help-13

[silver sphinx]

cos3x?

[lusty grotto]

Is that all

[silver sphinx]

so 2sin3x cos3x

[lusty grotto]

No I mean

But yeah this is what you shud be doing

The derivative of sin3x is not just cos3x tho

Notice cos3x is again a function of a function, the outer function is cos, the inner one is 3x

[silver sphinx]

(cos3x)3+3x-sinx

like this?

[lusty grotto]

What is the +3x-sinx

The derivative of sin3x is just (cos3x)*3

[silver sphinx]

oh okay

[lusty grotto]

Is it clear why

[silver sphinx]

no i didn't understand

[lusty grotto]

In sin3x, your outer function is sin, and inner function is 3x

Is that clear

[silver sphinx]

yeah

[lusty grotto]

Like what you have is sine OF 3x, so you first find 3x (inner function), and pass it as input to sine (outer function)

sin(3x)

Okay now the derivative of the outer function is the cos function right?

The derivative of the inner function is just 3

Is that clear?

[silver sphinx]

yeah

i got it

[lusty grotto]

Okay so cos OF 3x, multiplied by 3

Derivative of outerfunction evaluated on the inner function = cos(3x)

Multiplied by derivative of inner function = 3

Clear?

[silver sphinx]

yep

ohhhh okay

so we derivative for both sine and 3x

okay i got it

it's like there's two part right?

[lusty grotto]

Yes, this is a function of a function of a function

So you must use the chain rule twice

[silver sphinx]

okayokay

so it's like 6 sin3x cos3x

[lusty grotto]

Yes

[silver sphinx]

thank you soooooo much

and i have another question

can u help as well?

[lusty grotto]

Right

What is it

[silver sphinx]

the first part i can do it

but second idk

[lusty grotto]

What partial fractions did you find

[silver sphinx]

i'm doing it now

cuz i did it wrong before

[lusty grotto]

Okay can you send the right fractions after you find them then

[silver sphinx]

alright

okay so i got

5/x-3 - 2/3x+2

[lusty grotto]

$\frac{5}{x-3}-\frac{2}{3x+2}$

Right

[wraith dagger]

𝕾ilver𝕾oldier

[silver sphinx]

excatly

[lusty grotto]

Use parentheses though

Anyway

[silver sphinx]

alright

[lusty grotto]

So then $\der{}{x}\frac{13x+16}{(x-3)(3x+2)}=\der{}{x}\left(\frac{5}{x-3}-\frac{2}{3x+2}\right)$

[wraith dagger]

𝕾ilver𝕾oldier

[lusty grotto]

Right?

[silver sphinx]

yeah

[lusty grotto]

Now you can split the derivative on the right hand up into two parts

[silver sphinx]

wdym

[lusty grotto]

$\der{}{x}\left(\frac{5}{x-3}-\frac{2}{3x+2}\right)=\der{}{x}\frac{5}{x-3}-\der{}{x}\frac{2}{3x+2}$

[wraith dagger]

𝕾ilver𝕾oldier

[silver sphinx]

oh okay

[lusty grotto]

Coz the derivative of a sum or difference of functions is just the same as the sum or difference of their derivatives

Right now you just have to differentiate these functions

Can you find $\der{}{x},\frac{5}{x-3}$ ?

[wraith dagger]

𝕾ilver𝕾oldier

[silver sphinx]

okay i'll try

so i use y'=(vu'-uv')/v^2

and u is 5 v is x-3

[lusty grotto]

Yeah u can use the quotient rule

[silver sphinx]

okay

i do it now

how can i do for x-3

[lusty grotto]

The derivative of x-3

Shud be simple enug ig

[silver sphinx]

so just 1

[lusty grotto]

Yes

[silver sphinx]

okay~

and u'=0

[lusty grotto]

Yes

[silver sphinx]

so -5/(x-3)^2

[lusty grotto]

Yes

[silver sphinx]

that's the first part right?

[lusty grotto]

Yes

Now you need to differentiate the other one

[silver sphinx]

okay let me try

so

-6/(3x+2)^2

[lusty grotto]

Yes

[lusty grotto]

Now just put them in there

[silver sphinx]

so -5/(x-3)^2+6/(3x+2)^2

[lusty grotto]

Yes

[silver sphinx]

niceeee omg

[lusty grotto]

Jsyk, you could also have treated $\frac{5}{x-3}$ as $5(x-3)^{-1}$, and you could have used the chain rule to differentiate it..sometimes this is easier to do

[wraith dagger]

𝕾ilver𝕾oldier

[lusty grotto]

And similarly the other fraction

[silver sphinx]

true!

it's better

[lusty grotto]

[silver sphinx]

how can i close this

[lusty grotto]

Type .close

[silver sphinx]

okay

btw can i add u as friend

.close

[cedar kiln]

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[sinful lark]

how to verify a solution.

