help-17

[normal lynx]

its triangle proportional theorem

[junior forum]

Can you find the proportionality factor between the big triangle and the small triangle?

[vocal sleet]

@normal lynx Has your question been resolved?

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[vast shale]

i need help[

[vast shale]

i need help someone

[lyric fjord]

[vast shale]

ok

my brain is kinda not working rn

x = 45+(97x5)+54-1

i need to solve for x

[warped junco]

97*5 first

[vast shale]

485?

i think

[warped junco]

Add 45

[vast shale]

530

[warped junco]

Add 54

[vast shale]

584

[warped junco]

-1

[vast shale]

583

[warped junco]

Wow congrats

[vast shale]

!

ty

[warped junco]

[vast shale]

i have more thp

[loud walrus]

Also, don’t use x as a product symbol

[vast shale]

?

[loud walrus]

And even less if you have the x for other purpose, like in the equation

[vast shale]

so what do iuse

[loud walrus]

Here

Use a dot

•

Middle dot, not the other

. •

[vast shale]

oh

ohhh ok

now i get it

woopds

there

i need help with these

but i gtg

ill msg when i get back got a class

ummmm

anyone there

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

<@&286206848099549185>

can i have some help please

[candid otter]

!status

[vocal sleet]

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

[vast shale]

all of them pls

hello?

<@&286206848099549185>

dude i rlly need help

i gtg soon

and this is due soon

ok

[cursive turret]

stop spampinging

[vast shale]

sorry

[vocal sleet]

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[vocal sleet]

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[fossil goblet]

sparse means not populated
rife- a description of something harmful or negative
inexorab;e- impossible to stop
gainsay- to deny a staement
odious- worthy of hate
amiable -friendly and pleasent
misanthrope- a perosn who hates people and does not want there company at that time
odious- worthy of hate?... wwow
sully- to damage the reputatioin
cessation- an end or interuption
decree nisi- - end o f a marraige paper
scofflaw- someone who commits minor crime with a hope that they will escape punishment
deed-poll changing your name

[vocal sleet]

@fossil goblet Has your question been resolved?

[mild flower]

what?

[vocal sleet]

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[minor atlas]

Quick question regarding matrices.
If I have $UAU = Da$, how can I alter it to get A by itself on the LHS?
Like this? $A = UDaU$

[twin meteor]

Chizz

[minor atlas]

sorry not sure how to do the notation right with TeXit

But U* represents the conjugate transpose

[hard kraken]

don't you need to multipy the inverse of U twice

on the right sid

[minor atlas]

Like $U^(*-1) Da U^(-1)$

[hard kraken]

I think so but I'm just a dumb 9th grader that just learned matrices like a week ago

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[cursive crane]

hello, i'm not sure on how to solve this question

[vocal sleet]

@cursive crane Has your question been resolved?

[sweet flower]

!status

[vocal sleet]

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

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[cursive crane]

not sure how to find the delta X of this problem

[vocal sleet]

@cursive crane Has your question been resolved?

[median harbor]

@cursive crane you can close it now

helped you in other server

[cursive crane]

perfect timing because

[median harbor]

you got another one?

[cursive crane]

for some reason its suppose to be easy but my answer no work

i did 5(sum of all the midpoints)

[median harbor]

you know what midpoint riemann sum is right

okay

first find all the midpoints

[cursive crane]

its already listed

[median harbor]

i meant like find all the bounds

[median harbor]

ok nvm you said this that means you found the bounds already

[cursive crane]

my answer was 1632.5 by the way

[median harbor]

is that not right

WAIT HOLD UP

[median harbor]

THE UNITS

THE UNITS

YOU GET WHAT I MEAN

[cursive crane]

no

what do you mean the units

they are all in km

[median harbor]

its 19.5km/h for 5 seconds

think about it

how would 0 —> 75km/h in 30 seconds walk for 1632.5km

[cursive crane]

but the graph disagrees

[median harbor]

what

wdym

its 19.5km/h for second 0 to second 5

so it walks for 19.5km/h for 5 seconds

[cursive crane]

unless im understanding you wrong

at 5 its very clearly over 25km/h

[median harbor]

but we’re using the midpoint riemann sum

[cursive crane]

right

[cursive crane]

yes thats the first midpoint

[median harbor]

yeah so for seconds 0 to second 5

[cursive crane]

i already have it in my equation

[median harbor]

ok wait do you get why for seconds 0 to seconds 5, the car is walking at 19.5km/h

[cursive crane]

not really

i just memorize formulas

[median harbor]

do you understand this?

[cursive crane]

yes those are the mid points

well half drawn rectangles

[median harbor]

and we want the area of all those rectangles right?

[cursive crane]

yes

[median harbor]

so what’s the width of the first rectangle in seconds

and what’s the height of the first rectangle in km/h

[cursive crane]

5

[median harbor]

5 seconds yep

[median harbor]

and this?

