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[glass jewel]

How should I think about these curves to figure out which is which?

something about functions/derivatives changing the - or + of their slope?

[jolly kernel]

How does the derivative affect the power of X?

[glass jewel]

brings it down 1

2x^3 becomes 6x^2 for ex

[jolly kernel]

What shows the difference between the two in a graph

How can you tell them apart

[glass jewel]

I'm not sure how to tell visually

[jolly kernel]

Try graphing these on desmos and tell me what stands out

[glass jewel]

hmm, it's not obvious to me why the derivative does that

[glass jewel]

I'm guessing a is the original function

[jolly kernel]

Sorry I'm not sure how I could explain better

If someone doesn't help in a bit I recommend pinging the helper role

[glass jewel]

ok

[pearl pond]

@glass jewel Has your question been resolved?

[glass jewel]

<@&286206848099549185>

[severe token]

Hi Smart! Hope you are well. A is original, B is derivative, c is second derivative.

[glass jewel]

I was looking to get an explanation if you can

try to avoid giving answers

[severe token]

When a is reaching its max, b is becoming zero. When b is reaching its max then c is becoming zero

see derivative function is equal to zero when original function is reaching local max or local min

so it can be very good criteria to understand what is original function and what is derivative

please let me know if you have questions

[glass jewel]

are you able to draw lines on my picture to show what you mean about the max and min and derivative = 0 ?

@severe token

[severe token]

@glass jewel Hi! See when a is max b is crossing x, do you see that?

@glass jewel when b is max c is crossing x, do you see that?

this is very good criteria

[glass jewel]

did I correctly point out max on my picture?

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[copper kestrel]

why did you multiply the slopes?

[compact wing]

To see if it’s perpendicular or not

M1 x m2 = -1

[compact wing]

Or we are not supposed to?

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[gentle acorn]

hello, i need some help with multivariable calc
problem is a given variable w = wind in mph t = temperator in F
wind chill is a result of combination of both, f(w,T)

[gentle acorn]

i need to estimate f_w(10,30) and give practical interpretation of that, not sure how to start

[pearl pond]

@gentle acorn Has your question been resolved?

[gentle acorn]

<@&286206848099549185>

[severe token]

Hi! f(10,30) is 21. Just check intersection of 10 and 30. It means that 10 mph an 30 degree of Farenghait gives 21 wind chill

Hope this he;ps

[gentle acorn]

im looking for the derivative of the function at 10/30

the answer isnt 21

its f_w at f(10,30)

[gentle acorn]

<@&286206848099549185>

[pearl pond]

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[gentle acorn]

<@&286206848099549185>

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[kind mural]

Topic: Real Analysis

When function is differentiable on an interval I : [a,b], does that mean that for each point in the interval, there exist a derivative?

[rose rock]

Yes

the limit expression for the derivative exists

for every point in the interval

[kind mural]

Thank you

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[glass jewel]

How does this want me to set this up?

[pearl pond]

@glass jewel Has your question been resolved?

[pearl pond]

@glass jewel Has your question been resolved?

[glass jewel]

<@&286206848099549185>

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[worthy belfry]

Hello I need help on this question

[vestal tapir]

you know this

you can do it

i's hard to say what it even tests for

it's like the topic is confidence

[plush bramble]

,calc (22)^4.5

[jolly parrot]

Result:

1.0987580342332e+6

[worthy belfry]

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[dense sphinx]

A coin is thrown upwards up to a height of 50 cm from the launch point: what speed did it have at the beginning of the upward motion? (approximate the acceleration of gravity to 10 m/s^2)?

[dense sphinx]

The answer is 3m/s

the coin travels 0.5 meters so the initial velocity must be 1 m/s right?

[viscid sierra]

v^2=u^2+2as

[dense sphinx]

?

any other solution rather than formula

[viscid sierra]

is the answer 3m/s only?

[dense sphinx]

It is about 3m/s

[viscid sierra]

yes not exact

use formula

easier

[dense sphinx]

but non logical

[viscid sierra]

at max height velocity=0

[dense sphinx]

cant use this formula for every question. All questions has a unique formulas

[viscid sierra]

There are only 3 formulas
v=u+at
s=ut+1/2at^2
v^2=u^2+2as

[dense sphinx]

I do not feel like doing I am math when I use em

[autumn fossil]

Would you like to derive at least some of the formulas by yourself?

