help-17

[vocal sleet]

@waxen breach Has your question been resolved?

[vast shale]

divide through by 2π and then square both sides and multiply up through by 9.81

[vocal sleet]

@waxen breach Has your question been resolved?

[waxen breach]

Thanks but how would i solve this one

[vast shale]

idk much about pendulum motion sorry

[knotty pendant]

You might be able to use the equation $T = 2\pi \sqrt{\frac{L}{g}}$

[twin meteor]

dragonbreath

[knotty pendant]

Then once you solve for L, multiply by two and plug it back into this equation

[vocal sleet]

@waxen breach Has your question been resolved?

[vocal sleet]

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[ivory sonnet]

I am not really sure what this question is asking me... Am I supposed to be inputting T=1 or finding the slope? Can someone clarify what I am supposed to be doing here and if I am on the right track with my shown work?

[subtle helm]

f'(x) = that limit of difference quotient, that's just the slope of f at a particular x value

so part a) using the definition you just need to evaluate the derivative at the point they're asking you to

didn't check if your work was correct, but assuming you didn't make any mistakes, yes you just evaluate it at t=1 now

[ivory sonnet]

ok thank you, Im still a bit lost - I got 2t+4 as the slope, am I supposed to be using this? Or is the question just asking me inputting t=1 for the h(t)=t^2+4t?

[subtle helm]

2t+4 is the slope for a general value of t

you're being asked for what is happening at t = 1

you want to evalutate t = 1 for the derivative, no the original function

if oyu eval h(1), that's just a point on the graph of h, (1, 5)

[ivory sonnet]

sorry I am still a bit lost, What would my derivative be?

[unique sonnet]

it may help you if you put into words what a derivative is

[ivory sonnet]

'the rate of change of a quantity y with respect to another quantity x' - Would my derivative be 2/t? 2/1 after inputting t=1?

[unique sonnet]

so if this is your function:

what should your derivative be?

[ivory sonnet]

2t+4?

[unique sonnet]

very good. this means that the derivative may change if we plug in different values for t, correct?

[ivory sonnet]

I think so - the deriv changing do you mean inputting t values into the orginal h(t)= or the 2t+4?

[ivory sonnet]

so if t=3 then itd be 2(3)+4?

[unique sonnet]

yep

[ivory sonnet]

ok thank you. So this question is basically just asking me to find the derivative and input t=1, 2(1)+4, and show my work for the different methods? For example, part B, I would start with the original h(t)=t^2+4t but they want me to show my work/use the power rule instead?

[unique sonnet]

yep

[ivory sonnet]

Both my answers should be the same but they just want me practicing the different methods?

ok thank you

Very helpful

Last thing, my answer would just be 6?

[unique sonnet]

looks like it

[ivory sonnet]

ok thank you

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[viscid mural]

Is there some trick to this or do I just have to sit through a horrible computation?

[vocal sleet]

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@viscid mural Has your question been resolved?

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[green gulch]

Hi can i please get help with this question

[green gulch]

it says simplify

[tawdry canopy]

the original form = $3\sqrt{11}(2\sqrt{2}\sqrt{11} - 4\sqrt{3}\sqrt{11})$

[twin meteor]

e_waste

[tawdry canopy]

multiply them and you can get $6\sqrt{2}\times 11 - 12\sqrt{3} \times 11$

[twin meteor]

e_waste

[green gulch]

ohh right because the root11s cancel each other out

and make

11

[tawdry canopy]

so the answer is $66\sqrt{2} - 132\sqrt{3}$ i guess

[twin meteor]

e_waste

[green gulch]

i understand now ty

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[feral scroll]

How exactly do you do this conversion/simplification? Is it some sort of substitution?

[heavy yoke]

we substituted x with a new indexing variable (confusingly also called x), which is one less than the previous one

[feral scroll]

I see

so like doing u sub but when its over just replace it with x?

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[vocal sleet]

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[frail torrent]

$\int_{-1}^{1} \arctan{e^x} , dx$

[vast shale]

open a new channel this one is already closing

[frail torrent]

okay

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[minor blade]

yesterday i asked for help with this function, the question was to find its asymptotes and the possible answers were:
a) it only has a horizontal asymptote
b) it has an oblique asymptote
c) it doesn't have asymptotes
d) it has a horizontal asymptote and a vertical one (the right one)

[minor blade]

to do that i had to find its domain but there wasnt an accurate way to do it without a calculator so i found graphically a point where cosx = 2x which implied the existence of a discontinuity in the function

so because of that discontinuity i figured out that it was implied the existence of a vertical asymptote as well

but now im thinking that its not really correct to say that the existence of a vertical asymptote is implied by a discontinuity

like for example this function has a discontinuity for x = 1 but it doesnt have a vertical asymptote

[minor blade]

so im wondering what's the correct way to come to the conclusion that this function has a vertical asymptote if calculators arent allowed

since i cant find the exact solution for 2x = cosx im not able to study the limit as the function approaches that point

[vocal sleet]

@minor blade Has your question been resolved?

[vast shale]

f(x) is a a/0 form

lnx/x-1 is a 0/0 form

[minor blade]

ohh that makes sense lol

thank you =)

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[opal raft]

Can these three circles make a triangle in some mathematical world?

The circles are the sides of an equilateral triangle.

[hollow sinew]

Prob not according to our definition of a triangle, which is made up of line segments
Also the circles are not even joined tho? a triangle is a closed figure

[opal raft]

How do you define a line segment?

