help-23

[broken yew]

its 25 times 9

[broken yew]

understand why?

[solar cradle]

Uhhh

Because the exponent applies to each number?

[broken yew]

yes it distributes

or if we write it out

,,(5\times3)^2=(5\times3)\times(5\times3) \= 5\times3\times5\times3 = 5\times5\times3\times3 = 5^2\times 3^2

[solar cradle]

Yeah I get that but then what exactly is 5x?? Like I thought there’s something in the x

[dense prism]

x can be any number

[broken yew]

5x means 5 times x

where x is any number yes

[solar cradle]

Umm yeah so we don’t know or

[broken yew]

when working with algebra, we no longer use the multiplication symbol x

by the way

we write things next to each other to indicate multiplication

[solar cradle]

That’s what I thought brackets are for

[broken yew]

its not we dont know, its saying it can be anything

brackets work too.

For example:

,,2(x + 1) = 2x + 2

[broken yew]

You understand what this is saying?

The point of working with x, an unknown

[solar cradle]

Yes

[broken yew]

is this equality applies for any x

we choose

[solar cradle]

Okay but see how the 2 multiplied into the x

[broken yew]

if we plug in x = 5, it works.

yes.

[solar cradle]

When u I have (5x) and then it has (5x)^2 I thought by multiply the 2 onto the 5 I’m already adding to the x

[broken yew]

no thats not what that notation means

^2 means squared

or ^n means exponentiation by n

which is multiplying the thing by itself n times

,,(8x)^3 = (8x)(8x)(8x)

[broken yew]

there are three 8's and three x's multiplied together.

If we change the order that gets us 8^3 times x^3

[solar cradle]

Hmmmmm so I’m multiplying 2 different things???

[broken yew]

,,(8x)^3 = (8x)(8x)(8x)= 8^3x^3

[solar cradle]

I’m multiplying the x and the number next to it

[broken yew]

well yes 8x means 8 times x

[broken yew]

do you understand this working fully?

[solar cradle]

But 8 times 3 x’S?

I think so

[broken yew]

???

the 3 is exponentiation

not times

8 times x all to the power of 3

[solar cradle]

Is that 3 the same as the one outside the bracket

[broken yew]

i am talking about that 3 yes

[solar cradle]

Exponention is the same@to the power of yeah?

[broken yew]

yes

By the way, I really think you should try having someone in real life - probably your teacher go through this with you

rather than us teaching you the entire thing from scratch

[solar cradle]

Sorry

[broken yew]

it's ok

[solar cradle]

I’m not good at learning

[broken yew]

its just harder on here

and im finding it hard to make sure youre definitely following everything im saying

[solar cradle]

I think I do so far

[broken yew]

But yeah this is what you werent doing before basically.

[solar cradle]

Okay so if I applied it to the xs then it’s okay so far what I did right?

[broken yew]

the last bracket

-3 to the power of -3 is not -27

-3 to the power of 3 is -27.

[solar cradle]

But I thought the were 3

[dense prism]

-3≠3

[solar cradle]

-3*-3 +9*-3 -27

[broken yew]

huh........

do you know what negative exponentiation entails

I dont think you do.

[solar cradle]

I’m multiplying-3 to the power of 3 . 3 times

[broken yew]

-3 times.

[solar cradle]

Like -3 times -3

Times -3

[broken yew]

,,(-3)^3 = -27

[broken yew]

youre not being asked this

youre being asked

[dense prism]

$a^x≠x\times a$

[flat frigate]

radiation1

[broken yew]

,,(-3)^{-3}

[broken yew]

Yeah I feel like theres all sorts of confusion going on here and it really would be best to clear it up with your teacher

[solar cradle]

Ummmm

[broken yew]

You need to review the topic of exponentiation in algebra

[solar cradle]

I thought it was an exponent

[broken yew]

A youtube video might also help if anyones got a recc

[solar cradle]

The 3 outside the bracket

[dense prism]

-3 is the exponent

[broken yew]

its -3 not 3

[dense prism]

Not 3

[solar cradle]

Yeah that’s what I mean

[broken yew]

in here

[broken yew]

do you know what this is

[solar cradle]

-2

[broken yew]

no

its 1/2.

[dense prism]

.

[solar cradle]

How

[broken yew]

well im not sure if thats the issue

just to check

what did you think

[broken yew]

this is

[solar cradle]

It’s 3 multiplied by -3

[dense prism]

-9 or -6?

[broken yew]

its not

thats not how negative exponentiation is defined

unfortunately

[solar cradle]

-9

[broken yew]

this is how it is defined

[solar cradle]

Hmmm

[dense prism]

Yeah you should really learn about exponents first

And then attempt problems

[solar cradle]

So that’s why I flip some of those negative upside down to make them positive?

