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ipn
7 days ago
5

the volume of a cone is 3πx3 cubic units and its height is x units. which expression represents the radius of the cone’s base, i

n units? 3x 6x 3πx2 9πx2
Mathematics
2 answers:
Inessa [3.9K]7 days ago
7 0
The volume provided is 3Pi(x^3) with a radius of x. To determine the volume of a cone, the formula used is V= [1/3]Pi(r^2)*height. By substituting, we get [1/3]Pi(r^2)x = 3Pi(x^3). This simplifies to (r^2)x = 9(x^3). Eventually, we find that r^2 = 9x^2, which leads to r = sqrt[9x^2] = 3x. <span>Answer: r = 3x</span>
Zina [3.9K]7 days ago
6 0
The response to this question is 3x.
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There was 5/8 of a pie left in the fridge. Daniel are 1/4 of the left over pie. How much of a pie did he eat
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Step-by-step explanation:

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1/4 x 2 = 2/8

2. Deduct 2/8 from 5/8.

5/8 - 2/8 = 3/8

Daniel consumed 2/8 of the remaining pie, and now there are 3/8 left.

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8 days ago
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Find the point on the circle x^2+y^2 = 16900 which is closest to the interior point (30,40)
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Response-

(78,104) represents the point closest to the interior.

Explanation-

The equation defining the circle,

\Rightarrow x^2+y^2 = 16900

\Rightarrow y^2 = 16900-x^2

\Rightarrow y = \sqrt{16900-x^2}

Since the point lies on the circle, its coordinates must be,

(x,\sqrt{16900-x^2})

The distance "d" from the point to (30,40) can be calculated as,

=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

=\sqrt{(x-30)^2+(\sqrt{16900-x^2}-40)^2}

=\sqrt{x^2+900-60x+16900-x^2+1600-80\sqrt{16900-x^2}}

=\sqrt{9400-60x-80\sqrt{16900-x^2}}

Next, we need to determine the value of x for which d is minimized. The minimum distance occurs when 9400-60x-80\sqrt{16900-x^2} is at its lowest value.

Let’s set up the equation,

\Rightarrow f(x)=9400-60x-80\sqrt{16900-x^2}

\Rightarrow f'(x)=-60+80\dfrac{x}{\sqrt{16900-x^2}}

\Rightarrow f''(x)=\dfrac{1352000}{\left(16900-x^2\right)\sqrt{16900-x^2}}

We find the critical points,

\Rightarrow f'(x)=0

\Rightarrow-60+80\dfrac{x}{\sqrt{16900-x^2}}=0

\Rightarrow 80\dfrac{x}{\sqrt{16900-x^2}}=60

\Rightarrow 80x=60\sqrt{16900-x^2}

\Rightarrow 80^2x^2=60^2(16900-x^2)

\Rightarrow 6400x^2=3600(16900-x^2)

\Rightarrow \dfrac{16}{9}x^2=16900-x^2

\Rightarrow \dfrac{25}{9}x^2=16900

\Rightarrow x=\sqrt{\dfrac{16900\times 9}{25}}=78

\Rightarrow x=78

Then,

\Rightarrow f''(78)=\dfrac{1352000}{\left(16900-78^2\right)\sqrt{16900-78^2}}=\dfrac{125}{104}=1.2

Since f''(x) is positive, the function f(x) achieves its minimum at x=78

When x is set to 78, the corresponding y value will be

\Rightarrow y = \sqrt{16900-x^2}=\sqrt{16900-78^2}=104

This leads us to conclude that the closest point is (78,104)

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