[hollow sinew]

A part of a line ig, bounded by end points

[opal raft]

What is a line in this definition?

Sorry, I am a bit confused about these concepts, basic as they are

[hollow sinew]

An object with no width, depth or curvature, just length

[opal raft]

I am looking up the definitions also, in the meantime

[opal raft]

Are there objects with only width and no length?

[hollow sinew]

That would just be a line ig

[opal raft]

I think an idea that relates is that we can draw a triangle on a sphere, and it will be a circle

[cobalt crypt]

what a vague definition

what is "width", "depth", "curvature"

what is "length"

[hollow sinew]

Don’t think it would become a circle tho?

[opal raft]

I don't know

[wary mantle]

Yeah, though I heard that's how Euclid defined these in the elements

[cobalt crypt]

eons ago, before mathematics was formalised

[hollow sinew]

It would be a spherical triangle but not really a circle

[cobalt crypt]

what is your goal here?

[opal raft]

It is something that occured to me while trying to solve a geometry exercise. As I was relaxing conditions, this "triangle" appeared in my mind, and wondered if it can in fact be a triangle.

[cobalt crypt]

how do you imagine to be a triangle though

[opal raft]

Then, I wondered how to define a triangle, and what a line was, and found that I could not yet define it.\

[cobalt crypt]

i mean, if you would like to define a triangle in the plane

you could define it as the boundary of the convex hull of 3 points

[opal raft]

A hull is a structure?

[cobalt crypt]

no it's a way of defining a region in space

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

[opal raft]

Awesome, thanks!

[cobalt crypt]

a filled in triangle is the convex hull of 3 points

[opal raft]

So a line is a convex hull of 2 points!

With no curvature

[cobalt crypt]

yes

a line segment

[opal raft]

An infinite convex hull

[cobalt crypt]

a line would typically be described by an equation

like in the plane

ax + by + c = 0 describes a line

[opal raft]

How do we describe the plane?

[cobalt crypt]

the plane is constructed as R^2

pairs of real numbers (x, y)

[opal raft]

Thank you @cobalt crypt I will ponder this! Thank you @hollow sinew

[vocal sleet]

@opal raft Has your question been resolved?

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[neat zephyr]

need help to find the value of 'a'

[neat zephyr]

gradient for first line= a+1

gradient for second line= 2

[hallow plover]

What's the condition under which two lines are perpendicular?

[neat zephyr]

perpendicular lines have different gradients ( the reciprocal is flipped and turns into the opposite)

[livid horizon]

Yes

[neat zephyr]

yes

[livid horizon]

Or in other words, product of their slopes is -1

[hallow plover]

Phrased weirdly but yeah

[neat zephyr]

what is the next step

[hallow plover]

Well if the slope of the first is m and the slope of the second is n then you have m * n = -1 right?

So plug in m and n

[neat zephyr]

is that true?

[hallow plover]

yup

[livid horizon]

You can check it out yourself

[neat zephyr]

2a+2=-1 , a=-3/2

[livid horizon]

You said taking the reciprocal and changing the sign

[livid horizon]

Now try multiplying this with the original slope, you will get -1

[neat zephyr]

alr I get it onw ty

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[barren kiln]

Could someone pls help me with this

[barren kiln]

Nvm

.close

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[lethal silo]

i dont really know where to go after i get Px+Qy+Rz

[stone gazelle]

Setup a standard surface integral over a vector field

[vocal sleet]

@lethal silo Has your question been resolved?

[lethal silo]

Like this?

forgot the jacobian but still

[stone gazelle]

You won’t be able to apply Div thm because the upper hemisphere is not a closed surface

Use the same parametrisation, compute the normal vector field and the vector field over the parametrisation, then flux will be given by the double integral of the dot of the vector field with the normal vector field

[lethal silo]

What’s the normal vector field

@stone gazelle

[stone gazelle]

The normals to the surface at any given point

Use the parametrisation
[ \Psi (\theta,\varphi) = ( \rho\cos\theta\sin\varphi, \rho\sin\theta\sin\varphi, \rho\cos\varphi) ]

[twin meteor]

shsgd

[stone gazelle]

And you know rho to be 2

Now compute $\Psi_\theta \cross \Psi_\varphi$

[twin meteor]

shsgd

[stone gazelle]

That would give you your normal vector field

[ \text{Flux} = \iint_\Psi ( \textbf{L}(\Psi (\theta,\varphi)) \cdot \Psi_\theta \cross \Psi_\varphi) \rho^2 \sin\varphi , \dd \varphi , \dd \theta ]

[twin meteor]

shsgd

[stone gazelle]

@lethal silo

[lethal silo]

are these the general steps for when you dont have an enclosed surface and need the flux

[stone gazelle]

Yes

If the surface is enclosed then you should apply div thm

[lethal silo]

I don’t think I’m doing this correctly

[stone gazelle]

Your partials look wrong

[vocal sleet]

@lethal silo Has your question been resolved?

[vocal sleet]

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[hybrid oyster]

dude im sorry but im really bad at math and i have no diea what the hell is going on so uhh starting from umber 1 a i litterally do not know what to do

[scenic ravine]

ok, so let $cos(x)= \xi $, I'm just chosing $\xi$ because it's fancy, feel free to chose any letter

[twin meteor]

ƒ(Why am. I here)=I don't Know

[hybrid oyster]

ok

[scenic ravine]

so can you re-write the equation for me

[hybrid oyster]

so it would be using x bc its easy
x + rt3 = -x

[scenic ravine]

cool

so solve for $\xi$ or in your case $x$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[hybrid oyster]

it would be -(rt3/2) yes

but its asking in degrees

OHHH

ok i get it so i put that in cos?

