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Korolek [52]
3 years ago
10

Find the diameter of a circle whose equation is

Mathematics
1 answer:
Olin [163]3 years ago
7 0

Answer:

D) 20

Step-by-step explanation:

x^2 + (y - 2)^2=100

This can be rewritten as

(x-0)^2 + (y - 2)^2=10^2

This is in the form

(x-h)^2 + (y - k)^2=r^2

The radius is 10

The diameter is 2 times the radius

d = 2*r = 2*10 = 20

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Formula for the area of a circle: A = pi * r^2

A semi-circle is half of a circle. So, we will need to divide the area of the total circle by 2 to find the area of a semi-circle.

r = d / 2

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Find the area between the following curves.<br> y=x^3-x^2+x+8 ; y=5x^2-7x+8
solmaris [256]

Find where the two curves meet.

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The area between the curves is

\displaystyle \int_0^4 \left|\left(x^3-x^2+x+8\right) - \left(5x^2 - 7x + 8\right)\right| \, dx = \int_0^4 \left|x(x-2)(x-4)\right| \, dx

When x is between 0 and 2, x(x-2)(x-4) is positive; when x is between 2 and 4, x(x-2)(x-4) is negative. So we split the integral at x=2 to get

\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_2^4 x(x-2)(x-4)\,dx

In the second integral, substitute y=x-2 to get

\displaystyle \int_0^2 x(x-2)(x-4) \, dx - \int_0^2 (y+2)y(y-2)\,dy

\displaystyle \int_0^2 x(x-2) \bigg((x-4) - (x+2)\bigg) \, dx

\displaystyle -6 \int_0^2 x(x-2) \, dx

\displaystyle 6 \int_0^2 \left(2x - x^2\right) \, dx

\displaystyle 6 \left(x^2 - \frac13x^3\right)\bigg|_0^2 = \boxed{8}

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2 years ago
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harina [27]

Answer:

$16

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If 3 rides cost $28 and 6 rides cost 40, each ride costs $4. 28 - (3 x 4) = $16 admission fee

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