[broken yew]

You should really ask your teacher or maybe someone here can recommend a video

[solar cradle]

Wait so your saying

I can’t even multiply -3x)^-3 because it’s negative

I have to put it under a table?

[dense prism]

No offense but do you actually call that a table? Or are you unaware of what that sign means

[solar cradle]

It’s a fraction

[dense prism]

Yes

[solar cradle]

I’m just trying to be like you guys.

(-3)(-3)(-3)

-3*-3 is positive 9*-3 is -27

Sorry everybody. Im just no good

.close

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[safe radish]

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[little cipher]

how to solve x?

[desert pasture]

!status

[safe radish]

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

[little cipher]

3

[pseudo scroll]

Cosine rule

[little cipher]

o

$a^2=b^2+c^2-2bc*cosA $or something?

[pseudo scroll]

yes

[little cipher]

a

[little cipher]

so trig ratios only work on right triangle?

[pseudo scroll]

yeah the opposite/hypotenuse stuff only works with right angled triangle

But the functions usually transcend right angled triangles and are used everywhere

You only need cos(26°)

[little cipher]

was side with length 15, a?

[pseudo scroll]

15 or 19, doesn't matter

[little cipher]

oh ok

[little cipher]

$15^2=19^2+x^2-38xcos26$

[flat frigate]

紅卫兵

[little cipher]

$15^2-19^2+38xcos26=x^2$

[flat frigate]

紅卫兵

[little cipher]

$-136+38xcos26=x^2$

[flat frigate]

紅卫兵

[little cipher]

$\sqrt{38xcos26-136}=x$

[flat frigate]

紅卫兵

[little cipher]

$$\sqrt{38x\frac{9}{10}-136}=x$$

[flat frigate]

紅卫兵

[little cipher]

$38x\frac{9}{10}-136=x^2$

[flat frigate]

紅卫兵

[little cipher]

$38x*9-1360=10x^2$

[flat frigate]

紅卫兵

[rustic goblet]

try starting with $x^2=19^2+15^2-570cos26$

[flat frigate]

ING

[rustic goblet]

cosine law works for all three sides

but in this case

you only have the angle opposite to x

[little cipher]

why is $2bc=570?$

[flat frigate]

紅卫兵

[little cipher]

wait

where did i get 38

[rustic goblet]

you did 2 * 19 *x

[little cipher]

is that even correct

[rustic goblet]

cosine law is c^2 = a^2 + b^2 - 2ab*cosC

angle C in this case is 26 degrees

since its opposite to the side we dont know

[little cipher]

ah

[rustic goblet]

like this image

the angles are denoted with capitals and are always opposite to the sides, which are denoted with lowercase letters

so in our case, the angle 26 is C

and the unknown side is c

[little cipher]

so a can be c?

[rustic goblet]

it could, yes

you can keep it as x though

c is just the general way of writing it

[little cipher]

okay

around 8.6?

thx

.close

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[frigid spruce]

hello hello

[lean otter]

What's your question ?

[frigid spruce]

its not like a ques

like

i want to ask

[lean otter]

Umhmm

[frigid spruce]

or can someone provide me with

properties of logs

[lean otter]

Just google it

[frigid spruce]

there are much complicated in google

[rustic goblet]

like this?

[lean otter]

Oh right then wait

[frigid spruce]

yes

but there are a lot more

[lean otter]

All I have ever learnt

And in the way I did

[frigid spruce]

this is sufficient right?

[lean otter]

Like tbh take these and write it yourself somewhere

[lean otter]

Till 12th yeah ig

You might also need a few other things like Base changing theorem and all but that's for your teachers to teach

[lean otter]

I assure you 80% questions with the right use of all these

[frigid spruce]

there was a chapter like functions and realtions

in which

[lean otter]

Other 20% need different topics combined

[lean otter]

11th?

[frigid spruce]

many functions like, mod, gif, log, where given

[lean otter]

Yeah go on

[frigid spruce]

yes

so my teacher

like according to our book

it was given about the

thier domain, range and graph

[lean otter]

Yeah

[frigid spruce]

and some basic properties

[lean otter]

Umhmm

[frigid spruce]

but when i reached

quadratic eqs

there were a lot of ques

in which other properties were getting used

thats why i was aking

askinhU8

[lean otter]

These are sufficient then

For 11th grade

I'm damn sure

And one more thing

Don't ever confuse log² x with log x²

And always put a Mod when you take root x²

[frigid spruce]

like both positive and negative values?

[lean otter]

Yeahh

[frigid spruce]

okay

[lean otter]

And I assure you log questions will be a 100% done for you

[frigid spruce]

but what is the, base property

[lean otter]

Lmao don't go into it rn if you're in 11th, you'll study in 12th

[frigid spruce]

so, will there be any chapter only dedicated to logs?