[scenic ravine]

when you solve for $\xi$ you get cos(x)

[twin meteor]

ƒ(Why am. I here)=I don't Know

[hybrid oyster]

that would mean its in either quad 2 or 3 making it 150 or 240?

ohhh right

[scenic ravine]

yes

use the same logic for the other equations

[hybrid oyster]

alright ty

does it work till number 3?

[scenic ravine]

yes

[hybrid oyster]

oh wait for b theres 2x in the sin

[scenic ravine]

yes

[hybrid oyster]

does that meani need to divide the degree by 2?

[scenic ravine]

yes

[scenic ravine]

see, you're better at maths then you thought

[hybrid oyster]

alright ty very much!!!! is it fine if like i get stuck agian ill come back?

[hybrid oyster]

all the memory related math stuff im good at but counting and algebra is my issue

im a biologist biologist need good memory :)!!!!

agian ty!

[scenic ravine]

cool!

[scenic ravine]

yes, just message here

[hybrid oyster]

oh on c

the sin is squared

what does that mean exactly?

[scenic ravine]

$sin(x) \cross sin(x)$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[hybrid oyster]

uhhh ic

so on c im at

[slow horizon]

Hey guys idk if this is correct channel but please can someone help@im stuck

[scenic ravine]

!occupied

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hybrid oyster]

sin^2 x = (3/4)

[scenic ravine]

cool

so let $sin^2(x)=u$

now what

[twin meteor]

ƒ(Why am. I here)=I don't Know

[hybrid oyster]

knowing that its sin x*sinx = (3/4)

oh

uhhh

ok

am i allowed to rt(u)?

[strange crater]

of course

[hybrid oyster]

then right after i sin

oh alright

OHHHH YEAH I tyy

i got it

ok i kinda continued after c but

with D

i am left with tan= (rt3/2)

at the end

rt3/2 isnt a perfect angle (idk if its called that in english i learn math in my native tongue)

for tna

[scenic ravine]

it is

[hybrid oyster]

so uhh

what do i do with tan= (rt3/2)

is there omething im missing?

OHHH

AM I SUPPOSE TO TRANSFORM THE TAN INTO SOMETHING ELSE?

into

uhh

sin/cos was it?

hm nope uhh

was there a miss calculation on m part

[strange crater]

,w calc arctan(sqrt(3) / 2)

[twin meteor]

Results provided by WolframAlpha

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Requested by cwatson

[hybrid oyster]

7tanθ=3rt(3)+tanθ
6tanθ=3rt(3)
tanθ=(rt(3)/2) ```

[hybrid oyster]

this question is supposed o be solveable calculator less so uhh sorry i dont hthin this is it

uhh maybe we can move on so

for sinx to = 1

it could only be 90degrees right

180+90 wouldnt make sin x = 1 right?

is there a sure fire way of figuring out a different instance of like a cos/sin/tan value that is equal to the first x?

like im pretty sure theres 2 instances of cos +(1/2)

but i jsut dont know where

and how to find

same with sin

there should be 2 isntances of +(1/2)

well uhh

without just

remembering the entire trigonometric perfect angles

[vocal sleet]

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[echo tree]

Find the derivative of f(x) using the chain rule

[echo tree]

This is where I’m at for the first chain rule but I’m not very familiar with simplifying radicals so I don’t think this is correct

[scenic ravine]

this works but IMO might be a better idea to square it first

$y^2=\sqrt{x+\sqrt{x+\sqrt{x}}}$

$y^2=x+\sqrt{x+\sqrt{x}}$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

$\left(y^2-x\right)^2=x+\sqrt{x}$

[twin meteor]

ƒ(Why am. I here)=I don't Know

[scenic ravine]

should be easy from here

[echo tree]

Oh I see that makes sense, I think my teacher wanted me specifically to do a chain rule inside a chain rule but that’s much faster

[scenic ravine]

here you'll have to use the chain rule and product rule on the LHS anyway

[echo tree]

What’s an LHS?

[scenic ravine]

Left hand side

[echo tree]

Oh ok thanks

[scenic ravine]

wait

this may mess up the domain though

nah, shouldn't change much, just remember the domain remains the same

[echo tree]

Ok the domain in this case would just be x >= 0 right?

[scenic ravine]

$x\geq0$ yes

[twin meteor]

ƒ(Why am. I here)=I don't Know

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@echo tree Has your question been resolved?

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[acoustic patio]

which one is funcion graphic?

[round garnet]

I don't understand your question. Do you mean which picture resembles a function?

[acoustic patio]

yes ig

[round garnet]

Ok so the definition of a function is smth like:
Every x-value gets ONE y- value but never more

[acoustic patio]

so the right side should be a funtion then?

[round garnet]

Yes exactly

[acoustic patio]

thank you

[round garnet]

But what about the left one?

[acoustic patio]

no idea

[round garnet]

Well it's not as you figured

But let me tell you why

So I take the x-value x=2

Then I have one y-value which is y=6
But I also have y=-2
(And technically all the ones in between)
So I don't have only ONE y-value but "a lot"

Therefore it isn't a function!