[crimson sedge]

wdym

[sinful lark]

verifying solutions of slope

so like

[crimson sedge]

just show what the original problem is, your solution to it, and your doubt

[sinful lark]

is it gray

[crimson sedge]

what are lines 3 and 4

[crimson sedge]

the text? it is readable

[sinful lark]

okay

line 3 is :

y=-1/3x - 4

line 4 :

y=2x+11

[crimson sedge]

yeah okay

so what is your main confusion rn

[sinful lark]

how to like

solve it

because this is the only thing idk how to do

[crimson sedge]

so for part a i assume u found the intersection graphically?

[sinful lark]

yes

[crimson sedge]

i dont believe thats correct but i guess u wouldn't have been able to get it right completely from just the graph

anyways like

do u know how to solve a system of two equations like

[sinful lark]

substitution and elimination

i know how to do

[crimson sedge]

yuh

so it is the same idea for [
y = -\f13 x - 4\
y = 2x + 11]

[sinful lark]

ohhhhh

OMG

so we need to use substitution

let me see

[crimson sedge]

actually

you r very lucky

because the equation is very nice the way it is given rn

because like

[sinful lark]

🙏🙏🙏

[crimson sedge]

do u see how both equations are y =

it means that if we want to find the intersection between both, the y values have to be the same

so you can immediately find the x value by setting the two equations equal to each other

i hope this made sense?

[sinful lark]

yes

[crimson sedge]

did u understand what i meant though

[sinful lark]

yes i did

[crimson sedge]

okay uh

ur algebra

isnt right

[sinful lark]

wdym

[crimson sedge]

[
-\f13 x -2x {\c r {}\neq{}} -\f53 x
]

[sinful lark]

oh

oopsie

[crimson sedge]

lmaoo

[sinful lark]

it’s 7/3

[crimson sedge]

just 7/3?

[sinful lark]

yes

i think

[crimson sedge]

nop

[sinful lark]

unless i can simplify it

[crimson sedge]

it is not positive 7/3

[sinful lark]

-7/3

[crimson sedge]

yee

okay

go from there

[sinful lark]

will it change my answer though

[crimson sedge]

yes

i mean

entirely different number so yes

[sinful lark]

o

i got

45/7

[crimson sedge]

uh

ur sign is wrong

[sinful lark]

-45/7

[crimson sedge]

yes

there you go

[sinful lark]

but

it didn’t prove that the solution is true

[crimson sedge]

it does?

[sinful lark]

how

[crimson sedge]

those are the calculations?

[sinful lark]

-45/7 x = 15

[crimson sedge]

what?

[sinful lark]

oh wait

no

-45/7 = x is the answer but it isn’t equal to -9?

[crimson sedge]

like if you want to prove it holds compute what
[
y= -\f13\parens{-\f{45}{3}}-4 \
y = 2\parens{-\f{45}{3}} +11
]
is for both of them, u will find out they will be the same value

[sinful lark]

this is confusing me

omg

wait

[crimson sedge]

the coordinates of the intersection u wrote in part a, (9,7), are wrong

[sinful lark]

Omgg

it isn’t wrong

i accidentally made the slope negative

i wrote it as pisitive on my ipad

;-;

[crimson sedge]

oh

what do u mean

like the equations themselves are wrong?

[sinful lark]

it would be this

but without the negative slope

of 1/3

[crimson sedge]

ah so are the equations [
y = \f13 x - 4\
y = 2x + 11]

[sinful lark]

yes

[crimson sedge]

ok uh

ur answer in part a is stil wrong

[sinful lark]

how

[crimson sedge]

idk why u wrote -9 in the equations but 9 in the box

u should fix that

[sinful lark]

i did

[crimson sedge]

because its not (9,7)

[sinful lark]

i fixed both of them

[crimson sedge]

its (-9,-7)

oh okay

good

anyways

[sinful lark]

[crimson sedge]

do the same thing we just did lmaoo

[sinful lark]

okay

[crimson sedge]

yeah okay good job

woooooo

[sinful lark]

thank you

so much

.close

[cedar kiln]

Channel closed

Closed by @sinful lark

Use .reopen if this was a mistake.

[crimson sedge]

lmaoo good luck

[cedar kiln]

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[sinful lark]

back again Lol

[cedar kiln]

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

[sinful lark]

1

[silk thistle]

What do you know about subsitution

[sinful lark]

they are equal equations

idk

[silk thistle]

Substition is the method you have to use in order to find x and y

Heere you have 2 equations

[sinful lark]

yes

[silk thistle]

You know that y = 3x

[sinful lark]

yes

[silk thistle]

So the substition method means that you can plug y=3x into the second equation

[sinful lark]

so

y=3x+x=-32?

no wait

[silk thistle]

Yes 3x + x =-32

Because x + y =-32 and y=3x

[sinful lark]

but

[silk thistle]

There is no y= in what you wrote

[sinful lark]

actually nvm

okay so 3x+x=-32

[silk thistle]

So you have 4x = -32

[sinful lark]

which is -8 when divided

[silk thistle]

Yes

So what do you conclude ?

