The volume of gas a balloon can hold varies directly as the cube of its radius. imagine a balloon with a radius of 3 inches can hold 81 cubic inches of gas. how many cubic inches of gas must be released to reduce the radius down to 2 inches?

Answers

table a is wierd

step-by-step explanation:

6*f

\begin {array}{c|c}\underline {\quad x\quad}& \underline{\quad y\quad}\\ -2& 4\\-1& 2\\0& 0\\1& 2\\ 2& 4\\\end{array}

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

57 in³

Step-by-step explanation:

1. You know that the volume (V) of gas a balloon can hold varies directly as the cube of its radius (r). Then, you can write the following expression:

V=kr^{3}

Where k is the constant of proportionality.

2. You can calculate k as following:

k=\frac{V_1}{r^{3}}\\k=\frac{81in^{3}}{(3in)^{3}}\\k=3

3. Then, to calculate the cubic inches that must be released to reduce the radius down to 2 inches:

- Solve for the volume as following:

V_2=3(2in)^{3}=24in^{3}

- Make the subtraction of the two volumes shown above. Then, the result is:

V_3=81in^{3}-24in^{3}=57in^{3}



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