[lean otter]

Idk about that

[frigid spruce]

okay

[lean otter]

But yeah

[frigid spruce]

okay

[lean otter]

You talking about base changing theorem right

[frigid spruce]

yeah ig

and the properties

[lean otter]

Yeah you'll learn next year

[lean otter]

The ones which I sent, sufficient

[frigid spruce]

okay

thnks sir

.close

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[waxen moon]

Hi, I got sin2x(sinx+cosx)
Is it correct?

[lean otter]

I believe it to be correct

I mean yeah it is

Oh wait no

Nvm idk

[rustic goblet]

yes

[waxen moon]

Alr, I have another question, is the following correct?

[rustic goblet]

looks good to me

[waxen moon]

Is the following question correct:

[fast musk]

how did sinx-1 turn into -cosx?

[waxen moon]

-cosx = sinx- 1

[fast musk]

that's not correct

[waxen moon]

From sinx + cosx = 1

Oh

[fast musk]

no, sin^2x+cos^2x = 1

the 'squared' is important

try replacing the 'cos^2x' in your equation instead

[waxen moon]

I get sinx = 0, sinx = 1

[fast musk]

I think that's also wrong?

you have cos^2x = 1-sin^2x

so if you replace that you get 2(1-sin^2x)+sinx-1=0

then solve

[waxen moon]

I'm not getting 1 - sin^2'

Sin^2x

On the LS

[fast musk]

cos^2x+sin^2x=1. subtract sin^2x from both sides

[waxen moon]

sin^2(x) is not on both sides tho

[fast musk]

I don't know what you mean

if you have x=y, then x-sin^2x = y-sin^2x

[fast musk]

re-read what I wrote

"try replacing the 'cos^2x' in your equation instead"

[waxen moon]

I did

It became 1 - sin^2(x)

[fast musk]

using cos^2x = 1-sin^2x

Oh

idk why you moved sinx to the RHS

[waxen moon]

Alr

[fast musk]

also you didn't distribute the '2' to the sin^2x correctly

[waxen moon]

I got sinx = -1/2

Sinx = 1

[fast musk]

yeah

[waxen moon]

What would sinx = -1/2 be?

For an exact pi value

[fast musk]

Do you know sin/cos for the basic angles? 30,45,60?

[waxen moon]

Yeah

Sin30 is 1/2

[fast musk]

well draw the circle and the line y=-1/2

[waxen moon]

So sin -pi/6

[fast musk]

yes, what else?

[waxen moon]

-5pi/6

[fast musk]

Yeah

[waxen moon]

That's it

For that rante

[fast musk]

Yep

[waxen moon]

Range

So then

The answer is:
X = pi /2, -pi/2, -pi/6, -5pi/6?

[fast musk]

-pi/2?

what's sin(-pi/2)?

[waxen moon]

Oh nvm

Just pi/2

So excluding that, those are all the values?

[fast musk]

yep

[waxen moon]

Alr, thx for the help

I have another question

I'm having trouble with this

For the left side

[fast musk]

expand, then simplify

let me know what you get

[waxen moon]

Is that correct?

[fast musk]

no

3cos^2x+sin^2x does not = 3

you could expand it to cos^2x + cos^2x + cos^2x + sin^2x and then replace cos^2x + sin^2x =1 to get: cos^2x + cos^2x + 1

or 2cos^2x+1

[waxen moon]

I'm not getting the final answer from there tho

[safe radish]

@waxen moon Has your question been resolved?

[safe radish]

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[edgy latch]

hey yall can someoe check if my answer is right for this problem

[lean otter]

What problem?

[edgy latch]

it says

determine the common ratio. then find a geometric growth model:

64, 48, 36, 27, 20.25

my growth model is y=64(4/3)^x

is that right

[mossy fractal]

do you mean you want a formula to find the nth term?

[edgy latch]

yes i think

[lean otter]

This is incorrect

[mossy fractal]

then that would be correct if x was n-1

actually

no

i read wrong

the 4/3 is wrong aswell

[edgy latch]

oh boy

here wait is it ok if i keep this channel open and search how to do it on yt

i dont wanna waste ur guys's time lol

[mossy fractal]

what is the common ratio you have?

[edgy latch]

or should i close it and come back to it

4/3

[mossy fractal]

how do you get from 64 to 48, then from 48 to 36 and so on

what do you multiply by

[edgy latch]

4/3

or 3/4

[mossy fractal]

thats greater than 1

[mossy fractal]

yeah

if you multiply 64 by 3/4 you get 48

so the common ratio is 3/4

[edgy latch]

i see

how do i write a growth model

is there a formula i can use or somethinf

[lean otter]

$$P(X) = P(0)e^{kt}$$
$$P(x)=\frac{256}{3}e^{\ln(\frac{3}{4})x}$$

[mossy fractal]

this is the formula for a geometric growth model

where a is the first term

[edgy latch]

where did the negative one come from

[mossy fractal]

r is the common ratio

think of it, if you want the first term

[edgy latch]

is a the first number then?

