L=16
W=21
Set up a systems of equations:
x=length
y=width
xy=336
x+5=y
Use substitution to solve:
x(x+5)=336
x^2+5x=336
Solve using factoring:
x^2+5x-336=0
(x-16)(x+21)=0
x=16 and x= -21
Since length can't be negative, l=16
To find width, plug length into the first equation:
(16)y=336
y=21
So...
L=16
<span>W=21</span>
We will use the right Riemann sum. We can break this integral in two parts.

We take the interval and we divide it n times:

The area of the i-th rectangle in the right Riemann sum is:

For the first part of our integral we have:

For the second part we have:

We can now put it all together:
![\sum_{i=1}^{i=n} [(\Delta x)^4 i^3-6(\Delta x)^2i]\\\sum_{i=1}^{i=n}[ (\frac{3}{n})^4 i^3-6(\frac{3}{n})^2i]\\ \sum_{i=1}^{i=n}(\frac{3}{n})^2i[(\frac{3}{n})^2 i^2-6]](https://tex.z-dn.net/?f=%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%20%5B%28%5CDelta%20x%29%5E4%20i%5E3-6%28%5CDelta%20x%29%5E2i%5D%5C%5C%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%5B%20%28%5Cfrac%7B3%7D%7Bn%7D%29%5E4%20i%5E3-6%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5D%5C%5C%0A%5Csum_%7Bi%3D1%7D%5E%7Bi%3Dn%7D%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2i%5B%28%5Cfrac%7B3%7D%7Bn%7D%29%5E2%20i%5E2-6%5D)
We can also write n-th partial sum:
Answer:
(2 * 6) - 20
Step-by-step explanation:
<u>Step 1: Convert words into an expression
</u>
Multiply two and six then subtract 20
<em>(2 * 6) - 20
</em>
Answer: (2 * 6) - 20
Answer: 3 hours and 20 minutes
Step-by-step explanation:
First , you convert your fraction to a decimal.
6/5 = 1.2
Second , you subtract 4 minus 1.2 to see how much of the work shift does she have left ( since 1.2 hours is what she worked).
4 - 1.2 = 2.8 = 2 hours and 80 minutes (60 minutes is an hour) = 3 hours and 20 minutes