help-10

[craggy sundial]

so it makes sense to get a number from that?

[wary vigil]

yes

[craggy sundial]

so if i was asked to find dH/dq, this equation would be my answer. and if they gave me a coordinates, i would follow the steps u gave me and the answer to that is the value at that certain point?

[wary vigil]

well it depends on whether they want it in xyz and pqr or only in pqr

[craggy sundial]

meaning?

[wary vigil]

so yes that equation is the one

but they may want you to fill x, y, and z in

[craggy sundial]

with the values we got from plugging in p q and r yea?

[wary vigil]

no the x = p r ln(q)

fill that into the equation

to get an equation only depending on p, q, and r

[craggy sundial]

oh thats gonna be a fat equation then presimplification

[wary vigil]

but they'll probably tell you if they want that

[craggy sundial]

okay cool

thanks! is it cool if i dm u if i need more help? its hard to get noticed in this server

[obtuse pebble]

@craggy sundial Has your question been resolved?

[obtuse pebble]

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[green shore]

hi i need help doing this fourier questions. im not usre how to approach these questions

[upbeat gazelle]

Which one?

[green shore]

first one

even though i dont undestand either >,<

i thought they r closely related so i just sent htem together nayway

[upbeat gazelle]

Do you remember the expression of the coefficients of the Fourier series of a periodic function?

[green shore]

this is the answer its in terms of pi

i remember but i didnt know how to put it and hcnage the limits

what do i put the limits as since it gives me 2 on the t axis on in terms of pi

[upbeat gazelle]

Here it would be $\frac{1}{4}\underset{-2}{\overset{2}{\displaystyle\int}}f(t)\cos\left( \frac{\pi}{2} t \right)\mathrm{d}t$

[green shore]

and the same for sin ?

[upbeat gazelle]

Yes

[green shore]

why is it 1/2

[upbeat gazelle]

Because the period is 2

Oh sorry

It is 4

[green shore]

i dont understand why my uni teaches me like this

[upbeat gazelle]

So the pi is Divided by 2

[warm shale]

black_couscous

[upbeat gazelle]

Because usually you work on $2\pi$periodic functions

[warm shale]

black_couscous

[green shore]

bu t then wouldnt the period be

1/2pi

so why is it 1/pi

[green shore]

why would this be pi/2 t in the bracket

[upbeat gazelle]

Because they appear to have renormalized it

[obtuse pebble]

@green shore Has your question been resolved?

[obtuse pebble]

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[onyx hearth]

is this the correct null hypothesis

[onyx hearth]

or am I hypothesizing incorrectly

okay dope thankss

[tame narwhal]

no

[onyx hearth]

oh

[tame narwhal]

that would be the alternate hypothesis

[onyx hearth]

why?

[tame narwhal]

because you want to "prove" that is true

[onyx hearth]

oh the alternative is always the one you're trying to prove?

[tame narwhal]

generally

[onyx hearth]

okay cool

thanks a ton

.close

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[timid silo]

Please help

[sage geode]

Generally it's good to first figure out the equations that represent the outlines of the region given

[sage geode]

Here we have a region between y = x and x = 4, right?

[timid silo]

Yes

[sage geode]

Right

First you can notice that the entire region is located left to x = 4, so we say x <= 4

[timid silo]

Ye

[sage geode]

And it is also below the line y = x

So the other inequality would be y < x

[timid silo]

Ok

[sage geode]

Note that it's not y <= x because the diagonal line is plotted with dots

[timid silo]

Thank you I managed to do it👍

.close

[obtuse pebble]

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[ancient stag]

Im confused as to why this diverges

[ancient stag]

.close

[obtuse pebble]

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[quasi harness]

Hello

[quasi harness]

1. Show that you can graph the reciprocal of a quadratic function by graphing

, where f(x) is any quadratic expression of your choice with 2 distinct zeros. Be sure to show all appropriate details for a reciprocal graph as explained in our video lesson.  Your work should be done on a single, LARGE axes system and show both the graph of y = g(x) and y = f(x).

[glossy basalt]

...

didn't you posted this yesterday with no reply?

i mean i replied, and you did follow up

but then you're suddenly gone 🥲

[obtuse pebble]

@quasi harness Has your question been resolved?

[glossy basalt]

🥲

[obtuse pebble]

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[quasi harness]

.reopen

[obtuse pebble]

✅

[quasi harness]

@glossy basalt

are you still there

[glossy basalt]

?

[quasi harness]

hey

How are you today

can i please post a question

[tame narwhal]

you have posted this same thing 2 days in a row

[quasi harness]

@tame narwhal I didn’t receive clarification that’s all

[tame narwhal]

I remember yesterday someone asked you where does "A(x)/x" come from

[quasi harness]

Yes

And that’s what I seem to be stuck upon

[nocturne minnow]

#help-8 message

And it doesn't help if you ghost people helping eithet

[quasi harness]

@nocturne minnow I didn’t receive a response so I left

[glossy basalt]

erm....

