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emmainna [20.7K]

2 years ago

7

Find the volume of a right circular cone that has a height of 12.5 cm and a base with a circumference of 5.8 cm. Round your answ

er to the nearest tenth of a cubic centimeter.

1 answer:

BaLLatris [955]2 years ago

4 0

Answer:

$V = 11.2 cm^3$

Step-by-step explanation:

Given

Shape: Right Circular cone

Height, h = 12.5 cm

Base circumference, C = 5.8cm

Required:

Calculate Volume

The volume of a cone is calculated as thus;

Volume, $V = \frac{1}{3}\pi r^2h$

Where V, r and h represent the volume, the radius and the height of the cone respectively

But first we need to calculate the radius of the cone.

Using formula of circumference.

$C = 2\pi r$

We have C to be 5.8cm

By substituting this value;

$5.8 = 2\pi r$

Divide through by $2\pi$

$\frac{5.8}{2\pi} = \frac{2\pi r}{2\pi}$

$r = \frac{5.8}{2\pi}$

$r = \frac{2.9}{\pi}$

Now, we can calculate the volume of the cone

By substituting $r = \frac{2.9}{\pi}$ and h = 12.5 cm;

$V = \frac{1}{3}\pi r^2h$ becomes

$V = \frac{1}{3}\pi * (\frac{2.9}{\pi})^2 * 12.5$

$V = \frac{1}{3}\pi * \frac{8.41}{\pi^2} * 12.5$

$V = \frac{1}{3} * \frac{8.41}{\pi} * 12.5$

Take $\pi$ as 3.14

$V = \frac{1}{3} * \frac{8.41}{3.14} * 12.5$

$V = 11.15976$

$V = 11.2 cm^3$ ---- Approximated

Hence, the volume of the cone is; $V = 11.2 cm^3$

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Answer:

(i). $y = 2\, x^2 - 4\, x + 1 = 2\, (x - 1)^2 - 1$ . Point $A$ is at $(1, \, -1)$ .

(ii). Point $Q$ is at $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ .

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Step-by-step explanation:

<h3>Coordinates of the Extrema</h3>

Note, that when $a(x + b)^2 + c$ is expanded, the expression would become $a\, x^2 + 2\, a\, b\, x + a\, b^2 + c$ .

Compare this expression to the original $2\, x^2 - 4\, x + 1$ . In particular, try to match the coefficients of the $x^2$ terms and the $x$ terms, as well as the constant terms.

For the $x^2$ coefficients: $a = 2$ .
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For the constant terms: $a \, b^2 + c = 1$ . Since $a = 2$ and $b = -1$ , solving for $c$ gives $c =-1$ .

Hence, the original expression for the parabola is equivalent to $y = 2\, (x - 1)^2 - 1$ .

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Assume $(m,\, n)$ is an intersection of the graphs of the two functions $y = 2\, x^2- 4\, x + 1$ and $x -y + 4 = 0$ . Setting $x$ to $m$ , and $y$ to $n$ should make sure that both equations still hold. That is:

$\displaystyle \left\lbrace \begin{aligned}& n = 2\, m^2 - 4\, m + 1 \\ & m - n + 4 = 0\end{aligned}\right.$ .

Take the sum of these two equations to eliminate the variable $n$ :

$n + (m - n + 4) = 2\, m^2 - 4\, m + 1$ .

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For point $A$ , the $x$ -component is $1$ while the $y$ -component is $(-1)$ .
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Similarly, the $y$ -component of point $M$ would be $((-1) + 7) / 2 = 3$ , the average of the $y\!$ -components of point $A$ and point $P$ .

Hence, the midpoint of segment $AP$ would be at $(2,\, 3)$ .

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$\displaystyle \left( y - \frac{7}{2}\right) = \frac{1}{5}\, \left(x - \frac{1}{2} \right)$ .

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