[mossy fractal]

that is a

yeah

[edgy latch]

ok

so my thing would be y=64(3/4)^-1

[mossy fractal]

eg if you want the 2nd term which is 48, you want to do 64(3/4)^2-1

[mossy fractal]

^n-1

[edgy latch]

yeah

i just dont get where the negative one comes from

is there any way u can explain it v simply lol im not rlly familiar w all the math terminology

[mossy fractal]

try to use that formula to find the next terms for that sequence

the first term being 64

[edgy latch]

do i replace the 64 with the terms that come after

or do i just put the number of the terms order for n

sorry if this is confusing

[mossy fractal]

no keep the 64

since its a common ratio you can use the first term

[edgy latch]

i just dont understand the ^n-1

oh omg nevermind

i get it

i see it now

tysm this was v helpful

[mossy fractal]

np

[edgy latch]

.close

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[flat frigate]

Brandon H

[slow lily]

I need help

[slow lily]

I have no idea how to solve this, there is not enough information

the question is about parabolas

I made an assumption that the x intercepts are 0 and 50, and the axis of symmetry is 25 but other than that I don't know what to do next

[safe radish]

@slow lily Has your question been resolved?

[safe radish]

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[cold aurora]

It clearly has insufficient information.

We need to know the leading coefficient of parabola at least to find the height.

[safe radish]

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[split fulcrum]

I’m trying to put everything into one fraction, but is what I have correct so far?

[safe radish]

@split fulcrum Has your question been resolved?

[safe radish]

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[hearty fractal]

im confused on this

[safe radish]

@hearty fractal Has your question been resolved?

[hearty fractal]

.close

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[lean otter]

Why is the second question No.?

[lean otter]

is it becasue two points have the same minimum?

1 & 3

[sharp tundra]

not quite. the point near 3 is lower than the one near 1

the problem comes from it being an open dot

since that endpoint is not included, we could get closer and closer to that open circle (ie, approach x=3 from the left) but never hit it, and no matter where we stopped, we could have actually gotten a little closer, meaning no value is actually a minimum

"that endpoint" meaning the endpoint of the middle line segment btw

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

i see

[safe radish]

@lean otter Has your question been resolved?

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[forest plaza]

The Eurovision Song Contest attracted performers from 100 countries, one from each country. Each performer sings a song of three verses and leaves immediately. The organizers, after reading the lyrics, find that each verse is offensive to exactly to exactly one participating country. Prove that it is possible to schedule performances so that each performer hears no more than 3 insults.

[forest plaza]

.close

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[dreamy verge]

supposed to simplify this

[dreamy verge]

wait one sec

.close

[safe radish]

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[dreamy verge]

.reopen

[safe radish]

✅

[dreamy verge]

supposed to simplify this

these are the possible correct options

not sure what to do after that step

[calm bridge]

in that last step how did you get 10^-21?

[dreamy verge]

ah its -20, you're right

anyways im still stuck

[calm bridge]

what will be 20 - 1/9

[dreamy verge]

179/9

[calm bridge]

correct

[dreamy verge]

take 1/9 common out of bracket?

[calm bridge]

yep

and you are done

[dreamy verge]

yep got it

thanks!

[frigid spruce]

old guy

[dreamy verge]

.close

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[cold aurora]

Do you know what being congruent means?

Yes.

[lusty folio]

Two line segments are said to be congruent to each other if they have equal lengths. They may or may not align at the same angle with an axis or position in the plane.

[cold aurora]

So what makes you think they are same ?

[safe radish]

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[lusty folio]

If KL = 30, LM = 18, MN = 13, then NO = (Blank 1). (Round your answer to one decimal place, if necessary.)

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[lean otter]

What is the best source to study straight lines (in general, coordinate geometry) from?

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.close

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[tight meteor]

I am learning about the taylor series, but I am struggling to calculate zeta. Wikipedia states that zeta must be between a and x, but I don't know how to do that. As you can see in the screenshot of desmos, I get a zeta value of roughly 1.585 when trying to approximate sin(0.1) from 0 and with n = 5. This would mean that zeta would have to be between 0 and 0.1, but I don't know how to get that. The value I get for zeta does seem to work as shown by the third last calculation in desmos.

[harsh cargo]

Help me pls

Hw

I don't know

[tight meteor]

@harsh cargo This is my channel. You can create a new help channel in #help-2

[primal granite]

This channel is occupied create your own. See #❓how-to-get-help for directions

[tight meteor]

(Just reposting so people don't get confused)
I am learning about the taylor series, but I am struggling to calculate zeta. Wikipedia states that zeta must be between a and x, but I don't know how to do that. As you can see in the screenshot of desmos, I get a zeta value of roughly 1.585 when trying to approximate sin(0.1) from 0 and with n = 5. This would mean that zeta would have to be between 0 and 0.1, but I don't know how to get that. The value I get for zeta does seem to work as shown by the third last calculation in desmos.