I hope that makes the definition a little clearer

[acoustic patio]

yes, you helped a lot thank you

.close

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[north sky]

any ideas on how to solve these kinda problems

[vocal sleet]

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@north sky Has your question been resolved?

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[warped sleet]

what's |AD|?

Yes the side length of AD

what is it?

?

yes

what do you know about the side length of a general square

anything "interesting"?

okay so what's C?

yes

no

huh

what's the y coordinate of C?

why?

C is some units above D

it doesn't make sense for C and D to have the same y coordinate

no

look at the x coordinate of D

and compare it to C

?

x coordinate

does it move to the left or right?

How do you go from D to C?

Describe in words?

If you had to walk from D to C

by how many units?

?????

up 7 units

there is no right or left

so decide your coordinate point for C

with that observation

it doesn't move left/right

only up 7 units

bro

ye

sure

Lol

[strange crater]

take it to #discussion

<@&268886789983436800> spam

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[umbral ember]

Part a:
let R be the region in the first quadrant enclosed by the graphs of x^(1/2) and x/3

part B
write but dont evaluate an expression involving one or more integrals that give the volume of the solid generated when R is revolved around the horizontal line y=4
only need help with B

[chrome raptor]

What have you tried?

Do you know the method of discs and the method of cylinders?

[umbral ember]

i know discs and washers

nothing else

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@umbral ember Has your question been resolved?

[wraith python]

What have you tried?

[vocal sleet]

@umbral ember Has your question been resolved?

[umbral ember]

washers, does R=4-x^(1/2) and r=4-x/3 sound correct at all for those?

[vocal sleet]

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[crystal lava]

I need help

Write the (x, y) coordinate of the y-intercept of this graph.

[upbeat spindle]

y intercept when x = 0

[tight sonnet]

do you know what y-intercept is?

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[vocal sleet]

Please don't occupy multiple help channels.

[thin vale]

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[thin vale]

Stick to #help-5

[vast shale]

ah okay

my bad

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[mellow marsh]

Can someone please help with how to go about this question?

[sturdy vortex]

bruteforce should be fine no? the number has to be a natural number < 20 if in not mistaken

like x^3 * (x+3) is approx x^4 and 4th root of 100000 is less than 18

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[lone linden]

exponential decay reflected over the x axis ig

[dire mason]

okay ty

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.close

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[dull depot]

need help with b don’t know how to do it

[simple plank]

V8 and manual are independent events, you mulitply

[dull depot]

so it would be ⅓ * 1/2?

[simple plank]

yes

[dull depot]

for the first part of b is it correct

already wrote down 12/36

[simple plank]

simply 1/3. no matter what other choices are, only consider those three choices

[dull depot]

so i’m correct

need help on the next problem tho this one is more confusing

[clear stirrup]

How did you get 36 as your sample space?

[dull depot]

i manually counted the combinations. i used fundamental counting principle first doing 2* 3* 2 but i only got 12

after counting the different combinations myself i got 36

[clear stirrup]

12 is the total combinations

[dull depot]

but how

wait no i’m dumb

it is 12

still don’t know how to do 3 tho

[clear stirrup]

Are you referring to c?

[dull depot]

yes

i am completely stuck

i don’t know how to take into account the limitations when using fundamental counting principle

i have 2x3x2x8x11

[clear stirrup]

So it's still using the product rule except we have to consider the restriction on the color

[dull depot]

and idk how

do you just subtract 3

[clear stirrup]

Use the sum rule to find the weighted average

[dull depot]

what

this isn’t probability it’s asking to find the number of combinations there are

[clear stirrup]

$\frac{1}{3} * 11 + \frac{1}{3} * 11 + \frac{1}{3} * 8$

[twin meteor]

Akuma

[dull depot]

why are we multiplying by fractions

[clear stirrup]

For the engine choices

Or you can separate into 2 cases

(2x2x2x8x11) + (2x1x2x8x8)

Does that make sense

[dull depot]

this makes more sense

[clear stirrup]

👌

[dull depot]

thanks

.close

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[rain yarrow]

Given 4 groups of 4 objects in a 4x4 square, what is the probability that at least one of the groups has 1 item in every row?

[rain yarrow]

<@&286206848099549185>

i know it >14% but not much more

<@&286206848099549185>

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[coarse wren]

Hey there, I'm needing help with some trigonometry. I have a given of 18 ft and need to figure out if 20, 25, or 40 ft would create a 75 degree angle. I'm just looking to get a pointer on how I could do this. Thanks in advance! :)

[woeful cobalt]

ok so just making sure are the sides making a angle that is not right

or is this triangle not a right angle

[lime abyss]

are the values 20 25 40 hypotenues values ??

[coarse wren]

I think 20, 25, and 40 are the hypotenuse values yeah. The problem is basically the roof is 18 ft from the ground, what ladder can you use (20, 25, 40) that will make a 75 degree angle (from the ladder and ground)

[lime abyss]

got it

see the 75 degrees is with the ground

we know one side is 18

[coarse wren]

I'm thinking that 18 is the opposite or adjacent while the ladders are the hypotenuse, making this a tangent problem, wasn't totally sure though

[lime abyss]

using the sin we can find the hypotenuse

[coarse wren]

Gotcha, what indicates that 18 is the opposite and not the adjacent?

