help-10

[timid silo]

and if u have

0.03

and 0.07

u need

0.03x=3

x=100

so u know

100x=35

x=35/100

0.35

any program can solve this for you

i can write you one right now

in java

[uncut girder]

What's missing then?

from making a program that just spits out average when I feed it the two frequencies?

[timid silo]

u can make a program

now that finds

the times

and then in the end

sum and divide

u know the cycle is on 35

so u take a loop

5x

u need 2 loops actually

one to loop through 5

and one to loop through 7

do u want me to help u with this

u know the cycle is 35 for 7 and 5

5 loops from 1 to 7

7 loops from 1 to 5

5x

for x = 1

5x1=5

now we loop through 7

which number multiplied by 7

gives us a number

for which

7 multiplied by this number - 5

gives us something bigger than 0

well no one

loop stops

go back at original loop

5x

for x = 2 now

10

what number multiplied by 7

that 5x-7 is bigger than 0

well

1

7x1=7

10-7=3

3 is bigger than 0

wow

for passenger 2

3 minutes

then u store this minutes

as as equence

and in the end

u sum and divide by the length

of the sequence

and u get the average

[uncut girder]

yep I wanted to give someone who can code instruction like this to buld me a calculator

[timid silo]

i can make u the program if u want

in java

[uncut girder]

the real purpose btw. is calculating latency in video games

you have framerate and monitor refreshrate

I translated that to passangers and trains xD

[timid silo]

ah xdd

do u need this for a school project

?

[uncut girder]

If you can make a program like this it will be awesome

[timid silo]

i can make it in java

[uncut girder]

Ok awesome

maybe I can pay you a little bit

I don't know how to compile or anything though you would need to explain like im five xD

[timid silo]

let me make it first and show u how it works

[uncut girder]

you da best

[timid silo]

first is first

u introduce 2 numbers

so ask the user to introduce 2 numbers

ill make it 1 sec

ill make these 2 steps

the scanner keyboard

is to start the scanner that will detect our inputs from keyboard

when we ask it

whats next

to ask for 2 numbers

train and passenger

gives a message to the user

that asks him how many minutes it takes for a train to pass by

[uncut girder]

ok

[timid silo]

he will type in a number

we detect it

mintrain = next keyboard number input

now we ask how many for a passenger

now we must check if they are integers or no

we need to make a bolean method

that returns true if they are integers

or false if they arent

ill call the method

isInteger

this means the method is boolean (returns true or false)

and it does it based on a arbitrary number double x

[uncut girder]

yup I kinda understand what ur doing so far

[timid silo]

if x is a integer x

then we say its true

else

its not an integer

ill try the method to make sure its good

we are trying with 1

true its a integer

ill try with 1.3

[uncut girder]

epic fail

[timid silo]

false

its not an integer

so the method works

1 is a integer

1.3 is not a integer

now well have to pass both our numbers through the method

if they are both integers

we write this

if the minutes of train

&& means and

if the minutes of train and the minutes of passenger are integers

im making them pass through the method isInteger

then

to get the cycle

we just multiply them

so far we have this

we ask to introduce minutes of passenger and minutes if train

and a method that tells us if they are integer

if they are both integers then cycle = mintrain*minpassenger

[uncut girder]

what we do if they are not?

[timid silo]

we have to write that

lets write if they are both integers first

then we jump on if they are not

[uncut girder]

yup

[timid silo]

so we need a loop

loops in java are written as

for

[uncut girder]

Is what we will do now kind of like a simulation?

[timid silo]

now we are doing it for only integers

[uncut girder]

k

[timid silo]

let me eat my chicken first

10-15 min

[uncut girder]

absolutely

I need quick break too

[uncut girder]

I'll add you incase the thread gets closed

[timid silo]

i accepted you

im working on it

[obtuse pebble]

@uncut girder Has your question been resolved?

[timid silo]

i stopped working on it ill get back to it in 20 minutes

its just one of the loops

is faulty

[obtuse pebble]

@uncut girder Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[fathom bridge]

how do you do 2^​x*​3^​x

[fathom bridge]

i feel like you cant

atleast not at my level

[rigid lintel]

you want to write $2^x \cdot 3^x$ as one exponent?