[safe radish]

@tight meteor Has your question been resolved?

[tight meteor]

<@&286206848099549185>

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@tight meteor Has your question been resolved?

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@tight meteor Has your question been resolved?

[plucky elk]

doesn't look like your function is 1-1 so your arccos can give one, but not the same one from lagrange's formula

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[lean otter]

Find the critial points(zeros), test the ranges

[mossy fractal]

so then by looking at that graph, when is the graph below the horizontal line

[lean otter]

what?

[mossy fractal]

yes

that is the answer

[lean otter]

That is not the answer

you have to take into account the other parabola as well

[mossy fractal]

oh i thought they were 2 seperate parts

https://tenor.com/view/clash-royale-gif-5535732

me when i spread misinformation

just do the exact same thing for the other equation and find the parts where domain will overlap

ie find where it equals 0 and then from there see where it is below the horizontal line

[wild ridge]

Best emote

[mossy fractal]

you have 3 diagrams you drew on the right hand side

you want the top one

so if the values at the horizontal line are 1 and 4, when is the curve going to be below the line?

but its not both less than 4 and less than 1

yeah so when x is more than 4, the curve is below the line

but its not less than 4

when x is between 1 and 4, the curve is clearly greater than 0

yeah

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[quasi bison]

ab < 15 does not imply any combination of "a<15" and "b<15".

that's what's wrong

i mean even for equalities it doesn't work

from x(x+2)=15 you would NOT be able to deduce that x=15 or x+2=15

"cross method"?

no, i am not american.

this appears to be some kind of shortcut or visual for factorization

you've correctly reduced the inequality to x^2 + 2x - 15 < 0

this factors as (x+5)(x-3) < 0

i personally do not see much use in this cross visual. it seems to require you to already know what numbers to put.

dunno i just used vieta's formulas i guess

and/or the related but slightly different in terms of sign

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

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[twilit torrent]

@hard crest could u clarify what you meant by this?

you said something but i was afk for a bit so i couldn't get back to that channel

for inequalities with 2 variables, we're just dealing with graphing mostly right?

i guess random sketches should work?

[twilit torrent]

okay fair enough

what would my specification mean though

are they assuming the same thing?

simultaneous inequalities aka solving by graping?

[hard crest]

i think so?? but also things like uh

,, \begin{align*}
3x + 4 < 17 \
7x - 3 \geq -1
\end{align*}

[flat frigate]

Hayley

[hard crest]

where it's just one variable and you have to combine them together

[twilit torrent]

oh yeah the number line thing

[hard crest]

yeah

[twilit torrent]

okay fair

i know both of these

i didn't know the algebra way to solve

the multi variable inequality though

so i was just checking in case

[hard crest]

there isn't really one

[twilit torrent]

o

[hard crest]

most of it is finding the intersection point

[twilit torrent]

okay fair enough, thanks

.close

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[safe radish]

@wise sky Has your question been resolved?

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[cold edge]

how can $O((\log_2 n)^7) \subset O(0.1n)$ be true?

[flat frigate]

severed_toast

[cold edge]

im supposed to order all these functions in an increasing order with respect to the big O notation

[potent osprey]

See https://math.stackexchange.com/questions/2857391/how-can-i-prove-that-logkn-on-epsilon for example

[cold edge]

i dont think im understanding the question properly

[cold edge]

i tried to order them by growth rate but what is this answer key doing?

i.e. i tried $\lim_{n \rightarrow \inf} \frac{g(n)}{f(n)}=0$ for each

[flat frigate]

severed_toast

[potent osprey]

Yes that's a fine definition

It holds for all of them

[cold edge]

im confused

[cold edge]

plotting these gives

which means 0.1n grows much slower than (log_2 n )^7

[potent osprey]

Initially yes

But the behaviour as n approaches infinity is wanted in big O

,wolf limit as n approaches infinity of (log2(n)^7)/(0.1*n)

[flat frigate]

Results provided by WolframAlpha

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Requested by _daili

[cold edge]

alright ig

even with n = 10000 or something large 0.1n is still much smaller]

[potent osprey]

I mean yeah that seventh power is putting in some work

,wolf log2(n)^7 < 0.1*n

[flat frigate]

Results provided by WolframAlpha

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

Requested by _daili

[cold edge]

didnt you mean to evaluate it the other way?

[potent osprey]

alternatively, you could for example also look at 10^5 n instead of 0.1n

[cold edge]

ah wait nvm

okay so n needs to be 1.87 * 10^12

makes sense

[potent osprey]

yeah if you take C to be 1 in the definition

but as i said you could also use C = 10^6

and then it will be a lot earlier

where the linear term overtakes the logarithmic power

[cold edge]

what do you mean by C?

