help-13

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[crimson sedge]

Kevin, Lenny and Mike are three workers who can do a certain job. If Lenny and Mike work together they will do it twice as fast as Kevin alone. If Kevin and Mike work together they will be three times faster than Lenny alone. How much faster will Kevin and Lenny be relative to Mike alone?

[crimson sedge]

My solution so far:

We can let K, L and M be the time it takes Kevin, Lenny and Mike to do a certain job. We can set up the system of equations
L+M=2K
K+M=3L
Now we can eliminate L to see how fast Mike is relevant to Larry, and then we can do the same for Kevin.
To eliminate L:
-2K+M=-L
K+M=3L

        6K-3M=3L
        K+M=3L
    Subtract them:
        5K-4M=0
        
        5K=4M
        K=4M/5
    Now we can now eliminate K:
        -2K+M=-L
        K+M=3L

        -2K+M=-L
        2K+2M=6L
    Add them:    
        3M=5L
        L=3M/5    
    Now we have 
        K=4M/5
        L=3M/5

how close am i to the end?

[jaunty mural]

We discussed this

'time' is wrong with those equations

Other than that, write down in algebra what you need to answer the question.

[crimson sedge]

hmm

dont i have my answers

[dawn jetty]

you are very close to the end

but M L K would represent rate here

[crimson sedge]

what should i write instead

[crimson sedge]

rate ah

[dawn jetty]

rate of work or amount of work per second \ any unit of time

however the reason you can equate them like that is since Work would be determined by rate*time

but here L*t +M*t = K*2t

therefore t, the time is eliminated from the equation

but besides that you pretty much got everything you need

[dawn jetty]

^

[crimson sedge]

hows this We can let K, L, and M be the rate of work per some unit of time.

[jaunty mural]

'some' isn't necessary

'per unit of time' is perfectly fine

Also say that it is for Kevin, Lenny, Mike respectively.

[crimson sedge]

cool

why respectively?

[crimson sedge]

K=some stuff with M
L=some stuff with M

i have those already

[dawn jetty]

Because K - kevin L-Lenny M- Mike

Let K,L,M be the rate of work per unit of time for Kevin, Lenny, Mike respectively..

[dawn jetty]

but you're not required to compute the rate ratio between each worker

[crimson sedge]

wdym

isnt

K=4M/5
L=3M/5

my answer?

[fallen heath]

Kenny, Lenny both slower than Mike.

[crimson sedge]

are they?

i probably have one more step

[fallen heath]

no you're done

[jaunty mural]

In algebra maybe, but write out the final conclusion in words after that.

To complete the answer.

[crimson sedge]

just say that those are the speeds?

[jaunty mural]

'How much faster will Kevin and Lenny be relative to Mike alone?'

Answer this.

That is the question, remember.

[crimson sedge]

i gotta add them together

We can let K, L, and M be the rate of work per unit of time it takes Kevin, Lenny, and Mike respectively. We can set up the system of equations 
        L+M=2K
            K+M=3L
        Now we can eliminate L to see how fast Mike is relevant to Larry, and then we can do the same for Kevin.
        To eliminate L:
            -2K+M=-L
            K+M=3L
            
            6K-3M=3L
            K+M=3L
        Subtract them:
            5K-4M=0
            
            5K=4M
            K=4M/5
        Now we can now eliminate K:
            -2K+M=-L
            K+M=3L

            -2K+M=-L
            2K+2M=6L
        Add them:    
            3M=5L
            L=3M/5    
        Now we have 
            K=4M/5
            L=3M/5
        Now that we have our individual values of K and L in terms of M, we can add them together to get our final result. This makes K+L=7M/5. 

there

[fallen heath]

'How much faster will Kevin and Lenny be relative to Mike alone?'

[crimson sedge]

and

[fallen heath]

Kinda confusing in that, does it want "how does Kenny and Lenny combined together fair against Mike working alone" vs "how does kenny and lenny fair against mike working alone"

[dawn jetty]

isn't it supposed to be K+L = xM ?

huh

it is

[fallen heath]

it is... which?

[dawn jetty]

confusing

[fallen heath]

XD

Well, I'll believe it's the latter tbh

[dawn jetty]

well.. K+L is more in the spirit of the question

[fallen heath]

And the conclusion would just be: concluding with K = 4M/5, L = 3M/5

[dawn jetty]

I suppose you could write both

[crimson sedge]

sure

does

Now that we have our individual values of K and L in terms of M, we can add them together to get our final result. This makes K+L=7M/5. 

cover both?

[jaunty mural]

Can you write in words

A final sentence in a manner that would actually answer the question

'How much faster will Kevin and Lenny be relative to Mike alone?'

If I asked you this irl and you answered with a bunch of numbers/symbols/letters, I would be wtf.

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

words

[jaunty mural]

'How many more oranges did you buy than apples'

'6O = A'

^ please tell me you don't answer people like this.

'I bought 6 times more apples than oranges' would be an appropriate answer.

