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Lerok [7]
3 years ago
7

One of the legs of a right triangle measures 1 cm and its hypotenuse measures 4 cm. Find the measure of the other leg. If necess

ary, round to the nearest tenth.
Mathematics
1 answer:
Ivan3 years ago
8 0

Answer:

3.9 cm

Step-by-step explanation:

One of the legs of a right triangle measures 1 cm and its hypotenuse measures 4 cm. Find the measure of the other leg. If necessary, round to the nearest tenth.

We can solve the above question using Pythagoras Theorem

Hypotenuse ² = Opposite ² + Adjacent ²

From the above question

Hypotenuse = 4 cm

1 leg = Opposite = 1 cm

Other leg = Adjacent = x

Hence

4² = 1 ² + x²

x² = 4² - 1²

x = √ 16 - 1

x = √15

x = 3.8729833462 cm

Approximately to the nearest tenth = 3.9 cm

The measure of the other leg is 3.9 cm

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Assume that it costs Apple approximately E(x) 25,600 + 100x + 0.012 dollars to manufacture x 32GB iPods in a day. (a) The averag
uranmaximum [27]

Answer:

(a)C'(x)=\dfrac{x^2-2560000}{x^2}

(b)x=1600, Minimum Average Cost Per iPod=$132

(c)C''(x)=\dfrac{5120000}{x^3}

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

Step-by-step explanation:

Given that it costs Apple approximately $ C(x) to manufacture x 32GB iPods in a day, where:

C(x)=25,600+100x+0.01x^2

(a)The average cost per iPod when they manufacture x iPods in a day is given by:

Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}

The average cost per iPod is therefore:

C'(x)=\dfrac{x^2-2560000}{x^2}

(b)To minimize average cost of x iPods per day, we set the average cost per iPod=0 and solve for x.

C'(x)=\dfrac{x^2-2560000}{x^2}=0\\x^2-2560000=0\\x^2=2560000\\x=\sqrt{2560000}=1600

The resulting minimum average cost (at x=1600) is given as:

Cost \:Per \:iPod=\dfrac{C(x)}{x} =\dfrac{25,600+100x+0.01x^2}{x}\\\dfrac{25,600+100(1600)+0.01(1600)^2}{1600}\\=\$132

<u>Second derivative test</u>

(c)The answer above is a critical point for the average cost function. To show it is a minimum, we calculate the second derivative of the average cost function.

C''(x)=\dfrac{5120000}{x^3}

At the critical point,  x=1600

C''(1600)=\dfrac{5120000}{1600^3}=0.00125

The result, C''(1600) is positive, which means that the average cost is Concave up at the critical point, and the critical point is a minimum.

3 0
3 years ago
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