[dense sphinx]

Yes I can but it takes so much time

[autumn fossil]

Yeah, solving problems without using formulas made by someone else usually takes some time.

h=1/2at^2 + ut, is this formula logical for you? u stands for initial velocity, a for accel, h for height at given time t

[dense sphinx]

not really but once I tried to reach this.

from v+v/2 something

but it took like 10 min

[autumn fossil]

yeah, it takes some time. Are you ok with using that formula to solve this problem?

together with v = u + at where u is initial velocity? This one comes from definition of accel

You dont need to remember the last formula sent by beard, but those 2 are necessary

[dense sphinx]

lets see...

[viscid sierra]

bead

[dense sphinx]

what about time

[autumn fossil]

sorry

[dense sphinx]

we need time to find initial v

[viscid sierra]

isn't it ut+1/2at^2

instead of vt

[autumn fossil]

it's the same, our physics teacher used v

[viscid sierra]

v is generally used for final velocity while u is initial velocity

[autumn fossil]

I added "v stands for initial velocity" for clarity

[viscid sierra]

oh then ok

[autumn fossil]

aight, lets use u

[autumn fossil]

Yep, so we will start by setting up few equations

what can you tell about the velocity at the point where it reaches maximum height?

[dense sphinx]

0

[viscid sierra]

Sorry for the interference but using those 2 equations instead of v^2=u^2+2as makes it unnecessarily long and complicated

[autumn fossil]

So 0 = u - 10t

[autumn fossil]

Is this one even taught?

[viscid sierra]

to us it was

3 equations of kinematics

You may continue

with your process

[autumn fossil]

Well with this one, you would just do 0=u^2 - 2*10*0.5

Do you know this one @dense sphinx ?

[dense sphinx]

yes

but as I said before it is just remembering

the formulas

[viscid sierra]

The other 2 are also formulas

[dense sphinx]

yes

[autumn fossil]

Unlike this one, they make sense (at least to me)

this one seems to be derived algebraicaly from the other 2

[viscid sierra]

Yeah continue with your process I won't interfare

[autumn fossil]

You're actually right

[dense sphinx]

nevermind u equals to 10t

[autumn fossil]

0 = u - 10t
u = 10t
then use second formula to get
h=ut+1/2at^2
0.5=10t*t-1/2 * 10 * t^2 (0.5 meters is the height, 10t was substituted for u)

simplifying it's
0.5=10t^2-5t^2=5t^2
Which means
0.5/5 = 5t^2/5
1/10 = t^2
t=sqrt(1/10)

I didnt

u=10t, so u=10*sqrt(1/10)

sqrt(1/10)=1/sqrt(10)

[dense sphinx]

isn't t equals to sqrt1/30

[autumn fossil]

why should it be

[dense sphinx]

5t^2 + 10t^2 = 0,5

[autumn fossil]

the acceleration pushes the object down

which means it has negative sign

[dense sphinx]

ow yeah true...

in this case answer will be sqrt10

[autumn fossil]

,w calc sqrt10

[jolly parrot]

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Requested by MathIsAlwaysRight#4079

[autumn fossil]

around 3

[dense sphinx]

okay thanks

also @viscid sierra thanks

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[pearl pond]

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[warm echo]

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[worthy cairn]

what's the status?

[midnight haven]

I need explanation

[worthy cairn]

well 3* of the answer can be eliminated by simple substitution

[midnight haven]

put n=1,2,3.. and observe

[worthy cairn]

do you have any concrete question here?

[narrow hinge]

Are you trying to find solutions of n for each question?

[midnight haven]

No

[midnight haven]

Actually, I already have solutions... Need a better explanation

[worthy cairn]

If there are no questions, there is nothing to explain

[midnight haven]

ans is a

put values of n na

so easy qsn

don't make it tough

[narrow hinge]

Well 4^n + 1 cannot be even for any natural value of n. Do you need true or false answers for each question?

