help-10

[astral tulip]

just wondering if any could solve this with some steps...having a hard time working backwards

[versed cave]

working backwards? What do you mean?

[astral tulip]

im not sure how to work it out, im just thinking i would have to work backwards or something like that

[versed cave]

What's the function that tells us the gradient at (almost*) each point of a function?

[astral tulip]

not sure

[versed cave]

The gradient is the slope, right?

[astral tulip]

oh yes

[versed cave]

Derivatives

Do you know derivatives?

[astral tulip]

yes

[versed cave]

what does the derivative tell us?

[astral tulip]

the slope of the tangent line

i think

[versed cave]

Yes nice

So you can find for what points the derivative = 8

What's the derivative of x²?

[astral tulip]

2x

[versed cave]

Yeah

[astral tulip]

hey i just went to the answer sheet and found 4 , 16. just wondering how i get to that from 2x.

[versed cave]

.

[astral tulip]

lol

[versed cave]

[The derivative] = [The slope you need]
That's how to find at which points the tangent has that slope

But you should get 4 only, I don't know why 16 is there

Maybe there's a part of the question you didn't post?

[astral tulip]

this is the full question, it asks for coordinates

[versed cave]

oh, it 16 is the y value of the point

[versed cave]

anyways, do this

[astral tulip]

ok ty

[versed cave]

Once you found the x value of the point, it's easy to find the y

[astral tulip]

yeah ty

.close

[obtuse pebble]

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[clever nacelle]

I need help understanding this question, more particularly a more concrete definition of a relation and graph

[warm canopy]

hint: a function cannot send an element in the domain to two different elements in the codomain

think of a relation as a set of (x,y) points

take {(0,1),(1,1)} this sends 0 to 1, and it sends 1 to 1

but if we had something like {(5,6),(5,7)}. IF this were a relation for a function, then this function would send 5 to 6 and ALSO send 5 to 7?? not possible so it must not represent a function

[obtuse pebble]

@clever nacelle Has your question been resolved?

[clever nacelle]

Okay so a relation is basically the cartesian product of the sets and the (x,y) ordered pairs show what the function gets for every x

Ok understood

Ty

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[velvet spade]

try manipulating the right equation so the the denominator is the same as on the left

...if you can do it, then why not do it?

do you know how to get the denominator of the right hand side the same as on the left?

for example,
[\frac{A}{x + 5} = \frac{A (3 x - 1)^2}{(x + 5) (3 x - 1)^2}]

[warm shale]

Samsamson33

[velvet spade]

see how the denominator is (x + 5) (3 x - 1)², like on the left?

do that to every term then add them up

on the left, you have [\frac{39 x^2 + 2x + 59}{(x + 5) (3 x - 1)^2}]

[warm shale]

Samsamson33

[velvet spade]

so the denominator is (x + 5) (3 x - 1)²

you want every fraction to have that denominator

[velvet spade]

I already did one for you here

use the same idea for the other two

yep

now all the fractions have the same denominator, so you can add them up

yes, you would

you're given that the left equation equals the right equation

so you set them equal to each other and solve for A, B, and C

[warm shale]

Samsamson33

[velvet spade]

huh, that get cut off

you need to find A, B, and C

the denominators are all the same, so you can multiply everything by (x + 5) (3 x - 1)²

yep

now expand the right hand side

exactly

idk, but it's not going to be helpful

try expanding the right hand side here

A (3x - 1)² + B (x + 5) (3x - 1) + C (x + 5)

ok, now collect the terms for each of the powers of x

for example, A x² - 5 + 3 x - B x² would become (A - B) x² + 3 x - 5

ah, ok

so they wanted you to use the zeroes to your advantage to solve the last bit

[obtuse pebble]

@timid silo Has your question been resolved?

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[timid silo]

I need help with it

[turbid sail]

What would you guess

[timid silo]

Ummm

[turbid sail]

Any ideas

[timid silo]

I think A or B

Idk tho

[turbid sail]

Ok are you familiar with absolute value

|-1|=1

[trail pewter]

i can't see the letters

[timid silo]

Well yeah but not with the equation thing

[turbid sail]

Plug in some points

Try x=-1

[timid silo]

alright

[turbid sail]

y=-|-1|

y=?

[timid silo]

uhhh

1 right…..

idk

[turbid sail]

it would be -1

[timid silo]

Ohh

[turbid sail]

So the absolute value of |-1| is =1

And then the negative applies

After that

[timid silo]

Alright

[turbid sail]

can you see that ?

[timid silo]

A little I’m just confused

Is C wrong?

[turbid sail]

I’m not sure what c is

Are you doing vertically

Or horizontally

[timid silo]

Ummm

[turbid sail]

Bottom left?

Top right?

[timid silo]

Well um bottom left I think

[turbid sail]

Yes it’s wrong

[timid silo]

Ohh

[turbid sail]

There can be no values

[timid silo]

Wouldn’t D be wrong as well

[turbid sail]

Where this function is positive

D is wrong as well

[hexed agate]

I suggest you review graph transformations, would be much easier

[turbid sail]

Why?

[timid silo]

Is B wrong?

[turbid sail]

I mean what do you think

Yes or no

And explain why

[timid silo]

Ummmm I think no idk

[turbid sail]

It’s correct

[timid silo]

Oh

[turbid sail]

I feel like you need to understand that any value

Will be negative

So absolute value

Returns only positive values

If you take that to be negative

Than we can only have negative values

[timid silo]

Ohhh

I got 6 more problems that I need help with 🥲

[turbid sail]

Do you understand that though?