[cursive crane]

5\cdot 19.5

5 * 19.5

[median harbor]

no

dont omit the units

leave them in

5 seconds x 19.5km/h

do you see where it went wrong now

[cursive crane]

o

so do i divide 19.5 by 3600

[median harbor]

yep

ok but think a bit more first

what does 19.5km/h * 3600 mean

like whaat is it evaluating

in terms of the car

[cursive crane]

• or /

[median harbor]

?

what

[cursive crane]

ok yeah

this was the answer

[median harbor]

yep

[cursive crane]

the devil really is in the details

[median harbor]

that’s right

yeah

[cursive crane]

ok thanks

.close

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[median harbor]

np

[vocal sleet]

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[strong thorn]

an i doing number 5 right?

[flat whale]

Yes

,w diff (log(x))^2 / x

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by rie.mann

[strong thorn]

alr so the problem is, we’re given the answers and we’re supposed to show our work

the answer is e^2 but i keep getting 1

i don’t see how it would equal 0 at e^2

[edgy gulch]

2-ln(x) = 0

x = ?

[strong thorn]

oh

right

ohhh

there we go

thank you

.close

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[waxen hawk]

uh

[waxen hawk]

ayy

u know

ayy

ayy

ayy

[flat whale]

ayy

[waxen hawk]

ay

[flat whale]

Ayyyyyyy

[waxen hawk]

its lit

you know

ayy

[round ridge]

ayyyyyyy

[waxen hawk]

ayy

[round ridge]

ayyy

[waxen hawk]

So here's a

scenario

So I think it is a physics-related scenarios

and were trying to get the T represented by these constant

m1,m2,g etc..

[waxen hawk]

but when I manipulate these shit myself

I got a different T

Am I tripping or the teacher

[merry shuttle]

third line, it will be (m1 + m2 - m1 + m2)

[vocal sleet]

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[ripe ivy]

I'm trying to design a little game using playing cards and am wondering this.

Say you have a 5x5 grid of playing cards, ranging from values 1-10 (including 1 and 10). and placed randomly.

Now, imagine a scanner that looks at the grid horizontally and vertically.
Like XXX XOX
OOO XOX
XXX XOX

The scanned grid's value is summed together.

Across the whole grid, what are the chances of the sum being 18 or over?

Also, if you could be so kind, what would the probability of it being 15, 16, 17, 19, 20, 21 or 22 be? (optional)

[ripe ivy]

Thank you in advance :D

i figured out that theres 30 different spaces possible for the "scanner" to scan.

( horizontally, 3 possibilites for 1 row. times by 5 = 15
do the same vertically, 15x2=30)

i love probability questions, id love to know how to do these on my own.

.close

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[muted nexus]

these are my answers

other than the last one, im unsure which ones i have wrong

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[leaden jackal]

What I do wrong

[wicked sonnet]

What is the question?

Find the area?

[leaden jackal]

Surface area

Formula is 2(lw+lh+wh)

[mild flower]

that's for cuboids

this is not one

[leaden jackal]

It’s for a prism

[wicked sonnet]

Wait where'd you get the formula from?

[leaden jackal]

From the course

[wicked sonnet]

This doesn't look like the formula for it

[leaden jackal]

It’s the formula for a prism

[mild flower]

it's the formula for a prism with a rectangular base

which this isn't

just consider each side and add them up

[leaden jackal]

Hmmm

[wicked sonnet]

You could do some smart work by seeing that the shape is symmetric so like, find the area of the triangular side once and then double it

[leaden jackal]

So I’ll just add up all the sides and it will give me the surface area ?

[wicked sonnet]

Yes

That's what surface area is, isn't it

Noo wait.

[leaden jackal]

Idk

[wicked sonnet]

She meant, area of sides.

[leaden jackal]

Area of sides?, so the surface area for each side?

[wicked sonnet]

Do you know what surface area is?

[leaden jackal]

Yea

Idk what area of sides is

[wicked sonnet]

What is surface area?

[leaden jackal]

Is squared measurement of a side

[wicked sonnet]

Wdym squared

[leaden jackal]

Wel for a square it’s side *side

It means how many little squares you can fit

[wicked sonnet]

That's not what surface area is

[leaden jackal]

…

[wicked sonnet]

Its used for solid figures, or three dimensional objects

For two dimensional figures, or plane figures, it's just called area

[leaden jackal]

That’s the same thing

Just one surface

[wicked sonnet]

I'm talking about terminology

[leaden jackal]

Yeah

I’m talking about math

[wicked sonnet]

Okay so, for a solid figure

The (total) surface area is just the total area of it's surface

It's implied by the word itself.

[mild flower]

surface area is how much paint you'd need to color it red

or wrapping paper maybe

that's better

[leaden jackal]

Okaaay

[mild flower]

so like for your shape

you have the triangle in front, the triangle in back, the rectangle on the bottom, and the two rectangles on the top

[leaden jackal]

Okay

[mild flower]

those are all faces (2d shapes), and you can find the area of each one individually

add all those areas up to get the total surface area

[leaden jackal]

Ahh okay

[mild flower]

,tex .plane geom

[twin meteor]

hayley

[mild flower]

here are some tools to help you :D

[leaden jackal]

Okay im going to save this

I@know some of these

This triangle one I know

And the rectangle

[mild flower]

that's all you need in this case

[leaden jackal]

And the parrlellagram

[mild flower]

but it is a useful image to have

[leaden jackal]

Yeah

Okay well

I’ll

Go and solve this then.