[lime abyss]

oppposite

use this image

so letting the hypotenus be x

xsin75 = 18

sin75 is $(sqrt6+sqrt4)frac/4$

[coarse wren]

Gotcha, so I believe the 20ft latter would suffice then, as it's the closest to 18 and the problem states to use the smallest ladder to accomplish what's needed

[twin meteor]

the_legend

[lime abyss]

did you calculate or an gusse

[coarse wren]

I did 20sin75, 25sin75, and 40sin75, and 20sin75 comes out to be 19.319 (rounded), being closest to 18. The problem basically just says the smallest ladder should be used

[lime abyss]

the answer is 25

[lime abyss]

no

[coarse wren]

huh, I must have did something wrong, let me see

[lime abyss]

bro my bad its 20

ft

sry

[coarse wren]

ohhh no problem

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[earnest harness]

Hello

[vocal sleet]

Please don't occupy multiple help channels.

[earnest harness]

Can I get help with this

I’ve waited and entire hour just to get help

And I need just the answer I don’t need any steps to solve this

It dosent ask for it

<@&286206848099549185>

[vocal sleet]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[sly bolt]

bro tryna run away

just hear me out brother

I understand it doesn't ask for it

but you need to understand the material

so you can get the answers yourself

this server helps you understand

[earnest harness]

Do u not understand that I’m late turning in my assignment

[sly bolt]

should have done it earlier

[earnest harness]

Stop stalking me like a weirdo

[sly bolt]

i got tagged

[earnest harness]

Let somebody else respond

[sly bolt]

you @ ed helpers

[sly bolt]

he did

[earnest harness]

And for ur info most people who helped me just gave me the answers

[sly bolt]

then go to those people

[earnest harness]

The David kid keep asking for answers kick him

.close

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[earnest harness]

.close

[potent finch]

If this is geometry let me help u here #geometry-and-trigonometry

[earnest harness]

Ok

Thank u

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[hybrid oyster]

dude this is real quick

2.d

[thin vale]

why real quick?

[hybrid oyster]

as mentioned above the trigometric function needs to be 0-2pi

[thin vale]

that's the domain for x

[hybrid oyster]

does that mean the x has to be between 0-2pi

[thin vale]

yes that's exactly what it means

[hybrid oyster]

so theere should be 4 answers?

[thin vale]

it means 0 <= x < 2pi

Yes 4 solutions

[hybrid oyster]

alright ty thats all

how do i end this

[thin vale]

!done

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.close

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[vast shale]

what should i do after this

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

[tawdry canopy]

it just reqires you to make choices

so you can just calculate f(0) g(0) f(1) g(1) and verify all the choices

[vast shale]

i did but im wrong

[final otter]

idk if i'm going insane but all of the equations on that sheet are wrong

e.g. g(f(0)) = g(max(1 + sin(0), 1 - cos(0)) = g(max(1,0)) = g(1) = max(1, |1-1|) = max(1,0) = 1

[river kettle]

Yajat

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[fresh wasp]

Hello

[fresh wasp]

i actually have no clue how to attempt this 😄

Please help

[sharp frost]

Hey

[fresh wasp]

Hello

[sharp frost]

There is like a trick for this

Sin nx period where n is a number is 2pi/n

Generally

For f(nx) where n is a number

Period will [period of f(x)]/n

So here it'll be 2 pi/n

So the period of each term will be 2pi/2 , 2pi/3, 2pi/5 and 2pi/7

[vocal sleet]

@fresh wasp Has your question been resolved?

[fresh wasp]

Hm

But how did i find the total period then?

Add them together?

[sharp frost]

Common multiple

Least common multiple

[fresh wasp]

Ah

Okay thank you for your help

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[orchid pier]

why is 18*18 not equal to (10+8)(10+8)?

[steep finch]

you made a mistake when expanding the brackets

8*8

[orchid pier]

oh shi

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[hoary talon]

differentiate, is my answer and work correct?

[flint idol]

yeah looks good

[hoary talon]

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[plush summit]

Hi I need to answer these questions, are my solutions correct?
Let the vectors u, v and w be elements of IR3. Judge whether the following statements are true or false. Please give reasons for your answer in each case.
a) If v is orthogonal to x, y and z, then also to 2x − 5y + z.
b) If u is orthogonal to v and w, then v and w are parallel.
c) There is always a linear combination cv + w, c ∈ IR that is orthogonal to w.

[plush summit]

so for a)
if x,y,z are orthogonal to v it means v^T * x = 0
v^T * y = 0
v^T * z = 0

v^T * (2x - 5y + z) = 2(v^T * x) - 5(v^T * y) + (v^T * z)
Every scalar product is 0 so the result is 0. This means the statement is true, it can be orthogonal.

b)
u^T * v = 0
u^T * w = 0

If they are orthogonally it doesn't mean that they are parallel because the vectors can be orthogonal in another dimension. For example they lay in different layers, one in x1 and x3 and one in x1 and X2

c)
(cv + w)^T * w = 0
cv^T * w + w^T * w = 0
c(v^T * w) + | |w²| | = 0

It's orthogonal because | |w²| | is always positive?

But still unsure...

[vast shale]

yup u were going insane all of them were right i got it checked by my teacher (just ignore this skenox)

[final otter]

huh???

[final otter]

what was wrong with this calculation

i'm gonna help skenox instantly as an apology for my interruption

[final otter]

a) and b) look good, and for c) you're very close

in the equation "c(v^T * w) + | |w²| | = 0", can you choose c in such a way that the equation becomes true?