[nocturne minnow]

#help-8 message
That says otherwise

[quasi harness]

Yah I see now

Well I’m sorry @glossy basalt

[glossy basalt]

i am being neglected...😢

I'm sad

[quasi harness]

No never

We appreciate your service

[nocturne minnow]

I don't

[glossy basalt]

hehe

[quasi harness]

Ok are you available now

[glossy basalt]

nah, imma about to sleep

[quasi harness]

It’ll be quick

[glossy basalt]

start

....

I don't believe this will be quick then

[quasi harness]

Graph the function with equation

its sending

bear with me

@glossy basalt its sent

@glossy basalt are you still there

[glossy basalt]

yes and nope

yes i am reading

[tame narwhal]

there should be 2 vertical asymptotes

[glossy basalt]

nope i don't have the brain power to read.

[quasi harness]

@tame narwhal can you please help me

yes I stated y= x-2 & x=2

[tame narwhal]

two vertical asymptotes

"is one right" what does this mean

[quasi harness]

a vertical asymptote

nevermind

@tame narwhal is their anything else Im missing

[tame narwhal]

sorry I meant there are 2 places where f(x) is indeterminate

I see now you wrote that

[quasi harness]

you got me confused

[tame narwhal]

your graph looks "flipped" over the x-axis compared to what wolfram alpha plots

[quasi harness]

what does the wolfram alpha plots look like

[tame narwhal]

,w plot (x^3 + 4x +3x)/ (x^2-2x)

[quasi harness]

its not -4x

its +

[warm shale]

Results provided by WolframAlpha

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

Requested by cwatson#5438

[quasi harness]

ok

is my solving portion correct though

does this graph have a vertical asymptote or is it only HA

[tame narwhal]

it has a vertical asymptote at x=2 yes

what you wrote looks OK, but it's been a long time since I've done such problems

[quasi harness]

Is the HA x-2

@tame narwhal are you still there

.close

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[sweet edge]

For this problem, I tried finding the particular solution using the steps in the image but I'm not sure where it went wrong

[sweet edge]

I didn't fully write everything out but the steps should be clear enough

[obtuse pebble]

@sweet edge Has your question been resolved?

[obtuse pebble]

@sweet edge Has your question been resolved?

[obtuse pebble]

@sweet edge Has your question been resolved?

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[woeful cliff]

How should I go about solving this system of equations? I’ve completely exhausted my options and have made little progress, so I would like to know how to tackle these problems in the future

[obtuse pebble]

@woeful cliff Has your question been resolved?

[woeful cliff]

<@&286206848099549185>

[timid silo]

yes?

[woeful cliff]

After trying for an hour I’m still not sure how I should proceed with this, could you point me into the right direction?

[timid silo]

I can't help you, I'm sorry

[solemn jacinth]

proceed with what

[woeful cliff]

@solemn jacinth Finding solutions of (x,y,z) over R in this system of equations

[solemn jacinth]

hmm lemme see

[solemn jacinth]

solvin it rn

[solemn jacinth]

what does over R mean??

[woeful cliff]

For all real numbers x,y,z

[solemn jacinth]

ok i am able to solve

[solemn jacinth]

hooked at some parts
btw which chapter/category is this sum from

[woeful cliff]

My country’s national olympiad

It’s listed as an algebra question

[solemn jacinth]

will be back in a while bro

[obtuse pebble]

@woeful cliff Has your question been resolved?

[obtuse pebble]

@woeful cliff Has your question been resolved?

[obtuse pebble]

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[brave moon]

does anyone know what the symbols mean

i just learned about relations in discreet math

like x=y

is reflexive symetic and transitive

but i dont know how to do it in this context

[dark stirrup]

~ is just a binary operation that is either true or false

[brave moon]

so what would that mean like if x doessnt =y

[dark stirrup]

x~y is true if and only if x-y is odd

= is never used anywhere in your problem

[brave moon]

wait sorry i dont really understand how would you pronounce x~y

[dark stirrup]

"X tilde y", "the relation of x to y", "tilde of x, y".

[brave moon]

so the relation of x to y iff x-y is odd

[dark stirrup]

If you prove that ~ is an equivalence relation, you can also say "x is similar to y", or "x is equivalent to y"

[brave moon]

is it a equivalence relation if it it is reflective symetric and transitive

[dark stirrup]

I'd probably say "x is related to y iff..."

[brave moon]

ok that makes more sense

so then it would be reflective and not symetric?

because x-x=0

oh wait

it would be symetric

[dark stirrup]

But not symmetric reflexive

[brave moon]

hm wait why if 3-6 is odd then 6-3 is also odd

oh okok because 0 is even

ok and its symetric because x-y = -(y-x)

so if its odd it stays odd

just the negative version

what does x|y mean

[obtuse pebble]

@brave moon Has your question been resolved?

