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2 years ago

Find the area of a circle with a circumference of 6 pi

2 answers:

N76 [4]2 years ago

7 0

To find the circumference of a circle you use the formula C=pi*(radius*radius). If you know that the cuircumference is 6 pi, you first plug that in to solve for the radius.
6pi=pi*(r*r)
6pi/pi=(r*r)
Square root of(6pi/pi)=r
2.45=r
Then to find the area you use the formula A=pi*(r*r) which is
A=Pi*(2.45*2.45)
A=6pi or 18.85

elena-14-01-66 [18.8K]2 years ago

3 0

The circumference formula is 2 $\pi$ r and if the circumference is 6pi then the radius is 3. apply that to the area fromula ( $\pi$ r^2 ) and you get 9 $\pi$

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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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3 years ago

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Mamont248 [21]

Answer:

(x+8,y-6)

Step-by-step explanation:

it's the number of values that x and y differ

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