3.70+1.25+1.48= 6.43.
20-6.43=13.57
You would get $13.57
Answer:
Hello,
Step-by-step explanation:
We divide the interval [a b] in n equal parts.



9514 1404 393
Answer:
20 square units
Step-by-step explanation:
The area is given by the formula ...
A = bh
where b is the base of the parallelogram, and h is the perpendicular distance between the parallel bases. Using the numbers from the figure, we have ...
A = 5·4 = 20 . . . . square units
_____
<em>Additional comment</em>
You will notice this is the same formula as is used for a rectangle. You can consider the parallelogram to have the same area as a rectangle of these dimensions if you look at what happens when you cut the right triangle from the right end and add it to the left end. The result is a rectangle 5 units wide and 4 units high.
The same formula applies even if the skew is so great that the bases do not overlap. (That much skew would not result in a triangle of the kind shown in this diagram.)