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.
Answer:
C.
Step-by-step explanation:
If you are distributing, you must distribute the outside to both terms in the inside.
Answer:
616 in²
Step-by-step explanation:
14 × 14 × 22/7
=> 616
Step-by-step explanation:
- 3x+5y=7b equation 1
- 2x+4y=1
2x=-4y+1
- x=(-4y+1)/2
- x=(-4y+1)/2 in equation 1
3×(-4y+1)/2+5y=7b
(-12y+3)/2+5y=7b
(-12y+3+10y)/2=7b
(-12y+3)/2=7b
-12y+3=14b
-12y=14b-3
- y=3-14b/12
- y=3-14b/12 in x=(-4y+1)/2
x=(-4×{3-14b/12}+1)/2
x=(-3+14b)/3+1/2
x=14b/3×1/2