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Svetradugi [14.3K]
3 years ago
13

What is the base area of the cone?

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

Step-by-step explanation:

In order to find the area of the circle base, we need the radius, and we don't have it. We'll use the volume to find the radius, as follows:

V=\frac{1}{3}\pi r^2h and fill in the givens:

75=\frac{1}{3}\pi r^2(5) and solve that for r;

r=\sqrt{\frac{3(75)}{5\pi} } so

r = 3.784698783 and plug that in for the radius to find the area of the circle base:

A=\pi r^2 so

A=\pi (3.784698783)^2 gives you that

A = 45 meters squared, exactly. No decimal.

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3 years ago
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Find the roots of h(t) = (139kt)^2 − 69t + 80
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Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

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The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

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