[timid silo]

Not much

I suck at math

Do I send the other problem here

[turbid sail]

Ok do you know what these symbols mean

Let’s break it down

|x|

[timid silo]

Yes

[turbid sail]

what does that mean

[timid silo]

Positive right 🥲

[timid silo]

Did I click the right one

[turbid sail]

it returns the positive value yeah

try some examples of this

What is |3|

[timid silo]

3

[turbid sail]

Yes

What is |-6|

[timid silo]

6

[turbid sail]

okay good

Now we add a negative to this

so think of it as

-(6)

[timid silo]

Okok

[turbid sail]

for -|-6|

So it is

[timid silo]

-6?..

[turbid sail]

Yes

[timid silo]

okok

[turbid sail]

Good

[timid silo]

Question

Did I click the right answer for this

[timid silo]

This one

[turbid sail]

No

Ahh

[timid silo]

🥲

[turbid sail]

Okay so work through it

[timid silo]

I’ll try

[turbid sail]

n(-1) = -|-1|

What is that =

[timid silo]

Uhhh

[turbid sail]

when x=-1

The function returns what value

[timid silo]

Is it -1

[turbid sail]

Yes

Good

Okay so we now know that

When x=-1

y=-1

Right?

[timid silo]

Yeah

[turbid sail]

So we are basically done here

Because we only have one graph

That satisfies that

[timid silo]

Ohhh

[turbid sail]

But it’s important to know how these behave

Just intuitively

[timid silo]

True

Ummm so what is the answer for the graph

lemme label brb

[turbid sail]

Which one where x=-1 y=-1

[timid silo]

Ummm

A… idk

[turbid sail]

Uhhhhhhhhhhh

[timid silo]

😭😭 idk

[turbid sail]

Ok you need to review how graphs are

[timid silo]

Ik 😭😭

[turbid sail]

Look at a

[timid silo]

Is it C

[turbid sail]

Look at x=-1

Just look at a

[timid silo]

Okok

[turbid sail]

When the x coordinate is =-1

Where is the y value at that point

[timid silo]

Uhhh

[turbid sail]

Just think don’t just type

Take a second if you need

[timid silo]

Idk

Wouldn’t it be diff or the same

[turbid sail]

Look at when x=-1

for n

[timid silo]

Oh..

[turbid sail]

What is the ordered pair

(-1,?)

[timid silo]

0..?

[turbid sail]

YES

So that can’t be true

[timid silo]

So is it C

[turbid sail]

Because -|x| when x=-1 …….. is -1

……………

[timid silo]

Ohhhh

Ummm do I send the next problem or

[turbid sail]

Have you figured out the first

[timid silo]

Think so

[turbid sail]

What is it

[timid silo]

Ummmmmmmm

Definitely not a

Or b

😭😭

[turbid sail]

Maybe I’m just not explaining it well enough I gotta go

<@&286206848099549185>

[timid silo]

☠️☠️

[trail pewter]

it's b

[obtuse pebble]

@ripe gale Has your question been resolved?

[paper socket]

I think he was trying to explain it without telling him the answer

[trail pewter]

Sorry

[obtuse pebble]

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[pliant reef]

\

[pliant reef]

can someone help explain this probelm to me!

[dusk portal]

@pliant reef you know the total volume this shape should have, 600 cube inches

you know all but one required measurement. So, set up an equation equal to 600 with your one unknown (height of the rectangular portion) and solve for it

[pliant reef]

would 600=10x10xH work?

[dusk portal]

well 600 is for the whole thing

so 600 = (volume of rectangular portion) + (volume of cylinder)

[pliant reef]

so would i have to solve for the cylinder volume first?

[dusk portal]

you can already work out the cylinder's volume

600 = 10*10*H + (volume of cylinder)

[pliant reef]

for the volume, would it be pi(2.5)^2 x 7?

[dusk portal]

yup

[pliant reef]

so 600=100h + 137.375

but once i solve for h i get 2.5 inches

which doesn’t match the correct answer

[dusk portal]

you're right with 137.375 but how did u get 2.5?

600 - 137.375 = 462.625 = 100H

so H = 4.63

[pliant reef]

ohhh i added the 137.375 to the 100h by accident

[obtuse pebble]

@pliant reef Has your question been resolved?

[obtuse pebble]

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[delicate edge]

i dont get this

[delicate edge]

domain:f\left(x\right)=\frac{\sqrt{x+7}}{x+1}

[chrome mesa]

$$domain:f\left(x\right)=\frac{\sqrt{x+7}}{x+1}$$

[warm shale]

Tronsi

[delicate edge]

the answer is [-7,-1) u (-1, infinity) but i dont get why you wouldnt include (-infinity, -7] too for the domain

x+1 would just be x cannot = -1

which would mean that it would be like (-infinity, -1) u (-1, infintiy) right?

[kind hawk]

remember that you can't take the sqrt of a negative number

[delicate edge]

ohhhhhhh

yeah you right

okay well would x^6-\left|x\right| be an even function

x^6-\left|x\right|

t!cat

[sinful light]

Here is your random cat! 🐱

[delicate edge]

$x^6-\left|x\right|$

[warm shale]

BlueCookiez

[delicate edge]

is this an even function or a neither function?

<@&286206848099549185>

[obtuse pebble]

@delicate edge Has your question been resolved?

[velvet spade]

what's the definition of an even function?

[obtuse pebble]

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[errant harness]

hello i have a question about this formula

[timid silo]

yes

[errant harness]

can i simplify it to this?