[brave moon]

<@&286206848099549185>

[stable thistle]

Here

Aright what you need?

I can try to help.

[wintry swift]

x is a divisor of y.

[brave moon]

hey sorry ok thank you so what would that mean in the context of the question

[wintry swift]

x is in relation to y if x is a divisor of y.

[brave moon]

so a is a divisor of a

so its reflective

3 goes into 9 but 9 doesnt go into 3 so its not symmetric?

ok awesome but what would it mean for it to be transitive

like a|b b|c a|c

how do i know if thats works or not

[wintry swift]

if a is a divisor of b and b is a divisor of c. what can you say about a anc c? is a a divisor of c?

[brave moon]

this might be a stupid question but does divisor mean it can go into somthing and return an interger?

[tardy vigil]

No, that's not what divisor means. A divisor is a number that divides another number exactly (with no remainder). For example, the divisors of 12 are 1, 2, 3, 4, 6, and 12, because each of these numbers divides 12 exactly

[obtuse pebble]

@brave moon Has your question been resolved?

[brave moon]

ok thank you

i get it

.close

[obtuse pebble]

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[ruby elm]

The question is written in black.

[ruby elm]

The information given is as follows--
Consider a solid sphere is charge Q and radius R. At a distance R/2 from its center, a cross section is cut. Find the net flux crossing due to this surface due to charges present on both side of surface.

[obtuse pebble]

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[white glade]

I am doing problem M2 from Ch 13 in Artin's Algebra 2nd Edition, and it says that "A subset S of Z^2 that is closed under addition and contains the zero vector is an example of a semigroup." What does Z^2 denote here? I thought it might be the ordered pairs of integers, but the problem uses a "normal" alphabetical Z, and not a doublestruck Z.

[serene fossil]

are you sure its from CH13 artin 2nd edition? I just looked and 13.M.2 is different in my version

also its most likely the integers

without further context i would assume as such

[white glade]

my bad... I meant to write ch 12

[serene fossil]

Is this the question? in my copy the Z is clearly mathbb font

[hollow jacinth]

Hey guys I need help

[serene fossil]

#❓how-to-get-help

[hollow jacinth]

I know some of you know soh cah toa

And their reciprocals

[serene fossil]

get your own chat

[hollow jacinth]

But does anyone know this question?

[white glade]

oh for some reason our problem is slightly different? it's the exact same problem, just phrased slightly differently. But yea I guess it is the integers, thanks!

[serene fossil]

interesting must be the publisher or smth

[obtuse pebble]

@white glade Has your question been resolved?

[obtuse pebble]

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[pearl charm]

I don't know how to solve this porblem

[royal basin]

if the points here had their rectangular coordinates given, would you be able to do the problem?

[obtuse pebble]

@pearl charm Has your question been resolved?

[pearl charm]

maybe

as in if they were in like (x,y) form?

probably

<@&286206848099549185>

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[placid cedar]

two part question

[placid cedar]

to solve the integral $\int \frac{1}{\sqrt{25-x^2}} \dd x$, you can just use $u = \frac x 5$. I realize that's basically $u = \frac x a$ where $a = \sqrt{25}$. where does this intuition of subbing this exact value come from?

[warm shale]

@placid cedar

[placid cedar]

also, that's the standard way to do it. i did it this way which, after putting the final line into integral calculator, was leading me elsewhere. where did i go wrong here?

[fluid compass]

i'm sorry i do not understand both the ways of solving this as I would've directly used a given formula $\int \frac{1}{\sqrt{25-x^2}} \dd x$ = asin(x/5)

[warm shale]

akhsunaaa

[fluid compass]

the formula being $\int \frac{1}{\sqrt{a^2-x^2}} \dd x$ = asin(x/a)

[warm shale]

akhsunaaa

[obtuse pebble]

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[oak elm]

a^log b to the base c = b^log a to the base c, is this correct?

[obtuse pebble]

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[gritty elk]

I dont know the answer please confirm

I am getting 2880

[royal basin]

Representatives of 7 different countries are invited to attend a summit in India. The representative from China does not want to sit besides either the US or the UK. In how many ways can they be seated?

how are you getting 2880?

[gritty elk]

wait I am sending my working

my working

[royal basin]

china, us, uk together can only happen in 2 ways not 6

US-CN-UK or UK-CN-US

otherwise China is only adjacent to one

[gritty elk]

uk us china

but there are 4 other countries also

what answer are you getting

[royal basin]

,calc 7! - 6! * 2 - 6! * 2 + 5! * 2

[warm shale]

Result:

2400

[royal basin]

yes, and? does that contradict my point?