[vocal sleet]

@plush summit Has your question been resolved?

[plush summit]

If w is the zero space then it can't be orthogonal but if it's not the zero space then c is element of IR?

or no then there is no value for c but I can be orthogonal if w is the zero space

[final otter]

yeah but in any case you're seeing that the statement isn't always true, right?

so just describe a counterexample and you're done

[plush summit]

okay hm 1 * (2 * 3) + 2 = 0
8 ≠ 0
You mean like that? So c = 1 for instance wouldn't work, but the vectors are also different hm

[final otter]

oh no, you need an example where no value of c works

because to disprove "there is always a value of c so that...", you need to show that there are examples where no value of c fulfils the statement

[plush summit]

-1(v^T * w) → (-v^T * -w) which isn't necessarily orthogonal, so this is a counterexample, when c = -1 or < 0

hmm

I'm a bit lost xD

[final otter]

To find a solution, you need to find a c that fulfils c(v^T * w) + | |w²| | = 0, right?

how can you solve for that c?

[plush summit]

is it c = -| |w²| | / v^T * w?

[final otter]

Sure! yeah

so is the statement always true, because you can find c like this?

[plush summit]

no the c can be not found and therefore the statement would be false when there is for example a zero as a result of v^T * w?

[final otter]

Yeah when v^T w = 0, then the equation just becomes | |w^2| | = 0, and if w isn't the zero vector, that's always false

[plush summit]

Okay great, thank you for the help!

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[plush summit]

Hi is the statement true or false? The following term delivers a number in IR.

I think it's true because for example u= (1,0,0) and v = (1,1,0)
u * v^T = 1
and then for the sum it is
0 * 1 = 0
1 * 1 = 1
2 * 1 = 2
3 * 1 = 3
4 * 1 = 4

I think there is no counterexample for this. It could be only zero but still it's IR

[plush summit]

\sum_{i=0}^{4} i \cdot (u \cdot v^T)

aaah okay I post a picture instead

[upbeat spindle]

is this i the imgaginary i

[sharp lynx]

It's the index of summation

[plush summit]

Oh and u and v are supposed to be vectors in IR3

[sharp lynx]

The dot product returns a real number, so you're summing the product of integers and reals

[plush summit]

And is an integer and real number a real number?

[sharp lynx]

What

[plush summit]

think that was just confusing

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[sage dew]

e^-1 is not negative

[hallow plover]

The range of a function is the values y can be, not the values that x can be

[sage dew]

x is the input

aka domain

you analyze the behavior of the function and take note of any discontinuities or asymptotes that it might have

e^x doesn't approach zero for increasing x

then yes

as x approaches - infinity e^x tends to zero but never becomes zero

[river minnow]

Approaches 0 <-/-> Is positive

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✅

[kind wadi]

intuitively you can see that theres no way for e^x to be negative

[fading plume]

2e^x is close to zero for negative values of x
and you should know the range of x is R

[kind wadi]

2e^x - x can be negative if 2e^x < x

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[urban laurel]

how do i prove that this function is continuous over R²\{(0,0)}

[twin meteor]

Adam Ch.

[hybrid flicker]

well R^2\{(0,0)} is open

so around any (x,y) in that set, f(x',y') will equal the first form

and you're left to find $\lim_{(x',y')\to (x,y)}\frac{x'y'^3}{x'^2+y'^2}$

[urban laurel]

what does this mean ??

why do i have to evaluate that limit ?

[hybrid flicker]

the definition of continuity

you have to show $\lim_{(x',y')\to (x,y)}f(x',y') = f(x,y)$ no?

[twin meteor]

rafilou2003

[hybrid flicker]

for any (x,y) in R^2\{(0,0)}

[urban laurel]

what's x' and y' ??

[hybrid flicker]

?

(x',y') is a point that approaches (x,y)

when (x',y') approaches (x,y), f(x',y') approaches f(x,y)

that's the definition of continuity

[urban laurel]

shouldn't the limit be : $\lim_{(x',y')\to (x,y)}f(x,y) = f(x',y')$ ?

[twin meteor]

Adam Ch.

[hybrid flicker]

no

f(x,y) is constant when you fix (x,y)

there is no point in taking the limit of a constant

plus (x',y') is not defined on the RHS of the equality

[urban laurel]

ok how do i evaluate this ?

[twin meteor]

rafilou2003

[hybrid flicker]

right sorry I edited

I see what you meant

well (x',y') -> (x,y)

so x'^2+y'^2 converges to...

and x'y'^3 converges to...

[urban laurel]

.close

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[urban laurel]

got it ty @hybrid flicker

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[tight monolith]

can someone help me with my math performance task?? it’s about coding and decoding matrices

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[vast shale]

Could you help me in the 3rd one

[vast shale]

.close

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[glossy sorrel]

can someone help me

and tell me how to do this

[mellow oyster]

do you know about the trig ratios, like sin and cos and tan?

soh cah toa stuff?

[glossy sorrel]

yeah like

opposite

and adjacent

and hypotenudse

[mellow oyster]

perfect, that's what you want to use here

[tender socket]

Write out SOH CAH TOA

it should help you out

[glossy sorrel]

but how

i dont have the number

[tender socket]

Mhm, but you have the angle and hyp

[mellow oyster]

so those ratios are all relative to an angle, the only useful angle that they've given you in your triangle is 61 deg

and yeah you have the length of the hypotenuse

[mellow oyster]

relative to the 61 deg angle, is HJ the opposite or the adjacent?