[mild flower]

oki

[leaden jackal]

Okay thanks bye!

.close

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[vast shale]

There is a party with normally 250 people and each person must pay 6 dollars. Per each 1 dollar increase 25 people leave what’s the max revenue they can make

[quartz snow]

where do you need help in?

[vast shale]

I just did it on a. Test

I’m just curious if

1600 is correct

[quartz snow]

the maximum value happens when x=8

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@vast shale Has your question been resolved?

[vast shale]

You mean x=2

[vocal sleet]

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[snow light]

this is the assignment

[snow light]

this is as far as i got

[desert hornet]

is that $x^2\cdot \sqrt{x-1}$ or $x\cdot \sqrt[2]{x-1}$

[twin meteor]

kheerii

[desert hornet]

probably the former

[snow light]

i thing it is the second one

think

[desert hornet]

well, they wouldn't specify the 2 there

i dont think?

[snow light]

i was correcting myself

i wrote “thing” instead of “think”

[desert hornet]

no i meant

specifying the 2 on the radical seems a bit ridiculous

it's probably the former

anyway, both of them have the same solution

[snow light]

it should be x> 1

[snow light]

but now i cannot gather anything (if it’s called gather in english)

i’m so dumb

[snow light]

i didn’t change the sign on the 1 and -2x while bringing them on the other side of the disequation

now it results

.close

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[rain hollow]

i have a 30 degree sharp angle rombus there is a circle inscribed in it and in that circle there is a square i need to find the ratio of rombus area to squares area

[vocal sleet]

@rain hollow Has your question been resolved?

[rain hollow]

<@&286206848099549185>

[mystic remnant]

!question

[rain hollow]

i have a 30 degree sharp angle rombus there is a circle inscribed in it and in that circle there is a square i need to find the ratio of rombus area to squares area

@mystic remnant

[split wind]

did you try to draw it out?

[rain hollow]

yea but its a bad drawing

[split wind]

oh, I'll try to draw it out then, it'll be bad though

[rain hollow]

ok

[split wind]

let one side of the rhombus be a, then you can use the triangle to find the radius of the circle

[mystic remnant]

do you know the answer

@rain hollow

[rain hollow]

Yes

[mystic remnant]

can you tell pls

i jus want to verify a thing

[rain hollow]

The ratio is 4

But in the circle there also is a square

@split wind there’s also a square in the circle and rombus ses sharp angle is 30 degrees

[split wind]

that means that's it's 30° instead of 15°?

[rain hollow]

No i think you got it right but if 1 side is a how can i find the rest of the triangle

[split wind]

oh right

do you know sine law and cosine law?

[rain hollow]

Yes

[split wind]

i can give further hint, but the drawing will look messier

[rain hollow]

So the green line is just the side?

[split wind]

yea, the green lines are the diagonals of the rhombus

[rain hollow]

Cos 30 is sqrt 3/2 right

[split wind]

yea

[rain hollow]

Is thea 2-sqrt3/2?

[split wind]

?

[rain hollow]

1-sqrt3

Am i not supposed to sopve the green equation?

[split wind]

hmmm, I'll draw more

[split wind]

sorry, got a typo

[rain hollow]

wait

the sides are a right?

[split wind]

DA should be a/4

Added: 2(Area)÷base=height

[rain hollow]

So the answer is indeed 4 right?

This is quite complicated can you exlain from the beginning?

@split wind

[vocal sleet]

@rain hollow Has your question been resolved?

[vocal sleet]

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[spiral compass]

what happened here

im not sure

specifically this part

here is the whole problem for those who need

[opaque hill]

looks like it used L'Hopitals rule at first glance

[crude arrow]

it's right btw

[spiral compass]

oh yes ik, this is the answer key of my mock exam

however

[spiral compass]

oh okayyy

i get it

nice nice

[crude arrow]

oh they just make sure it's indeterminate

and used l hopital

in that case

[tribal moss]

i do not want to disturb, but read this:

$\text{we know that:}\text{ }\lim_{x \to 0}\left( 1+x \right)^{\frac{1}{x}}=e\text{ }\text{ then}\\\lim_{x \to 0}\left( 1+ax \right)^{\frac{1}{x}}=\lim_{x \to 0}\left[ \left( 1+ax \right)^{\frac{1}{ax}} \right]^{a}=e^{a}\text{, where }a\neq 0$

[twin meteor]

Joanna Angel

[spiral compass]

oh wow

thats a pretty aadvanced theorem

[tribal moss]

no 🙂 but very useful one

[spiral compass]

oh sorry advanced for me rather HAHHAHAHAHHA

[tribal moss]

it's ok )

[spiral compass]

im still doing that manually using $x=e^{lnx}$

[twin meteor]

adumb

[spiral compass]

another question

the originial question would be (a)

I was under the impression that implicit differential was to be used here

i will write my final answer pls wait

i am not sure why they ln both sides

[crude arrow]

I preferred the first handwriting

[crude arrow]

it makes it easier to differentiate

https://en.wikipedia.org/wiki/Logarithmic_differentiation#:~:text=In calculus%2C logarithmic differentiation or,rather than the function itself.

refer to this

[spiral compass]

sorry sorry for the terrible handwriting

[spiral compass]

i thought logarithmic differentiation was only used when there was a variable raised to another variable present within the equation

<@&286206848099549185>

[vocal sleet]

@spiral compass Has your question been resolved?