[crimson sedge]

lol

ok

[shut zenith]

L + ratio

[crimson sedge]

?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[jaunty mural]

???

[crimson sedge]

how bout this

This means that even is 4 times as fast as Mike divided by 5, and Lenny is 3 times as fast as Mike divided by 5. Their combined speed it 7 times faster than Mike divided by 5.

[crimson sedge]

i had a math class, but i wanted to keep the channel open

[dawn jetty]

I don't think Mike would appreciate being divided by 5 much..

if you're talking about ratios and values, mention the variables you assigned for their rates K,L,M

that would sound much better

[jaunty mural]

When I said words, I was referring only to the final conclusion

Q: 'How much faster will Kevin and Lenny be relative to Mike alone?'

A: 'Kevin and Lenny would be ??? times faster than Mike alone'

Would perfectly suffice to conclude it.

[crimson sedge]

4/5

This means that even is ⅘ times as fast as Mike, and Lenny is 3/5 times fast as Mike. Their combined speed it 7/5 times as fast as Mike.

.close

[cedar kiln]

Channel closed

Closed by @little wave

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[raw sentinel]

Hi guys does anyone know how to do 1c?

[raw sentinel]

im stuck

[jaunty mural]

how are you stuck

continue and isolate for y?

[raw sentinel]

i isolated y alr

[jaunty mural]

y = ???

you have e^y = ???

[raw sentinel]

x^2+c

do i divide e^y both side

[jaunty mural]

How do you solve $$e^x = 3$$

[wraith dagger]

Shuri2060

[jaunty mural]

for x.

[raw sentinel]

i dont rly know

[vernal palm]

do you know what a logarithm is

[jaunty mural]

$$\ln(e^x)=\ln3$$

[raw sentinel]

yea

[wraith dagger]

Shuri2060

[jaunty mural]

I applied the ln function to both sides.

Now we can isolate x because

$$x = \ln(e^x)=\ln3$$

[wraith dagger]

Shuri2060

[jaunty mural]

make sense?

[raw sentinel]

yes

[jaunty mural]

Same thing for the original, then.

[raw sentinel]

is it like this

[jaunty mural]

yh

[raw sentinel]

oh so ln(e^y)=y

?

[jaunty mural]

yh.

And thats it

[raw sentinel]

i see

thanks

is this correct?

[jaunty mural]

yh. you're doing fine.

[raw sentinel]

im stuck here

[jaunty mural]

heheh

Last time you had $$e^y = ???$$

[wraith dagger]

Shuri2060

[jaunty mural]

And you said you were stuck

Can we do something similar . . .

[raw sentinel]

log on both sides?

or times e on both sides

[jaunty mural]

If you log both sides you would get log log

doesnt sound great

multiplying by e doesn't quite work either...

[raw sentinel]

ln on both sides?

[jaunty mural]

You will end up with ln(ln(???))

[raw sentinel]

i have no idea

divide ln on both side?

[jaunty mural]

ln is a function

You need to undo it.

To undo a function, you need to apply its inverse function. . .

like to undo a square root, you square...

[raw sentinel]

is it e^1+y

= e^1=x + k

i used this

im not too sure

[jaunty mural]

$$\exp(x) = e^x$$

[wraith dagger]

Shuri2060

[jaunty mural]

Have you seen this notation before?

Consider applying the exp function to both sides

[raw sentinel]

is it like this

[jaunty mural]

not quite

Dont kill the modulus signs for no reason

and the right isn't quite right.

$$\ln|1+y| = \ln|1+x| + k$$
$\implies$
$$e^{\ln|1+y|} = e^{\ln|1+x| + k}$$

[wraith dagger]

Shuri2060

[jaunty mural]

The entire expression is exponentiated.

$\implies$
$$|1+y| = e^{k+\ln|1+x| }$$

Then you must use rules of exponents to hande the right

[wraith dagger]

Shuri2060

[raw sentinel]

ok

[raw sentinel]

is it like this

[jaunty mural]

yes, there's more to do.

[raw sentinel]

move the 1 over to the right side?

[jaunty mural]

Its inside the modulus sign. You cannot.

I am talking about your right-most term

[raw sentinel]

i am not sure

[jaunty mural]

you have e to the ln

[raw sentinel]

oh so its = 1+x modulus?

[jaunty mural]

yh.

but you're not done.

$$|1+y| = e^k|1+x|$$

[wraith dagger]

Shuri2060

[raw sentinel]

ok

[jaunty mural]

Tell me about $e^k$

[wraith dagger]

Shuri2060

[jaunty mural]

What is it.

[raw sentinel]

it can be any constant?

[jaunty mural]

k can be any real number.

However, what about $e^k$

[wraith dagger]

Shuri2060

[raw sentinel]

it cant be any number

[jaunty mural]

ok.

something more specific though

lol

what can it be

[raw sentinel]

not too sure

[jaunty mural]

If k can be any real number

and I'm asking you about what e^k can be

I'm essentially asking you about the range of the function e^x

[raw sentinel]

ohh

so e^k=e^x

[jaunty mural]

huh no.

no no

I am talking about the function $f(x) = e^x$

[wraith dagger]

Shuri2060

[jaunty mural]

Separately from this question.