[midnight haven]

oye

I didn't get it

[worthy cairn]

4^n is a mulitple of 4 so it must be an even number

if you take any even number and you add 1 you always get an odd number

[midnight haven]

u know addition subtraction multiplication or no

[midnight haven]

Hmm , i know

[midnight haven]

then just put values of n as 1,2,3,4.. and observe urself

What if we put n = even numbers answer is still odd

yes so option A is correct

wt is the problem

😞

Cool 😅

[pearl pond]

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[fossil lodge]

For q23

I found $\frac{dy}{dx} = \frac{3\sin(3x)}{\sin(x)}$

[jolly parrot]

KN

[vital estuary]

You sure its x?

[fossil lodge]

ok, it should be theta

[vital estuary]

dy/dx should be in terms of theta yes

[fossil lodge]

When I find the horizontal tangent, making the numerator = 0, I get $\theta = \frac{n\pi}{3}$ and the vertical tagent making denominator =0, I get $\theta = n\pi$

[jolly parrot]

KN

[old geyser]

You could turn it into a Cartesian equation

[vital estuary]

Meh

[old geyser]

It's too much work anyways

[vital estuary]

Ok, so when you look for vertical tangents, you notice the slope become 0/0

[fossil lodge]

Yeah, so those points are its multiple of pi

[old geyser]

You should know that $x \in [-1, 1]$ so really just find when $\dv{x}{\theta}$ and/or $\dv{y}{\theta}$ = 0$

[jolly parrot]

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

[vital estuary]

You got this umbra keep going

[old geyser]

Lemme cook!1!1!1!

[fossil lodge]

ok... so anyway. It turns out after taking limit of dy/dx at theta = pi, it gives 9 (and so multiples of pi will always give 9)

SO basically since verticle tangents are at theta = n pi

There is no vertical tangent

As for horizontal tangent, we still have theta values such as $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.

[jolly parrot]

KN

[fossil lodge]

SO this means the graph just have a bunch of horizontal tangents at npi/3 for integer n except when n = 0, 3, 6, etc?

but the answer says horizontal tagent only at pi/3 and 2pi/3

Im just wondering if there is technically infinitely many points

[pearl pond]

@fossil lodge Has your question been resolved?

[pearl pond]

@fossil lodge Has your question been resolved?

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[vague rock]

hi

[vague rock]

I only know that option 3 in wrong because its not cont at x=1

I don't know what to do with bracket

.close

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[sinful fern]

probablility and staticstics question?: If you have a uniformly random variable X~U(0,1). What is the CDF and PDF of Y~U(0,X)

[sinful fern]

This is all I’ve got?

[pearl pond]

@sinful fern Has your question been resolved?

[plush bramble]

looks fine.

[woven matrix]

does int_0^1 1/x dx converges tho, I don't think so

and such notation as P(Y = y) loses its meaning since Y is uniformly distributed on an interval

the joint density seems fine but after that it kinda goes wrong

$\text{f_Y should be} \int_0^x \dfrac{1}{x} dy$

oh gosh latex

https://cdn.discordapp.com/attachments/439153117997563915/1108494817404657724/249632691564183553.png

so 1

StaleDegree, you're mistaking f_Y and f_X,Y

[sinful fern]

That can't be

[pearl pond]

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[sinful fern]

.reopen

[pearl pond]

✅

[sinful fern]

the integral 1/x is imposible over 0 -> x

[woven matrix]

it is with respect to dy

[plush bramble]

@sinful fern read this

[woven matrix]

your mistake is here

mistaking density of Y for joint density of X,Y

(swap X and Y in your case obv)

[sinful fern]

what about this then

[woven matrix]

doesn't make sense, P(Y = y) doesn't make sense in the first place

[sinful fern]

P(Y=y|X=x) is 1/x for 0 < y< x

[woven matrix]

P(Y = y) = 0 since Y is uniformly distributed on an interval

[sinful fern]

so what is f_xy in this case

[woven matrix]

joint density of X and Y

[sinful fern]

I don't know how to get that

[woven matrix]

https://en.wikipedia.org/wiki/Conditional_probability_distribution

[sinful fern]

already been there

[woven matrix]

this is f_X,Y

[sinful fern]

I know what it is lol

I just don't know how to derrive it

[woven matrix]

you've already given f_X,Y

right here

[sinful fern]

thats f_y|x

[woven matrix]

f_x * f_y|x = f_x,y

[sinful fern]

oh no way I'm retarted

[woven matrix]

and then you've taken f_x,y as if it was f_y

[sinful fern]

so from there how do we get f_y

[pearl pond]

@sinful fern Has your question been resolved?

