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Alenkasestr [34]

2 years ago

14

Two cones are similar. The surface area of the larger cone is 65π square inches. The surface area of the smaller cone is 41.6π s

quare inches. The radius of the smaller cone is 6.4 inches. What is the radius of the larger cone?
8 inches
10 inches
11.52 inches
14.4 inches

2 answers:

vaieri [72.5K]2 years ago

6 0

8 inches hope it helps

3241004551 [841]2 years ago

5 0

Step $1$

<u>Find the scale factor</u>

we know that

$surface\ area\ of\ the\ larger\ cone=(scale\ factor^{2})* surface\ area\ of\ the\ smaller\ cone$

$(scale\ factor^{2})=surface\ area\ of\ the\ larger\ cone/ surface\ area\ of\ the\ smaller\ cone$

in this problem we have

$surface\ area\ of\ the\ larger\ cone=65\pi\ in^{2} \\surface\ area\ of\ the\ smaller\ cone=41.6\pi\ in^{2}$

Substitute the values

$(scale\ factor^{2})=\frac{65\pi }{41.6\pi } =1.5625\\ scale\ factor=1.25$

Step $2$

<u>Find the radius of the larger cone</u>

we know that

$the\ radius\ of\ the\ larger\ cone=(scale\ factor)* the\ radius\ of\ the\ smaller\ cone$

in this problem we have

$the\ radius\ of\ the\ smaller\ cone=6.4\ in$

substitute

$the\ radius\ of\ the\ larger\ cone=1.25* 6.4=8\ in$

therefore

<u>the answer is</u>

the radius of the larger cone is $8\ in$

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Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = $\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]$ $\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]$

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n = 36i + 20j + 45k

Equation of a plane is generally given by:

$a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0$

Then, replacing with point P and normal vector n:

$36(x-5) + 20(y-0) + 45(z-0) = 0$

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

$z = \frac{180-36x-20y}{45}$

When z = 0:

$y = 9 + \frac{-9x}{5}$

When z=0 and y=0:

x = 5

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$\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx$

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$\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx$

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$\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}} \, dy } \, dx$

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1 year ago

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Answer:

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We know that equation of an object's height t seconds after the launch is in form $s(t)=-gt^2+v_0t+h_0$ , where

g = Force of gravity,

$v_0$ = Initial velocity,

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For our given scenario $g=-9.8$ , $v_0=49$ and $h_0=58.8$ . Upon substituting these values in object's height function, we will get:

$s(t)=-9.8t^2+49t+58.8$

Therefore, the function for the height of the arrow would be $s(t)=-9.8t^2+49t+58.8$ .

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2 years ago