[warm shale]

heavy

[fathom bridge]

right

[rigid lintel]

alright

since the exponents are equal

you can write $2^x \cdot 3^x = (2\cdot 3)^x = 6^x$

[warm shale]

heavy

[fathom bridge]

ah okay so you factor out x

[rigid lintel]

well i wouldnt call it that hahah

[fathom bridge]

for some reason i thought the base had to be the same in order to multiply

exponents

[rigid lintel]

usually that is the way

but thats kind of different from this

this only works because the exponents are equal

they are both to the same power

[fathom bridge]

okay that makes sense

thanks

.close

[obtuse pebble]

Channel closed

Closed by @fathom bridge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[covert bison]

hello

[grizzled shore]

What’s your question?

[obtuse pebble]

@covert bison Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[mossy pewter]

When only given M, do you use the slope to find two points to create a function?

[mossy pewter]

I was asked to create a function with a slope of m = -3

Teacher explained how to create a function given points and a slope, or just points, but not just the slope

[high lily]

pick any random point you want

[mossy pewter]

given the parameters of the slope?

[high lily]

if all they want is something with a slope of - 3

[mossy pewter]

yep thats all, just a graph with the slope of -3

so I could just use

(-6, -6)?

[high lily]

()

(-6,-6) or any other point

[mossy pewter]

literally any other point?

doesnt matter what it is?

[high lily]

yes

[mossy pewter]

wow

ok

thanks

[high lily]

literally anything is thats all in the question wants

I mean to make this easier,
start with slope intercept form
y=mx+b
and choose any value you want for b

[mossy pewter]

yeah I just plugged it into point slope formula

M(x - a) = y - b

[obtuse pebble]

@mossy pewter Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[rigid yacht]

mornin USAAAA

[rigid yacht]

is this correc

they didn give answer key 💔💔💔

[wise talon]

is that the original question

[rigid yacht]

yessirr

[wise talon]

k is a constant

yes

[rigid yacht]

ok so it’s correc

lesgoooooo

[wise talon]

probably didn't look carefully

[rigid yacht]

huh

[wise talon]

assuming you solved it correctly yes

your methodology is fine

[rigid yacht]

aightt

https://tenor.com/view/ebarbs-kiss-clash-royale-clash-royale-kiss-emote-gif-26331316

have a kiss

anyway 2nd question

i tried long divisioning it

but I can’t get rid of the x in it

and I can’t use quotient/product rule

hold up

I can separate the ln

into 2 different

ln

ohhhh

that’s how I was supposed to approach the question

right?

[wise talon]

yes

[rigid yacht]

alright

[wise talon]

that is eye watering

[wise talon]

oh yeah the

bad notation there

[rigid yacht]

a?

oh

[wise talon]

do not leave the operator there after you already differentiated

finish everything in 1 step

[thorn stag]

dont attempt to derive a fractional log

just use log identities to break it into subtracting and then distributive the derivative operator

[wise talon]

,w differentiate e^sqrt(x)

[warm shale]

Results provided by WolframAlpha

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

Requested by DerpZ#9141

[rigid yacht]

could you circle where I left the operator

im confused

[wise talon]

3rd line

[wise talon]

this is simply wrong

[rigid yacht]

ohhh

so I was supposed to differentiate them all at one go

[wise talon]

yes

[thorn stag]

No it is not

[rigid yacht]

wha abou dis

[thorn stag]

$ln \left ( \dfrac{3x}{x^2+5} \right ) = ln(3x)-ln(x^2+5)$

[warm shale]

SirMonkey

[thorn stag]

$\dfrac{d}{dx} ln \left ( \dfrac{3x}{x^2+5} \right ) = \dfrac{d}{dx}ln(3x)-\dfrac{d}{dx}ln(x^2+5)$

[warm shale]

SirMonkey

[thorn stag]

$\dfrac{d}{dx}ln(u) = \dfrac{u'}{u}$

[warm shale]

SirMonkey

[thorn stag]

$\dfrac{d}{dx}ln(3x) = \dfrac{3}{3x}= \dfrac{1}{x}$

[warm shale]

SirMonkey

[thorn stag]

$\dfrac{d}{dx}ln(x^2+5)= \dfrac{2x}{x^2+5}$

[warm shale]

SirMonkey

[thorn stag]

$\dfrac{1}{x}-\dfrac{2x}{x^2+5}$ is the derivative of the log but obviously you simplify

[warm shale]

SirMonkey

[thorn stag]

Plug it into any derivative calculator and you'll get that, it makes it so easier than dealing with fractions and using quotient rule.