[timid silo]

sadly not, no :/

[errant harness]

is it a radicals rule

[timid silo]

for radicals this only applies for multiplication:
√(x*y) = √x * √y

never ever do this with addition

[errant harness]

oooooh ok thank you

[timid silo]

like, keep a simple example in mind, is the square root of 16+9 the same as the square root of 16 plus the square root of 9?

[errant harness]

no

ok that makes sense

thanks for the help! i hope you have a great rest of your day bye bye

.close

[obtuse pebble]

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[obtuse pebble]

@quiet cradle Has your question been resolved?

[quiet cradle]

I figured it out

.close

[obtuse pebble]

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[real galleon]

I am attempting to evaluate the trig relationships in a right triangle. We are given the adjacent and opposite but not the hypotenuse . I calculated the hypotenuse but the answer sheet for my problem shows me having a different hypotenuse.

Right now the opposite is 2, the adjacent is 3, therefore the hypotenuse is sqrt(13), however the answer key shows ig as having 3 sqrt(13)

[real galleon]

here is a photo of my work so far

[slow fulcrum]

you should be right... if the side lengths of the legs are 2 and 3 then the hypothenuse should be sqrt(13)

[real galleon]

here is what the answer key says

here is the original question

[obtuse pebble]

@real galleon Has your question been resolved?

[timid silo]

@real galleon what's your problem?

[real galleon]

.

[timid silo]

what you did was right

simply the answer key is simplified

$\frac{3\sqrt{13}}{13}=\frac{3}{\sqrt{13}}$

[warm shale]

Al3dium

[trail musk]

Sqrt(x)/x = 1/sqrt(x)

This is science

[real galleon]

oh I see

[spice chasm]

do x^-.5?

[obtuse pebble]

@real galleon Has your question been resolved?

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[obtuse pebble]

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[fallen atlas]

Hi, I need help with this problem: The bell tower of a clock takes (m+1) seconds to ring m² chimes, how many chimes will it ring in four seconds?

[gilded needle]

would love to understand the physics behind such a bell tower 😆

[fallen atlas]

idk, im solving a form and there's no alternative

[fallen atlas]

It is a topic that I take in class, it is simple, only a bit tedious to write, and I see that problem as a bit confusing, so I came to ask 😅

[gilded needle]

yeah it was just a humorous observation that the bells ring faster and faster if you give it more seconds, for example if m = 10 then that means in 11 seconds it rings 100 times. if m = 100 then it means that in 101 seconds it rings 10,000 times

[fallen atlas]

If you think about it like that, it gets even more confusing, but it makes sense

[gilded needle]

10,000 rings in 101 seconds is more than 100 rings per second

you wouldn't even be able to capture it with 60Hz video

[fallen atlas]

yea, I understand

[fallen atlas]

xd

these are the alternatives that's why I discard the idea that m = 3

[gilded needle]

well one of them does in fact equal 9 when m=3

[fallen atlas]

wich?

[gilded needle]

plug in m=3 to each one and see

[fallen atlas]

oke, let me see

oh

its A right?

[gilded needle]

yep

[fallen atlas]

cuz, 4m-3 = 4(3)-3 = 12-3 => 9

[gilded needle]

very weird question

but A is correct

[fallen atlas]

I know

[fallen atlas]

ty :D

.close

[obtuse pebble]

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[wintry cedar]

this is the question

using the exponent rule thing, we know that 1/x^9 is the same as x^-9

and x^8 is x^8/1

I'm stuck here

yes, which would be x^9, right

so then 1/x^9 + x^8/x^9?

which is just this

tf

how would that work here tho

why wud that be the denom

[warm shale]

donut

[wintry cedar]

lol

I think i got it tho

haha

[warm shale]

donut

[wintry cedar]

ok wait so I got it

no

thass not it i dont think

wait a sec

ah

i got it

answer is this

wait

answer

ez W

[wintry cedar]

this is also correct

just looks uglier 😛

.close

[obtuse pebble]

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[wise imp]

could you guys help me

[graceful oyster]

what's the question

[obtuse pebble]

@wise imp Has your question been resolved?

[upbeat plinth]

@wise imp pls post ur question and what u tried, also only use 1 channel

[obtuse pebble]

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[timid silo]

which one do you want help with ?

you just have to use identities in all of them

[opaque cairn]

Oh so is it cos/sin=tan?

[urban frost]

more like sin/cos lol

[opaque cairn]

Sorry god i havent looked at these for ages 😳

[robust sleet]

Loll

[urban frost]

AHAHA

[timid silo]

that's not the only identity in existence

[robust sleet]

Sadge

[opaque cairn]

yes i know

sorry im lost could you give me the first answer please

[timid silo]

which identity would you use for the first one

sin2x = 2sinxcosx

[robust sleet]

Get a list of identities and play puzzle

[timid silo]

use this

[opaque cairn]

oh god thank you

[timid silo]

sin2x= 2sinxcosx
cos2x= 1 - 2sin^2(x) = 2cos^2(x)-1 = cos^2x-sin^2x
tan2x = 2tanx/(1-tan^2x)

these are all the identities which are required

[opaque cairn]

Thank you so much!

.close

[obtuse pebble]

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[proper basin]

I’m just confused how to start

[timid silo]

what is the general equation of a line in point-slope form

[proper basin]

Y-y1=m(x-x1)

[devout sable]

Okay so where are you stuck?

