Answer:
(a) 120 square units (underestimate)
(b) 248 square units
Step-by-step explanation:
<u>(a) left sum</u>
See the attachment for a diagram of the areas being summed (in orange). This is the sum of the first 4 table values for f(x), each multiplied by 2 (the width of the rectangle). Quite clearly, the curve is above the rectangle for the entire interval, so the rectangle area underestimates the area under the curve.
left sum = 2(1 + 5 + 17 + 37) = 2(60) = 120 . . . . square units
<u>(b) right sum</u>
The right sum is the sum of the last 4 table values for f(x), each multiplied by 2 (the width of the rectangle). This sum is ...
right sum = 2(5 +17 + 37 +65) = 2(124) = 248 . . . . square units
Answer:
You can draw three rectangles:
1- 28 * 1.
2- 14*2
3- 7*4
Step-by-step explanation:
We have a total area of 28. If we use only integer numbers, we can find all the divisors of 28. The possible rectangles will be organized with them, taking into account that the product of them, which means the area should be 28.
28 = 1*2*2*7
We can organize them as follows:
R1: 28 = 1*28
R2: 28 = 2*14
R3: 28 = 4*7
Finally, we can conclude that there are only three possibilities
1- 28 by 1.
2- 14 by 2
3- 7 by4
The perimeters will be:
Perimeter 1 = 2x1 + 2x28 = 58
Perimeter 2 = 2x2 + 2x14 = 32
Perimeter 3 = 2x4+2x7 = 22
Answer:
5540.340
Step-by-step explanation:
That's My Answer For It