1. is 450ft, and #2. is 64ft.
9514 1404 393
Answer:
(c) 1.649
Step-by-step explanation:
For a lot of these summation problems it is worthwhile to learn to use a calculator or spreadsheet to do the arithmetic. Here, the ends of the intervals are 1 unit apart, so we only need to evaluate the function for integer values of x.
Almost any of these numerical integration methods involve some sort of weighted sum. For <em>trapezoidal</em> integration, the weights of all of the middle function values are 1. The weights of the first and last function values are 1/2. The weighted sum is multiplied by the interval width, which is 1 for this problem.
The area by trapezoidal integration is about 1.649 square units.
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In the attached, we have shown the calculation both by computing the area of each trapezoid (f1 does that), and by creating the weighted sum of function values.
The shortest leg=2*hypotenuse/3
hypotenuse^2=4*hypotenuse^2/9 + 5hypotenuse^2/9
the longest leg is V5/3* hypotenuse
Answer:

Step-by-step explanation:
Problems like this require that you recognize that the denominator of the right term is a factor of the denominator of the left term. That is, you're supposed to know how to recognize and factor the difference of two squares.

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