A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 86m long and 62m wide.

Mathematics

1 answer:

-Dominant- [34]3 years ago

6 0

The area of the rectangle is 5,332.

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Find the Average Rate of Change between the values of [0, 1]

e-lub [12.9K]

Answer:

The average rate of change cover the given interval is -6

Step-by-step explanation:

The average rate of change of a function over a range can be calculated by;

f(b) - f(a)/b-a

f(b) and f(a) refers to the respective y values of the x values a and b

Since we have a and b, we can get f(a) and f(b) from the plot

From the graph, since a = 0, f(a) is the y-intercept = 6

b = 1

f(b) = 0

So we have the slope as

= (0-6)/(1-0) = -6/1 = -6

3 0

3 years ago

Given the tables for f & g below, find the following: x f ( x ) g ( x ) 0 9 7 1 4 8 2 3 6 3 5 1 4 7 5 5 1 2 6 6 0 7 0 4 8 2

d1i1m1o1n [39]

$\\(i)\\
\text{We know that the average rate of change of a function f(x) on inerval [a,b] is}\\
\\
f_{avg}=\frac{f(b)-f(a)}{b-a}\\
\\
\text{so using this the average rate of chang of g for x on interval }[3,7] \text{ is}\\
\\
g_{avg}=\frac{g(7)-g(3)}{7-3}\\
\\
=\frac{4-1}{4}\\
\\
=\frac{3}{4}\\
\\
\text{so average rate of change of g on interval [3,7] is }\frac{3}{4}$

$\\(ii)\\ \text{Average rate of chang of f for x on interval }[3,7] \text{ is}\\
\\
f_{avg}=\frac{f(7)-f(3)}{7-3}\\
\\
=\frac{0-5}{4}\\
\\
=\frac{-5}{4}\\
\\
\text{so average rate of change of g on interval [3,7] is }\frac{-5}{4}$

$\\(iii)\\
\text{Since g(4)=5, so we have}\\
\\
f(g(4))=f(5)\\
\\
\text{since the value of f is 1 at x=5. so }\\
\\
f(g(4))=f(5)=1\\
\\
\text{hence we have, }f(g(4))=$ 1

$\\(iv)\\
\text{Since f(3)=5, so we have}\\
\\
g(f(3))=g(5)\\
\\
\text{since the value of g is 2 at x=5. so }\\
\\
g(f(3))=g(5)=2\\
\\
\text{hence we have, }g(f(3))=$ 2