All Answers

#1

+46

+5

to eat PIE AND BACON..... NO, BACON PIE

PiscesNov 30, 2016

#1

0

10000

GuestNov 30, 2016

#1

+12

0

It depends. What is your question?

AnathemaNov 30, 2016

#1

+1491

0

I believe all this question is asking you is... what is 5% of 640.

5% being the margin of error.

640 being your total.

HighSchoolCalculusNov 30, 2016

#1

+37091

0

y=40x + 75 for 70 hours y = total fee = 40(70) + 75 =$ 2875.

ElectricPavlovNov 30, 2016

#2

+7

0

Are you sure that you are sure that he really exists because you may be asleep and dreaming abot him so are you sure that you are sure

GerardWayNov 30, 2016

#2

+1491

+5

I think he means... i as in the imaginary number.

First off you have to know that i = sqrt(-1)

Say if you have sqrt(-6):

Since there cannot be a negative rooted to any EVEN power then there has to be an i in the simplified form.

You would end up getting i * sqrt(6)

Iḿ sure you can tell what I mean at this point.

If you want to use i on the calculator on this site, just type in "i"

On a graphing calculator such as a TI-84, there should be an i key in "shift" mode.

Any other questions?! Just ask!

HighSchoolCalculusNov 30, 2016

#1

0

yes

GuestNov 30, 2016

#1

+7

0

632,432,423,567,975,654,357

Nailed it

GerardWayNov 30, 2016

#1

0

uhh yeah pretty sure you do

GuestNov 30, 2016

#3

+30

0

He goes to the moovies

einsteinofmathNov 30, 2016

#4

+1491

+10

Thanks guys... turns out the problem was not homogeneous at all.

You could not distribute "t" out of tan(tx + ty).

That being said it could not fufill f(tx,ty)= t^n(f(x,y))

HighSchoolCalculusNov 30, 2016

#1

0

19 go fucking kys you degenreate f**k dank master memer weeaboo

GuestNov 30, 2016

#1

+2489

+10

How large is the the mess?

$\large \int_{0}^{60} 70e^{-0.05t} = \\ \int \:70e^{-0.05t}dt \hspace {14pt} |\text {Calculate the indefinite integral }\\ =70\int \:e^{-0.05t}dt \hspace {14pt} | \text { (Move constant) } \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\ $

$=70\int \:-20e^udu \hspace {14pt}|\int f (x (g)) \cdot {g}'(x)dx =\int f(u)du, u=g(x)\\ \hspace {84pt} \Rightarrow \:du=-0.05dt\\ \hspace {84pt} \Rightarrow \:dt=\left(-20\right)du\\ \hspace {84pt}=\frac{d}{dt}\left(-0.05t\right) $

$=70(-20\int e^udu) |\text { Move constant }\\ =70(-20e^u) |\text { using} \int e^udu=e^u\\ =70(-20e^{(-0.05t)}| \:\text{for }u=-0.05t\\ =-1400e^{-\frac{t}{20}}\\ $

$-1400e^{-3} = -69.70 \hspace {14pt}| \text{ for t=60}\\ -1400e^{0} = -1400 \hspace {20pt}| \text{ for t=0}\\\\ -1400-(-69.70) = -1330.3\\ \hspace {1pt}\\ \text {1330 liters leak in the first hour}$

I’m glad I don’t have to mop this up.