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laila [671]
2 years ago
7

For the function f(x) = 2x - 11, what is the average rate of change over the interval -1 ≤ x ≤ 1? average rate of change:​

Mathematics
1 answer:
kobusy [5.1K]2 years ago
8 0

Answer:

Average rate of change is 2

Step-by-step explanation:

If a function f(x) is continuous over the interval [a,b], then the average rate of change over that interval is \displaystyle \frac{f(b)-f(a)}{b-a}:

\displaystyle \frac{f(b)-f(a)}{b-a}\\\\=\displaystyle \frac{f(1)-f(-1)}{1-(-1)}\\\\=\frac{(2(1)-11)-(2(-1)-11)}{1+1}\\ \\=\frac{(2-11)-(-2-11)}{2}\\ \\=\frac{-9-(-13)}{2}\\ \\=\frac{-9+13}{2}\\ \\=\frac{4}{2}\\ \\=2

Thus, the average rate of  change over the interval [-1,1] for the function f(x)=2x-11 is 2.

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The radius of Circle A is 6 mm. The radius of Circle B is 4 mm greater than the radius of Circle A. The radius of Circle C is 2
labwork [276]

Answer:

Area of circle A =113.14 mm²

Area of circle b = 314.29 mm²

Area of circle C = 452.57 mm²

Area of circle A = 254.57 mm²

2.25 times

Step-by-step explanation:

Area of a circle = nr²

where n = 22/7

r = radius

Circle A's radius = 6mm

Circle B's radius = 6mm + 4mm = 10mm

Circle C's radius = 10mm + 2mm = 12mm

Circle D's radius = 12mm - 3mm = 9mm

Area of circle A = (22/7) x 6² = 113.14 mm²

Area of circle b = (22/7) x 10² = 314.29 mm²

Area of circle C = (22/7) x 12² = 452.57 mm²

Area of circle A = (22/7) x 9² = 254.57 mm²

Number of times the area of circle D is greater than that circle A = Area of circle D / Area of circle A

254.57 mm² / 113.14 mm² = 2.25 times

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