They’ve given (x1,y1) = (1,2) and m=3

[proper basin]

What do I use as y

[timid silo]

y is variable (x,y) represents all points on the line, writing an equation of a line essentially is writing the constraint relations of x and y, the condition which is supposed to be met for (x,y) to be a point on that line

[obtuse pebble]

@proper basin Has your question been resolved?

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[bright pumice]

But I don't understand why

Like I can understand it's sqrt(69120/30)

Which makes 48 but why?

[obtuse pebble]

@bright pumice Has your question been resolved?

[bright pumice]

<@&286206848099549185>

[obtuse pebble]

@bright pumice Has your question been resolved?

[obtuse pebble]

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[timid silo]

I just wanted to check if this was mathematically sound or if I’m just making it up

[timid silo]

I was looking through the answer methodology and it looks like they’ve done this but I’m not sure if this is something you can actually do??

To clarify

[stable rain]

yea

[stable rain]

yea

[timid silo]

Thanks :))
I don’t fully get the process of why 3(4^x) becomes 2(4^2x)

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

would anyone be able to explain why 2(4^x) becomes 2(4^2x) i dont see the line of reasoning :v

[wild swallow]

multiply the top equation by 4^x

$2 \cdot 4^x + 3 \cdot 4^{-x} = 7 \implies 2 \cdot 4^{2x} + 3 = 7 \cdot 4^x$

[warm shale]

夢雪

[timid silo]

aaah thankyou

i feel like i was staring at that for years, i really appreciate the help! :))

.close

[obtuse pebble]

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[royal basin]

do you know what a linear combination is?

a linear combination of $\vec{a}_1, \vec{a}_2, \vec{a}_3$ is a vector of the form $c_1\vec{a}_1 + c_2\vec{a}_2 + c_3\vec{a}_3$ where the $c_i$ are scalars.

[warm shale]

Ann

[royal basin]

you need to find whether or not there exist such scalars that the linear combination equals b, and if they do, present them

[obtuse pebble]

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[timid silo]

multiplication of two numbers equal to their sum, quadruple multiplication of these numbers equal to the sum of the numbers to the power of two. you need to find all the pairs of numbers. ```
xy = x+y
4xy = x^2+y^2

[rigid agate]

You sure "quadruple multiplication" is 4xy? To me that would be "quadruple addition". Nontheless that exprrssion seems odd.

[timid silo]

sorry for my english

xy = x+y, 4xy = x^2+y^2 is right

[rigid agate]

Ok👍 No worries

Try multiplying the first equation by -4 and adding both equations

But I'm not sure actually if that leads to progress😅

Solve one equation for x or for y and substitute

[timid silo]

ive done this

my problem is xy=6

[rigid agate]

One solution of xy=6 would be x=2,y=3. But that doesn't satisfy xy=x+y, so something went wrong.

[timid silo]

well

perhaps (0;0) is the only solution

but i'm not sure

[rigid agate]

No, there is more🙂

3 pairs (x,y) satisfy both equations

(0,0) and two more

[timid silo]

why?

[rigid agate]

Magic calculator😅 But try solving it without one

[timid silo]

that's right

[rigid agate]

Try this

You will get a new equation

[timid silo]

i found\

[rigid agate]

Yes that looks good

(3-sqrt3 , 3+sqrt3)

[timid silo]

yes

it's 6

[rigid agate]

Now notice that the equations are symmetric in x and y, so for any solution (x,y) also (y,x) is a solution.

[timid silo]

yes

[timid silo]

thank you btw

[rigid agate]

yw

[timid silo]

.close

[obtuse pebble]

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[queen flume]

uhmm i need help

[queen flume]

i dont even get it:')

[obtuse pebble]

@queen flume Has your question been resolved?

[river meteor]

.close

kimm just type that

@queen flume

[queen flume]

.close

[obtuse pebble]

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[queen flume]

thankssss

[obtuse pebble]

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[final radish]

I dont quite understand this partial fractions question 🤔

[royal basin]

what exactly do you not understand about this one in particular?

[final radish]

mb

this one

[royal basin]

my question still stands

what is causing you issues in understanding?

[final radish]

So I get this @royal basin

but now I am stuck

I can let x = 0 to get C but then what

[royal basin]

you are dodging my simple question of "what is it that you don't quite understand?"

also, no, taking x=0 will not give you C

also, you are missing a plus sign

[final radish]

how I solve for A, B and C

that is what I dont understand

I dont know how I would solve for them in this situation

I will add the plus sign now

[royal basin]

well there are many ways

you can get C by taking x=-3/2 to zero out the A and B terms.

[final radish]

ah ok lemme try ti

[final radish]

ann I got a sus answer I think

i show now

[final radish]

thats what I got so far

[royal basin]

and you consider this suspicious?

[final radish]

ohh ok ok ok ok!

i think this might be right

i get -2

thanks @royal basin i also solved for B cuz x = 0 ( i got 2)

thanks

.close

[obtuse pebble]

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[royal basin]

still do not appreciate how you do not answer questions plainly

[final radish]

I just wanted to first put out all my working so you can see where I went wrong

[royal basin]

when you are asked a yes-no question you should reply with "yes" or "no", not weasel your way out

[obtuse pebble]

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[violet compass]

I know that x=6 is the correct

By graphing the sqrt(5x+6)=x

[crystal moss]

If you take a square root of a number the answer has to be 0 or positive

[violet compass]

Oh because x can be positive or negative since sqrt would give us an answer in +/-

And we can’t do sqrt(-1)

(Well yet at this level)

[crystal moss]

the +- sign is always written outside the square root

Square root it self can’t give you negative value

[crystal moss]

(I mean imaginary numbers aren’t negative either

[violet compass]

(Not but at this level we not touching it :))

[timid silo]

Ig x should be positive becoz root can never be neg

For real nos

[crystal moss]

Haha yeah

[clever nacelle]

I'm pretty sure both are correct

wait nope

[violet compass]

howw?