[sinful lark]

x=-8

[silk thistle]

Yes

And what is y then ?

[sinful lark]

um

-32?

[silk thistle]

..

What did you do to find -32 ?

[sinful lark]

wait

i have to

use x to find y

[silk thistle]

Yes

Exactly

[sinful lark]

-24

[silk thistle]

Yes

Because of y = 3x

So the answer you have is (x,y)=(-8,-24)

[sinful lark]

play

okay

can we try one more

[silk thistle]

Hope you understood

Yes

[silk thistle]

I let you do it

Write what you would do if you were alone

[sinful lark]

okay

because y=2

we can do

2x+2=8

and we do inverse operations

so we get rid of the 2

which gives us 2x=6

and we divide it by 2

to get 3

and since we already have our y because y=2

we have our solution

(3,2)

[silk thistle]

Yes !

You did it alone, great !

[sinful lark]

yay

okay

i need help on one more

[crimson sedge]

hihi again

[sinful lark]

hello

[crimson sedge]

so same old same old

sub or eliminate

[sinful lark]

sub

[crimson sedge]

okay like

did u

try that

lmaoo

[sinful lark]

no idk what to do

[crimson sedge]

you see the second equation?

solve for y in it

[sinful lark]

i’m

um

[crimson sedge]

yea?

[sinful lark]

[crimson sedge]

[sinful lark]

[crimson sedge]

whats the problem?

[sinful lark]

i don’t know what to do

[errant vector]

Plug 2x + 3 from the first equation into the second equation for y

[sinful lark]

Oh

OHHHHHHHH

wait

[errant vector]

Same system

[crimson sedge]

that or solve for y in the second equation and plug it in in the first

[sinful lark]

x-2x+3=-2?

[errant vector]

So if y = — in the first equation, that’s what it equals in the second equation too

x - (2x-3) = -2

[sinful lark]

why parenthesis

[errant vector]

I just do that always

[sinful lark]

oh

[errant vector]

When I’m substituting a variable for an expression

Now distribute the -1

[sinful lark]

-2x+3=-2

?

[errant vector]

Now solve for x

[sinful lark]

-2x=-5

[errant vector]

x = 5/2

Right

-5/-2

[sinful lark]

yes

[errant vector]

So now you know your x value

Plug that into any of the equations to get your y value

The first equation would be easier since it’s already solved for y

[errant vector]

After you get your answer do it this way so your more comfortable solving systems in different ways @sinful lark

[sinful lark]

okay

i got y=8

[errant vector]

Nice check your answer with the other way

[sinful lark]

❔

[errant vector]

y = ~

y = -

• = ~

[sinful lark]

Wha

it says it’s wrong on edulastic

[crimson sedge]

whats wrong

what was the thing and what did u get

[sinful lark]

it says -1 , 1

[sinful lark]

5/2, 8

i think

idk

[crimson sedge]

what was the equation

(s)

[sinful lark]

this

[crimson sedge]

oh right

lets see

uh

yeah (-1,1) is correct

you must have fucked up ur algebra

show what u did

[sinful lark]

the other person helping me said that x=5/2

so that’s what i put

[crimson sedge]

they probably did not confirm...

and just thought u got it right lol

well

its fine, just means you shouldn't blindly believe what others say

[sinful lark]

okay

[crimson sedge]

anyways do u still want to know how to get (-1,1) or

[sinful lark]

yes

[crimson sedge]

okay

show what u did to get (5/2,8) and we can parse our way through that

[sinful lark]

okay hold up i’ll have to get my ipad out

actually no

[crimson sedge]

lmaoo

[sinful lark]

give me one sec

[crimson sedge]

alright just

ping me

because i might just forget this chat

[sinful lark]

5/2

y=2x+3

y=2(5/2)+3

y=5+3

y=8

@crimson sedge

[crimson sedge]

no but like

5/2 is wrong to begin with

so tell me how u found that

[sinful lark]

umm

they said to plug what y= in

so

y=2x+3

[sinful lark]

x-y=2x+3-2?

[crimson sedge]

okay

so

you have

y = 3 + 2x
x - y = -2

in the second equation, i want you to solve for y

such that the equation is like y =

[sinful lark]

i’m not sure how to do it

[crimson sedge]

okay lets do it together

so we need to make that x on the left be on the right

so what do you do to make it be on the right?