[thorn stag]

3c is correct but you might find it easier to just rip the constant e out and just deal with the exponential with the variable

[covert bison]

help me

https://cdn.artofproblemsolving.com/attachments/b/a/26f2119e4d1d603c222cc3ee64c4e07ba8bb37.pdf

no. 2

can someone pls help me?

[rigid yacht]

alrighty thanks y’all

.close

[obtuse pebble]

Channel closed

Closed by @rigid yacht

Use .reopen if this was a mistake.

[covert bison]

wow

hello?

[thorn stag]

12

[covert bison]

the answer is 10

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[lost violet]

solve for x

💀

[wooden cipher]

!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

[timid silo]

anyone know how to use a graph calc

[fierce lagoon]

Let u = 5^x

Make a substitution

[lost violet]

I love you

.close

[obtuse pebble]

Channel closed

Closed by @lost violet

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

Just looking to see if awsner is right

Order of operations

I can show work i needed

[teal turret]

Could u send a clearer pic

[timid silo]

looks pretty clear to me but alright

Exponent is 2 btw

looks like an 7

[tardy epoch]

Can you use a calculator to check your work

,calc (5/4+2/3)^2

[warm shale]

Result:

3.6736111111111

Stephen

Stephen

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Sorry im back

so i have simplifiy before using exponent??

[warm shale]

Stephen

[teal turret]

But that’s not necessary here

Just simplify the inside first then go on

[timid silo]

Huh?

[tardy epoch]

Just do this

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[lunar hedge]

Let V be a vector space over R. Let u, v, and w be distinct vectors in V. Prove that {u, v, w} is
linearly independent if and only if {u + v, u + w, v + w} is linearly independent.

[lunar hedge]

nvm i solved it

.close

[obtuse pebble]

Channel closed

Closed by @lunar hedge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

would the degree of 12x^2y^3 be 5 or would it be 6

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[timid silo]

The lengths of the sides of a right triangle are consecutive even integers. Find these lengths. (Hint: Use the Pythagorean theorem.)

[trail cloak]

Let the triangle have legs: n and n+2

And hypotenuse n+4

And use pythag to find the n that satisfies those constraints

[timid silo]

would it be n+1 for both?

[trail cloak]

Thanks

[trail cloak]

[timid silo]

wait I think im confusing myself

[trail cloak]

Let it be n, n+2, n+4

[timid silo]

okay thank you

[trail cloak]

Pythag and solve for n

Then find n, n+2, and n+4

[timid silo]

alright tha k you consecutive integers are confusing me alot rn

[trim portal]

This thing is true because they must be consecutive even numbers, even numbers are away from each other exactly 2 units (2,4,6,8...) If you let n be the first one, then the second side must be n+2, and the hypothenuse (greatest side) has to be 2 units away from the n+2, so hypothenuse is n+4

[trim portal]

triple of consecutive integers is eg: 1,2,3 or 6,7,8. But they must be also even. So it can be e.g. 2, 4, 6. or 10, 12, 14

[timid silo]

.close

[obtuse pebble]

Channel closed

Closed by @slim jasper

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[warm kelp]

Why is the area of this square 16r not 16r^2 as 4r X 4r = 16r^2

[warm kelp]

So solution error

@timid silo

[obtuse pebble]

@warm kelp Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[hushed raft]

how do you find the derivative of ln(x) given using the fact that e^x and ln(x) are inverses?

[hushed raft]

basically f(x) = e^x and g(x) = ln(x) and f(x) = g^-1 (x)

I thought that all you do is 1/ d/dx(g^-1(x))

in other words, 1/ d/dx(f'(x))

which is = to 1/ e^x

but this isn't the derivative of g(x)

because g'(x) = 1/x not 1/e^x

[hollow portal]

do you know how to do implicit differentiation

[daring rock]

I think you're missing a piece of this formula

$\frac{d}{dx} f^{-1}(x) = \frac{1}{f'[f^{-1}(x)]}$

[hushed raft]

ye

[warm shale]

tatpoj

[daring rock]

you were just doing d/dx f in the denominator

[hushed raft]

wont that turn out to be 1/f'(e^x) which is 1/e^x?

bceause f'(x) = e^x

[daring rock]

wait, f(x) = e^x and f^-1(x) = lnx right?

so the denominator will be e^(ln x)

= x

[hushed raft]

OH

so its

1/f'(f^-1(x))

which is 1/f'(ln(x))

which is 1/e^ln(x)

and e^ln(x) cancels out to

just x

so its 1/x?