[potent osprey]

what is the definition of big O you have

[cold edge]

oh you mean in this definition

[potent osprey]

thats little o?

big O, that 'for all c' is a 'there exists a c'

[cold edge]

right my bad

so instead of solving that limit

[potent osprey]

so for example choose C = 10^6 then you have that log2(n)^7 <= C * 0.1n for very small n

that limit one is an easy way to do this stuff

the definition with C is harder to proof the stuff with

since you need to show that the inequality holds for ALL n bigger than a n0

[cold edge]

and n0 would be 1.87 * 10^12 right

[potent osprey]

if you take c=1 yes

but you can't just rely on wolfram for this stuff

if you want a proof the limit definition is the way to go for most cases

[cold edge]

okay great

thanks for clearing all that up for me

appreciate it

.close

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[lean otter]

can I complete the squares here

[pseudo scroll]

what's your goal?

[sly field]

yes

[lean otter]

to graph the circle

[pseudo scroll]

yes then complete the square

[lean otter]

I'm guessing it's 0.25?

[autumn owl]

you definitely can and that's how you usually bring the circle equation to it's proper form

[sly field]

👍👍

[lean otter]

is this correct then?

[sly field]

yes

[lean otter]

alright tyyy

.close

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[quartz parrot]

10 mL of sulfuric acid is pippetted out (with an error of 0.02 mL) and diluted to 50 mL using a volumetric flask (with an error of 0.06 mL). The volume/volume concentration of the resulting solution is 20% ±

the answer I got is 0.06 but it is showing 0.05

could anyone pleasee correct me

[safe radish]

Please read #❓how-to-get-help

[safe radish]

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[sudden shoal]

How do I do part b?

Also can someone check my answer to a, i think ive gone wrong somewhere

[safe radish]

@sudden shoal Has your question been resolved?

[sudden shoal]

<@&286206848099549185>

[sudden shoal]

can you help me w this

[fresh nebula]

<@&286206848099549185>

[safe radish]

@sudden shoal Has your question been resolved?

[sudden shoal]

Can someone pls help lol

[safe radish]

@sudden shoal Has your question been resolved?

[sudden shoal]

sadly not : (

[cold aurora]

This is correct.

[cold aurora]

This as well.

[sudden shoal]

tysm

[sudden shoal]

how do I solve part (b)

<@&286206848099549185>

[sudden shoal]

I GOT IT!

.close

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[verbal vault]

So im told at least one of these is wrong by my prof but he didnt give me any clues

Although Im sure K4 and K5 are right

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@verbal vault Has your question been resolved?

[verbal vault]

<@&286206848099549185>

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@verbal vault Has your question been resolved?

[safe radish]

@verbal vault Has your question been resolved?

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[languid rover]

I'm taking SAT Test Prep on Khan, and came across this q. After using the hint, I got this answer, but I still haven't a clue how to decipher the question. I have taken a college algebra class from WVU, but that was several months ago, so I'm kinda out of the math loop... 😅

[peak siren]

The biggest value of R(q) would be at the vertex. This is easy to find as the question is in vertex form. Just (h,k)

[languid rover]

The context of the question confuses, me I guess... I'm interpreting it as: "What's the most this company makes in a given month, based on q sales?"

But I kinda forgot about the vertex form in general,,,

[peak siren]

Yep, q is the amount of sales and R(q) is the revenue

[languid rover]

So for what reason would the revenue drop to 0 in the graph?

I don't understand how more sales results in less revenue

[peak siren]

In the real world, I’m not sure

[languid rover]

So in the case that I run into this again, just use math logic instead of question logic>

?

(so I don't confuse myself again)

[peak siren]

So if it is vertex form or y = a(x - h)^2 + k

We get the vertex at (h,k)

[languid rover]

And a is

[peak siren]

In this problem it is -.31

[languid rover]

Well generally in the context of a vertex equation, what does a represent?

[peak siren]

it is a vertical stretch

[languid rover]

Sweet

[peak siren]

And if negative it is a reflection over the x axis

[languid rover]

Thanks! I think I'm gonna go dig out my pre cal notebook

.close

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[spark ocean]

Can someone

[spark ocean]

Help with a physics doubt

[calm bridge]

,rotate

[spark ocean]

I can't approach it at all

I have tried [e-mail]

[calm bridge]

v = dx/dt

where x is displacement

[spark ocean]

Yes

[calm bridge]

so vdt = dx

[spark ocean]

Lemme do side by side

Yes

[calm bridge]

integrate with suitable limits according to the intervals

[spark ocean]

How to?

[calm bridge]

you dont know how to integrate?

[spark ocean]

Can u help me solve by doing it on a notebook I swear I have been trying from an hr and i can't proceed

[spark ocean]

We have been taught kinematics yesterday itself

Along with basic maths

So I m not much good

[calm bridge]

so you know how to integrate?