[gritty elk]

but why are we not subtracting us uk china

[royal basin]

as in, when they sit in the order US-UK-CN?

also. you're replying to the wrong message and that annoys me

[gritty elk]

oh sorry

[gritty elk]

yes

[royal basin]

these cases are already accounted for in the UK-CN adjacency and not in the US-CN adjacency

so you don't need to do anything else with those

[gritty elk]

ig I got it

thnx a lot

really

.close

[obtuse pebble]

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[jagged kernel]

.close

[obtuse pebble]

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[earnest field]

On my way to school

[earnest field]

How do I find the circumference 💔

[prisma hatch]

,rccw

[graceful pollen]

pi × r + 2r

[earnest field]

?

[graceful pollen]

yea

[earnest field]

How do I find the height if I have the bases of a trapezoid

[prisma hatch]

do u mean trapezium

[earnest field]

No

[prisma hatch]

whats trapezoid

[earnest field]

Don’t judge how it looks I drew it on the bus 🫂

[prisma hatch]

thats trapezium

[graceful pollen]

so a trapezium-

[earnest field]

It said trapezoid

💔

[graceful pollen]

same thing i suppose

[earnest field]

How do I find the height

[prisma hatch]

area of trapezium = height * (sum of bases)/2

[earnest field]

Man I have a quiz as soon as I get to school I totally failed lmaooo

[graceful pollen]

rearrange the formula to isolate the height

[prisma hatch]

which formula

[graceful pollen]

the formula u wrote above

[prisma hatch]

you mean equation?

[graceful pollen]

same thing?

[obtuse pebble]

@earnest field Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[next spruce]

help

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[next spruce]

1

[teal turret]

If they’re similar, the ratios of their corresponding side lengths will be proportional

For example

[warm shale]

Stephen

[next spruce]

Ty!

[teal turret]

Did u get it?

[next spruce]

yes

tysm

[teal turret]

Alr cool

Np

[next spruce]

.close

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[iron eagle]

I have tried using substitution for indefinite integrals, but im not sure how to use them in definite ones, does the range have to change? or do I have to these some other way?

[oak plover]

u = xt
1/x (cosu du)
1/x -sin(xt)
Then evaluate
1/x (-sin(x))

[obtuse pebble]

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[hearty wigeon]

Hello i need small help:
In a triangle ABC with side BC of length 2 cm, point K is the centre of side AB. The points L and M divide the side AC into three congruent lines. Triangle KLM is isosceles with right angle at vertex K.

Determine the lengths of the sides AB, AC of triangle ABC.

[obtuse pebble]

@hearty wigeon Has your question been resolved?

[hearty wigeon]

<@&286206848099549185> im sorry to bother you, couldn´t you gave me small help

[reef pollen]

It's okay

[obtuse pebble]

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[timber island]

Have u tried drawing a figure?

Aight lemme see if i can draw one

U aren't given any angles?

U found the height?

Of ABC

Getting a very irrational quantity

[obtuse pebble]

@wooden blade Has your question been resolved?

[timber island]

can u try to find height?

also since DE is minimum, I think we can assuming (in ur figure) DEA is right-angled

since if it is obtuse/acute, the DE becomes longer

well i amma see if mathway can help solve the quadratic for the height

yeah its a very irrational height

[warm shale]

ItzKraken2

[timber island]

set up a quadratic.

1 sec

use pythagoras then

yeah

u could use trig, but that needs the angles in the 1st place

and i forgot the trig wayyyy soo i had to setup this

how...? and which one again?

the one i drew?

how? its sides are 2,3 and sqrt(7)

1 sec

AB = 2, AC = 3, BC = sqrt(7)

and height is AO

havent drawn it drawing it

1 secc

Yeah

[night plaza]

the angle near point A

yes

[obtuse pebble]

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[smoky sedge]

I have a test in a few hours and I don’t know how to do the content that will be in the test could someone help me learn how to do them

[leaden ginkgo]

do you know the formula for calculating arc length?

[smoky sedge]

I do not

I don’t know anything to determine the length of an arc, area in a circle, or anything similar

[leaden ginkgo]

iirc $\frac {\theta}{360} \cdot 2\pi r$

[warm shale]

! saad

[smoky sedge]

How do I • it by 2pi

And what is the r

[leaden ginkgo]

r is the radius

[leaden ginkgo]

wdym

[smoky sedge]

How would I • it by two pi’s

I’m guessing that’s what 2pi means

[leaden ginkgo]

using a calculator?

[smoky sedge]

I mean like

2

Would I just • by pi twice

[leaden ginkgo]

,w (45/360) * 2 * pi * 11

[warm shale]

Results provided by WolframAlpha

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

Requested by ! saad#3616

[smoky sedge]

Ohh

How do I find arc degrees

How do I find which one is intercepted arc and which one is inscribed arc

[nocturne minnow]

That's the one you need

What is the arc you are looking for

[smoky sedge]

CD in P

Would it be 63?