[spiral compass]

so whenever there is a log, we use log diff

that makes so much sense omg bruh

[vocal sleet]

@spiral compass Has your question been resolved?

[vocal sleet]

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[dense wharf]

hi , is anyone here good with compound angles?

[vocal sleet]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[dense wharf]

For the first part of this question, do i just use algebra to combine the two equations?

The first equation in algebra would be Xsin(wt+theta)

just not sure how to combine the two equations together

i1= Xsin(wt+theta)

i2= Ysin(wt)

and i basically need to combine them but im not too sure as to how i do it

both wts are different values

i think i need to use compound angles? but i dont know how

<@&286206848099549185> ?

[vocal sleet]

@dense wharf Has your question been resolved?

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[sage panther]

(Skip to next paragraph for direct question) For a bit of contextual background, I'm not a student currently, I'm just self-learning to prepare for when I do get back into university. When I was in high school, I had pretty limited trigonometry taught to me, so working with questions like this are completely unknown to me

That said, I'm actually perfectly fine with deriving, so I'm mainly just wondering what to replace the variables with in this question. I'm not too sure what the relationship is between a, b, c, and cosine, and just wondering how to set this up properly

[sage panther]

I'm assuming I'm letting cos(theta) become cos(A-B), but what do I do with a, b, and c?

Also like I said, I'm not asking for the full solution to the question; Just what trig formulas I have to use to replace the variables with, if that's even what I'm supposed to be doing

[torpid talon]

you know two of the sides

they are radii

[sage panther]

Right. In the end, my full equation is supposed to look like cos(A-B) = cos(A)cos(B) + sin(A)sin(B), and I just don't know how to get there from side lengths. Do I replace the 'a' and 'b' with the identity sin^2a + cos^2a and sin^2b + cos^2b? Or am I thinking about this completely wrong?

Also sorry if my message comes in late, my internet's being very faulty right now

[vocal sleet]

@sage panther Has your question been resolved?

[vocal sleet]

@sage panther Has your question been resolved?

[sage panther]

Just for a bit of a cleanup on this question, I'm just asking about the first step in my post. I don't need to know how to directly derive the equation, I just need to know what I'm replacing a, b, and c with

[vocal sleet]

@sage panther Has your question been resolved?

[sage panther]

<@&286206848099549185>

[vocal sleet]

@sage panther Has your question been resolved?

[spiral inlet]

@sage panther in order to use the law of cosines, you need to have expressions for all three sides and an angle

In terms of A and B.

Do you have everything you need? If not, what are you missing?

[sage panther]

I guess first of all, am I right to assume sin^2A + cos^2A and sin^2B + cos^2B take on the variables 'a' and 'b' in the law of cosines?

Or am I headed in a completely incorrect direction with that

[spiral inlet]

those are both just equal to 1

[sage panther]

Oh, literally just implementing 1 for those?

[spiral inlet]

I guess I'm a bit unsure how you got to those expressions in the first place

[sage panther]

What do I do with the 'c' then?

[spiral inlet]

Well hang on, I want to make sure we're on the same page

[spiral inlet]

when you say this, are you referring to the two sides of the triangle which are radii of the circle?

[sage panther]

Yeah. Because I need to derive the law of cosines into the formula for cos(A-B)

I'm fine doing the deriving by myself, I think, but I'm not sure how to start it, since I know I have to turn a, b, and c into different expressions

Or at least, that's how I'm interpretting it

[spiral inlet]

well, in the law of cosines, a, b, and c refer to the three sides of the triangle

two of the sides of your triangle are equal to 1, so yes you could use just a=1 and b=1

[sage panther]

Oh, I see. I thought I was supposed to plug in sin and cos into those. So I can just use the numbers themselves, gotcha. What about c?

[spiral inlet]

c would be the length of the third side

Remember the points on a unit circle have coordinates which are the cosine and sine of the corresponding angle in standard position

so these two points are (cosB, sinB) and (cosA, sinA)

Using the distance formula, you can find an expression for the third side

[sage panther]

Ahh gotcha, I see now. Thanks

[spiral inlet]

No problem 👍

[sage panther]

.close

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[thin vale]

For a question like this, is it supposed to be obvious that "the set of all polynomials of degree greater than n" means polynomials with coefficients in a field....? Like I might just be overthinking it
P(x) is supposed to be P(F) and that's the set of all polynomials with coeffs. in F

so I think it is probably implicitly saying like, "the set of all polynomials, with coeffs. in F, of degree greater than n", but I also wonder if it might be a trick question

[urban edge]

Uh oh

Party time

[thin vale]

the party is dead

[urban edge]

D:

[hushed pewter]

What is n?