[glossy sorrel]

opposite

[mellow oyster]

good, so you want to solve for the opposite and you have an angle and the hypotenuse

which trig relation related opposite and hypotenuse

[glossy sorrel]

so how would i set that up

[mellow oyster]

^

[glossy sorrel]

sine

[mellow oyster]

so sine of an angle = opposite / hypotenuse, your angle is 61 degrees, hypotenuse = 8

which means sin61 = opposite / 8, so opposite = 8 * sin(61 deg)

which is your answer that you'd plug into a calculator

[glossy sorrel]

sin 61= x/8?

[mellow oyster]

yes

in general for these right triangle problems you first figure out the angle they gave you, and then figure out what side lengths you have and are solving for

[glossy sorrel]

so how would i solve

[mellow oyster]

how would you solve for x?

[glossy sorrel]

idk

😭

[mellow oyster]

sin61 = x / 8, to isolate for x, you can just multiply both sides by 8

[glossy sorrel]

OH

[hexed ember]

for the reference
sin if:
60 <a<70 ≃ 0.9
70 <a<80 ≃ 0.95

80<a<90 ≃ 1

[glossy sorrel]

so would i multiply the 61 x 8

wait what

im so lost

[mellow oyster]

no sorry

sin(61) is just a number

you can call that number a if it helps you think about it

then a = x / 8, so x = 8 * a

you don't need to bring the 8 inside the sine function

is the problem that x = 8 * sin(61) doesn't look like an answer? all of that works out to a number, it's approximately 7

if you're asked to write the exact answer, you would put 8*sin(61), otherwise if you're trying to approximate the answer you'd plug that into a calculator

[glossy sorrel]

ohhhhh

ok

thank u sir

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[prime moon]

@mellow oyster hey so im still a little bit confused

[mellow oyster]

confused about

[prime moon]

how can i write equation

like with the bot

so i can better explain

[mellow oyster]

oh i'm sorry i can't help you there, i don't use discord much so idk how the bot works

[prime moon]

oh ok ok

[mellow oyster]

it looks like it just uses latex formatting if you know that

[prime moon]

yeah yeah i think you'll understand

so right now i have the equation set up like this

sinx/sinx+cosx=(sinx/cosx) (cosx/sinx)

i wrote brackets just so u understand its multiplication

how can i get the sides to equal same thing

[tight wasp]

That multiplication just cancels out 😳

[prime moon]

yeah ik but my goal is to prove that left side equals right side

[dreamy torrent]

i think it is written as
$\frac{sin(x)}{sin(x)+cos(x)} = \frac{sin(x)}{cos(x)} x \frac{cos(x)}{sin(x)}$

[twin meteor]

Macbeth

[tight wasp]

Its not true for all x tho

[dreamy torrent]

multiplication not x but yea

[dreamy torrent]

not possible
anyways the answer should be cos(x) = 0

[prime moon]

let me show you the question from the start

maybe i made a mistake

question d)

@dreamy torrent @tight wasp

[dreamy torrent]

do you know what tan(x) is equal to?

[prime moon]

yes its sinx/cosx

[dreamy torrent]

you made a mistake here

try transforming 1 to $\frac{cos(x)}{cos(x)}$

[twin meteor]

Macbeth

[tight wasp]

You just made the +1 vanish

[prime moon]

wait so what would the right side look like

[dreamy torrent]

would be a nightmare to try to type it in latex

lol

[dreamy torrent]

but it would be (sinx/cosx) / ((cosx+sinx)/(cosx))

try simplifying this

[prime moon]

done

tysm

/close

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[cunning saddle]

Help

[thin vale]

Have you tried drawing it

[cunning saddle]

Okay let me a second , can help me ?

[thin vale]

What do you need help with

[cunning saddle]

with all

[thin vale]

did you draw this?

[cunning saddle]

yes

is beatiful

Okay lets go to 2

istn complete?

[thin vale]

Do you know what a complete graph is

[cunning saddle]

Connected: True
Complete : False

[thin vale]

Yes

[cunning saddle]

Is it when a vertex is connected to every vertex in the set of V?

[cunning saddle]

Im alr?

[thin vale]

Yes

[cunning saddle]

Okay now this part

how i can find cycles

[thin vale]

What's a cycle

[cunning saddle]

It is when a vertex travels a path and returns same vertex?

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[bitter bison]

The sides of a square are 1 cm long. If we consider all vertices, how many points in the plane are exactly 1 cm away from any two vertices?

[tight wasp]

You need the distance to work for all pairs of vertices or any one of the pairs

[mild flower]

i feel like you can solve this graphically

[bitter bison]

i think for all pairs of vertices

[subtle blaze]

Yeah I was gonna say it's probably easiest to just draw it and sketch circles around vertices

Seems like a constructions q

Well, it could be

[bitter bison]

I tried to make a square bigger than the initial one that had all its sides 1cm away from the initial one but I got stuck. How else can it be graphed???

[mild flower]

is that what the question asked you to do?

[bitter bison]

Not really... But that's how i tried to approach it

[bitter bison]

This is the question

[mild flower]

how would you find all points in the plane that are 1cm away from the first vertex? (call it A)

[bitter bison]

making a circle of radius 1 cm where the vertex is the central point? but that would only work for each vertex individually, right?