[nocturne minnow]

No, because you are applying the wrong concept. You need to use the central angle one

So you need to find the measure of the central angle that intercepts arc CD

[smoky sedge]

54?

360 - 126 - 90 - 90

[nocturne minnow]

Yes

[smoky sedge]

Epic

What is x referring to in This

The line, area, arc? and how would I find it

[nocturne minnow]

The chord

[smoky sedge]

Is that the long black line

[nocturne minnow]

If you don't know your circle terms, I suggest reviewing them

[smoky sedge]

Ok

[nocturne minnow]

For this look into congruent chord theorem

[smoky sedge]

I’ll check that out

[obtuse pebble]

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[sage geode]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@dapper abyss Has your question been resolved?

[obtuse pebble]

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[fluid snow]

A similar question has been asked on Math StackExchange: https://math.stackexchange.com/questions/1064165/find-lim-x-to-0-frac-arcsin-xx

[obtuse pebble]

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[warm palm]

for f'(x) would i use first part of the FTC? so i just need to sub x so i would have f'(x) = 2x+1/(sqrt(4+x^4)

[warm canopy]

yes exactly

[obtuse pebble]

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[alpine heath]

How could I rationalize this?

[strange trellis]

just multiply it by (1-2^1/3)/(1-2^1/3) no?

[alpine heath]

doesn't it work just for square roots?

[strange trellis]

no?

i don't think so

maybe, idk

[alpine heath]

It wouldn't cancel the root

it would stay like this

[strange trellis]

wait

could you

multiply it by (2^2/3-1)/(2^2/3-1)

[alpine heath]

2^{2/3-1}?

[strange trellis]

no nvm

[alpine heath]

maybe there's no way to rationalize it

[strange trellis]

there is

what's the question

that makes you need to rationalize it?

[alpine heath]

none

I just wanna do it out of curiosity

[strange trellis]

where did you find it?

like in a book on your course etc

[alpine heath]

I created it

'cause I just saw examples with square roots

so I wondered how could I do it with any nth root

[strange trellis]

let me send you a picture on how to do it

[alpine heath]

Ok

[strange trellis]

just give me a minute to write it out

okay

if you really want the answer then you can continue my expansion

[alpine heath]

Ok

[strange trellis]

,r

,rotate

[strange trellis]

ignore the fact i wrote it on a test paper lmao

[alpine heath]

lol

[marsh geyser]

Use cube sum

[alpine heath]

I makes sanse to use this

but it would have to be like that

or not?

[marsh geyser]

no

like this

so you have 1+2 in the denominator

[alpine heath]

isn't it the same?

[marsh geyser]

you are not razionalitazing

you have this

[alpine heath]

It would happen right after it

[marsh geyser]

1/a+b * (a+b)(a^2-ab+b^2)/(a+b)(a^2-ab+b^2)

but you already have a+b in the denominator

so you don't need to multiply with a+b

1/a+b * (a^2-ab+b^2)/(a^2-ab+b^2)

[alpine heath]

yeah

[marsh geyser]

so you use this one

which is better

[alpine heath]

It was not my solution

[marsh geyser]

to eliminate the roots

[alpine heath]

I put the red marks just to comment on it

But I understood how to it know

though

[marsh geyser]

nice

[alpine heath]

How

1 sec

What about this expression?

[marsh geyser]

is the same, but now you don't use cubic

[alpine heath]

oooh

wait

Could I use (a + b)(a - b) = a^2 - b^2 somehow?

[marsh geyser]

everytime you have a "1" you can do probably whatever you need

cause it's unchanged no matter the power

[alpine heath]

what about now?

I wanna know if there's always a way to rationalize

[marsh geyser]

2 = a
2^(1/4) = b

there is always a solution

but not all of them are casual

[alpine heath]

Nice

Interesting

Thank you

[marsh geyser]

you're welcome

[alpine heath]

!close

.close

[obtuse pebble]

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[obtuse pebble]

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[static mason]

how tf do i solve this

[static mason]

can someone explain what the bracket mean

[daring rock]

It's defining the set W. W contains all polynomials of the form a+bx+cx^2 such that 2a+c > 3

Presumably a,b,c are real numbers

[static mason]

what to do

what did i benefit from 2a+c>3

[daring rock]

Well without that restriction, W would just be equal to P2

P2 is the set of all degree 2 polynomials

Use the definition of a subspace

to check if W is a subspace of P2

Check if W fits the definition

[static mason]

what is the definition of subspace

a vector inside a vector (space)

what now

W= a+bx+cx^2 and p2 = 2a +c >3?

[obtuse pebble]

@static mason Has your question been resolved?

[static mason]

<@&286206848099549185>

[obtuse pebble]

@static mason Has your question been resolved?