[daring rock]

yep

[hollow portal]

yes

[daring rock]

What Nav was saying is a good point though. You don't really need to memorize this formula, it's just implicit differentiation in a different package

[hushed raft]

thank you 🙏🙏🙏

[hollow portal]

yeah at first I thought something like y=lnx and then do $e^y=e^{lnx}$

whoops

[warm shale]

ℕαv

[hushed raft]

@hollow portal @daring rock closing this, thanks for the help again

[hollow portal]

and then differentiate from there

[hushed raft]

.close

[obtuse pebble]

Channel closed

Closed by @hushed raft

Use .reopen if this was a mistake.

[daring rock]

yep

[daring rock]

Np 👍

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[whole temple]

how to solve sin(2theta) = 0.882?

[whole temple]

problem asks for two angles

I understand to isolate using arcsin, but this only produces one angle, how would I get another angle

[obtuse pebble]

@whole temple Has your question been resolved?

[spring sandal]

whats the domain

for theta

[whole temple]

hmm not sure, it's a physics problem so I would assume first quadrant

[spring sandal]

in degrees ?

[whole temple]

yes in degrees

[spring sandal]

ok

So its in first quadrant

Meaning its between 0 and 90

correct ?

[whole temple]

yes

[spring sandal]

But we are talking about 2 theta

meaning we need to double the domain

so its between 0 and 180

so what was the degrees u got ?

[whole temple]

I got 30.942

for one value of theta

[spring sandal]

OK

Value for 2theta is 61.88 ?

correct ?

in first quad

[whole temple]

yes

[spring sandal]

OK

So do you know the rule that the second quadrant sin is positive

[whole temple]

because y is positive?

[spring sandal]

No

ASTC rule

[whole temple]

oh wow

now I do haha thanks

[spring sandal]

ok

so Sin is positive

and there is a rule where sintheta = sin(180-theta)

so your second option should be sin(180-61.88)

so the angle should be wat

118.12

So your two angles r 61.88, 118.12 (but this is still 2theta)

so u need to half both angle

then your done

[whole temple]

gotcha, it wouldn't involve arcsin? to isolate theta? that's how I solved the first one

[spring sandal]

Yea

u use arcsin

to find the initial angle of 61.88

but u need to find the angle in the second quadrant too

because the domain is 2theta not theta

[whole temple]

okay cool, so the other question it's asking is what additional data would I need to know which angle is the correct one

I'm assuming that it's talking about restricting the domain

[spring sandal]

whats the context of the question ?

like was that the firsrt part

[whole temple]

here I will upload the question

[spring sandal]

ok thanks

[spring sandal]

wouldn't it be the range and max height ?

not too sure

[whole temple]

that's true the range is provided, but if we get the max height would that determine which angle to use?

[spring sandal]

Well the greater the angle woudln't it reach a greater height

because the vertical component of the velocity will be greater

[whole temple]

yes but it wouldn't go as far right?

[spring sandal]

yea
exactly

hence range and height

[whole temple]

ah, but height isnt given

so we would need to know the height?

[spring sandal]

no

true

just range

if height was includded then that would be too

but for this question range shoiuld be sufficient

since the smaller angle will have a greater horizontal component

[whole temple]

I'm confused, so the other angle would be arcsin(180-61.88) / 2?

[spring sandal]

ok

here i explain

so

if we are talking about two quadrants

like first two

the angles are theta and 180 -theta

i can write a solution if u want

my explanation is probabl ypoor

[whole temple]

so need to solve for theta

so this what I did for the first angle,

sin(2theta) = 0.882
2theta = arcsin(0.882)
theta = 1/2 * arcsin(0.882)

so theta = 30.942 degrees

[spring sandal]

i did with a mouse

[whole temple]

I get domain error when I plug it into calculator

plug 180 - 61.88

[spring sandal]

what do u mean

[whole temple]

I try to do arcsin(118.12)

[spring sandal]

u dont ahve to

the angle is 118.12

not the ratio

[whole temple]

oh I see

it is implied from the other angle we solved for?

[spring sandal]

yes

u need to know that second quadrant angle is represented by 180 - theta

where theta is the angle in the first quadrant

[whole temple]

so for this problem these are the two angles

so for that last question what additional information would allow you to determine definitely which of the two angles was used to produce the data in the table?