I need more explanations

[clever nacelle]

if you substitute the answers into the equation given we will get sqrt(1) = -1

and sqrt(1) is 1 not -1

[violet compass]

is it not +/-

istg it can be

[clever nacelle]

substitute x=-1 into:

[violet compass]

yh it would get to sqrt(1)

[clever nacelle]

exactly

[violet compass]

oh and -1 != 1

[clever nacelle]

yeah

I would only use that +- thing for completing the square or using the quadratic formula

we don't know the range of the function in this case so just use what they have given and don't over complicate it

[violet compass]

.close

[obtuse pebble]

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[warm shale]

Shihab

Hi, "If $z_{1},z_{2}$ are complex numbers, $z_{1} = h+i^{27}$ and $Arg(z)=\frac{5\pi}{4}$ and $z_{2}=(k+7\sqrt{3}i^{2019})z_{1}$ and Mod $z_{2}=14$ Mod $z_{1}$, find $h,k \in R$" I tried to solve this and i got $k=0$ and $h=1$ is that right ? i did first calculate Mod $z_{1}$ = \sqrt{h^{2}+1}
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.57 ...i did first calculate Mod $z_{1}$ = \sqrt{
                                                  h^{2}+1}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```

[obtuse pebble]

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[timid silo]

.close

?

[obtuse pebble]

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[warm shale]

I can't DM you! Do you have DMs disabled?

Aug 31, 2022, 2:15 PM

I can't DM you! Do you have DMs disabled?

Aug 31, 2022, 2:17 PM

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

`tex` command documentation. (Aliases `,`, `mtex`, `align`, `latex`, `texsp`, `texw`, `tikz`.)

Render LaTeX code.

Usage

,, <equations>
,tex <code>
,align <align block>
,texsp <code>
,texw <code>
,tikz <code>

Click to show more information.

[pearl aurora]

,tex <The price of a kg of cake flour is 2 Pounds more expensive than the price of a kg of regular flour. 5 kg of cake flour is more expensive at
8 Pounds from the price of 6 kg of regular flour. What is the price per kg of flour of each type?>

[warm shale]

Confutatis Maledictis

[pearl aurora]

How to solve?

[obtuse pebble]

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[obtuse pebble]

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[hollow mason]

Can someone verify my answers? I’m not sure if I did them correctly

[weary delta]

Looks good to me

[hollow mason]

appreciate it, I get intimidated once I don’t see multiple choice 😂

.close

[obtuse pebble]

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[feral warren]

If roots of ax^2 + bx + c are imaginary, find sign of (a + 2b + 4c)(4a + 2b + c)

[feral warren]

<@&286206848099549185> anyone?

[weary delta]

If the roots of a quadratic are imaginary, what does that say about it’s discriminant?

[feral warren]

D < 0

@weary delta uh

[weary delta]

Sorry, got distracted. I thought I saw something there but it doesn’t pan out

[feral warren]

F

[weary delta]

If you expand that mess of a product, you’ll get a lot of terms that are trivially positive due to either being squares or the discriminant. However you get stuck with a 10b(a+c) that’s not easy to show to be positive

[feral warren]

oh I see, thanks for your input

.close

[obtuse pebble]

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[open nimbus]

The median for this would be 2-3.9 right ?

[obtuse pebble]

@open nimbus Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[subtle eagle]

hello , im relearning calculus, i can't really remeber how to do this excercise

[full cloud]

The first order derivatives are ∂f/∂x and ∂f/∂y while finding the partial derivative with respect to x, you treat y as a constant and similarly for y as well

So here ∂f/∂x here would be 12x^2 + 4y

[subtle eagle]

and for y would be 4x - 6y

yes

[full cloud]

Yeah

[subtle eagle]

i got the answers but how to get there it's the thing

[full cloud]

Well I told it

Do you know the normal derivative rules?

[subtle eagle]

oh damn i did not read that carefully

[subtle eagle]

yeah i know most of them

[full cloud]

Ah ok

[subtle eagle]

thank you i got some direction where i'll look into

.close

[full cloud]

👍

[obtuse pebble]

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[hexed thunder]

I need help understanding this relations and functions

[hexed thunder]

So what I got for the domain was

nvm

.close

[obtuse pebble]

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[timid silo]

can someone help me? i dont know which side is x and y for both quadrant 3 and 4, i can find the coordinates, i just don't know which side is x and y

[pulsar merlin]

Y is the line going from top to bottom

X is the side going from left to right

[timid silo]

okok thank you so much!!

.close

[obtuse pebble]

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[timid silo]

ok hi again uhhhhh so i'm just gonna type this (im using < as the angle symbol) <RQT is a straight angle. What are m<RQS and m<TQS?

[royal basin]

have you made any progress on this so far?

[timid silo]

i tried to but i think im doing it wrong

[royal basin]

show your work

you may well have been on the right track

[timid silo]

i dont have it anymore 😭 im sorry

[royal basin]

so you made an attempt, deemed it wrong, and threw it out?

[timid silo]

i erased it and the paper ripped

[royal basin]

well now you'll have to redo it from scratch and then show it here.

