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3 years ago

A cable hangs between two poles of equal height and 3737 feet apart.

Mathematics

1 answer:

geniusboy [140]3 years ago

6 0

Answer:

15450.4 pounds

Step-by-step explanation:

Given the distance between the poles is D = 3737 feet, and

f(x) = 10 + 0.1x^3/2

= 10 + 1/10 x^3/2

We need to find the arc length of the cable and this is given by:

L = integral of √(1 + [f'(x)]²) dx, between D=0 and D/2

Where f'(x) = d(f(x))/DX

f'(x) = 3/2 × 1/10x^1/2 = 3/20 √x

Hence, L = integral of (√(1 + [3/20√x]²))dx, between D=0 and D/2

L = integral(√(1 + (9/400)x))dx; D=0 and D/2

L = integral(√(1 +9x/400))dx; D=0 and D/2

Let u= 1+ (9x/400) ; du/dx = 9/400

dx = (400/9)du

L= integral (√u) × +(400/9)du; btw D=0 and D/2

L = (400/9) × integral (√u)du

L= (400/9) × (u^3/2)÷(3/2)

L = (800/27)[u^3/2]

Now we replace 1 + (9x/400)

and evaluate between D=0 and D/2 = 1868.5 ft

L= (800/27)[(1 + (9× 1868.5/400)-(1 + 0)]

L=( 800/27) × (9× 1868.5/400)

L = (800 27) × 42.04

L=1246 feet (approximately)

The weight = 12.4 × 1246

Weight = 15450.4 pounds

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3 years ago

Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the

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Answer:

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Step-by step explanation:

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$M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)$

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$M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)$

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