[sinful fern]

f_y is not 1 it's -ln(y)

how could f_y be 1

the number closer to 0 are def more likely to occur

[woven matrix]

yeah indeed

I realised it and thinking about where is the mistake

probably when I assume that derivative of F_Y|X is f_Y|X if I have to guess

no ok the problem is F_Y|X itself I think

[pearl pond]

@sinful fern Has your question been resolved?

[sinful fern]

<@&286206848099549185>

[plush bramble]

how do you know this

[sinful fern]

Someone gave me the answer without explaining. And I just thought it was a beautiful problem

So I want to know why thats the probability

[plush bramble]

go ask them then

[sinful fern]

It was a while ago

and I did but they didnt' respond

[woven matrix]

ok I found it

it is indeed -ln(y)

since

$\int_y^1 \dfrac{1}{x} dx = -ln(y)$

[jolly parrot]

Mélo

[woven matrix]

F_Y|X = y/x on 0 < y < x, 0 or 1 elsewhere
so f_y|x = 1/x on 0 < y < x, 0 elsewhere
so f_x,y = 1/x on y < x < 1, 0 elsewhere
since f_x = 1 on 0 < x < 1

@sinful fern

[sinful fern]

so where do you get teh integral from y -> 1

[woven matrix]

from f_x,y
f_x,y = 1/x * indicator of 0 < y < x * indicator of 0 < x < 1
so integral of f_x,y dx from -inf to +inf = integral of 1/x on [y, 1]

because when x is in [0, y] the density is simply 0

it implies x < y

and the indicator is then 0

[sinful fern]

what is indicator

[woven matrix]

a function that is 0 when its condition is not respected

1 when it is

indicator of 0 < y < x = 1, if 0 < y < x
= 0 elsewhere

[sinful fern]

right

[sinful fern]

here you mean f_y|x

?

[woven matrix]

no

f_x = indicator of 0 < x < 1 since X is uniform on [0, 1]
f_y|x = 1/x * indicator of 0 < y < x

and f_x,y is the product of the two

so f_x,y = 1/x * product of the indicators

then you integrate over R with respect to x, and because of the indicators, the integral is non zero only on the interval where 0 < y < x < 1 ie when x is in [y, 1]

[sinful fern]

Wowwwww

I get it now

You are actually a legend

Thank you for helping me see it

.close

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[hidden marsh]

this graph is hamiltonian, but what strategy would you use to find a spanning cycle without completely randomly guessing?

[limber oasis]

Start at the top. Do 4/5 of the outer layer, jump in the inner layer, do 4/5, take the middle and finish the cycle.
So many edges means a bit of strategy and gluttony work well together

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[autumn fossil]

show that n^7 - n is divisible by 42. Where do I start?

[autumn fossil]

for every positive integer n

[midnight haven]

can u not factor and prove by exhaustion

.

[autumn fossil]

$n\left(n-1\right)\left(n^{2}+n+1\right)\left(n+1\right)\left(n^{2}-n+1\right)$ I got this when I factored it

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

lemme look up what does exhaustion mean...

oh brute force basically...

[autumn fossil]

This proves divisibility by 2 and 3, the last thing I need to prove is divisibility by 7

[midnight haven]

kind of ;; 7 terms only tho

like 7p , 7p + 1, 7p +2 ,... 7p+6

[autumn fossil]

alright, thanks

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[cursive flare]

idk if im hallucinating, but i can't find the error ,since x=y=z=pi/3 doesn't work

[cursive flare]

oh wait

the first line

[slender viper]

yeah

$\cos(z)=-\cos(x+y)$

[jolly parrot]

Element118

[cursive flare]

thanks ogganesson

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[pearl pond]

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[split knoll]

I saw the word maximum so I did the derivative function thing and set it to 0

but now idk what to do next

[brazen heath]

You got the value of x on substitute it in the equation for volume you got before differentiation

[split knoll]

cheers!

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[wild fable]

When it says on the domain [0,2pi], that means the maximum x length it goes to is 2pi right

[toxic lichen]

it means that the x-axis on your graph goes from 0 to 2pi

[wild fable]

does this work?

[wild fable]

and y=-f(x) its just reflected across the x axis ?