[timid silo]

okok

[timid silo]

im confused because im supposed to be getting 2 answers

[royal basin]

you have found the value of x

and ignoring the notational mishap near the end, the value of x is correct

now you need to find the values of 12x+2 and of 7x+7 with that value of x

[timid silo]

ohh okay i will do that rn

so i got 110 and 70

hey something came up so im gonna go ahead and close this

.close

[obtuse pebble]

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[timid silo]

How am I supposed to do the last part?

[timid silo]

substitute y = 4x

$\lim_{(x, 4x) \to (0, 0)} \frac{4x^2 + 64x^3}{x^2 + 16x^2}$

[warm shale]

Aude

[timid silo]

notice that you can reduce the limit

@timid silo

[timid silo]

Then doesnt it just become 4?

no

[timid silo]

Then how do I do this?

you look at what I wrote

$\lim_{(x, 4x) \to (0, 0)} \frac{4x^2 + 64x^3}{x^2 + 16x^2} = \lim_{(x, 4x) \to (0, 0)} \frac{4 + 64x}{17}$

[warm shale]

Aude

[flat rune]

why is x^2 = 1?

[timid silo]

Got it. Thank you

.close

[obtuse pebble]

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[tardy plaza]

Why does the limit not exist; shouldn’t it be -3 because the point (-1,-3) is on the graph?

Or maybe, when it doesn’t have the - or + exponent on the x->-1 thing, it means we only look at the left side/graph (approaching -1 from the left)?

Thanks in advance and please ping me

[rich sparrow]

A limit, unless specifically stated, exists for both sides of the approach.

If L(left) != L(right), there is no limit.

[rich sparrow]

.

You will see a tiny arrow beneath the limit (above the number) that shows a direction when a limit is one-sided

[spice basalt]

The exponents you are talking about are what typically denote direction. $\lim_{x\to a^-}$ means that we approach from the left, $\lim_{x\to a^+}$ from the right. As pointed out, both sided limits must exist and be equal for $\lim_{x\to a}$ to exist

[warm shale]

Tymelord14

[tardy plaza]

Tyvm, that makes a ton of sense now!

.close

[obtuse pebble]

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[maiden radish]

abc are positive reals; proof that $$\dfrac{a}{a+\sqrt{(a+b)(a+c)}}+\dfrac{b}{b+\sqrt{(b+c)(b+a)}}+\dfrac{c}{c+\sqrt{(c+a)(c+b)}} \leqslant 1$$

[warm shale]

what do i live for v2.0

[maiden radish]

@compact shadow

[compact shadow]

<=1?

[maiden radish]

yes

[timid silo]

have you tried multiplying the fractions with a-sqrt(...), b-sqrt(...), c-sqrt(...)?

[slate zephyr]

I suggest canceling a,b and c first.

[maiden radish]

no ?? i will try

[tardy epoch]

$\frac{1}{1 + \sqrt{(1+\tfrac{b}{a})(1+\tfrac{c}{a})}}$

[slate zephyr]

[warm shale]

riemann (eric tao for honorable)

[tardy epoch]

the first term can be written like this (similar expressions for the other two. and maybe this is less than 1/3?

[slate zephyr]

unfortunately no

[timid silo]

if you multiply it like this it should give them common denominators, but havent checked myself

[slate zephyr]

Summing those fractions in the only way I can see.

[obtuse pebble]

@maiden radish Has your question been resolved?

[timid silo]

it should give (x+y)(x-y) in the denominators, which is x²-y², cancelling the square roots y

and then a²-(a+b)(a+c)=-ac-bc-ab which will be the common denominator

[compact shadow]

Done but complicated

First

[obtuse pebble]

@maiden radish Has your question been resolved?

[compact shadow]

Let $\begin{pmatrix}0&1&1\1&0&1\1&1&0\end{pmatrix}\begin{pmatrix}a\b\c\end{pmatrix}=\begin{pmatrix}u^{2}\v^{2}\w^{2}\end{pmatrix}$, therefore $\begin{pmatrix}a\b\c\end{pmatrix}=\frac{1}{2}\begin{pmatrix}-1&1&1\1&-1&1\1&1&-1\end{pmatrix}\begin{pmatrix}u^{2}\v^{2}\w^{2}\end{pmatrix}$
Now the original inequality is equivalent to $\frac{vw}{(v+w)^{2}-u^{2}}+ \frac{wu}{(w+u)^{2}-v^{2}}+\frac{uv}{(u+v)^{2}-w^{2}} \geq 1$
Since they are homogeneous we assume that $u+v+w=1$, it becomes $\frac{vw}{v+w-u}+ \frac{wu}{w+u-v}+\frac{uv}{u+v-w} \geq 1$
Again, let $\begin{pmatrix}0&1&1\1&0&1\1&1&0\end{pmatrix}\begin{pmatrix}x\y\z\end{pmatrix}=\begin{pmatrix}u\v\w\end{pmatrix}$, we have $\begin{pmatrix}x\y\z\end{pmatrix}=\frac{1}{2}\begin{pmatrix}-1&1&1\1&-1&1\1&1&-1\end{pmatrix}\begin{pmatrix}u\v\w\end{pmatrix}$ the inequality becomes $\frac{(x+y)(x+z)}{x}+ \frac{(y+z)(y+x)}{y}+ \frac{(z+x)(z+y)}{z} \geq 2$
By $x+y+z=\frac{1}{2}$, it’s equivalent to $\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z} \geq \frac{1}{2}$,the last one is trivial I think, if you can’t prove it I will work it out for you

[warm shale]

Cogwheels of the mind

[compact shadow]

Oh the last one is trivial indeed:

Wait said something wrong

Fixing

[maiden radish]

ok

[compact shadow]

Oh the last one is proved:

$\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}=\frac{y^{2}z^{2}+z^{2}x^{2}+x^{2}y^{2}}{xyz}= \frac{\frac{y^{2}z^{2}+z^{2}x^{2}}{2}+\frac{z^{2}x^{2}+x^{2}y^{2}}{2}+\frac{y^{2}z^{2}+x^{2}y^{2}}{2}}{xyz} \geq \frac{yzx^{2}+zyx^{2}+zxy^{2}}{xyz}=x+y+z=\frac{1}{2}$

[warm shale]

Cogwheels of the mind

[slate zephyr]

The quivalent to does not make sense to me.