[obtuse pebble]

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[north anchor]

someone can explain me this how i can solve this

[tame narwhal]

which part of the solution are you having trouble with

[north anchor]

this

[tame narwhal]

the part in square brackets is the same as x

[north anchor]

how

[tame narwhal]

compute it, you'll see

[north anchor]

how do i calculate it\

[violet sentinel]

(x - 2) + 2 = x - 2 + 2 = x

[north anchor]

where did it come from

[tame narwhal]

what do you mean where? that is how they were able to separate integrals

since they subtracted and added 2, it doesn't change the equation. but it made it easier to compute the integral

[north anchor]

Did you mean it's like dividing by 1? is an equivalent expression

?

[north anchor]

why you cant use :

(x+2)-2

[tame narwhal]

so you can combine exponents, which was done on line 2 of the solution

[north anchor]

Now I noticed where the 3/2 came from?

[tame narwhal]

yes

[north anchor]

i dont understand why is 3/2

[tame narwhal]

they are multiplying $(x - 2)\sqrt{x - 2}$, add exponents

[warm shale]

cwatson

[north anchor]

wow it's my first time playing this integral, so you mean if there is a similar expression I can do that?

[tame narwhal]

possibly

[north anchor]

If I understand correctly, you noticed that the radical inside is similar to a substitution of x. Does this mean that there is a rule for integrals or something similar that requires this situation?

[north anchor]

In integrals using the substitution method, something similar happens if the derivative of the substitution is similar to the one outside, the substitution method can be applied.

I am right?

[tame narwhal]

they didn't use substitution in the typical way, just recognized that there was a straightforward way to "obtain" another (x - 2) to make the integral easier to solve

[north anchor]

If so, the internal function and the external function should be similar?

or multiplication

[tame narwhal]

I'd say it depends on the problem. there might be some "rules", I'm not sure

[north anchor]

oh dnt worry thx for help me

.close

[obtuse pebble]

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[obtuse pebble]

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[distant schooner]

(log_4 (125))/(log_4(25))

[distant schooner]

i used the log law and made it

log_25 (125)

is there a way to simplify this any further

[high lily]

yes

[distant schooner]

how please

[wooden marten]

25 to the what is 125

[high lily]

would you be able to simplify log(125)
and log(25)

[distant schooner]

none

cuz 25^1 is 25 and 25^2 is 625

[distant schooner]

yes

(log_4(5)^3)/(log_4(5)^2)

but i dont know what to do from there

[wooden marten]

those aren't cubed

the exponent comes down

into multiplication

[distant schooner]

i edited it

now it should make sense

[wooden marten]

ye

[distant schooner]

so i bring the power out

i was also thinknig of

bringint the power to the front of the log

like

[wooden marten]

exactly

[distant schooner]

(3log_4(5)/(2log_4(5)

[wooden marten]

exactly

[distant schooner]

is it possible to cancel out the log_4(5)

and get 3/2

or is that not allowed

[wooden marten]

well yes it is a number

[high lily]

no, notation bad

[distant schooner]

oh alr

[wooden marten]

also one more thing I want to point out
25^1.5 = 125

[distant schooner]

i dont know how to do the typing notations

[distant schooner]

i thought the power can only be a whole number

[high lily]

power is with the argument of the log

[wooden marten]

(5^2)^3/2
power multiplication
you get 5^3

[wooden marten]

not at all
in fact taking the square root is ^1/2

[high lily]

log_4(125) = log_4(5^3)
power of 3 is inside and applies to the 5

[distant schooner]

7es

[high lily]

as opposed to outside in the way you wrote it

[distant schooner]

ooh i see

[distant schooner]

oh right theres that as well

would they still accept my answer as log_25 (125)

or does it have to be 3/2

[wooden marten]

it has to be 3/2

simpler

[obtuse pebble]

@distant schooner Has your question been resolved?

[distant schooner]

alright thanks

[obtuse pebble]

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[timid silo]

hello

[timid silo]

how tto represent a parametric equation

?????

and how it works????????

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[wooden marten]

every value of t gives you a point

based on what (x,y) is

then you increase t and draw out the points

[obtuse pebble]

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[tepid talon]

I just don't know what do

[formal vigil]

the arc from F to H and the angle at G are supplementary angles

[tepid talon]

okay

major arc FH or minor arc FH

[formal vigil]

minor arc

[tepid talon]

👍

[formal vigil]

So using that you should be able to figure out the answer

[tepid talon]

so minor arc FH is 15 degrees?

wait

no I'm stupid

[formal vigil]

supplementary not complementary

[tepid talon]

it's 115

[formal vigil]

yes

[tepid talon]

so I subtract 115 from 360?

[formal vigil]

yes, that's the next step

[tepid talon]

okay so now I have 245

and do I divide by 17?

[formal vigil]

right

[tepid talon]

and that's it?

[formal vigil]

I think so but hold on one sec

I want to check something

[tepid talon]

I tried both 14 and 14.4 but it's marking them as wrong

let me try one more

[formal vigil]

how about just 245/17, is that an answer?