[thin vale]

I am also assuming it is a positive integer

or nonzero integer

but it isn't clear

this is the whole problem statement

[hushed pewter]

Sorry but what is P(x)?

[thin vale]

P(x), is supposed to be P(F), my professor sent an email correcting that

and P(F) is the set of all polynomials with coeffs. in F

[hushed pewter]

F being whatever field

[thin vale]

yes

reals or complexes for me

[hushed pewter]

Yes, I think this is the right way to see it

[thin vale]

alright

I thought so

either way the answer is no

ty SWR

.close

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[somber arch]

last time i ask for help fr im new to this stuff, what is going on 😭

[somber arch]

am i correct?

[outer warren]

a° is correct
b° is incoorect

[somber arch]

im confused on b

[outer warren]

what's the total angle in a revolution / around a pioint?

[somber arch]

?

like in a circle?

[buoyant notch]

yh

[somber arch]

360 degrees

[buoyant notch]

what’s a again

[somber arch]

62

[buoyant notch]

No

You said 118

[somber arch]

my bad 118

yes

[buoyant notch]

so b and a have to equal 360

[somber arch]

bro i didnt get taught this

[buoyant notch]

[somber arch]

its 242 then

[buoyant notch]

Yh

[somber arch]

alright thanks is there anyway i can get more information to study stuff

like the 360 thing

whats it called

[buoyant notch]

js study your angle theorem stuff

[somber arch]

ok thanks

.close

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[dusty escarp]

how do i deal with square roots here, i always find myself confused on removing the square roots cause i dont really know its properties

[proven knot]

Well square root in general only behaves nicely with multiplication and division

This problem is a bit more involved than most because you have 3 distinct pieces with the variable under the square root

I think you have no choice but to start by squaring both sides

Keep in mind that on the right side you must expand that out using distribution (FOIL)

[dusty escarp]

so i would square sqrt(x+3) and square sqrt(7-x) or would i square both under one like (sqrt(x+3) + sqrt(7-x))^2

[proven knot]

The latter

When you have an equation and you perform an action like squaring

It is done to the entire side of the equation

You can imagine that each side of the equation is wrapped in parentheses

Every time you perform an operation

[dusty escarp]

okay i see, but just for example to make sure, if the first side of the equation was sqrt(2x+4) + 3x = ... , would i also square the 3x?

[proven knot]

You would wrap the entire left side in parentheses

And then square that object

[dusty escarp]

wdym?

[proven knot]

Say you had

a + b = c + d + e

To square both sides

I would get

(a+b)^2 = (c + d + e)^2

[dusty escarp]

ohh i see

[proven knot]

The whole side is first put in parentheses before you perform the squaring

[dusty escarp]

but how is foiling this performed?

[proven knot]

Well you'll get something similar to

(a+b)^2 = a^2 + 2ab + b^2

[Sqrt(x+3)]^2 = x + 3

Similar for sqrt(7-x)

The middle term is

2*sqrt(x+3)*sqrt(7-x)

When you multiply square roots together you have a nice property for that

Sqrt(ab) = sqrt(a)*sqrt(b)

[dusty escarp]

so it becomes sqrt((x+3)(7-x))?

[proven knot]

So you can call.the middle term:

2sqrt[(x+3)(7-x)]

Yea

[dusty escarp]

and does the square root go away at some point?

[proven knot]

So what you'll have to do

Is move all.the non square root terms to one side

Leave the single square root term thats left alone on one side

Then square both sides again

You'll end up with a quadratic equation without any square roots

[dusty escarp]

ahh okay, but seeing that its 2sqrt[(x+3)(7-x)] should i divide the equation by 2 to keep only the square root or does the not really matter

[proven knot]

It doesnt really matter if you do that

Depending on the numbers it may be easier to not divide

But make sure you also square the 2

If you leave them together

After you finish solving, you have to check to make sure the solutions you get do not cause problems in the original problem

Because this started as several square root terms, you can sometimes find extraneous solutions

These would be answers that you reach through the algebra

That cause a negative square root in the original expression

Whatver answers you find, check whether they cause square root of a negative in the original problem

If so they must be discarded

[dusty escarp]

oh okay that makes sense ill try everything and see what i get

thank you so much

[proven knot]

Yw goodluck

[warped junco]

Remove square root symbol by squaring then solve

[vocal sleet]

@dusty escarp Has your question been resolved?

[dusty escarp]

.close

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[ionic otter]

can anyone please help me with few basic statistics questions such as identifying the data types in a question?

[paper depot]

http://dontasktoask.com/

[ionic otter]

huh

Do you want me to show my question?

[paper depot]

yes exactly

[sullen thunder]

don't ask to ask a question... just ask the question

[ionic otter]

oh okay my apologies

[paper depot]

and for future reference you can and SHOULD just open with the question itself

[ionic otter]

I have 10 questions like this and I am not sure about them, I have answered with my knowledge but not 100% sure if I am correct

<@&286206848099549185>

[paper depot]

why do they have point values attached to them btw

this sounds like you might be in a test

where you are supposed to do it yourself

[ionic otter]

oh no its our graded homework

[paper depot]

k

[ionic otter]

its a check in

if it was a test it would be monitored

[paper depot]

well i guess show all 10 questions and your answers?