[compact shadow]

QED

[compact shadow]

Which step

[slate zephyr]

"Now the original inequality..."

[compact shadow]

I use first term as an example for you, rest is similar:

a=(1/2)(v^2+w^2-u^2)

So first fraction is

(v^2+w^2-u^2)/(v^2+w^2+2wv-u^2)

The numerator

You plus 2vw then minus 2vw

It becomes 1-2vw/((v+w)^2-u^2)

So 3-Σ2vw/((v+w)^2-u^2))<=1

[slate zephyr]

ah

[compact shadow]

Okay

As long as you got it

[maiden radish]

wow rly complicated
i will take some time to get it
thnx very much

[compact shadow]

If it wasn’t complicated, it wouldn’t have taken me this long…😂

Np

[slate zephyr]

Has anyone tried just mtuliplaxying out?

[compact shadow]

Please leave this open for a moment, others want to check it for me

[maiden radish]

ok

[compact shadow]

Little mistake here, should be xyz^2 , the first term of numerator of the last fraction, won’t effect your reading though I suppose

[obtuse pebble]

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[jagged steppe]

someone help

[obtuse pebble]

Please don't occupy multiple help channels.

[jagged steppe]

.close

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[knotty goblet]

How would I do this problem?

[daring rock]

What's the directions? Factor it?

[knotty goblet]

yeah

[daring rock]

Expand it out first, combine all the like terms

Then you should be able to factor it

You might need more room than that lol

[knotty goblet]

so like unfactor it then factor it back?

[daring rock]

Yes

But actually

we can be more clever than that

I just noticed

There is a common factor of (2x-1)^2

in the first and second part

[knotty goblet]

I could divide (2x-1)^2 from both sides right?

[daring rock]

Don't just divide it away. Factor it out

$$(2x-1)^2(x-3) + (x+1)(2x-1)^3$$
$$= (2x-1)^2[(x-3)+(x+1)(2x-1)]$$

[warm shale]

tatpoj

[knotty goblet]

I think I understand it

[daring rock]

Yeah, you can expand the stuff in the brackets and factor that

[knotty goblet]

alright, thanks!

[daring rock]

sure thing

[knotty goblet]

.close

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[spice chasm]

h(x)=sqrt(x+4). h(x) is a composite function of f and g such that h(x)=f(g(x)). Give solutions for f and g.

[short spruce]

what do you know about composite functions

[spice chasm]

u put in what g(x) equals into f

[timid silo]

yeah so have you tried any f or g?

[spice chasm]

g=x+4
f=sqrtx

[timid silo]

that's a good one

how many are you asked?

[obtuse pebble]

@spice chasm Has your question been resolved?

[spice chasm]

1

[obtuse pebble]

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[rain trench]

inverse function of

[timid silo]

how about you subtract y, and then xy - y = (x-1)y

[rain trench]

i usually ended up here

where is the -1 from?

[timid silo]

xy - y = xy - 1y

[rain trench]

I see I see

so the next would be this

[obtuse pebble]

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@rain trench Has your question been resolved?

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[final grove]

me

[final grove]

i need help

[daring rock]

so

post your question

[crystal moss]

The shaded region is 1/2 of the rectangle ‘s area, the “trick “ is to see how both triangle have the same height (which is also the height of the rectangle) and thus let the height be h, the area of shaded region = ((BF x h)/2) + ((FC x h)/2) = (h(BF+FC) /2)

[obtuse pebble]

@final grove Has your question been resolved?

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[small rock]

You are eligible to be President of the U.S.A. only if you are at least 35 years old, were born in the U.S.A., or at the time of your birth both of your parents were citizens, and you have lived at least 14 years in the country.

Express your answer in terms of
e: “You are eligible to be President of the U.S.A.,”
a: “You are at least 35 years old,”
b: “You were born in the U.S.A.,”
p: “At the time of your birth, both of your parents were citizens,” and
r: “You have lived at least 14 years in the U.S.A.”

[small rock]

Oof that formatting

One sec

I believe it is e → ( a ∧ (b ∨ p) ∧ r)

But it also might be (a ∧ (b ∨ p) ∧ r) → e

This is discrete math btw

I mean, what is the difference between "If you are eligible for president then the following is true" and "If the following is true then you are eligible for president?"

<@&286206848099549185>

[obtuse pebble]

@small rock Has your question been resolved?

[small rock]

Sad face

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@small rock Has your question been resolved?