What is the range of this function

[raw sentinel]

is it x?

[jaunty mural]

Are you familiar with the words 'domain' and 'range'

[raw sentinel]

no

[jaunty mural]

ok.

This is the graph of e^x

The 'domain' of a function is the set of possible inputs. In this case, it is the real numbers.

[raw sentinel]

ok

[jaunty mural]

The 'range' of a function is the set of possible outputs.

In this case it is...?

[raw sentinel]

0,-5,-10 and -15?

[jaunty mural]

no

If x is allowed to be any real number

The entire continuous red line

represents the function

The graph is obviously zoomed in.

We need to know the entire possible range of outputs

[raw sentinel]

ohh

[jaunty mural]

?

no idea?

[raw sentinel]

it can be any number?

[jaunty mural]

No.

[jaunty mural]

Pictorially, do you know what the range is represented by

[raw sentinel]

until where the graph ends?

[jaunty mural]

The range is represented

By which y-values the graph reaches

You can imagine drawing horizontal lines

and asking which ones touch the graph

eg. y = -1 doesn't touch

y = 1 does touch

Tbh I'm surprised you haven't done range/domain since you're already solving differential equations

It is quite important

The domain of e^x is R (set of real numbers)

The range of e^x is R^+ (set of positive real numbers)

When you exponentiate, you can only get a positive number. not negative. not 0.

[raw sentinel]

Ok

[jaunty mural]

So going back to the question

$k$ is the constant of integration. It can be any real number

[wraith dagger]

Shuri2060

[jaunty mural]

$e^k$ can be any positive real number as a result

[wraith dagger]

Shuri2060

[raw sentinel]

yes

[jaunty mural]

A common technique when you have e^k

is to rename it to a new constant

called A

We let A = e^k

and say A is any positive real

$$|1+y| = e^k|1+x|\quad , k\in\bR$$

$$|1+y| = A|1+x|\quad , A\in\bR^+$$

[wraith dagger]

Shuri2060

Shuri2060

[jaunty mural]

Make sense?

[raw sentinel]

yea

[jaunty mural]

Next, we need to get rid of the modulus

$$1+y = \pm A(1+x)\quad , A\in\bR^+$$

[wraith dagger]

Shuri2060

[jaunty mural]

Can you see why this is true?

[raw sentinel]

yes

[jaunty mural]

Now notice, how A is any positive real

we can again define a new constant

$B = \pm A$

[wraith dagger]

Shuri2060

[jaunty mural]

Which set must B belong to?

[raw sentinel]

1+y

[jaunty mural]

no.

$$A\in\bR^+$$

[wraith dagger]

Shuri2060

[jaunty mural]

Have you seen this set notation before?

'A is in the set of positive real numbers'

[raw sentinel]

no

i have nvr seen this symbol before

[jaunty mural]

Ok

So the e

means 'in'

R means 'real numbers'

• means 'positive'

[raw sentinel]

ok

[jaunty mural]

If you're not familiar with it, then don't use it, no problems

'A is in the set of positive real numbers'

^ however, if this is true, what must B be in

$B = \pm A$

[wraith dagger]

Shuri2060

[raw sentinel]

bcan be negative or positive numbers

[jaunty mural]

In other words.

Any real number except 0

I am guessing your teacher does not need such a thorough solution...

but it is still best to understand

$$1+y = B(1+x)\quad , B\in\bR-{0}$$

$$y = B(1+x) - 1\quad , B\in\bR-{0}$$

[wraith dagger]

Shuri2060

Shuri2060

[jaunty mural]

Then this is almost done except

there is one thing you did earlier

[raw sentinel]

ok

[jaunty mural]

In this step you divided both sides by 1+y

This cannot be done if y = -1

That's dividing by 0.

Therefore, you have to consider this case separately

If y = -1, then dy/dx = 0, from your original ODE

This is consistent, so it turns out you have another solution

If that makes sense

[raw sentinel]

ok

[jaunty mural]

$$y = B(1+x) - 1\quad , B\in\bR-{0}$$

[wraith dagger]

Shuri2060

[jaunty mural]

But notice if we let B = 0

This is exactly that solution

$$y = B(1+x) - 1\quad , B\in\bR$$

[wraith dagger]

Shuri2060

[jaunty mural]

So this is the final solution to the ODE

Normally most teachers do not show these steps

And skip straight to this final line

From here.

[raw sentinel]

I see

[cedar kiln]

@raw sentinel Has your question been resolved?

Channel closed

Closed by @raw sentinel

Use .reopen if this was a mistake.

[cedar kiln]

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.

[crimson sedge]

Solve the equation $\log_3 (\log_2 x) = 2$.

[wraith dagger]

crabbo

[crimson sedge]

how does this work

not the log, but how do i write out the equation

@crimson sedge do you know how to convert a logarithm into an exponent?

[jaunty mural]

'write out the equation' ?