[sinful lark]

subtract it

so it would become

[crimson sedge]

yes

exactly

[sinful lark]

-y=-x-2

[crimson sedge]

yes

you are so right

okay

[sinful lark]

and then you divide by -1

[crimson sedge]

yes

[sinful lark]

so it will then become y=x+2

[crimson sedge]

yes

u got it, good job

so now you have
y = 3 + 2x
y = x +2

i want you to do the same trick we did

set them equal

[sinful lark]

yes

3+2x=x+2

[crimson sedge]

yes

so now

[sinful lark]

so we do

[crimson sedge]

lets solve for x

[sinful lark]

3x-x

which is 2

x

2x

[crimson sedge]

no

wait

pause

where did the 3x come from

[sinful lark]

no wait oops

sorry messed up

[crimson sedge]

no worries

[sinful lark]

So it’s 2x - x

which makes it x

so

[crimson sedge]

yeppie

[sinful lark]

3+x=2

[crimson sedge]

yes

[sinful lark]

and then we subtract 3

to get -1

[crimson sedge]

yes

yes

[sinful lark]

so x=-1

[crimson sedge]

ding ding

[crimson sedge]

from the second equation

you have

y = x + 2

plug in -1 in there

what do u get

[sinful lark]

y= 1(-1)+2

y= -1+2

y=1

[crimson sedge]

ding ding

[sinful lark]

yayay

[crimson sedge]

g'jobb

[sinful lark]

okay after that i need help with 2 more give me a second 🙏

i’ll try to solve them and if i get stuck i will ask

[crimson sedge]

alrighty

again just

ping me

[sinful lark]

kk

[cedar kiln]

@sinful lark Has your question been resolved?

[sinful lark]

ok idk if i did this right but

@crimson sedge

[cedar kiln]

@sinful lark Has your question been resolved?

[sinful lark]

;-;

[errant vector]

;-;

I wasn’t here for most of the convo so

[sinful lark]

that’s fine

[sinful lark]

<@&286206848099549185> 🙏🙏

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[fleet terrace]

hi

[fleet terrace]

I am trying to create a very complex problem with an answer of pi
the answer here is currently -pi
so I'm looking for a hard question with an answer of -1
do you have any?

and not e^ipi

$i^2$

[wraith dagger]

skittle

[bright juniper]

Let me rephrase that ain't -π

[fleet terrace]

I don't get it

[fleet terrace]

the answer to this is negative pi

[bright juniper]

Not really

[fleet terrace]

why not?

oh i messed up the variables

you gotta change the n in the first limit to another letter

then it is correct

[cedar kiln]

@fleet terrace Has your question been resolved?

[fleet terrace]

$\arctan x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}x^{2n+1}$

[wraith dagger]

skittle
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fleet terrace]

$$\sum_{k = 1}^{+\infty} \frac{(-1)^k \times (-1)}{2k+1}$$

[wraith dagger]

skittle

[fleet terrace]

$$\sum_{k = 1}^{+\infty} \frac{(-1)^k \times (-1)^{2k+1}}{2k+1}$$

[wraith dagger]

skittle

[fleet terrace]

$\arctan x&=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}x^{2k+1}$

[wraith dagger]

skittle
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fleet terrace]

$-2$

[wraith dagger]

skittle

[fleet terrace]

$\left(\tan\left(\theta\right)^2+1\left(\cos\left(\theta\right)^2-2\right)\right)$

[wraith dagger]

skittle

[fleet terrace]

$\left(\tan\left(\theta\right)^2+1\left(\cos\left(\theta\right)^2\right)-2\right)$

[wraith dagger]

skittle

[fleet terrace]

$\left(\left(\tan\left(\theta\right)^2+1\right)\cos\left(\theta\right)^2-2\right)$

[wraith dagger]

skittle

[cedar kiln]

@fleet terrace Has your question been resolved?

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[foggy steppe]

How did that get to this? Am I missing a crucial rule?

[soft owl]

what is s ?

[soft owl]

they already gives you the rule

$\frac{\dd}{\dd x} \qty(15000 e^{-0.0004x}) =15000* \frac{\dd}{\dd x} \qty(e^{-0.0004x})$

[wraith dagger]

Mehdi_Moulati

[foggy steppe]

but it ended up being a negative 15000, no?

oh

nvm

I am dumb thanks

.close

[cedar kiln]

Channel closed

Closed by @foggy steppe

Use .reopen if this was a mistake.

[soft owl]

anytime

it's happened

[cedar kiln]

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[crimson sedge]

anyone who can direct me w the first step ?

[ornate hearth]

what are you looking for?

[crimson sedge]

need to solve for bottom theta but dont know where to start

[ornate hearth]

was 40° given?