[mild flower]

sure, but if you look at vertices A and B, what are the points that are 1cm away from both of those?

[vocal sleet]

@bitter bison Has your question been resolved?

[bitter bison]

I'm sorry im not sure

[mild flower]

the circle around A is the points 1cm away from A

the circle around B is the points 1cm away from B

what are the points 1cm away from A and 1cm away from B?

[bitter bison]

the diameter of the circle?

wait no, sorry, i think that doesn't makes a lot of sense. Could it be like the points where the two circles intersect?

[bitter bison]

Wait i think i get it. The circles of each vertex intersect 3 times, right? and one of those times it's also a vertex?? does it make sense?

[mild flower]

yes this

[bitter bison]

Yess!!! thank uuu

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[elder talon]

solve for x

i had this question on my test on friday

it was the only problem i struggled with

but im not sure if its possible as when solving i end up with someting thats not quadratic

i asked my teacher if the denominator for the first term was supposed to be x+2 instead but she said no

but i think i was able to solve the question either way , it was -5.75184 but I don't think my teacher was looking for taht

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[haughty tulip]

1 - see if you can factor and then send a pic of what you did

[elder talon]

that picture was the question

that was on my test

the goal was to isolate x

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[sinful hound]

could someone explain the change of variables in the second-last step

[sinful hound]

i get why they did it but i don't understand why it's 1 to t

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[sinful hound]

<@&286206848099549185>

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[ivory sonnet]

Is the blue graph correct?

[pallid forge]

no

[hybrid flicker]

f'(1) = 0...

[sharp frost]

No

[hybrid flicker]

this wouldn't look like that unfortunately

[pallid forge]

first bullet you got, i'm not sure about any of the rest

second bullet, the slope of the graph should be negative when x is less than 1 or greater than 4

[hybrid flicker]

it looks more like red graph is correct

[ivory sonnet]

ok thank you everyone

.close

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[slim hill]

I'm having trouble with these types of questions, I know how to use the formula but I dont know how im supposed to get to using that said formula, in this case I dont know how I was supposed to know how to get 290 (145x2)

[pallid forge]

so 145 going at 58 miles per hour

$\frac{58 miles}{1 hour}=\frac{145 miles}{x hours}$

[twin meteor]

Wumbo

[pallid forge]

apologies for the poor spacing

[slim hill]

ohh so I was supposed to use proportions

[pallid forge]

cross multiply and solve for x

[slim hill]

but my concern is how am I supposed to know to use proportion or any other formula

[pallid forge]

practice

[slim hill]

do you have any resources where I can practice these types of questions?

[pallid forge]

my first instinct is khan academy, but let me google it

wdym by "these types of questions"

like

dimensional analysis?

or specifically stuff to do with calculation speed, distance, and time

[slim hill]

like I can't tell exactly what to do so I'm trying to figure out the trick to knowing that

i couldnt tell I was supposed to use proportions

[pallid forge]

well

what else were you going to do

math isn't exact , there are sometimes lots of ways to go about a problem

[slim hill]

didnt know what to do at all

I have trouble with analyzing how questions are worded

[pallid forge]

well, now you do!

i think it really just comes down to practice and exposure

next time you come across a problem like this one, maybe you'll remember how to approach it

[slim hill]

I know how to solve this type of problem but I have trouble piecing together what kind of problem it is from how the question is worded

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[vast shale]

The function $h : \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ is defined by $h(<a, b>) = 2^a(2b + 1) - 1$. How do I show that the function is bijective?

[twin meteor]

ecoproducts

[vast shale]

First, was wondering for the injective part.

I suppose that $h(<a_1, b_1>) = h(<a_2, b_2>)$.

[twin meteor]

ecoproducts

[vast shale]

Then, is it simply algebra?

By equivalence of equations, we have: $2^{a_1} (2b_1 + 1) = 2^{a_2} (2b_2 + 1)$.

[twin meteor]

ecoproducts

[vast shale]

Do I need to log and figure something out?

Not sure.

[brittle minnow]

This follows from the fundamental theorem of arithmetic

Although I am not sure why a -1 is there

Ah okay, nvm mind the -1 makes sense

@vast shale what have you shown so far

[vast shale]

The equation above. $2^{a_1} (2b_1 + 1) = 2^{a_2} (2b_2 + 1)$.

[twin meteor]

ecoproducts

[vast shale]

For two pairs whose respective outputs by h are set to be equal.

[brittle minnow]

I mean have you shown either injection or subjection?

[vast shale]

No, I originally was given a map h and I simply showed that it was a well defined function.

Now I'm trying to show that h is injective.

Next, I'll move on to it being surjective.

[brittle minnow]

Alright, do you see why the function is a subjection

Hint: use the fundamental theorem of arithmetic

[vast shale]

I'm not supposed to use anything we haven't seen in class... This is for discrete math by the way. No need to use a theorem to show that h is surjective I think. I have it figured out though.

[brittle minnow]

Fundamental theorem of arithmetic is very basic. It's literally the fact that you can do prime factorisation

[vast shale]

I used prime factorization to show that h will always yield a natural number of the form 2k - 1 ≥ 0.

Yeah, well, there we go.

[brittle minnow]

Alright

[vast shale]

Anyways, surjection seems more straightforward than the injection part which I need help on.

[brittle minnow]

So for injection you can do the opposite route. Show that if x not equal to y then f(x) is not equal to f( y)