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2 years ago

Suppose that circles R and S have a central angle measuring 80°. Additionally, the measure of the sector for circle R is

Mathematics

2 answers:

GarryVolchara [31]2 years ago

7 0

Answer

B) 9 m

Step-by-step explanation:

18π

x = 9

When circles have the same central angle measure, the ratio of measure of the sectors is the same as the ratio of the radii squared.

riadik2000 [5.3K]2 years ago

6 0

The correct answer is B) 9 m.

The measure of the sector of circle R is 32π/9 m. The measure of the central angle is 80°. This means that the sector is 80/360 = 2/9 of the circle. The area of a circle is given by A=πr², so the area of the sector is A=2/9πr². To verify this, 2/9π(4²) = 2/9π(16) = 32π/9.

Using this same formula for circle S, we will work backward to find the radius:

18π = 2/9πr²

Multiply both sides by 9:
18*9π = 2πr²
162π = 2πr²

Divide both sides by 2π:
162π/2π = 2πr²/2π
81 = r²

Take the square root of both sides:
√81 = √r²
9 = r

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Answer:

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1 year ago

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1 year ago

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2 years ago

Solve the following by using mathematical Induction. For >/ 1

cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

$1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

1. When $n=1,$ we have

in left part $1^2=1;$

in right part $\dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.$

2. Assume that for all $k$ following equality is true

$1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)$

3. Prove that for $k+1$ the following equality is true too.

$1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)$

Consider left part:

$1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}$

Consider right part:

$\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)$

We get the same left and right parts, so the equality is true for $k+1.$

By mathematical induction, this equality is true for all n.

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