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DENIUS [597]
3 years ago
11

A cylinder has a diameter of 6 ft and height of 9 ft. What is the volume of this cylinder? A) 54 ft2 B) 169.8 ft3 C) 254.3 ft3 D

) 1,017 ft3
Mathematics
2 answers:
Margarita [4]3 years ago
7 0
<span>In order to find the volume of a cylinder, we must find the area of its circular face and multiply that value by its height. The area of a circle is (pi)*r^2. The diameter is twice the radius, so the cylinder has a radius of 6/2, or 3 ft. The area of the circular face of the cylinder is therefore 9(pi) or approximately 28.27 sq. ft. Multiplying that area by the height of 9 ft. results in a volume of approximately 254.3 cubic ft., which is answer C.</span>
Alexeev081 [22]3 years ago
7 0
C) 254.3

1. 3.14 * 3^2  
2. 28.26 * 9
3. 254.3
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Answer:

Choice b.

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The highest power of the variable x in this polynomial is 2. In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to 0.)

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Apply the quadratic formula to find the two roots that would set this quadratic polynomial to 0. The discriminant of this polynomial is (5^{2} - 4 \times 1 \times 6) = 1.

\begin{aligned}x_{1} &= \frac{-5 + \sqrt{1}}{2\times 1} \\ &= \frac{-5 + 1}{2} \\ &= -2\end{aligned}.

Similarly:

\begin{aligned}x_{2} &= \frac{-5 - \sqrt{1}}{2\times 1} \\ &= \frac{-5 - 1}{2} \\ &= -3\end{aligned}.

By the Factor Theorem, if x = x_{0} is a root of a polynomial, then (x - x_0) would be a factor of that polynomial. Note the minus sign between x and x_{0}.

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Verify that (x + 2)\, (x + 3) indeed expands to the original polynomial:

\begin{aligned}& (x + 2)\, (x + 3) \\ =\; & x^{2} + 2\, x + 3\, x + 6 \\ =\; & x^{2} + 5\, x + 6\end{aligned}.

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Mashutka [201]

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the question is asking us to add 0 to the given value of y. the given value of y is -11, so place -11 into 0 + y and you get 0 + -11. since you are adding zero, you are making no changes to the y value, so it stays at -11.

<em>hope this helps and have a great day!</em>

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