[toxic lichen]

yes to both

[wild fable]

ok cool

for tan graph

would it be any different drawing it assuming dom is still [0,2pi]

how would i draw it

[toxic lichen]

i mean... do you know what the graph of y=tan(x) looks like

[wild fable]

yes

but as in

i know the pattern of periods for sin and cos

but i dont get the period for tan graph

every 180 degrees theres a tan graph?

[toxic lichen]

the period of tan is 1pi, yes

[wild fable]

ok.

also

is there a difference between compression and dilation

[toxic lichen]

compression is squeezing things in, dilation is spreading them out...?

[wild fable]

i dont understand why the tan graph looks like this

did the 2 do anything

i know it got reflected

because of the negative

but what did the 2 do

oh

[toxic lichen]

to state the obvious, 1-2tan(x) ≠ tan(2x)

[pearl pond]

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.reopen

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[clear summit]

how da hell do i do this

Im in confusion

[coarse dawn]

Hi "in confusion", I'm dad

[clear summit]

xd

well be back

class wnded

[nova tendon]

Work out the area of the front panel, its 2 triangles and a square

then multiply that by 11 to get the volume.

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[pearl pond]

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

[dense sphinx]

I thought it was 1 due to the continuous lines on the left images

but the answer says that it was 4 somehow

[haughty gull]

A live for me to study 📚

And to be monitored

To not waste my time i hope it works

[dense sphinx]

what is this advertisement?

[haughty gull]

No just me

To not waste time but i feel uncomfortable

Do i have to turn it off

[old geyser]

<@&268886789983436800> this guy seems to be advertising something?

[haughty gull]

I swear I don’t

Just enter if you want to check

It is just my face and my notebook

[old geyser]

So how is that relevant to the OP's problem?

[haughty gull]

This is the proof

No excuse me to delete this bad photo

Now*

[bronze quiver]

Could you please translate the problem?

[dense sphinx]

Given the two orthogonal projections of the object (top view and front view), identify the corresponding third view (side view)

@bronze quiver

[pearl pond]

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[midnight haven]

Does anyone have a nice video or tutorial

[midnight haven]

on where to learn these

and know what theya re

Im not sure what the topic is called and cant find videos teaching all of them

[pearl pond]

@midnight haven Has your question been resolved?

[west sapphire]

i haven't watched this video, but the title sounds promising: "Recognizing functions from graphs"

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:functions/x2f8bb11595b61c86:recognizing-functions/v/graphical-relations-and-functions

[pearl pond]

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[edgy gazelle]

@midnight haven it's A

@midnight haven wait, what are the answers E and F?

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[frozen sandal]

Help

[frozen sandal]

I am not sure where to start

I was thinking of using a pythagorean identity

like

yes I do. Can't we do sin^2theta=1-cos^2theta?

also is there a way to add theta on discord? like with a bot

would help a lot with this chapter

[jolly parrot]

antrep

[frozen sandal]

thank you

okay let me try this one more time and show my work if i'm stuck

okay I'm confused on this:

$$cos^2theta=-1/7$$

[jolly parrot]

dasa

[frozen sandal]

how would I square root the negative?

what am I doing wrong?

ignore the top 2 lines

thank you

i see it now

I got it right! thank you!!!

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[digital path]

Could you solve using commutative and distributive property

[dim tangle]

what do you need help on specifically

[digital path]

to solve it by like regrouping

[dim tangle]

well at first glance i can see two common factors out of three terms. can you see the same?

[digital path]

yes, 3/7 ?

[dim tangle]

good, whats the third term?

[digital path]

what is it

[dim tangle]

well you saw 3/7 in two of them. whats the other term that doesn't have 3/7?

[digital path]

2/5

[dim tangle]

no, remember PEMDAS. due to multiplication, this already has two "groups," or terms.

there's another term that doesn't have multiplication, what is it

[digital path]

-1/14

[dim tangle]

good. so to regroup this i would figure out a way to factor out 3/7 from 1/14

what level math is this

[digital path]

8th grade

ncert

[dim tangle]

algebra?

[digital path]

rational numbers

[dim tangle]

have you covered algebra at all yet?

[digital path]

yea

[dim tangle]

so if i gave you something like 1/2x = 5 you would know how to find x?