[spring sandal]

range

because the two different angles will produce two different ranges

[whole temple]

but range is already given in the table

[spring sandal]

yes

so u test both angles and see which it amtches

[whole temple]

and what would be different for the two angles

the horizontal component of velocity, correct?

[spring sandal]

yes

[whole temple]

I believe I understand, it wouldnt be the vertical component because we didn't raise the cannon or anything

[spring sandal]

ok

nice

[whole temple]

thanks! just need to practice more

[spring sandal]

yep

[whole temple]

could you also figure it out if you knew time?

like flight time

which angle to pick I mean

[spring sandal]

not too sure

[obtuse pebble]

@whole temple Has your question been resolved?

Channel closed

Closed by @whole temple

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[bleak skiff]

.close

[obtuse pebble]

Channel closed

Closed by @bleak skiff

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[solid igloo]

Help with 3c

[quaint glen]

,rccw

[solid igloo]

My current working out has lead to nothing

The textbook answer is

,rccw

[obtuse pebble]

@solid igloo Has your question been resolved?

[solid igloo]

absofuckinglutely

[obtuse pebble]

Channel closed

Closed by @solid igloo

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[distant schooner]

.

a printing

[distant schooner]

a printing firm charges 35 for printing 600 sheets of headed notepaper and 47 for printing 800 sheets

find a formula, assuming the relationship is linear, for the charge, C, in terms of number of sheets printed, n

so i tried to find out the cost of each headed notepaper

so i did 600/35

which came out to be a decimal around 17

and did the same with 800/47, came out to be a decimal around 17

but those two decimal values were different

so i was kinda stumped there

[lost tree]

i remember seeing probably this very problem once before lol

[distant schooner]

its unlikely

lol

im using a 2023 book

[lost tree]

linear does not mean that those ratios must be the same

draw a graph

on the x axis, mark 35 and 47

[distant schooner]

ok

[lost tree]

on the y axis mark 600 and 800 above 35 and 47

draw a line passing through them

find the equation of that line

the thing is though that this way, ud see that the firm charges some amount even when u r not printing anything

but i guess u r to assume that n >= 1

[distant schooner]

alright

fuck i forgot how to find the linear equation

ok so

rise over run

800/47

[lost tree]

uhm no

[distant schooner]

wat

[lost tree]

yes rise over run

but the run is not 47 when the rise is 800

[distant schooner]

what

it is

but then it cant be 800/35

[lost tree]

you are gonna need to find two points on the line like this

like A, B

what two points are u using to find 800/47

[distant schooner]

oh shit

i was thinking 47 of the length

mbmb

so

800/35

thats the two points

[lost tree]

what are

[lost tree]

this is also wrong

tell me the coordinates of two points that u know are on the graph

[distant schooner]

or wait

800,47

wait no

47,800

35,600

[lost tree]

yes one point is (47, 800)

[lost tree]

yeah the other (35, 600)

the rise would be the difference in their y coordinates

[distant schooner]

yes

[lost tree]

and run wud be the difference in their x coordinates

[distant schooner]

yea

[lost tree]

if u consider these two points

so can u tell me then what rise/run shud be

[distant schooner]

47-35/800-600

12/200

so that simplifies to 3/50

[lost tree]

u found run/rise

not rise/run

[distant schooner]

oh woops

so

800-600/47-35

[lost tree]

yeah u just turn this around

[distant schooner]

so

thats the gradient

[lost tree]

yeah now find the intercept

u know ur line now has the equation y = (50/3)x + b

find b

[distant schooner]

yeah

ok

[lost tree]

using the coordinates of a point u know lies on the graph

[distant schooner]

b is also 50/3

[lost tree]

yeah that sounds right

so now u have it

[distant schooner]

so

now

i have

y= 50/3x + 50/3

[lost tree]

replace the x with n, coz we marked the number of sheets along the x axis

[distant schooner]

so thats my equation

ok yeah

[lost tree]

the x coordinate represents a number of sheets

and replace the y with C

[distant schooner]

also

what if i mark the number of sheets along the y axis

same answer?

[lost tree]

y = (50/3)x + (50/3)

[distant schooner]

C = (50/3)n + (50/3)

oh is that the answer

[lost tree]

well if u do that and proceed in exactly the same way we did now, u wud find an equation for n in terms of C

the problem is asking you for an equation for C in terms of n

[lost tree]

if u do this, you can rearrange the equation you get to find C in terms of n

[distant schooner]

oh

alright

thanks man

[obtuse pebble]

@distant schooner Has your question been resolved?