[paper depot]

yeah ok just making sure. we get cheaters sometimes and it's against server rules

[ionic otter]

i dont have questions about all 10, but ill show the ones that i have questions about

thats all i have questions about

[ionic otter]

should i send all?

[paper depot]

send the ones you want checked

[ionic otter]

i did above

[paper depot]

ok

[ionic otter]

thank you

[paper depot]

Q7 and Q8 seem ok. Q6 i would also tick "discrete" probably.

[ionic otter]

in the homework description it says that categorical only has two answers and same for numerical

discrete also can't be categorical as it is numerical

.close

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[heady moat]

Can someone help me solving this ?

[heady moat]

The first this I remember is the limit of sin(x)/x when x goes to 0 is 1

So I divide both of the numerator and denominator with x

But the rest, I dunno

(Sorry if you didn't get it, I didn't do English maths)

[vast shale]

what tools can you use

[vast shale]

Hint: its in 0/0 form so use L-hospital rule

are Taylor and l'hopital allowed or not?

it becomes straightforward with either

[heady moat]

What ?

[vast shale]

read my second message

[heady moat]

L-Hospital ?

[merry shuttle]

for numerator, (sqrt(x+1) - sqrt(1-x))/x , you can rationalise

[vast shale]

you can't make the approximation sin(x) = x without knowing the Taylor series. Unless this was a typo

[merry shuttle]

they divided by x

i was just talking about numerator

they've done the denominator sinx/x

[heady moat]

I can even show you the correction

The fact is I dunno what he did

[merry shuttle]

yea

he rationalised the numerator

or de-rationalise it

[viral copper]

best to just rationalize you're gonna directly get the sinx/x limit

[merry shuttle]

multiply and divide by sqrt(1+x) + sqrt(1-x) in the numerator

[heady moat]

Okay

But when did the 2x spawned ?

[merry shuttle]

multiplying the original thing with this

do you know the identity of (a+b)(a-b)

[heady moat]

Yeah

It's something like a^2-b^2

[merry shuttle]

a is sqrt(1+x) and b is sqrt(1-x) in this case

[merry shuttle]

yea

[merry shuttle]

apply the identity to this

[heady moat]

But isn't that a to the power of 2 instead of a to the power of 1/2 ?

[merry shuttle]

a is not 1+x
its the whole sqrt(1+x)

[heady moat]

But the sqrt of a to the power of 2 is a, isn't it ?

[merry shuttle]

you're mistaking a for 1+x

but yea, sqrt(1+x) to the power of 2 is just 1+x

[heady moat]

He didn't disappeared the sqrt on this one

[merry shuttle]

it disappeares on the top
the bottom would still be x(sqrt(1+x) + sqrt(1-x))

you're multiplying and dividing

the bottom stays

[heady moat]

Sooooo

How the 2x spawned?

Because of the power ?

[merry shuttle]

do (a+b)(a-b)=a²-b² for a=sqrt(1+x) and b=sqrt(1-x)

[heady moat]

Even tho, why did he reversed the a (square roots) with a some denominator

[merry shuttle]

you'll get 2x

[heady moat]

HUH UH

[merry shuttle]

what do you mean by reverse a

[heady moat]

x becomes 1/x

[merry shuttle]

theres no x becoming 1/x?

sorry im not getting you

okay so

$ \frac{\sqrt{1+x} - \sqrt{1-x}}{x} $

[heady moat]

There is sqrt(1+x) - sqrt(1-x) becomes 2x/sqrt(1+x) - sqrt(1-x)

[merry shuttle]

that would be a +

in the denominator

[heady moat]

$\frac{\sqrt{1+x} - \sqrt{1-x}}{\sin({x})}$

[merry shuttle]

yea

then they multiplied

[twin meteor]

Meyland

[merry shuttle]

consider the numerator first

after you divide by x

since you've done the sinx/x part

$\frac{\sqrt{1+x} - \sqrt{1-x}}{x}$

[twin meteor]

Fab

[merry shuttle]

this is what needs to be done now right

[heady moat]

Yeah I guess

[merry shuttle]

$\frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$

[twin meteor]

Fab

[merry shuttle]

do you understand this

[heady moat]

Hm, not really

[merry shuttle]

do you know rationalisation

how you would simplify something like 1/(2+ sqrt(3)) for example

[heady moat]

I can't

[merry shuttle]

oh

okay so

$\frac{1}{2+\sqrt{3}}$

[twin meteor]

Fab

[merry shuttle]

consider this fraction

rationalisation is when you turn the denominator from an irrational number to a rational number

for that you generally multiply and divide by the 'conjugate'

conjugate is basically flipping the sign
conjugate of a+b would be a-b for example

what do you think the conjugate for 2+sqrt(3) would be

[heady moat]

2-sqrt(3)?