[crimson sedge]

Yes sorry

[jaunty mural]

I'm asking crabbo

[crimson sedge]

Oh

[mellow dirge]

pls

[crimson sedge]

This channel is occupied.

[mellow dirge]

srry

[crimson sedge]

yes

Can you please show

but a logarithm is an equation

isnt it

Equation...?

Equation and logarithm are two separate concepts

[jaunty mural]

this 'log' thing

is a function.

[crimson sedge]

Yes ^

[crimson sedge]

$\log_3 (\log_2 x)$
$3^x=\log_2 x$

[wraith dagger]

crabbo

[crimson sedge]

It accepts an input and returns a value

[crimson sedge]

right

[jaunty mural]

Like a function f(x) kinda function.

[crimson sedge]

Are you sure this is correct

yes

thats how it works

You have $log_a(b) = c$, how would you convert it into an exponent?

[wraith dagger]

Pencil

[crimson sedge]

wdym

Exponent??

$log_a(b) = c$

[wraith dagger]

crabbo

[crimson sedge]

Yes now

lhs is a^x=b

Yes

[jaunty mural]

c

[crimson sedge]

So how would you get rid of log_3?

[jaunty mural]

But to be clear, you should do this in 2 steps.

[crimson sedge]

a^x=b=c?

No

[crimson sedge]

idk

[crimson sedge]

c

[jaunty mural]

$$\log_ab = c$$
$$a^{\log_ab} = a^c$$
$$b = a^c$$

[crimson sedge]

Yes ^

[wraith dagger]

Shuri2060

[crimson sedge]

Basically you get rid of a logarithm by raising itself to its own base

[jaunty mural]

You apply the exp function (raising to a power) to both sides of the equation. In this case it is exp base a

[crimson sedge]

Yes ^

[crimson sedge]

i dont understand

[jaunty mural]

which part.

[crimson sedge]

why are you making both sides the exponent of a?

To get rid of the logarithm

[violet hull]

What about taking antilog?

To finish log?

[crimson sedge]

how is $a^{\log_ab}=b$?

[wraith dagger]

crabbo

[crimson sedge]

Because the base is getting canceled

That's why

[crimson sedge]

Anti logarithm is equivalent to this @violet hull

$b = antilog_a(c)$ is equivalent of saying $b = a^c$

[wraith dagger]

Pencil

[jaunty mural]

This is the most important rule of logarithms.

[crimson sedge]

why?

what exactly is happening

[jaunty mural]

wait did i make a typo

no ok.

$$a^{\log_ax} = \log_a a^x = x$$

[wraith dagger]

Shuri2060

[jaunty mural]

This is how the log function is defined.

It reverses what exponentiation does.

[crimson sedge]

Exponents and logarithms are inverse functions of each other.

[crimson sedge]

right

[crimson sedge]

ah

$\log_3 (\log_2 x)$

[wraith dagger]

crabbo

[crimson sedge]

$3^{\log_3 (\log_2 x)}$?

[wraith dagger]

crabbo

[crimson sedge]

is this right?

[jaunty mural]

?

youve written 2 expressions down, no idea.

[crimson sedge]

Yes but the same will reflect on the RHS too

$3^{\log_3 (\log_2 x)}=3^c$?

now?

[wraith dagger]

crabbo

[jaunty mural]

No c in the original.

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

i meant to write

$3^{\log_3 (\log_2 x)}=3^2$

[wraith dagger]

crabbo

[crimson sedge]

what next?

[crimson sedge]

See this

If you look closely, the logarithm and the base cancels out

The inside part of the logarithm remains intact, don't do anything to that

[crimson sedge]

right

$\log_2x=3^2$

[wraith dagger]

crabbo

[crimson sedge]

Correct

Now how would you get rid of log_2?

$2^{\log_2x}=2^{3^2}$

[wraith dagger]

crabbo

[crimson sedge]

Correct

Do the required cancelation

And simplification

$x=2^{3^2}$

[wraith dagger]

crabbo

[crimson sedge]

Simplify the RHS

And yes correct

Hint: do not multiply 3 and 2

512

Correct

You got your answer 👍

cool :_

.close

[cedar kiln]

Channel closed

Closed by @little wave

Use .reopen if this was a mistake.

[cedar kiln]

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.

[crimson sedge]

why can i do this?

how can i be certain it will always work

why?

.close

[cedar kiln]

Channel closed

Closed by @little wave

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.

[crimson sedge]

help

[crimson sedge]

can somebody explain how to calculate the rank of a mtrix ik how to solve systems in ef and ref but i need to know how to calculate it

lets say i wanna calculate b

[upper garnet]

rref it

then the rank is the amount of non zero rows

[cedar kiln]

@crimson sedge Has your question been resolved?

[severe oyster]

X=64

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

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.

[lethal kernel]

Hello, how do you compute easily 100^2 + 500^2

[lethal kernel]

to get 100 * sqrt(26)

[jaunty mural]

how are the 2 related ?