Channel closed

Closed by @distant schooner

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[sage dagger]

How do you simplify the surd sqrt(125)

[near comet]

can you spot any square factors of 125

if it helps you can prime factorise it

[sage dagger]

25, 5, 15?

[earnest elk]

No 15

Prime factorise 125 like DylanPi said

[near comet]

if you can spot 25 as a factor, then you can do this

[sage dagger]

i dont understand the 5sqrt5 part

[near comet]

the square root of 25 is 5 right

[sage dagger]

yes

[near comet]

so root(25) times root(5)

is just 5 times root(5)

or 5 root(5)

[sage dagger]

ohh ok

how about

2sqrt(8) - 5sqrt(27)

[near comet]

just follow the same steps as above - try to split up 8 and 27 into a two factors, one of which is a square

then simplify after, just like we did above

[sage dagger]

2root(4) ?

[royal basin]

no, $2 \sqrt{8} \neq 2 \sqrt{4}$.

[warm shale]

Ann

[sage dagger]

or is it the other way around

[analog robin]

2 root 2

Times by 2

So it would be 4 root 2

8 has factors 4 and 2 so u can root the 4 which is two and then keep the factor of two in the root

And multiply everything by two

[sage dagger]

hm

and for the 5root27

factors for 27 are 9,3

[sage dagger]

the solution says 15sqrt(3)

i dont get where the 15 came from

.close

[obtuse pebble]

Channel closed

Closed by @sage dagger

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

Channel closed

Closed due to the original message being deleted

[dense adder]

.close

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[proven zephyr]

i need help again
$10a^{\log(a)} = 2a^2$

[proven zephyr]

$10a^{\log(a)} = a^2$

what am i supposed to do here

[warm shale]

help

[proven zephyr]

change of base?

,,log(a) = \frac{\log_a(a)}{\log_a(10)} = \frac{1}{\log_a(10)}

[warm shale]

help

[proven zephyr]

nah it's just making the question worse

[proven zephyr]

help this

multiple choice btw

original equation

[daring rock]

Take log of both sides, you should get a quadratic in log(a), right?

[proven zephyr]

hmm

[daring rock]

log(a)log(10a) = 2log(a)

[proven zephyr]

oh yeah

[daring rock]

log(a)[log(10)+log(a)] = 2log(a)

[proven zephyr]

the power is only on a tho

wait... am i doing it wrong..

[daring rock]

Do you see how we can get here?

[proven zephyr]

no...

[proven zephyr]

wait is log(a) log base a or log base 10 of a?

oh wait i think i get it

ohh we take log_10 from both sides

ok thanks

i get it now

ok another question

what does the | mean?

[tranquil quiver]

just notation

[proven zephyr]

x|x = 0

[tranquil quiver]

its

a line

[proven zephyr]

oh

ok

[tranquil quiver]

seperating them

so cool

[proven zephyr]

ok next question

wth am i supposed to do here

[tranquil quiver]

find radius and circumference in terms of A and B ig

[proven zephyr]

this is way too advanced for me 🥲

[tranquil quiver]

well do u know how to find radius of circle from equation

like x^2 + y^2 = 25

do you know the radius of the circle that this graph will make

[proven zephyr]

tbh no

i haven't learned what to do with ax and by

[tranquil quiver]

without ax and by

[tranquil quiver]

for this

yk?

whats radius

[proven zephyr]

it just

25

right?

[tranquil quiver]

❌

[proven zephyr]

huh

[tranquil quiver]

is 5

sqrt(25)

[proven zephyr]

oh

wait

[tranquil quiver]

x^2 + y^2 = r^2 yk

[proven zephyr]

i've never noticed that

yeah

[tranquil quiver]

what about

[proven zephyr]

i've just learned the equation like last week...

[tranquil quiver]

(x-3)^2 + (y+2)^2 = 25

whats radius

[proven zephyr]

5 still

or no?

[tranquil quiver]

yeah

5 still

so put the equation you have

in that form

[proven zephyr]

cuz
(x - h)^2 + (y - k)^2 = 25

oh ok

[tranquil quiver]

ye

do you know complete the square

[proven zephyr]

yes

ohhhh

[tranquil quiver]

you can use it and put that equation in that form (x-h)^2 + (y-k)^2 = r^2

just complete the square for x and y

seperately

[proven zephyr]

(x - 1/2 a)^2 + (y - 1/2 b)^2 - 1/4 a^2 - 1/4 b^2 + 10 = 0

kinda messy ima just put it on latex

[warm shale]

help

[proven zephyr]

$(x - \frac12 a)^2 + (y - \frac12 b)^2 = \frac14 a^2 + \frac14 b^2 + 10$

[warm shale]

help

[proven zephyr]

like this?

