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Lynna [10]
2 years ago
11

What is the volume of a sphere with a diameter of 3.2 m, rounded to the nearest tenth

Mathematics
1 answer:
frez [133]2 years ago
4 0

Answer:

Volume of a sphere = 17.164 m³ (Approx)

Step-by-step explanation:

Given:

Diameter of a sphere = 3.2 m

Radius of a sphere = Diameter / 2 = 3.2 / 2 = 1.6 m

Find:

Volume of a sphere = ?

Computation:

Volume\ of\ a\ sphere =\frac{4}{3}\pi \times r^3 \\\\Volume\ of\ a\ sphere =\frac{4}{3}\times \frac{22}{7}  \times (1.6)^3 \\\\Volume\ of\ a\ sphere =\frac{4}{3}\times \frac{22}{7}  \times 4.096\\\\Volume\ of\ a\ sphere =17.164

Volume of a sphere = 17.164 m³ (Approx)

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To prove:

$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1

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$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}

Multiply first term by \frac{1+cos x}{1+cos x} and second term by \frac{1-cos x}{1-cos x}.

        $= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}

Using the identity: (a-b)(a+b)=(a^2-b^2)

        $= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}

Denominators are same, you can subtract the fractions.

       $= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}

Using the identity: 1-\cos ^{2}(x)=\sin ^{2}(x)

       $= \frac{1+\cos^2 x}{\sin^2x}

Using the identity: 1=\cos ^{2}(x)+\sin ^{2}(x)

       $=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}

       $=\frac{\sin ^{2}x+2 \cos ^{2}x}{\sin ^{2}x} ------------ (1)

RHS=2 \cot ^{2} x+1

Using the identity: \cot (x)=\frac{\cos (x)}{\sin (x)}

        $=1+2\left(\frac{\cos x}{\sin x}\right)^{2}

       $=1+2\frac{\cos^{2} x}{\sin^{2} x}

       $=\frac{\sin^2 x + 2\cos^{2} x}{\sin^2 x} ------------ (2)

Equation (1) = Equation (2)

LHS = RHS

$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1

Hence proved.

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