[crimson sedge]

42 is given

[ornate hearth]

oh yes 42°

what about trying to find the other angle?

[crimson sedge]

i found that the last angle of the whole triangle is 48

by just adding 90 and 42 and subtracting 180

[ornate hearth]

thats a good start

[crimson sedge]

where would i go from there, i assume that the inner obtuse would make the upper angle different

[ornate hearth]

maybe try doing some trig to it?

[crimson sedge]

only thing i could think of is trying sin^-1(sin(42)*9/16) but that doesnt make sense since 9 isnt the comeplete side

[ornate hearth]

lets label every angle in terms of theta

The obtuse angle

$180-(42-\theta+48)$

[wraith dagger]

CrEpasPmkinPie

[ornate hearth]

,calc 42+48

[wraith dagger]

Result:

90

[ornate hearth]

what

[crimson sedge]

?

[ornate hearth]

uh

something weird happened

lemme try again

[crimson sedge]

kk

[ornate hearth]

$\angle O = 180-((42-\theta)+48)$

[wraith dagger]

CrEpasPmkinPie

[ornate hearth]

$180-42+\theta-48$

,calc 180-42-49

[wraith dagger]

Result:

89

[ornate hearth]

90

$Obtuse = 90+\theta$

[wraith dagger]

CrEpasPmkinPie

[ornate hearth]

And acute is 90-Ø

so lets do trig to find the adj to Ø

$cos(\theta)=\frac{a}{16}$

[wraith dagger]

CrEpasPmkinPie

[crimson sedge]

wait, wouldnt we be trying to solve for the tinier acute angle?

[ornate hearth]

Theta or the new acute angle?

[crimson sedge]

theta

[ornate hearth]

we are

[crimson sedge]

ohh ok

would the law of sine help solve ? or are we using the above equation

[ornate hearth]

law of sine probably would help

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[restive ore]

Can anyone please help me with distibutive property?

[crimson sedge]

Ask your question

[jagged torrent]

My calculus teacher was gone today and I need help with these problems. Here is some of my attempts.

Allegedly all of these are supposed to come out to be integers but I have no idea how this is possible

<@&286206848099549185>

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[crimson sedge]

Question 31 I don’t understand the diagram at all, could someone help

[crimson sedge]

Got it nvm easy stuff

.close

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[lone mesa]

I need assistance graphing.

I’m trying to factor out a 2.

To find the horizontal shift and the B value for the period.

y = 2 Cos 2(x -?) - 1?

[cedar kiln]

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[lone mesa]

.close

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[vapid sand]

Consider the function f(x)= (x-1)^2

show that the normal also instersects f(x) at x = -1/2

[vapid sand]

So I already found the equation for normal which is

y = -1/2x + 2

[muted bear]

set them equal

[vapid sand]

(x-1)^2 = -1/2x +2

like that?

[muted bear]

indeed

[vapid sand]

because I tried that and somewhere along the line I couldn't get the x into 1 term

[muted bear]

expand the (x-1)^2

[vapid sand]

x^2-2x+1 yes?

[muted bear]

yup

[vapid sand]

so I went ahead a bit and now up to x^2- 3/2x = 1

do I factorise x?

[muted bear]

keep moving everything to the left

[vapid sand]

x^2 - 3/2x - 1 =0

then I put this into the quadratic formula I assume

[muted bear]

yup

[vapid sand]

ah yea that works

I got x= 2 and -1/2

alright perfect thank you garlic

[muted bear]

no problem

[vapid sand]

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[bronze pivot]

We were taught to follow this convention for circuit analysis:

• V +
  ---> low to high - voltage rise so +V

• V -
  ---> high to low - voltage drop so -V

but what if V = (-25)?
how do I write it in the equation and why?

[crimson sedge]

That means the polarity is flipped

Think of it like 2 batteries connected in series with this configuration (+v1-)(-v2+)

[bronze pivot]

lets say
X + X + ___ = 0

how do i write it?

since it's a voltage rise

I think it should be +25

but since there is a sign

I don't know

[bronze pivot]

I understand

it means

instead of
+
-25V

its actually:

25V
+

am I right?

@crimson sedge

[crimson sedge]

I believe so

But yes you should take a second opinion as last time i studied circuits was about an year ago

We are yet to study this chapter in the next grade which will start in about 2 months

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[jade silo]

Simple question am i getting it correctly

[tropic oxide]

the diagonals of a kite don't bisect each other

otherwise correct

[jade silo]

oh yeah thanks i forgot that bisect mean divide into half

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[crimson sedge]

Hello everyone 🙂 I have a time and work question

If I can mow a lawn in 8 hours and my brother can mow the same lawn in 6 hours. Working together, how long will it take us to mow that lawn.