[digital path]

yeah

wait lemme do it

[dim tangle]

thats not relevant i was just making sure you knew some algebra

[digital path]

oh

wait can i send the work i have done

the thing is

[dim tangle]

yes send

[digital path]

i have the answer

but i didnt understand how it was done

so im asking it here so that someone could do it and explain it

[digital path]

if i ping you will u come back here

[dim tangle]

idk. im here now.

but if i gave you this, could you solve for x?

[digital path]

yeah

x = 1/14 - 3/7

[dim tangle]

sorry it should be more clear. 3/7 is multiplied by x

[digital path]

can we get to this later

[dim tangle]

oh i see how they want you to do it

so what don't you understand about that picture

[digital path]

how did "+" come

sorry for handwriting

[dim tangle]

okay so lets look at those two terms with common factors

[digital path]

yes, 3/7

so we group it using commutativity right

oh yeah please dont mind the brackets

[dim tangle]

well, commutativity also applies to subtraction.

[digital path]

i put it for reference

it wasnt there.

[dim tangle]

you see how there's a - for the second term and a - on the 3 of 3/7 in the first?

[digital path]

yeah

[dim tangle]

so we can visualize these terms like this

[digital path]

wait isnt -3/7 and - 3/7 the same

[digital path]

isnt this the same

[dim tangle]

yes

because of the commutative property

[digital path]

alr..

[digital path]

why does -1 come

and where did the "+" come in the middle

[dim tangle]

anytime you see a negative number or subtraction, all it's really telling you is that there is a -1 as a factor of whatever number it is or what you are subtracting.

1 - 2 is the same as 1 + (-1)2

[digital path]

yes

[dim tangle]

so seeing a negative instantly means there's a -1 factor.

multiplication is commutative, so we can pull that factor out as we please.

so that's how i got the result in that picture

[digital path]

hmm

[digital path]

but how did + come

[dim tangle]

x - y = x + (-1)y

you see how that + appeared?

[digital path]

shouldnt it be - because - was before the 3/7

[dim tangle]

addition and subtraction are extremely similar. you are combining two values. subtraction just means you are combining a value with a -1 as a factor.

so when we separate that factor, subtraction becomes addition.

[digital path]

ok

and then

[dim tangle]

1/2 - 1/3 == 1/2 + (-1)(1/3)

its the same thing.

[digital path]

yea

[dim tangle]

so going back to our problem

you see how i was able to group these terms by factoring the negative?

i pulled the first negative from -3/7 and the second negative from the subtraction.

we can do the same with any factor

and these two terms in the group share 3/7 as you said earlier.

so we could pull 3/7 also

[digital path]

hmm

[dim tangle]

when i say "pull," on a technical level i am dividing the whole thing by that factor, and to not change our final answer i put that factor outside the grouping.

so if i were to pull 3/7 out, i have to divide the whole thing in the brackets by 3/7

[digital path]

hm

this is kinda har

d

[dim tangle]

at first it can be

but lets look at a quick example without fractions

[digital path]

the way u teaching and the way i was taught

[dim tangle]

we won't even use numbers this time

when two letters are together in algebra they are multiplying each other right?

xy means x * y okay

[digital path]

yea

[dim tangle]

so lets say we have xyz + xjk

could we group this?

[digital path]

hmmm

[dim tangle]

grouping by the commutative multiplication property means common factors

when we multiply, the terms are called factors

[digital path]

we can for just x?

[dim tangle]

yes!!

so can you think of a way to rewrite that by factoring x?

[digital path]

hol up

[dim tangle]

remember when we are grouping, we are taking out a factor from a group of terms. this is the opposite of multiplication, so we must divide.

[digital path]

u having a hard time explaining

because i still dont kinda get it

[dim tangle]

what is the answer

[digital path]

-3/2

[dim tangle]

its ( 2 * 5 ) / 2

[digital path]

isnt that subtraction sign

[dim tangle]

sorry its a dot

[digital path]

the one here

oh

so 5

[dim tangle]

right, because you mentally just cancelled the 2's right?

[digital path]

yeah

or i multiplied 5x2 then divided it by 2

[dim tangle]

think of cancelling, it's easier.

same thing

you can cancel factors by divisiojn

[digital path]

yes

it has been 20 minutes.

anyways

[dim tangle]

so?

back to the problem

i grouped these terms with -1

the -1 is going to multiply back into the group in the end.