,w (1^2 + 5^2)100^2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Shuri2060#9525

[jaunty mural]

its just 260k

[lethal kernel]

a^2 + b^2 = c^2

just a simple pythagore problem, i search hyppotenus = c

i got a = 500

b= 100

with calculator is easy, but how do you procede mentally to get 100 * sqrt(26)

[jaunty mural]

I don't understand how the 2 are related.

Or what you are doing.

[lethal kernel]

o kwait

[jaunty mural]

$$100\sqrt{26}$$

This answer is in exact form.

[wraith dagger]

Shuri2060

[jaunty mural]

There is no need to do anything else to it.

[glossy halo]

1^2+5^2 = 26, if you must get your sqrt 26

[jaunty mural]

This in metres is your answer. That's it.

[lethal kernel]

thanks you

[cedar kiln]

@lethal kernel Has your question been resolved?

[cedar kiln]

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[valid bolt]

I want to say x is a natural number ( so whole number greater than 0) in set notation

[valid bolt]

but x is also not 1, 2, or 617

How will I write this?

{x | x ∈ ℕ, x ∉ {1, 2, 617}} ?

{x | x ∈ ℕ & x ∉ {1, 2, 617}} ?

I'm not sure about the & or , sign, because x has to fulfil 2 conditions

[reef laurel]

$x>2, x\not = 617$

[wraith dagger]

Luca

[reef laurel]

For example

Your way works as well

First one

[valid bolt]

Thank you very much

.close

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[crimson sedge]

could someone pls show me how to convert sin 2 theta back to non-theta form?

[lime tangle]

sin(2x)=2sin(x)cos(x) is useful

[crimson sedge]

oh

forgot about that

thx

.close

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[crimson sedge]

could someone tell me if my answer's correct?

[wraith crypt]

,w integrate 1/(sqrt(1+9x^2)) dx

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Camilleone#1596

[crimson sedge]

we're not supposed to use sinh lol

it's a section on trig substitution

[wraith crypt]

lol

i'm a bit confused about your working

in particular what exactly your substitution is

[crimson sedge]

ok for which term?

[wraith crypt]

in the very first line

and moving to the third line

what exactly are you substituting?

[crimson sedge]

see the stuffs on the top right?

[wraith crypt]

i know, but please explain how you got the third line from that

[crimson sedge]

I don't understand

I just plugged everything in

[wraith crypt]

because it looks like you're trying to set $9 + 9x^2 = (9\tan \theta)^2$

[crimson sedge]

1+ 9x^2

[wraith dagger]

Camilleone

[crimson sedge]

oh shit

it's supposed to be 3

not 9

but that doesn't change anything except the 3 LOL

[wraith crypt]

so how in the world did you end up with $1 + \tan\theta$ in the third line?

[wraith dagger]

Camilleone

[wraith crypt]

explain that simplification to me, because i'm not seeing it

[crimson sedge]

it's from the formula sheet we got

1 sec

[wraith crypt]

okay maybe let's start again

according to your formula sheet, what is the substitution you should make?

recall that now you're dealing with a $\sqrt{1 + 9x^2}$ term, and the coefficient of $x$ is not 1

[wraith dagger]

Camilleone

[crimson sedge]

uhh theta = tan^-1(9x/1)

[wraith crypt]

are you sure?

[wraith crypt]

the form is right, but are you sure about the numbers?

[crimson sedge]

based on the triangle

oh my lord

it's 3

3x

https://tenor.com/view/dh9511dh-head-banging-on-the-wall-gif-13355825

[wraith crypt]

right, so if i apply tan to both sides to make it simpler, $\tan\theta = 3x$

[wraith dagger]

Camilleone

[crimson sedge]

right

[wraith crypt]

can you start from here and substitute?

[crimson sedge]

yep could I get back to you in 10 mins?

or 5

[wraith crypt]

ping me when you do

[crimson sedge]

👍

im gonna try to do this again from scratch

@wraith cryptok so that 3x^2 doesn't make sense because it should be (3x)^2 no?

from a^2 + x^2

[wraith crypt]

you have 9x^2 = (3x)^2, yes

what exactly is the problem?

[crimson sedge]

uhh 1 sec

lol

so is this correct?

i guess im a bit confused on what exactly happens to theta when the coefficient of x isn’t 1

does it become 3 theta?

i got an answer that’s differnt from the calculator but it looks like im on the right direction?

[cedar kiln]

@crimson sedge Has your question been resolved?

[wraith crypt]

one way you can think about it is $1 + a^2x^2 = a^2(\frac{1}{a^2} + x^2)$, so you factor out the $a^2$

[wraith dagger]

Camilleone

[fallen heath]

the integration was done right.. I'm concerned about the reverse substitution here

[crimson sedge]

what should i have done?