Am stuck because : i am getting a sum of their hours as an answer, that is, 14 hours. Working together, they should take less to mow the lawn.
Thanks

[tropic oxide]

Am stuck because : i am getting a sum of their hours as an answer, that is, 14 hours.
well yeah that is nonsense

[crimson delta]

Imagine that the lawn was lets say 48m^2. How much of the lawn do you finish in an hour and how much does your brother finish

[crimson sedge]

ok so you are saying i need more information to answer the question?

[crimson delta]

No

[crimson sedge]

i can rephrase the question...
I make 1 burger in 8 minutes and my brother makes the same burger in 6 minutes. If we work together, how fast can we make the burger?

I am getting 14 minutes the answer and that is nonsense :p

[crimson delta]

Well with burgers you can't easily split up the work like that but ok.

[tropic oxide]

no, you do not need more info to answer the question, and you do not need the burger rephrasing

[crimson delta]

Can you just answer my question

[tropic oxide]

^

[crimson sedge]

yeah if we have more information it makes sense then.

[crimson delta]

We don't have more information. The 48m^2 is something I made up but we will see that this number doesn't matter at all

[tropic oxide]

there isn't anything sacred about the area of the lawn being proclaimed as 48 m^2. it is just easier this way for calculation purposes.

[crimson sedge]

8/1 + 6/1 , why is method is wrong?

[crimson delta]

Why should it be right

[crimson sedge]

^ its wrong

i think becaue we already have 1 in denominator thats the problem

[tropic oxide]

no you're overthinking it

and you aren't letting us guide you either

are you willing to be guided through how to solve the problem correctly, or would you prefer to keep beating your head against the wall of "why is this wrong method wrong??"

[crimson delta]

Can you please just humor us and answer my question

[crimson sedge]

ok guys, let me type in the real question with lots of info . Hang on ...

[crimson delta]

no

[crimson sedge]

this is real question

George takes 8 hours to copy a 50-page manuscript while Sonia can copy the same manuscript in 6 hours. How many hours would it take them to copy a 100-page manuscript, if they work together ?

i was trying to simply it, but i failed

[tropic oxide]

okay well this problem actually has a difference from the one you had posted originally.

[crimson sedge]

So my thought process
Two ways to solve it :

1. Find number of manuscripts typed in 1 hour working together
2. Time it takes to type 1 manuscript

[tropic oxide]

i think you would be better off counting pages and not manuscripts here.

given that the work times are given for 50 pages but the job you're calculating is 100 pages.

so, yknow, it matters.

[crimson sedge]

8/50 + 6/ 50 = 175/12

then

100/175/12 = 48/7 hours or 6.8 hours

6.8 hours is the answer

But i have a problem

guys i have gone nuts, i'll be back

Thank you Ann and Dena 🙂

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[nova snow]

How do I do part C

[nova snow]

( I know part B is wrong ignore it I fixed it)

[zealous compass]

what have you tried

[tropic oxide]

aren't b and c almost the same problem just with different numbers?

@nova snow what is troubling you with part c?

[cedar kiln]

@nova snow Has your question been resolved?

[tropic oxide]

bruh what

[nova snow]

Oh is it

Oops I meant part D

[tropic oxide]

ok

do you know what it means for a point to lie on a graph

if you dont then just say you dont.

would appreciate not having to see the bot's "has your question been resolved" thing

[nova snow]

Sorry, I was eating dinner I was not anticipating a fast response

[tropic oxide]

could have said "brb food" or something

[nova snow]

My bad

[nova snow]

For a point to lie on a graph would be any point on the graph that touches the line?

[tropic oxide]

overthinking it

the graph of an equation consists of all those points, and only those points, which satisfy the equation.

do you understand this, yes or no

[nova snow]

Yes.

[tropic oxide]

so how do you check if a point lies on a graph

[nova snow]

Possibly input certain values into the equation?

To test if it works or not

[tropic oxide]

what "certain values"

you plug the POINT (i.e. its coords) in

and see if the equation is satisfied or not

it really is as simple as that

[nova snow]

So for example if we want to find (1,-2) we put those values in?

[tropic oxide]

if we want to CHECK whether (1, -2) lies on the graph of y = 3x^2 - 3x - 2, then yes, we plug x=1 and y=-2 into y = 3x^2 - 3x - 2

[nova snow]

If that’s the case, I’m not sure where I’m putting the y=-2 into

[tropic oxide]

y = 3x^2 - 3x - 2

[nova snow]

So 3x^2-3x -2 = -2

[tropic oxide]

you are once again overthinking this

so hard

you have your equation y = 3x^2 - 3x - 2
you replace x with 1 and y with -2

you see if what you get is true or false

[cedar kiln]

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[waxen kestrel]

let A be a 5x5 matrix with det(A)=2. Calculate det(3A^5). I found that det(kA^n)=k^n*det(A)^n. Is this correct? In that case my answer would be 3^5 * 2^5=7776.