[proven zephyr]

did i do it right?

[tranquil quiver]

idk

lets hope so

[proven zephyr]

$(x - \frac12 a)^2 + (y - \frac12 b)^2 = \frac14 a^2 + \frac14 b^2 - 20$

[warm shale]

help

[proven zephyr]

second part it basically this

[tranquil quiver]

sick

[proven zephyr]

so i just take the r^2

$\frac14 a^2 + \frac14 b^2 + 10 = 4(\frac14 a^2 + \frac14 b^2 - 20)$

[tranquil quiver]

neh

[proven zephyr]

ye

[warm shale]

help

[proven zephyr]

realized

[tranquil quiver]

but

its 2 times the length

is that

circumference

or what

[proven zephyr]

the radius

the radius of the circle .... is two times the length of

[tranquil quiver]

the other circle

idk what length refers to tho

[proven zephyr]

yes

should be the radius

$\frac14 a^2 + \frac14 b^2 + 10 = a^2 + b^2 - 80$

[tranquil quiver]

u sure

[proven zephyr]

now what

[warm shale]

help

[proven zephyr]

positive.

[tranquil quiver]

but it say length of the circle

[proven zephyr]

uhh

ok im doubting myself now

[tranquil quiver]

im doubting myself too

lets remain in doubt forever

[proven zephyr]

wait ima ask the teacher

[tranquil quiver]

W

[proven zephyr]

Teacher is always online!

Fast response too.

[tranquil quiver]

u should ask em for help then

👍 👍 👍

[proven zephyr]

how to do it if it's talking about the radius tho?

[tranquil quiver]

who knows

[proven zephyr]

...

okay

wait... it's a multiple choice

i can just check the answers one by one

wait but how...

oh nvm i think i can

[proven zephyr]

he's confused 👍 👍

[tranquil quiver]

great teacher

[proven zephyr]

this is a competition's past paper

so yeah

he said it's probably the radius but he's not sure

ok thanks btw i think i can do it now.

❤️

i can't prove it mathematically, but i can prove it with trial and error

[tranquil quiver]

$\sqrt{\frac{A^2}{4} + \frac{B^2}{4} + 10} = 2\sqrt{\frac{A^2}{4} + \frac{B^2}{4} -20}$

[warm shale]

dog.

[tranquil quiver]

sick

i feel like u need more info or im missing smth

[proven zephyr]

nvm i dont think i can do it

yeah

it failed

trial and error

for the first time

nvm

it's

the

CIRCUMFERENCE

[tranquil quiver]

egg

[proven zephyr]

my teacher said circumference

[tranquil quiver]

honestly idekhow that helps

[proven zephyr]

yeah

[tranquil quiver]

ig we know A^2 + B^2 > 80

so then maybe only one of those answers meet that restriction

[proven zephyr]

teacher also confused with the problem now

he said radius

again

[tranquil quiver]

this shit gotta get worked out first

come back to me when u got an answer

[obtuse pebble]

@proven zephyr Has your question been resolved?

[proven zephyr]

radius!

and

also

my teacher said use the radius formula?

ah nvm he was referring to the equation

he said: find A^2 + B^2

didn't say how....

all he said was simplify

[warm shale]

help

[proven zephyr]

and go back to this

ah yes

simplify

very easy to say

i don't get it

<@&286206848099549185>

[proven zephyr]

this is the question

[proven zephyr]

i got to the radius's equation

[tranquil quiver]

im an enjoyer of this idea

a^2/4 + b^2/4 + 10 =a^2 + b^2 - 20

30 =3(a^2 + b^2)/4

120 =3(a^2 + b^2)

a^2+b^2 = 40

[gloomy vector]

huh

[tranquil quiver]

rad = sqrt((a^2+b^2)/4 - 20)

= sqrt(40-20) = sqrt(20) = sqrt(4*5) = 2sqrt(5)

[proven zephyr]

oh yeahhh

[tranquil quiver]

i think i fucked smth u

[proven zephyr]

wait yeah

wait lemme try

[gloomy vector]

$\frac{a^2}{4}+\frac{b^2}{4}+10 =a^2+b^2-20$

[warm shale]

Skill_Issue

[gloomy vector]

^^

[tranquil quiver]

thanks for resending what we already had 👏

[proven zephyr]

a^2/4 + b^2/4 + 10 =a^2 + b^2 - 20
1/4(a^2 + b^2) + 10 = a^2 + b^2 - 20
30 = 3/4 (a^2 + b^2)

like this?

and then just divide

[tranquil quiver]

yeah

that gets you 40 as well

[proven zephyr]

40 = a^2 + b^2

should be correct no?