[merry shuttle]

yea

so you multiply and divide by 2-sqrt(3)

you can do this because multiplying and dividing by the same thing is basically multiplying by 1

$\frac{1}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}$

[twin meteor]

Fab

[merry shuttle]

do you understand

[heady moat]

$\frac{1}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}$

[merry shuttle]

it will be a minus on the top

it has to be same thing

thats when it'll be 1

[twin meteor]

Meyland

[merry shuttle]

yep

look at the denominator now

[heady moat]

But this gonna be 1

[merry shuttle]

do you see something

[heady moat]

Oh

[merry shuttle]

we're not simply gonna cancel it

do you see any pattern in the denominator

[heady moat]

$\frac{2-\sqrt{3}}{(2+\sqrt{3}) \cdot (2-\sqrt{3})}$

[merry shuttle]

you forgot the parenthesis

[heady moat]

Where ?

[merry shuttle]

$(2+\sqrt{3})(2-\sqrt{3})$

[twin meteor]

Fab

[heady moat]

OH YES

[merry shuttle]

this would be the correct way of writing the denominator

[heady moat]

Sorry I'm a bit dumb

Sometimes

[merry shuttle]

soo

[merry shuttle]

do you think you can use anything here

[twin meteor]

Meyland

[heady moat]

Identities

[merry shuttle]

which one

[heady moat]

$2\power{2}-\sqrt{3}\power{2}$

[twin meteor]

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

[heady moat]

Bruh

[merry shuttle]

yea so 2² - (sqrt(3)² basically

[heady moat]

Yeah, that

[merry shuttle]

4-3

thats 1

$\frac{2-\sqrt{3}}{1}$

[twin meteor]

Fab

[merry shuttle]

did you understand the whole process

[heady moat]

Yes, it's pretty easy

When did we use the formula tho ?

[merry shuttle]

continue from this then

[merry shuttle]

(a+b)(a-b) is a²-b²

you used that

[heady moat]

When we wanna make an irrational number a real number?

[heady moat]

About rationalization stuffs

[merry shuttle]

there's not exactly what we'll doing here

but

the process is the sams

[heady moat]

Sooo

[merry shuttle]

continue from here

[heady moat]

$(frac{\sqrt{1+x}-\sqrt{1-x}) \cdot (\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})}$

[merry shuttle]

forgot the x in the denominator

and the parenthesis

[heady moat]

Always forgot this shi

[merry shuttle]

$\frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{x(\sqrt{1+x} + \sqrt{1-x})}$

[twin meteor]

Fab

[heady moat]

As you did

Okay

[merry shuttle]

do you see anything in the numerator

[heady moat]

Soo

$\frac{(1+x)-(1-x)}{x(\sqrt{1+x} + \sqrt{1-x})}$

[merry shuttle]

yea

square of sqrt(1+x) is just 1+x

simplify the numerator

psst

parenthesis

[twin meteor]

Meyland

[heady moat]

My bad

[merry shuttle]

do you see the 2x now

[heady moat]

Wait

OH

OHHHH

I got it

[merry shuttle]

yea

you can also cancel the x now

[heady moat]

$\frac{2}{\sqrt{1+x}+\sqrt{1-x}}$

[merry shuttle]

there's no x

[twin meteor]

Meyland

[merry shuttle]

yep

[heady moat]

Almost Forgot

[merry shuttle]

so this

simplifies to

[merry shuttle]

this

you can plug x=0 to get the limit of the numerator now

[heady moat]

It's 1

[merry shuttle]

yep

[heady moat]

And sin(×)/× = 1 too

[merry shuttle]

yep

[heady moat]

Tysm dude

[merry shuttle]

so it's all 1

youre welcome

[heady moat]

You learned me something important

Even my teacher was struggling lol

[merry shuttle]

well you got it now

goodluck on future limits

[vocal sleet]

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[waxen hawk]

🐯

[waxen hawk]

Can i interpret Angular velocity as how fast the object spins.

It is basically spinning velocity

The hurricane has a angular velocity of 100pi

For instance

Why

Why you retrieve it

For sure

[paper depot]

missing units

[waxen hawk]

“Rad”

[paper depot]

rad**/s** you mean?

[waxen hawk]

Yes

[paper depot]

angular velocity is the angle by which your thing rotates in one time unit

[paper depot]

basically yes

[waxen hawk]

Got it thank you

.close

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[main knot]

Having trouble with these 3

[main knot]

Solve for x here

[main knot]

isolate x here

[main knot]

answer as a number to the power of 3 here

I don't know where to start in any of these 3 please help if you can

[dry yacht]

Starting with the first one: $$\frac{1}{3}-\frac{x-5}{3x-7}=\frac{2}{3x}$$ Find the common denominator.

[twin meteor]

Good

[main knot]

3x-7?

[dry yacht]

What does it mean with common denominator?

[main knot]

im not sure what the common denominator can be in this case

unless you mean factoring out 3?

[dry yacht]

Common denominator means something that all fractions have in common. In this case, 3, 3x-7 and 3x are included.

[main knot]

I forgot so much of the math I learned sorry

[dry yacht]

If 3x-7 were 9x, then the common denominator only included 3x and 3.