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[carmine wharf]

Let $a_{1} = 2$ and $a_{n+1} = \frac{1}{2} (a_{n} + \frac{36}{a_{n}})$. Given that $\lim_{n\to\infty} a_{n} = L$, then $L = ?$

[warm shale]

kidnamedfinger

[carmine wharf]

I don’t even know how to start this :/

[compact shadow]

Take limit both sides

x=(1/2)(x+36/x)

Quadratic

Sorry not x, L

[carmine wharf]

$L = \lim_{n\to\infty} a_{n} = \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} \frac{1}{2}(a_{n} + \frac{36}{a_{n}}) = \frac{1}{2} (L + \frac{36}{L})$
$\ \$
$L = \frac{1}{2} (L+\frac{36}{L}) \implies L = 6$

[warm shale]

kidnamedfinger

[carmine wharf]

do you mean like this?

[carmine wharf]

<@&286206848099549185> Is this right?

[compact shadow]

Yeah

[carmine wharf]

yes sirrrrrr

thanks for the help

.close

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[mental zephyr]

need help with this operation on function guys

[royal basin]

@mental zephyr have you made any progress or are you completely stuck not knowing how to begin?

[mental zephyr]

i am stuck yes

idk how will i write it out in the function rules

[royal basin]

perhaps your problem is that you are overthinking things somewhat

let's go through this problem step by step

[mental zephyr]

okay

i’ll listen

[royal basin]

let's say maria makes x ensaymadas and sells them all. how much revenue will she have?

i.e. how much money will she have in her cash register from that day's sales, before any expenses?

@charred ivy this channel is occupied, please move!

[charred ivy]

alright

[mental zephyr]

abt how muchh?

[royal basin]

sorry, i don't understand your question.

[mental zephyr]

i cant speak english much

srry

how can i start solution

[royal basin]

well i don't speak any of the languages of the philippines either, but i am TRYING to show you how to start solving your problem.

ok let's make it even more simple

[mental zephyr]

ok

thank you

[royal basin]

too early to thank me...

this one only needs reading and no math or thinking: how much does 1 ensaymada sell for?

[mental zephyr]

about php 18

[royal basin]

what do you mean "about"???

[mental zephyr]

1 ensaymada costs php 18 it says

[royal basin]

yes, the problem says it's EXACTLY 18 pesos! what is the word "about" doing there, as if it could be 17.99 or 18.01??

[mental zephyr]

my bad

srry for english comprehension

its exactly php 18

[royal basin]

also let's not confuse cost (how much maria has to pay to make 1 ensaymada) with sell price (how much a customer has to pay to buy 1 ensaymada)

1 ensaymada sells for 18 pesos.

ok

[mental zephyr]

yess

and it costs 6 pesos to make one

[royal basin]

we aren't at cost yet

so far we are working with sales only

[mental zephyr]

okay

[royal basin]

we will get there soon

so, how much do 2 ensaymadas sell for?

[mental zephyr]

36

pesos

[royal basin]

and how did you calculate that?

[mental zephyr]

1 ensaymada costs 18 pesos two would cost 36

right?

[royal basin]

that's not what i am asking you

i am not asking you to restate your result

i am asking you to tell how you arrived at it.

maybe you don't know how to multiply and so you had to add 18+18 instead of multiplying 18*2...

[mental zephyr]

i multiplied 18 by 2

[royal basin]

right

ok, so in the same token

if maria sells x ensaymadas, how much will she make in sales?

[mental zephyr]

x?

[royal basin]

so from selling x ensaymadas, she will make x pesos?

is that what you're saying?

[mental zephyr]

yeah

[royal basin]

you are wrong and you contradicted yourself from earlier

you yourself said that when maria sells 2 ensaymadas (ie if x=2) then she makes 36 pesos

but now you are saying that 2 ensaymadas sell for only 2 pesos

you were able to say, correctly, that the revenue from selling 2 ensaymadas is 18 * 2 pesos. why not apply the same logic here? x ensaymadas, at 18 pesos/ensaymada...

[mental zephyr]

i kinda don’t get what you are coming from with if maria sells x ensaymadas, how much will she make in sales?

what is it that u mean by x

[royal basin]

x is a variable

it stands for a number

i.e. the number of ensaymadas sold by maria

our eventual goal will be to build a rule to express maria's profit in terms of this x

[mental zephyr]

okay

then?

[royal basin]

.

[mental zephyr]

so 18x?

[royal basin]

yes!

ok

we're making progress

so maria's revenue is 18x

now we will need to subtract from it the costs of maria's business

there are two such costs: the fixed 720 pesos in operating costs, and the 6 pesos/ensaymada production costs

[mental zephyr]

okayy

18x - 720

that it?

@royal basin

P(x) = 18x - 720

[royal basin]

you've accounted for the operating costs but not for the production costs

remember maria still has to spend 6 pesos per ensaymada made

[mental zephyr]

then i’ll substract it also?

18x - 6?

[royal basin]

not 6

the 6 is not fixed, it's the cost for every ensaymada

it varies just like the revenue does

[mental zephyr]

6x

[royal basin]

right

so let's put it all together now

[mental zephyr]

ok

by multplication function rules right?

[royal basin]

no, stop overthinking it

[mental zephyr]

how then

[royal basin]

profit is income minus expenses

income is the sales money, in our case 18x

expenses are the operating costs (720) and production costs (6x)

[mental zephyr]

yess

right

[royal basin]

ok so can you write down the profit now

[mental zephyr]

what do i do at this point again

[royal basin]

.

[mental zephyr]

subtract income and expenses,

?

[royal basin]

i've spelled it all out for you, you just need to put all these things together

yes...

[mental zephyr]

18x(720-6x)

[royal basin]

no

[mental zephyr]

how

[royal basin]

here you go again overthinking it

it's just 18x - 6x - 720 !!!