[upper garnet]

3x=tan@ => arxtan(3x)=@ so for sec@ it becomes sec((arctan(3x))

[wraith crypt]

alternatively, if $\tan\theta = 3x$, this means you can find both $\sin\theta$ and $\cos\theta$ in terms of $x$ (using a triangle), and it also means you can find $\sec\theta$

[wraith dagger]

Camilleone

[crimson sedge]

ok but

I don't get why the answer has the sqrt root while my answer doesn;t

[upper garnet]

did you4use the triangle

[crimson sedge]

no

was I supposed to?

im trying not to

is there a way to solve this without using csc?

[crimson sedge]

.close

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[tiny marten]

for a regular pentagon

can you find the interior angle of it?

[glad cave]

108

540 in total

outer angle is 252

[tiny marten]

good

for a quadrilateral

can you find the sum of all interior angles?

[glad cave]

360

i think 🤔

[tiny marten]

yes

can you continue?

you should be able to solve it now

[glad cave]

ah ok got it now

.close

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[stray bramble]

HOW DO YOU DO THIS ONE

[stray bramble]

NUMBER 1

<@&286206848099549185>

[foggy merlin]

is it a test ?

[stray bramble]

NO LOL

it is a practice exam

and im going over it

had to skip question 1 💀

[sweet vapor]

What’s ur working out

What equations have you set up?

@stray bramble

[stray bramble]

here

it is

@sweet vapor hello ?

[cedar kiln]

@stray bramble Has your question been resolved?

[stray bramble]

<@&286206848099549185>

[azure monolith]

Your are almost done. Dont subtract the values each of the parties have payed but set them equal

So 180 = 420.

[cedar kiln]

@stray bramble Has your question been resolved?

[stray bramble]

wdym

180 = 420

how does that give you an answer

[azure monolith]

both of these values add up to 600 so you need to equal them out

[stray bramble]

they should both be 300

not 180

or 420

[azure monolith]

yes but you need to see those in realtion to the original question

Ron and Andy paid 180$ Robin and Steve paid 420$. The party costs 600 that doesnt change but if Ron and Andy paid 120$ Robin and Steve they would both have paid equally for the party

[stray bramble]

why 120?

@azure monolith

[azure monolith]

because if Ron and Andy pay 120$ to Steve and Robin they would both have paid 300$ each

[stray bramble]

180 + 120 = 300

but robin and steve = 420

so 300 + 420 = 720

not 600

[azure monolith]

why does your cost change ?

why are you changing the cost of the party

[stray bramble]

wdym

im not

[azure monolith]

the party costs 600 that is a fixed value. If one party Ron and Andy pay the other party a certain amount the others pay less for the party. The values 180 and 420 are in relation to the 600

both need to add up to 600

[stray bramble]

so the answer should be c 120

[cedar kiln]

@stray bramble Has your question been resolved?

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[charred tide]

Is there a way to manipulate the triangle inequality to prove this to be true

I thought it had something to do with multiplying both sides by -1

but i guess idk how to think about what happens inside the norm

[foggy merlin]

how is it >= and not <= tho

oh I see now...

|v|=|v-w+w| <= |v-w|+|w|
so |v|-|w|<= |v-w|

[charred tide]

where did the <= |v-w|+|w| part come from?

OOOH wait it's just triangle ineq w v = v - w and w = w?

[foggy merlin]

yes

[flint plinth]

Side note, in exactly the same way you can also show that $|v - w| \geq |w| - |v|$, so in fact (combining this with the inequality in the problem statement) you have $|v - w| \geq |\ |v| - |w|\ |$.

[wraith dagger]

OurBelovedBungo

[charred tide]

thanks!

.close

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[acoustic sail]

what would separate lower lefts equation with the rest

[cedar kiln]

@acoustic sail Has your question been resolved?

[acoustic sail]

<@&286206848099549185>

[dire geode]

Vague question

Do you know what the hole means

[acoustic sail]

The dot cannot be a solution

[cedar kiln]

@acoustic sail Has your question been resolved?

[dire geode]

that's a little sloppy language. you have to be very precise with language in math. the hole means the function y=f(x) is not defined at x=3, meaning 3 is not in the domain of f(x). can you write down the domain for all 4?

[acoustic sail]

what are the equations for each graph

[dire geode]

you don't need to know the function, just the domain

[acoustic sail]

are these the domains

[dire geode]

https://math.libretexts.org/Bookshelves/Algebra/Map%3A_College_Algebra_(OpenStax)/03%3A_Functions/3.03%3A_Domain_and_Range

read "Finding Domain and Range from Graphs"

[acoustic sail]

yes

[cedar kiln]

@acoustic sail Has your question been resolved?

[cedar kiln]

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[spare walrus]

quick question
(V ×V ) ∩V = V
is this true or false?

[modern compass]

pretty sure that's false, unless you're being very lax with notation

[brave bough]

i think that's always false isn't it?? V×V is a set of ordered pairs

[spare walrus]

yeah that's the problem i guess

[brave bough]

oh if V is a null set it would be true

[modern compass]

Yeah, I only bring up lax notation because V is bijective with Vx{v} and so if you were being lazy that's true-ish. But strictly speaking, it shouldn't generally work

[spare walrus]

thank you!

