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3 years ago

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Grade 7 Need help with proportions (PICTURE INCLUDED)

1 answer:

jeka57 [31]3 years ago

4 0

What you are doing with proportions most the time is solving for a variable which in this example I will use an x. You have to do something called cross multiplying. now look at number 21. when you cross multiply you multiply the number of the top part of the fraction on the left side to the bottom part of the fraction on the right side. In that example 1 × 36 = 36. now write 36 = 16x since you have to cross multiply the other parts of the fraction. Then you would divide both sides of the equation by 16. I will make a mixed fraction for this example. 2 4/16 which simplified equals 2 1/4

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The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

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