[spark ocean]

Yes but not with variables

[calm bridge]

i dont have a camera rn, sorry

[spark ocean]

It's tough for me

[spark ocean]

Umm can u help me in this chat itself then

What limits to put I can't figure it out

And what ae those 3 relations

[calm bridge]

u havent done a problem where u are given a function of velocity and you have to find displacement?

[spark ocean]

Nope

There are 3 functios in this

I have done really basic problem

[calm bridge]

you need to first practice some basic problems of this type

jumping straight on this wont help much

[spark ocean]

But this is a HW question and I need to do it cos tomm we have to submit

[frigid spruce]

lol, just say you dont know how to solve this, to the teacher

[spark ocean]

I can't understand doubts from this teacher

Even if I tell him

I won't understand his solution

That's why I started using discord again

[frigid spruce]

old guy should help you

[spark ocean]

Who's that

[frigid spruce]

lol

dissrupt

i also know how to solve these ques

[spark ocean]

Ah

[frigid spruce]

but i cant

understand this ques's language

[spark ocean]

@calm bridge can u help

[spark ocean]

Same

[calm bridge]

sorry, im not much of a teaching type but a helping one

[spark ocean]

Its fine

Imma w8 for someone else

[frigid spruce]

wait

give me some time

let me give it a try

[spark ocean]

Kk thnx

Allt

Alot

[safe radish]

@spark ocean Has your question been resolved?

[frigid spruce]

lmao

wtf

is this ques

ping helpers, lol

[spark ocean]

I solved it now

Lol

[frigid spruce]

yo

how

tell me also

i also have kinematics test

on sunday

just tell me the basic idea

ill try later fir myseld

f

[safe radish]

@spark ocean Has your question been resolved?

[safe radish]

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[round cypress]

plz help

[safe radish]

@round cypress Has your question been resolved?

[marsh walrus]

do you have an example of what exactly theyre expecting?

their suggestion is pretty vague

@round cypress

[round cypress]

i think they just expect T at the end

like a truth value

or no

A

[final halo]

write out the what the left set is in set builder notation

[round cypress]

{x,y: xEA n yEB}

[final halo]

that is not the left hand set in set buider notation

[round cypress]

the left u said right

that is the left

AnB = {x,y: xEA n yEB}

[final halo]

sorry i mean the whole set on the left of the equals sign

also not sure why you have both x and y in your set builder notation

[round cypress]

oh

thats wut u meant

im i suposd to only put x

@final halo im i suposd just put x

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[sinful basalt]

Wouldn't this be actually 4 separate integrals? why does two single integrals work here?

maybe i am being dumb about how the double integral was broken up

Well, let me try and justify this.

Seems the two integral method has merit. However I am still a little bit confused on how you can break up that double integral like that. Because there's no x term maybe?

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[stoic cargo]

I don’t understand how to evaluate piece wise functions as shown on the problem like f(0), f(4), and etc.

[spare walrus]

if it was not a piecewise function how would you evaluate it?

[stoic cargo]

I forgot how to evaluate function, my brain is all foggy throughout summer, I had essentially forgotten how to do most of these stuff.

[spare walrus]

okay so function notation is f(x)

[stoic cargo]

Yes

[spare walrus]

where x is the x coordinate that you're evaluating the function at, and the value of the function is typically the y-value

[stoic cargo]

mhmm ok

[spare walrus]

so take f(0) first

we look at where 0 is on the x axis, and then you can go up or down along the y axis to find the value of the function

[stoic cargo]

So would it be 2?

[spare walrus]

yeah!

[stoic cargo]

Ohh

Is that essentially all I had to do

[safe radish]

@stoic cargo Has your question been resolved?

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[gusty lodge]

Hi can someone help me how would I factorise something like this

[little sorrel]

look at the cubic ones

and cancel the ones out that dont work with the signs infront of it

[gusty lodge]

but is there a way to factorise it without process of elimination

incase I wasn’t given options

[little sorrel]

idk try bunch the common factors together

then factorise multiple parts

[timid pasture]

For questions like this the best way is to eliminate the wrong options

For example last option gets eliminated cause, look at x in each of the brackets. You should have odd number of negatives to get x^3

[gusty lodge]

alright I’ll stick with elimination for now then 👍

thanks for the help

[hazy elbow]

You could but that would be pain

[gusty lodge]

hopefully I won’t get asked to do that then

[hazy elbow]

like the manipulation to factorize would require a lot of intuition

It's actually the best option to eliminate

by comparing coefficients

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[lean otter]

im able to get the answer when i find for C but get something competely different wen finding the area under

[pulsar pecan]

can you show your work

[lean otter]

lemme re write it real quick

oh wait nvm

.close

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[wraith prism]

where is this interval came from

0<x<3?

did they common interval at the end or sum the interval?

i guess sum the interval becuase of interval doesn't inculde 3/2

[wind stream]

it is quite unnecessary here to solve for (3-x)/x>0, as the condition (3-x)/x>=1 is already enough to make both the square root and logarithm to be defined.
But anyway, they are taking the union of (0,3/2] and (0,3) like you said

[wraith prism]

.close

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[wraith prism]

Range of x^2/(1+x^2)

[wraith prism]

Y=x^2/1+x^2

(y-1)x^2+y=0

-4(y-1)y>=0

(y-1)y<=0

Y<=1 ?