[waxen kestrel]

And how can I prove that det(kA^n)=k^ndet(A)^n?

[tropic oxide]

inaccuracy there

you're lumping together two properties of det

[tame wraith]

why is it k^n

i feel like it should just be k

[tropic oxide]

$\det(kA) = k^n \det(A)$, where $n$ is the size of $A$.

and $\det(A^m) = (\det(A))^m$ no matter what size matrix $A$ is (so long as it is square)

[wraith dagger]

Ann

[tropic oxide]

your feelings deceive you; the determinant of kI is k^(order)

[heavy sorrel]

,calc 3^5 * 2^5

[wraith dagger]

Result:

7776

[tame wraith]

yh i just realised, i had thought of the determinant in the same way i do vector magnitudes, but its not the same, since it would be like ka(kd*kh - ke*kg) - kb..., for a 3 by 3 so k^3 factor

glad i used words like "i feel like" instead of "it is"

[waxen kestrel]

How does that differ from what I wrote? I'm sorry for probably being stupid right now

[tropic oxide]

well, you've kind of silently assumed that the exponent on the matrix happens to be the same as its size

which imo is a bit too unjustifiedly narrow

[waxen kestrel]

okay, ill just keep getting nowhere then, since i cant find anything similar online.

[cedar kiln]

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[modern meteor]

guys i need help here

[fast hazel]

i cant rly read your writing, does that say integral of sec(y) from 0 to 1/6?

[modern meteor]

the integral sec y dy from zero to one-sixth of pi is log to base e of the square root of three times the sixty-fourth power of (?)

[fast hazel]

ah ok

i think, without using weierstrass, easiest method is probably knowing that sec x is the same as secx (tanx+secx)/(tanx+secx)

which should hopefully tell you what you want to do next

[modern meteor]

ohh thanks

[modern meteor]

i came to "i"

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[obtuse frigate]

can someone help with these 2 qns

[cedar kiln]

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

[obtuse frigate]

1

<@&286206848099549185>

[crimson sedge]

!15m

[cedar kiln]

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

[obtuse frigate]

oh sorryy

[cedar kiln]

@obtuse frigate Has your question been resolved?

[obtuse frigate]

<@&286206848099549185>

[cedar kiln]

@obtuse frigate Has your question been resolved?

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[vagrant elbow]

Everything boils down to a power of 2

[wraith dagger]

Mehdi_Moulati

[soft owl]

use this rules

to find the answer

[cedar kiln]

@dense jungle Has your question been resolved?

[dense jungle]

cann u showw the workinh

workingggg

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[empty anvil]

how did the top become the bottom?

[vagrant elbow]

Probably factored and cancelled

(x - 1) is a factor of the denominator

[quartz frost]

im pretty sure they are not equal

[lyric narwhal]

ye

[empty anvil]

oh what

[lyric narwhal]

they aren't equal

[vagrant elbow]

Yeah it doesn't make sense because they reduced the degree on top by 2 but only by 1 on the bottom

[empty anvil]

my teacher messed up?

this is for partial fraction decomposition

[lyric narwhal]

ahh

that is different

[vagrant elbow]

Are there more fractions

[lyric narwhal]

yeah

there must be multiple integrals

[empty anvil]

this was all thats given

[lyric narwhal]

maybe they were solving a different integral?

[empty anvil]

oh yea

[lyric narwhal]

i think they were solving the 2x/(x^2-5x+6) integral and not the original

[empty anvil]

Ok sorry for the confusion!

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[unreal pelican]

so is this just asking for the arc length?

[tame wraith]

Yh

[unreal pelican]

so i get one and the answer is saying 2π. are those equivalent

[tame wraith]

Wdym

What is equivalent

[tropic oxide]

you get that the length of one turn is 1?

[unreal pelican]

yes

[tropic oxide]

that's a bit sus

[tame wraith]

Did you forget to integrate

[tropic oxide]

show your work.

[unreal pelican]

how can i integrate when i dont have a or b

[tropic oxide]

one turn

[tame wraith]

@unreal pelican what are the 2 values of t where it makes a turn

[unreal pelican]

0 to 1

right?

[tame wraith]

@unreal pelican put those 2 in

And see if it's correct

As in r(t)

Do you think t = 1 will return you back to your i and j coordinate

You had at t = 0

[unreal pelican]

r(1) = 1/2ocs(1)t + 1/2sin(1)t + √(3/4)

but for arc length arent you supposed to do integrate √(f'(t)^2 + g'(t)^2 + h'(t)^2) ?

