Answer:
Let's say you sold some goods to a customer. If the customer is not satisfied with the goods and returns it back to you, then it is called sales return.
A = price of the desktop
b = price of the laptop
now, we know the laptop was more expensive than the desktop by 250 bucks, thus
b = a + 250.
for "a", he paid 8.5% in interest, for "b", he paid 5% in interest.
how much is 8.5% of a? well (8.5/100) * a, or
0.085a.
how much is 5% of b? well, (5/100) * b, or
0.05b.
now, we know the total charges for financing were $296, that means the interest paid in total was 296, thus whatever "a" or "b" are, we know that 0.085a + 0.05b = 296.

how much did the laptop cost? well, b = a + 250.
Suppose R = {1,3,5,7,9,11,13,15,17} and D={3,6,9,12,15,18,21,24,27} r d
Free_Kalibri [48]
The intersection of sets R and D is give by the following set:
R ∩ D = {3, 9, 15}.
<h3>What is the missing information?</h3>
This problem is incomplete, but researching it on a search engine, we find that it asks the intersection of sets R and D.
<h3>What is the set that is the intersection of two sets?</h3>
The set that is the intersection of two sets is composed by the elements that belong to both sets.
For this problem, the sets are given as follows:
- R = {1,3,5,7,9,11,13,15,17}.
- D={3,6,9,12,15,18,21,24,27}
Hence the intersection is given by:
R ∩ D = {3, 9, 15}.
As the elements 3, 9 and 15 are the only ones that belong to both sets.
More can be learned about intersection of sets at brainly.com/question/11439924
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Answer:
The answer is 
Step-by-step explanation:
To calculate the volumen of the solid we solve the next double integral:

Solving:

![[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.](https://tex.z-dn.net/?f=%5B6x%5E%7B2%7D%20%5D%7B%7B1%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%2A%20%5B%5Cfrac%7By%5E%7B3%7D%7D%7B3%7D%5D%7B%7B1%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
Replacing the limits:

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.
Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ 
Solving the double integral with these new limits we have:

This part is a little bit tricky so let's solve the integral first for dy:
![\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%5Cfrac%7B1%7D%7Bm%7D_0%20%5B%7B12x%20%5Cfrac%7By%5E%7B3%7D%7D%7B3%7D%7D%5D%7B%7B1%7D%20%5Catop%20%7Bmx%7D%7D%20%5Cright.%5C%2C%20dx%20%3D%5Cint%5Climits%5E%5Cfrac%7B1%7D%7Bm%7D_0%20%5B%7B4x%20y%5E%7B3%20%7D%5D%7B%7B1%7D%20%5Catop%20%7Bmx%7D%7D%20%5Cright.%5C%2C%20dx)
Replacing the limits:

Solving now for dx:
![[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.](https://tex.z-dn.net/?f=%5B%7B%5Cfrac%7B4x%5E%7B2%7D%7D%7B2%7D%20-%5Cfrac%7B4m%5E%7B3%7D%20x%5E%7B5%7D%7D%7B5%7D%20%5D%7B%7B%5Cfrac%7B1%7D%7Bm%7D%20%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%3D%20%5B%7B2x%5E%7B2%7D%20-%5Cfrac%7B4m%5E%7B3%7D%20x%5E%7B5%7D%7D%7B5%7D%20%5D%7B%7B%5Cfrac%7B1%7D%7Bm%7D%20%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
Replacing the limits:

As I mentioned before, this volume is equal to 1, hence:

The 2 outside the brackets means that everything within the brackets needs to be multiplied by 2