[digital path]

our teacher told us that

when there are 2 common factors we take one out

so we took -3/7 out

(ignore brackets)

[dim tangle]

this is the exact same thing

[digital path]

wait we can do that with variables?

[dim tangle]

when you divide x by x, or 3/7 by 3/7, what do you get?

yes

variables are just placeholders for numbers

[digital path]

do u get this

the positioning

wait i think im confusing you

and myself

[dim tangle]

its confusing because this is actually simplified, there's a division happening

in my example with variables, let me show you what is really happening

[digital path]

OHHHHHHHH

NOW I GET IT

[dim tangle]

we put the x on the outside so we don't change our final answer. we are only rearranging things.

we put the x on the outside because we must undo our division later by multiplying.

x/x = 1, and 1 * anything is just itself.

[digital path]

wait can i send you another sum

just a min

[dim tangle]

when we really think about it this is just the commutative property of multiplication in reverse.

or the opposite of distribution

[dim tangle]

what do you have an issue with here

[digital path]

why did "-" come

[dim tangle]

how they wrote the second line isn't right.

[digital path]

its incorrect?

wait what

oh yeah ignore brackets

[digital path]

except third line ignore brackets

[dim tangle]

it looks like 5/2 is a factor of the second term, so it looks like they got confused or wrote the first line wrong.

[digital path]

thats weird.

[dim tangle]

ohhh i see it was written wrong

[digital path]

i had a doubt with my teacher regarding this exact problem

[dim tangle]

is this the original problem?

[digital path]

he told me that the "-" came from the "-" in the 2nd line

[digital path]

no brackets.

[dim tangle]

brackets don't change anything, they just help you visualize the grouping happening from the multiplication because of PEMDAS

[digital path]

oh alr then yes

[dim tangle]

let me write out the work with notes so you can follow easier

[digital path]

btw it is correct

i checked the answers of the textbook online

exact steps without bracket.

[dim tangle]

right

in that picture, what do you not understand? the - inside the parantheses?

[digital path]

no look

my teacher told me that

the "-" sign came because the common factor had "-"

so that way i did many problems and it was always correct

until the one i had a doubt with

[dim tangle]

okay, so lets say we have 1 x 2 - 3 x 4

group that for me

put some brackets or parantheses around what you need to

[digital path]

i need to go right now, could you message me about this later

my mother is angry i spent 40 mins on it

please.

[dim tangle]

you can't rush understanding the material. it takes time to fully grasp concepts. when you fully understand it you will solve these problems in 30 seconds

(1 x 2) - (2 x 3) == 2(1 - 3)

1 x 2 + 2 x 3 == 2(1 + 3)

[digital path]

can i dm you later i really need to go im really sorry

[dim tangle]

bye goodluck

[digital path]

can i dm tho

[dim tangle]

yes

[digital path]

thanks

[dim tangle]

.close

[pearl pond]

@digital path Has your question been resolved?

[pearl pond]

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[full plover]

hello

[full plover]

can someone please help me find how i would do part b, the answer to part a is 1.5

1.50 (3s.f)

vm

nvm

.close

[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[twin plinth]

Hello, so I have this problem:

[twin plinth]

What I have tried so far is:

But yeah in the last step when I substitute in the given values I get... Zero

I'm not sure why

I based my work on an example in the textbook, which used a fixed error amount of $\varepsilon$, so I changed it to be $(1.05 x)$ for each measurement

[jolly parrot]

Memikri

[pearl pond]

@twin plinth Has your question been resolved?

[pearl pond]

@twin plinth Has your question been resolved?

[pearl pond]

@twin plinth Has your question been resolved?

[twin plinth]

any help?

[mint scarab]

I think the way the eqn is setup is destined to give you 0 from your calculations

I'm not sure what relative error is referring to in this case

[pearl pond]

@twin plinth Has your question been resolved?

[pearl pond]

@twin plinth Has your question been resolved?

[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[whole willow]

If ratio A:B is 5:4 and B:C is 6:7. What is ratio A:B:C?

[boreal aspen]

A is the first term

so everything is in terms of A right

If A is 4 B's

and 6 B's is 7 C's