.close

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[pure spire]

hihi! could someone please help with this question? What is the period of this function

[pure spire]

since the period of tan x is pi, shouldnt the period of this be (5/3)pi??

please ping!!

[fallen heath]

@pure spire Yes it'd be 5/3 pi lol

[crimson sedge]

Lol

[cedar kiln]

@pure spire Has your question been resolved?

[pure spire]

could i also have help for this one?

im not really sure how to get the amplitude

since y is now 4 after the 1/4, and then move down by 4 units

the maximum is 0

and then minumum is -8 please ping!!

[cedar kiln]

@pure spire Has your question been resolved?

[versed kayak]

the eqn of a wave is Asin(kx-omegat)

so compare the eqn

1/4 is the amplitude

(i think)

[cedar kiln]

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[wraith dagger]

Kepe

[livid hound]

by definition

its literally how logs are defined

[wraith dagger]

Kepe

[gaunt hamlet]

Therefore if you raise a to that power, you get x by definition

[sand cradle]

thank you!

[cedar kiln]

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[jovial mango]

I need to find for what least value of n this happens, how can I do it without trial & error method

[sand cradle]

isnt it $log_\frac{93}{100}$?

[wraith dagger]

Kepe

[sand cradle]

then you get $log_\frac{93}{100}(\frac{1}{2}) <= n$

[wraith dagger]

Kepe

[jovial mango]

How does it help to solve though ?

[gaunt hamlet]

You found n. What else is there to do?

[jovial mango]

How do I find n from there ?

[sand cradle]

$log_\frac{93}{100}(1/2)$ evaluates to about $9.5513375094$

[wraith dagger]

Kepe

[sand cradle]

$9.5513375094 <= n$

[wraith dagger]

Kepe

[jovial mango]

Umm I was trying to find it without using calculator 😅

I thought it can be reduce to some easier form

[sand cradle]

oh

[gaunt hamlet]

Something isn't right. Setting n = 0 or 1 doesn't solve the inequality

[sand cradle]

wait...

[gaunt hamlet]

The sign flips. log base 93/100 is strictly decreasing, so the inequality flips

That's why

[jovial mango]

Wait maybe i have complicated it, let me send the question

[gaunt hamlet]

n is actually greater than or equal to that expression, not greater than or less

[jovial mango]

a machine depreciates at the rate of 7% of its value at the beginning of a year. if the machine was purchased for $8500, what is the minimum number of complete years at the end of which the worth of the machine will be less than or equal to half of its original cost price?"

[jovial mango]

I tried to solve in my way and ended up with this

[gaunt hamlet]

I think you got it right

Although I doubt you're gonna solve it without a calculator if you're not brute forcing it

[jovial mango]

Someone sent me to solve it, well i can't explain him it other than calculator then 😅 i was trying to find a reduced fraction

Thank you then, I'll close

.close

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[cedar kiln]

@grand quarry Has your question been resolved?

[fallen heath]

:p

$\sum_{k=1}^n k(x_k - l) = \sum_{k=1}^n (kx_k - kl) = \sum_{k=1}^n kx_k - nkl$

[wraith dagger]

Ansh

[fallen heath]

Say, $S = \sum_{k=1}^n kx_k$

[wraith dagger]

Ansh

[fallen heath]

waito

$\sum_{k=1}^n k(x_k - l) = \sum_{k=1}^n (kx_k - kl) = \sum_{k=1}^n kx_k - \sum_{k=1}^n kl$

[wraith dagger]

Ansh

[fallen heath]

$= \sum_{k=1}^n kx_k - \frac{l\cdot n(n+1)}{2}$

[wraith dagger]

Ansh

[fallen heath]

Now, back to our original question.

LHS

$\frac{2}{n(n+1)}\left|\sum_{k=1}^n k(x_k - l)\right|$

[wraith dagger]

Ansh

[fallen heath]

$=\frac{2}{n(n+1)} \left|\sum_{k=1}^n kx_k - \frac{l\cdot n(n+1)}{2} \right|$

[wraith dagger]

Ansh

[fallen heath]

Hmm?

$=\left|\frac{2}{n(n+1)} \sum_{k=1}^n kx_k - \frac{2}{n(n+1)} \cdot \frac{l\cdot n(n+1)}{2} \right|$

[wraith dagger]

Ansh

[fallen heath]

$=\left|\frac{2}{n(n+1)} \sum_{k=1}^n kx_k - l \right|$

[wraith dagger]

Ansh

[fallen heath]

@grand quarry ?

"n" is natural numbers.. right? 🥺

what is l*n(n+1)/2?

nope.. sum of k is the sum of all natural numbers from 1 to n

= n(n+1)/2

[cedar kiln]

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[pearl creek]

Hello what is 5 x 12

[pearl creek]

+9

[south tundra]

I think it'd be much more efficient to use a calculator rather than to ask here

[cedar kiln]

@pearl creek Has your question been resolved?

