Ask question

Ask question

All categories

3 years ago

12

If California has 6,480 residents and each resident produces about 182 cubic feet of garbage per year, how long can the landfill

stay open? Round to the nearest year

1 answer:

svetoff [14.1K]3 years ago

5 0

Answer:

0

Step-by-step explanation:

Just 4 people per year would overflow the landfill, how would it survive 6480 people?

You might be interested in

What is 1.3 repeating as a mixed number?

photoshop1234 [79]

The answer would
be 13/10

5 0

3 years ago

How long does it take for a simple with total number n to decay to 1/3n?

Vlada [557]

<span>Income earned over one year is principal * int rate so we have 0.12*A + 0.13*B +0.14*C = 400 and A + B + C = 3000 and A+B = C</span>

7 0

3 years ago

What would be the steps to solving this problem?<br><br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D" id="TexFormula

lidiya [134]

Isolate the variable by dividing each side by factors that don’t contain the variable.

X = -2

Hope this helps!
Have a great day!

3 0

2 years ago

Read 2 more answers

What is the measure of angle R? A. 30° B. 50° C. 100° D. 180°

PIT_PIT [208]

Answer:

D. 180 degres

Step-by-step explanation:

5 0

3 years ago

What is the square root of -2i?

Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots:

<u>When k=1</u>

<u /> $\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

$\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i$

<u>When k=0</u>

<u /> $\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i$

Thus, the square roots of -2i are 1-i and -1+i

4 0

1 year ago