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suter [353]
3 years ago
6

Suppose the area of a circle is 254.4696 square inches. Whats the radius of the circle (use pie=3.1416. A. 18 in. B. 81 in. C. 9

in. D. 9in 2
Mathematics
1 answer:
zalisa [80]3 years ago
6 0

Answer:

The radius of the circle is 9 in

Choice C is correct

Step-by-step explanation:

254.4696 square inches. Whats the radius of the circle (use pie=3.1416.

The area of a circle with radius r is given as;

A=\pi *r^{2}

We substitute the given area and the value of pi given to solve for the radius;

254.4696=3.1416*r^{2}\\\\r^{2}=\frac{254.4696}{3.1416}\\\\r^{2}=81\\\\r=\sqrt{81}=9\\\\r=9

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ella [17]

Answer:

85.5 reals

Step-by-step explanation:

Aqui, usando a função de análise, queremos saber a quantidade de dinheiro que deve ser investido para fornecer o número de visualizações.

A maneira como podemos resolver isso é através da substituição. Precisamos apenas substituir F (x) na equação e resolver x, o que nos dará a idéia da quantidade de dinheiro a ser investido.

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F (x) = 40x + 80

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8 0
3 years ago
1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat
brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

x = 1 is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

x = 2 is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) f(x) = 6x + 6/x

f\prime(x) = 6 - 6/x^2 = 0 and x \neq 0

So, the roots are x = -1 and x = 1

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum.

2) f(x)=6-4/x+2/x^2

f\prime(x)=4/x^2-4/x^3=0 and x \neq 0

So the root is x = 1

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

x = 1 is absolute minimum.

3) f(x) = 8x^2/(x-4)

f\prime(x) = (8x^2-64x)/(x-4)^2=0 and x \neq 4

So the roots are x = 0 and x = 8

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum.

4) f(x)=6(x-2)^{2/3} +4=0

f\prime(x) = 4/(x-2)^{1/3} has no solution and x = 2 is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

x = 2 is absolute minimum.

5) f(x)=8\sqrt x - 6x for x>0

f\prime(x) = (4/\sqrt x)-6 = 0

So the root is x = 4/9

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum.

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Answer:7

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Answer:

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