[tame wraith]

Yes

But firstly you need to know your t values

To integrate from

[tame wraith]

Is the cos(0)i + sin(0)j = cos(1)i + sin(1)j

Or is it more likely

That

Cos(0)i + sin(0)j = cos(2π)i + sin(2π)j

[unreal pelican]

the second

so the t values are from 0 to 2π

[tame wraith]

Yh

[unreal pelican]

because 0 & 2π are equivalent, thats why they are my t values

[tame wraith]

Yh

[unreal pelican]

i see

i greatly appreciate your help

.close

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[manic barn]

how do i go about this

[vernal charm]

if two side are equal in triangle then angle opposite to sides are equal..

[manic barn]

so 180-(68*2)?

[vernal charm]

ye

[manic barn]

okay

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[wary anchor]

how do i make T the subject

can i divide everything by 500

[long arrow]

take ln(.), yes you can

[wary anchor]

then natural log

[viral cloak]

You're trying to solve for t?

[wary anchor]

yh

[viral cloak]

Yes, you can divide by 500. Then use ln to bring down the exponent.

[wary anchor]

but if i divide everything by t i get $e^01.94t = 2e^(0.1386t)$ do i do $ln(2e^(0.1386t))$ or if i do 2(ln(e^(0.1386t))) idk which one is correct

oh god

[wraith dagger]

Modus

[long arrow]

1st one is correct

and division by 500, not t

[wary anchor]

yeah 500 i mean

lemme try

i got $0.194t = 0.234t$ surely this is not right

[wraith dagger]

yomiko

Modus

[wary anchor]

oh

log rules

i forgot

🤦‍♂️

thanks i got the answer right ;))))))

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[foggy mason]

Anyone have tips for test taking in math?

Recently had a midterm and lost points mostly due to constant small errors like forgetting negatives, sometimes I rush so half I forget numbers lol

[red pumice]

i always double check my work

after finishing the test

[foggy mason]

not always an option tho its pretty close on time

[royal loom]

Well, I would recommend becoming as familiarized with the material as possible. The better you are at the math, the less likely you are to make the small mistakes. Also, if you are better then you can do it quicker and have more time to double check for the small errors aswell.

[red pumice]

true

but if you practice enough youll have time left over

[foggy mason]

I suppose so yeah that's true

I just hate making mistakes like those and getting beamed

well, thank y'all

.close

[cedar kiln]

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[runic garnet]

This is truly great advice

One of the advice of all time

[royal loom]

it really is one of the advice of all time

[red pumice]

lol

sometimes the advice you dont want to hear is the best advice

[cedar kiln]

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[amber crow]

If I have 9 balls labeled 1 to 9 in a box. I choose one at random and remove it from the box and do not put it back. how do i calculate the average number of tries before I get the ball with the number i want?

[grand forge]

well the odds of picking your ball first time is 1/9

2nd time is 1/8

3rd 1/7

[amber crow]

yeah i know that

[cedar kiln]

@amber crow Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[elder elm]

Why is this incorrect? This is my process: mass = integral from 4 to 10 of (17/x) dx
= 17 ln|x| |_4^10
= 17 ln(10/4)
= 17 ln(5/2)

[dire geode]

try 14 instead of 10

[elder elm]

It worked..you added 4 to 10 but not sure if this is supposed to be intuitive...

[dire geode]

it isn't. the problem's poorly worded

[elder elm]

Alright, thanks!

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[cedar kiln]

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[jade stag]

these are composition functions calculus class slides, why on the left side they put parentheses (x+c) and treat it like multiplication to the exterior, but on the right side they dont do the same?

[crimson sedge]

wdym by treat it like multiplication?

[jade stag]

whenever you put parenthesis on something

its like a barrier

in order to get rid of you it you got to take something out or multiply it by something for the parenthesis to go away

thats what i mean

[crimson sedge]

I honestly dont see the difference between how the left and right are presented

your finding f(g(x)) on the left and g(g(x)) on the right

[jade stag]

two different problems

but yes

does this make sense

[crimson sedge]

not really lol

what bit is specifically confusing?

[jade stag]

the left is different from the right

THE ERD

RED

[crimson sedge]

its not just different functions?

[jade stag]

the right is the confusing part

[crimson sedge]

g(x) = x-1

right?

[jade stag]

yes

[crimson sedge]

g(g(x)) = g(x-1)

right?

[jade stag]

ok

[crimson sedge]

okay

lets say j = x-1

g(j) = j-1

do you agree?

[jade stag]

yes

[crimson sedge]

so g(x-1) = (x-1)-1

[jade stag]

yup

now distribute the -1

[crimson sedge]

x-2?

[jade stag]

to the inside of the parenthesis

-x+1

[crimson sedge]

what

theres no multiplying

[jade stag]

compare

it to the left side

if there wasnt multiplication