[tepid talon]

no, I tried that too

[formal vigil]

haha it's 105 not 115

arithmetic error

105 + 75 = 180

[tepid talon]

💀

I'm so smart

[formal vigil]

nah everyone does that

[tepid talon]

it's 15 💀

tysm

[formal vigil]

ok cool

[tepid talon]

!close

wait wrong thing

.close

[obtuse pebble]

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[misty zinc]

.reopen

[obtuse pebble]

✅

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[timid silo]

okay so how would i solve this

[timid silo]

i have never touched matrices before besides one without any variables

and it didnt = anything so i didnt have to worry about that

i just multiplied which was simple. but i didnt take matrix algebra and i need to know it for a circuit class.

[grizzled shore]

do you know about gaussian elimination

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

no

someone else said rref was the easiest way

i have no idea what any of this means

i have literally never touched matrices besides the most simple thing you can possibly imagine

[grizzled shore]

do you need to solve it

or do you need to learn to solve it

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

i need to learn how to solve it so i can solve ones similar on exams

[grizzled shore]

then you have a lot to catch up on

first by learning gaussian elimination

or rather, first understand what matrices are, what they represent, what they mean and how you can manipulate them

[obtuse pebble]

@timid silo Has your question been resolved?

[wooden marten]

or you can solve the corresponding linear system of equations with your desired method

[obtuse pebble]

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[timid silo]

;rotate

,rotate

[timid silo]

2nd one

<@&286206848099549185>

[prisma hatch]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[prisma hatch]

6n/(n+1)

[wooden marten]

recall simplification of sum of first few square

[wooden marten]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[prisma hatch]

wow that was right!
ig

[wooden marten]

idk if it was right

but still

[prisma hatch]

nice Username

[timid silo]

I tried it like this

[wooden marten]

yes

that's a telescoping sum

1/(n^2 + n) you can split it with partial fraction

[timid silo]

idk how to solve telescopic sum

[timid silo]

idk this either

[wooden marten]

1/(n)(n+1) = 1/n - 1/(n+1)
you can verify with desmos or algebra

then you can write out first few terms
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4

[timid silo]

partial fractions ok

nice I get it

[wooden marten]

alr

[timid silo]

so should watch lecture on telescopic sum and then try to solve this sheet

[wooden marten]

it's just this

[timid silo]

nvm theres only one problem based on telescopic sum in the sheet

[timid silo]

ive learned this before in junior classes

to solve series which symmetrically simply

so it's just that ?

[wooden marten]

yes
all terms go away except first and last

[timid silo]

yeah I have solved many like that

nice

[timid silo]

how do we split such expressions with partial fracrions

fractions

[wooden marten]

1/(n)(n+1) we will assume is equal to A/n + B/(n+1)
then cross multiply and you get A(n + 1) + B(n) = 1
solve for A and B

[timid silo]

yeah but there can be some ugly looking fractions

[wooden marten]

there are some ugly cases everywhere

[timid silo]

okay I just made one myself and tried

It's pretty easy if they get factored

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[sinful kraken]

i know you might think i am stupid but no matter what i do i just dont get the right result

[obtuse pebble]

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

[sinful kraken]

i am stuck at this (x+3)^2 - (x+2)^2

[high lily]

what's the full question

[sinful kraken]

i am supposed to calculate them and write the solution down

sorry for my english

[dark stirrup]

!show

[obtuse pebble]

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

[high lily]

did you attempt the problem

assuming they're asking you to ||(apply a factorisation identity to)|| simplify

[sinful kraken]

i did that

[leaden ginkgo]

show your working!

[sinful kraken]

one time i tried this one but it was wrong

and one time i tried this one

something has to be wrong with 4

[leaden ginkgo]

,rotate

[leaden ginkgo]

,rotate

[sinful kraken]

this one is right i think

[sinful kraken]

but the 4 has to get subtracted

[high lily]

wheres (x-2)(x-2) coming from

[sinful kraken]

i thought maybe because of the - in front of it but it didnt made any sense to me too

oh

i understand what i did wrong

[high lily]

don't do stuff with no mathematical justification

and also subtract the entire expression,
don't only apply - to whatever term happens to appear first

[wooden marten]

needs parentheses aroudn the (x-2) stuff

[sinful kraken]

i have to calculate the second thing and after that i have to do the opposite signs

[high lily]

should've had (x+2)(x+2)

[leaden ginkgo]

its $-(x^{2} + 4x + 4)$
not $(x-2)^{2}$

[warm shale]

! saad

[high lily]

also supposedly they want you to simplify by applying the factorisation identity for the difference of two squares

[sinful kraken]

,rotate

[high lily]

looks ok

[wooden marten]

gonna factorize 2x + 5 as (sqrt(2x) + isqrt(5))(sqrt(2x) - isqrt(5))

[obtuse pebble]

@sinful kraken Has your question been resolved?