[pearl creek]

I don’t own a calculator

[livid hound]

there are online calcs
also buy a scientific calculator
also you don't need a calculator for this

[pearl creek]

Yes I do

I’m the highest in my class and I can’t figure it out, we just started learning multiplication

[livid hound]

your initial messages indicate that you are already doing coordinate geometry

you should have learned/know multiplication well before that

[cedar kiln]

@pearl creek Has your question been resolved?

[woven path]

Troll?

[cedar kiln]

@pearl creek Has your question been resolved?

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[trim stump]

oops wrong one

I dont understand this

im guessing

I move a

so i get x-a = 2by^2

[crimson sedge]

Move all the terms from RHS to LHS except y.
Then you will get an equation of the form "something = y".

[trim stump]

whats rhs and lhs

oh

right hand side

ok ok

[crimson sedge]

Hmm

[trim stump]

so when i get

x-2

what do I do with the 2by^2

should I divide the x-2 by 2b?

or should I add it to both sides

[dull anchor]

Why you got „x-2“ where did that 2 come from

[trim stump]

i mean

a

x-a

my bad

[dull anchor]

😁

[trim stump]

so after that

[dull anchor]

That sounds good

[trim stump]

so

x-a/2b

= y^2

[dull anchor]

Yes

[trim stump]

should I square root the rhs?

[dull anchor]

Yes if you do that

You only got an y on the rhs

Thats the answer

[trim stump]

but can u help me out

the site im doing the work

on only accepts certain answers

[dull anchor]

If you square root the right side you have to square root the left side

[dull anchor]

So you have to square root this too

[dull anchor]

Wym

[trim stump]

is this wrong

[dull anchor]

Ahh you forgot something

[dull anchor]

Think about the minus of the „-2by^2

[dull anchor]

This is not right

[trim stump]

im confused

what would be left

if i get x-2/2b on the RHS

[dull anchor]

Yes

What is the lhs

[trim stump]

2^y

y^2

i mean

[dull anchor]

well if you divide 2b from -2by^2, what do you get

[trim stump]

im not sure

[dull anchor]

you forgot the minus

Its -y^2

And before rooting that you gotta change the sign

[trim stump]

so do i add

y^2

to both sides

[dull anchor]

you could do that but there is a way easier way to do that

[trim stump]

is there?

what is it

[dull anchor]

think about

[trim stump]

times by 2

?

[dull anchor]

-y^2 = -1*y^2

instead of adding the -y^2 you could just change the -1 to 1

[trim stump]

so add 1 to both sides

[dull anchor]

Not really

[trim stump]

alr

[dull anchor]

You have to divide on both sides with -1

[trim stump]

ohhh

[dull anchor]

because -1 / -1 = 1

[trim stump]

how should I write that out

with the square root and everything

[dull anchor]

So you got

[trim stump]

i got this so far

so

i add a minus to the 2b?

[dull anchor]

(x-a)/2b = -y^2

Nah dont skip bro

You made a mistake earlier

[dull anchor]

Go on from here

[dull anchor]

If you divde -1 on both sides what you get?

[trim stump]

positive y^2

right?

[dull anchor]

Yes

And on the other side?

[trim stump]

uh

im already dividing by 2b

on the right hand side

so

how would I

divide by -1

[dull anchor]

You just have to divide the complete left side with -1

It just changes the sign of the left side, nothing else

[dull anchor]

You can divide as much as possible

[trim stump]

so what would that look like

[dull anchor]

Wait lemme write that in

$$ -(x-a)/(2*b)$$

Yoo

[trim stump]

yo

oh I get it

then

i square root all that?

[wraith dagger]

Big xdddd

[dull anchor]

Yea

[trim stump]

so let me show u

still says its wrong

should I put brackets

fr

[dull anchor]

Hahah

[trim stump]

im down bad

[dull anchor]

Same

[trim stump]

help me out

[dull anchor]

No need for brackets

[trim stump]

so that is right?

[dull anchor]

That shit should be correct actually

[trim stump]

anythin I could change?

[dull anchor]

Wait try

[trim stump]

hm?

[dull anchor]

$$sqrt((a-x)/2*b)) $$

Write that

[trim stump]

can u show me

[wraith dagger]

Big xdddd

[trim stump]

oh ok

whats the outer brackets for

[dull anchor]

Its that everything is in the square root

[trim stump]

alr

so like this

[dull anchor]

Nah

Put those outer braxkets away

[trim stump]

away?

[dull anchor]

Bro

[dull anchor]

a-x

[trim stump]

OHHH

OK

SORRY

[dull anchor]

Yes i only did that in discord so the program shows it right

[trim stump]

so like this

[dull anchor]

BRO HAHAHA

[trim stump]

can u write it on like ms.paint

[dull anchor]

THE FUCK ARE YOU DOING

[trim stump]

or somn

and send me ss

its been a long day

[dull anchor]

Ok

Wait

[trim stump]

and what about the 2nd

way

u were telling me

[dull anchor]

Does this work?

[trim stump]

isnt there supposed to be

a minus?

bruh

[dull anchor]

No

Just try that

[trim stump]

nope

didnt work

[dull anchor]

You cant be serious