Y<=0

[cold aurora]

How did you get this ?

[wraith prism]

Determinant

B^2-4ac=0

B is 0

[cold aurora]

Till here, it's fine.

After that, you can just make cases, i think.

[wraith prism]

How?

[wraith prism]

Is this step correct?

[cold aurora]

For (y-1)(y) to be negative, what are the possibilities for (y-1) and y ?

[cold aurora]

Yeah

[cold aurora]

For ab to be negative, exactly one of a and b needs to be negative.

[wraith prism]

These are the possibilities

[wraith prism]

Ohh yes

If (y-1)<=0
y<=1, y >=0

y<=0
(Y-1)>=0

y>=1,y<=0

All the cases mean y=0

[cold aurora]

Uh. No

Come on. When (y-1) <= 0, that means that y<=1.
But since exactly one of them needs to be negative, y>=0

So, you get y belongs to [0,1]

Similarly, do for other case.

[wraith prism]

(y-1)>=0, y>=1
y<=0

No common parts

[cold aurora]

Yes

[cold aurora]

So, this is your answer.

[wraith prism]

But i have doubt

When i draw it

Desmos doesn't include 1

[cold aurora]

Yeah. We shouldn't include 1. I was just thinking that.

Actually, limit of given expression when x tends to infinity is 1.

So, that's one reason why desmos doesn't include it as its x-axis can never go to infinity.

[wraith prism]

Limit here?

[cold aurora]

Yes.

See the limit when x tends to infinity.

[wraith prism]

1

2x/1+2x

[cold aurora]

Yeah. So function approaches 1 as x gets bigger and bigger(magnitude wise) but i suppose since infinity isn't like any other number, we can't say that it actually reaches 1.

[cold aurora]

What's that ?

Oh. L'hopital?

[wraith prism]

Yup

[cold aurora]

Yeah. That's fine.

So you get why we shouldn't include 1 ?

[wraith prism]

i draw the graph in desmos

They are not including it so

I guess we should put y in x terms

[cold aurora]

Well, graph can never show a point at infinity, so.

[wraith prism]

Here

[cold aurora]

That's another way to go.

[wraith prism]

Right

[cold aurora]

There are so many ways to solve this.

[wraith prism]

I guess this is the genuine way of finding range

[wraith prism]

True

I am just learning about the ways

As you taught calculus use here

[cold aurora]

One prominent way is that you check the domain of inverse function. Inverse for this however will not exist.

So, this method is good enough.
Also, you can use estimation. Write numerator as (x^2 + 1 -1) and separate the terms.

[wraith prism]

1-1/(1+x^2)

What next?

[safe radish]

@wraith prism Has your question been resolved?

[cold aurora]

Now, you do estimation(I dunno if there's a standard term for it).
Basically, we start with one thing we know for x when x is real:

$$x^2 >= 0$$
$$x^2 + 1 >= 1$$
$$ 0< \frac{1}{1 + x^{2}}<= 1$$

[flat frigate]

Enemagneto

[cold aurora]

Notice how 0 doesn't come in range for 1/(1+x^2). That's why 1 doesn't come in final range.

[calm bridge]

you can now take a minus to this whole inequailty

and add 1

this is the easiest way ig

[cold aurora]

Yeah

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[cinder pagoda]

The red part is wrong, why?

[mossy fractal]

because its not + 180k

that only works for tan

[cinder pagoda]

No, it's a special solution that happens when cos(x) = 0

[mossy fractal]

oh yeah since its x = 90, 360 - 90 is the same as 90 + 180

so idk why its wrong

[cinder pagoda]

I wasn't supposed to reach this equation, you need to see the full exercise to check

[mossy fractal]

you werent supposed to get cosx = 0?

[cinder pagoda]

ya

[mossy fractal]

what does that .7 mean?

[cinder pagoda]

Exercise number 7

[mossy fractal]

cosx = 0 should be valid for this

[cinder pagoda]

why?

[mossy fractal]

because 2cos^3x - cosx = cosx(2cos^2x - 1)

which is what you wrote

[cinder pagoda]

And?

[safe radish]

@cinder pagoda Has your question been resolved?

[cinder pagoda]

<@&286206848099549185>

[finite yacht]

The red part is correct

[cinder pagoda]

But its not in the answers

[finite yacht]

Then the answers are wrong