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[sage dagger]

For part B, would I use the general equation of a semicircle?

[royal basin]

why a semicircle when you're asked for a circle

[sage dagger]

oh

i forgor it asked a circle

i would use the standard form right

did i plug in the values into the correct places?

or was I supposed to plug (-2,1) into h and k

[royal basin]

no y= at the start.

[sage dagger]

oh

[royal basin]

you were supposed to plug the coordinates of the center into the letters representing the coordinates of the center, yes. in your case, those would be h and k, yes.

[sage dagger]

guess i have to expand now

.close

[obtuse pebble]

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[quiet flume]

differentiation of x^x

[warm shale]

Pro_Hecker

[haughty coyote]

x^x = e^(x ln x)

[quiet flume]

.close

[obtuse pebble]

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[daring fiber]

Hello

[daring fiber]

Can anyone help me with number 1 here?

Like what do i have to do?

[royal basin]

do you know what the notation f(S) means?

[daring fiber]

Because it’s says real numbers so I can’t just replace with x every value from 1 to 3 right?

[daring fiber]

Basically i replace them inside the function

[royal basin]

f(S) = {f(x) : x ∈ S}

you might find it helpful to do some graphing here actually

graph the function floor(3x) - 1, highlight the part of it that lies over the interval [1,3] on the x-axis, and note the set of y-values covered by that part.

[daring fiber]

To graph it i just take random x values from the same interval?

And then look at y?

I haven’t graphed floor functions inside an equation before

[royal basin]

you do not take anything "random", no.

do you know what the graph of floor(x) looks like?

[daring fiber]

right side empty and discontinues right

[royal basin]

vague.

[daring fiber]

Got the graph

Now i just say f(s) lies from interval [2,7]?

[obtuse pebble]

@daring fiber Has your question been resolved?

[obtuse pebble]

@daring fiber Has your question been resolved?

[obtuse pebble]

@daring fiber Has your question been resolved?

[obtuse pebble]

@daring fiber Has your question been resolved?

[balmy stump]

it is kinda vague

but assuming its just asking for a range

[balmy stump]

2 cannot be inclusive here

it should be (2,7]

[daring fiber]

Got it thanks a lot

Can someone help me with these two?

I have to find sum

[balmy stump]

have you studied geometric progression?

@daring fiber

[daring fiber]

Yeah

[balmy stump]

yea so both are geometric progression with infinite terms

for infinitely long GP that the formula for the sum becomes a/(1-r) where a is the first term r is the common ratio and r must be less than 1

[daring fiber]

Which is the r in the first case?

Is r=1/2?

[balmy stump]

i would say you write out the first few terms of the series and try to find out

[balmy stump]

yes

[daring fiber]

Okay got it same formula for the second one?

R here must be (-1/2) right

[balmy stump]

yea but be careful about a

dont mess up the sign in that

[daring fiber]

Will a still be 1/2?

[balmy stump]

no

in first term

-1 to the power 2 becoms 1

so the first term is actually 1/4

[daring fiber]

Oh got it so i start from j=2

[balmy stump]

yea

[daring fiber]

So starting from a different point makes no other difference in the formula except a right?

[balmy stump]

yes

but if you start from say the 7th term

then your final answer would be missing the first 6 terms being added to it

[daring fiber]

Oh so i make another summation with 0 to 6?

[balmy stump]

why 0

[daring fiber]

Thought 0 would be the first team but actually is 1

[balmy stump]

thats why always write first few terms when it is a summation using the sigma notation

that way you can find some patterns as well

[daring fiber]

Got it,here second one i need to add the first term after using formula too right?

[balmy stump]

no

the first term is just when j = 2

so if you start from there then no need to add anything

[daring fiber]

What difference would there be starting from 7th for example and 2nd like in this case?

Because you said at 7 i have to add the other 6 terms right

[balmy stump]

first of all in the second we are not starting from 2nd term

what j doesnt always tell us the position of the term

j is just a variable whose value starts as 2 and is being incremented by 1

do not confuse the rank of the term with the index of the summation

[daring fiber]

Got it

[balmy stump]

the index of the summation can start at any value but the value given at bottom is the first term

say j were 11

in the expression

instead of j =2

then the first term would have been the expression -1^j * 1/2^j with j = 11

[daring fiber]

Okay so first term always depends on value of j

[balmy stump]

every term depends on j if j is in the expression

think of it like this

i have a series 4, 5, 6, 7, 8...

4 is not the fourth term here

its the first term

because it is the first in the series

[daring fiber]

Okay i got it,thanks so much

🙏

[balmy stump]

no prob

if you are done when close the requet
using .close

[daring fiber]

Okay

.close

[obtuse pebble]

Channel closed

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[reef egret]

I need help

[obtuse pebble]

Please don't occupy multiple help channels.

[reef egret]

I need help

<@&286206848099549185>

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.