[main knot]

hmm ok im confused

what do you mean if 3x-7 is 9x?

[dry yacht]

The point with common denominator is to get rid of the denominators.

So to achive that, we need to multiply every denominator that are involved.

[main knot]

so 1/3 * (x-7) for example?

and then put them together?

[dry yacht]

In this question, 3, 3x-7 and 3x are involved. So we need to multiply 3(3x)(3x-7) on both side, which is every expression.

[main knot]

ohhhh

ok wait

brb

soething like this

then put in quadratic formula?

[dry yacht]

You sure the middle expression is correct?

$\frac{x-5}{3x-7} \cdot 3(3x)(3x-7)=(x-5)(3x)$?

[twin meteor]

Good

[dry yacht]

On spot.

[main knot]

what middle expression

[dry yacht]

$\frac{x-5}{3x-7}$

[twin meteor]

Good

[dry yacht]

As I mentioned here.

[main knot]

I forgot to multiply by 3?

I was hurrying a bit didn't want you to wait too long

[dry yacht]

On spot.

[dry yacht]

Take your time.

If I were out of time, someone else would takeover.

[main knot]

yeah maybe

so like this im guessing

rest is easy enough so if this isn't wrong then we can move on to the next one

[dry yacht]

Move on to quadratic equation.

[main knot]

ok hold on

[dry yacht]

PS: simplifying before the equation would make it a bit easier.

[main knot]

into x^2 + x + 7?

I seem to be getting a negative number for b^2-4ac lol idk wtf

[dry yacht]

9x^2 - 9x^2 = 6x^2?

Look over the second last step.

[main knot]

This is why I hate math

god damnnn

ok so just solve for x then?

[dry yacht]

Yes.

[main knot]

x=-7 then

[dry yacht]

On spot.

[main knot]

lol

I just can't help myself

with these dumb mistakes

for no reason

[dry yacht]

$\frac{2x-1}{k} = \frac{x+1}{b}$

[twin meteor]

Good

[dry yacht]

When it involves this many terms, it's important to look over again.

[dry yacht]

This is just isolating x.

[main knot]

yea

maybe i can also do k*b on both sides here?

[dry yacht]

On spot. Usually this is called cross multiplication.

[main knot]

I forgot about this thing but now I know

[dry yacht]

Good news.

[dry yacht]

Looks good.

[dry yacht]

$\frac{6561^{\frac{2}{4}}}{2187^{-\frac{3}{7}}}$

[twin meteor]

Good

[dry yacht]

This is a bit heavy on first look, but it's something you could do with the numerator.

What would you get by 6561/2187?

[main knot]

3

[dry yacht]

So 6561 = 2187 * 3

[main knot]

yea

[dry yacht]

$=>\frac{(3*2187)^{\frac{1}{2}}}{2187^{-\frac{3}{7}}}$

[twin meteor]

Good

[dry yacht]

Any thoughts now?

[main knot]

can't see anything

[dry yacht]

(ab)^2 = ?

[main knot]

a^2 + 2ab + b^2?

[dry yacht]

That is (a+b)^2

[main knot]

oh

a^2b^2

[dry yacht]

Exactly, you apply the same on the numerator.

[main knot]

how exactly do you calculate to the power of fractions?

what is te term for it?

how do you calculate fractional exponents

[dry yacht]

$x^{\frac{a}{b}} = \sqrt[b]{x^a} = (\sqrt[b]{x})^a$

[twin meteor]

Good

[main knot]

ok so I get sqrt3*sqrt2187

[dry yacht]

$=>\frac{\sqrt{3} \sqrt{2187}}{2187^{-\frac{3}{7}}}$

[twin meteor]

Good

[dry yacht]

And you may do something with the 2187 in the numerator and denominator.

[dry yacht]

It's a bit much of latex, you might want to look for a picture.

[main knot]

ur explanation was ok for me

I'm so confused honestly

Not getting this at all

[dry yacht]

In this question, you'll need $\frac{x^a}{x^b} = x^{a-b}$

[twin meteor]

Good

[main knot]

soo sqrt2187/2187^-3/7 = 2187^1/2+3/7?

[dry yacht]

On spot.

[main knot]

ok then I want to add together the exponents

with a common denominator or?

[dry yacht]

Yeah, as the way you deal with regular fractions.

[main knot]

2187^13/14

[dry yacht]

Looks good.

[main knot]

how can I answer that then as a number to the power of 3

[dry yacht]

2187=3^7

[main knot]

ohhh

I thought it was supposed to be like x^3

but yea u are right

I still don't really get this this really made no sense to be the whole way through

[dry yacht]

It's not easy to find the exponent of a number.

[main knot]

I'll figure it out over time probably

thank you so much though you are great at teaching

[dry yacht]

Hope any of the explanations will be of help.

[main knot]

definitely helped a lot

I'm gonna go over the last one again see if it won't click

.close

[vocal sleet]

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[long canopy]

how can i find the domain and range of a quadratic from the factored form y=2(x-5)(x+3) and how could i explain it to someone in simple terms