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2 years ago

X 10/10=18/12 solve for x

1 answer:

8 0

Simplifying: x+1=18/12

Simplify: x+1=1.5

Combine Like Terms: 1x=1.5-1

Simplify: 1x=0.5

Divide: x=0.5

Hope I helped :)

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Find the numbers b such that the average value of f(x) = 7 + 10x − 9x2 on the interval [0, b] is equal to 8.

barxatty [35]

Answer:

The numbers $b$ such that the average value of $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ on the interval [0, b] is equal to 8 are $b_{1} \approx 1.434$ and $b_{2} \approx 0.232$ .

Step-by-step explanation:

The mean value of function within a given interval is given by the following integral:

$\bar f = \frac{1}{b-a}\cdot \int\limits^b_a {f(x)} \, dx$

If $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ , $a = 0$ , $b = b$ and $\bar f = 8$ , then:

$\frac{1}{b}\cdot \int\limits^b_0 {7+10\cdot x -9\cdot x^{2}} \, dx = 8$

$\frac{7}{b}\int\limits^b_0 \, dx + \frac{10}{b} \int\limits^b_0 {x}\, dx - \frac{9}{b} \int\limits^b_0 {x^{2}}\, dx = 8$

$\left(\frac{7}{b} \right)\cdot b + \left(\frac{10}{b} \right)\cdot \left(\frac{b^{2}}{2} \right)-\left(\frac{9}{b} \right)\cdot \left(\frac{b^{3}}{3} \right) = 8$

$7 + 5\cdot b - 3\cdot b^{2} = 8$

$3\cdot b^{2}-5\cdot b +1 = 0$

The roots of this polynomial are determined by the Quadratic Formula:

$b_{1} \approx 1.434$ and $b_{2} \approx 0.232$ .

The numbers $b$ such that the average value of $f(x) = 7 +10\cdot x - 9\cdot x^{2}$ on the interval [0, b] is equal to 8 are $b_{1} \approx 1.434$ and $b_{2} \approx 0.232$ .

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The answer is <span>1.082 x 10^<span>8
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