[gloomy vector]

in my oponion latex form is way easier to read and understand

[tranquil quiver]

no actually

thats impossible

caue a^2 + b^2 > 80 has to be true

[proven zephyr]

how so?

[tranquil quiver]

maybe its circumference after all..

[tranquil quiver]

otherwise u get negative in the sqrt

so that circle cant exist

[proven zephyr]

wait... lemme try

oh yeah

that circle

cant exist

[tranquil quiver]

yes

but circumference wont give a nice answer

UGHHh

OH SHIt

this shit

$\frac{a^2}{4}+\frac{b^2}{4}+10 =a^2+b^2-20$

is wrong

[warm shale]

dog.

[tranquil quiver]

its meant to be

$\frac{a^2}{4}+\frac{b^2}{4}+10 =a^2+b^2-80$

[warm shale]

dog.

[tranquil quiver]

@proven zephyr

80

cause al that shit on the rhs got multiplied by 4

[proven zephyr]

oh

yeah

[tranquil quiver]

so then a^2 + b^2 = 90 * 4/3 = 120

[proven zephyr]

so... we got it?

[tranquil quiver]

so then we get radius is sqrt(120/4 - 20) = sqrt(30-20) = sqrt(10)

epic

[proven zephyr]

yeyyy

At LASTT

thanks alot @tranquil quiver

t!rep @tranquil quiver

[sinful light]

MarveI has given @tranquil quiver a reputation point!

[proven zephyr]

.close

[obtuse pebble]

Channel closed

Closed by @proven zephyr

Use .reopen if this was a mistake.

[tranquil quiver]

i got my own satisfaction from solving this problem

it really shouldve of been difficult but the hardest part was the ambiguous wording of the question

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[frosty goblet]

Hello Guys!
How would one restructure the first term to get that solution?
I know that 5^-k = 1/5^k but I dont understand the first part.
Thanks for the help!

[latent walrus]

they just split it at all the +

[warm shale]

Max..

[frosty goblet]

And I can do 2^2 and 2^k instead of 2^2k?

Resulting in:

• 2^k / 4^k = 1/2 ^k
• 3/4^k = 3 x 1/4^k

[warm shale]

Max..

[slim lake]

yeah nice

[frosty goblet]

Ah lovely...sometimes I feel like an idiot when I stare at it for minutes and cant see it

Thanks guys!

[slim lake]

we have all been there bro

have a good day

[frosty goblet]

You too!

.close

[obtuse pebble]

Channel closed

Closed by @frosty goblet

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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>.
• Type the command .close to free the channel when you're done.
• 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.

[umbral nest]

any idea how they are converting 20grades 10/27×1000min = 20g 37' 3.7"

[umbral nest]

when i devide 1000 by 27 i get 37.037

how many minutes is 1 grade

or how many second is 1 grade?

[timid silo]

bruh those units are actually so obscure

okay so you want to turn $18^\circ 20'$ into radians if i understood correctly?

[warm shale]

Lixera

[umbral nest]

it says convert 18°20' into centesimal measure that is in grades
ive already done the converting to radians

[timid silo]

yeah haven't heard of those in my life surprisingly, i will look them up for a bit and see if i can help

[umbral nest]

yea am also lookin em up

thank you so much you are so nice 😭

[timid silo]

really not sure where you would ever use this stuff, maybe aviation or aeronautics?

[timid silo]

oh anytime pff

[umbral nest]

girl idk either.....

[timid silo]

pffft

[proven zephyr]

may i type something here? srry if im interrupting

[umbral nest]

yea but ig i found it

[proven zephyr]

BROO YOU PUT IT ON THE WRONG EQUATIONNN

[timid silo]

ah what is it

[proven zephyr]

IT SHOULD BE THE
SQUARE ROOT OF A^2 + B^2 + 10 ONEE

ok thanks

ima leave now

[timid silo]

really have never have heard of either of those

interesting

very interesting