If a varies jointly as b and c, and a = -10, when b = 12 and c = 2, how do you find a when b = 8 and c = ½?

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If a varies jointly as b and c, and a = -10, when b = 12 and c = 2, how do you find a when b = 8 and c = ½?

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Answer 1 · 2017-04-27T16:47:13

$a = - \frac{5}{3} = - 1 \frac{2}{3}$ $a = -5/3 = -1 2/3$

Explanation:

There are two ways to find the answer, finding the constant of proportionality or just setting up a proportion.

Finding the constant of proportionality :

$a = k b c$ $a = kbc$

Plug in the first set of numbers to find $k$ $k$ :

$- 10 = k \cdot 12 \cdot 2$ $-10 = k * 12 * 2$

$- 10 = 24 k$ $-10 = 24k$

$k = - \frac{10}{24} = - \frac{5}{12}$ $k = -10/24 = -5/12$

Now use $k$ $k$ to find $a$ $a$ :

$a = - \frac{5}{12} \cdot \frac{8}{1} \cdot \frac{1}{2} = - \frac{40}{24} = - \frac{5}{3} = - 1 \frac{2}{3}$ $a = -5/12 *8/1 *1/2 = -40/24 = -5/3 = -1 2/3$

Using proportions:

$\frac{a_{1}}{a_{2}} = \frac{b_{1} c_{1}}{b_{2} c_{2}}$ $(a_1)/(a_2) = (b_1 c_1)/(b_2 c_2)$

$\frac{- 10}{a_{2}} = \frac{12 \cdot 2}{\frac{8}{1} \cdot \frac{1}{2}}$ $(-10)/(a_2) = (12*2)/(8/1*1/2)$

$\frac{- 10}{a_{2}} = \frac{24}{4}$ $(-10)/(a_2) = 24/4$

Use the cross product $\frac{a}{b} = \frac{c}{d}$ $a/b = c/d$ : $a \cdot d = b \cdot c$ $" "a*d = b*c$

$- 10 \cdot 4 = 24 \cdot a_{2}$ $-10 * 4 = 24*a_2$

$- 40 = 24 \cdot a_{2}$ $-40 = 24 * a_2$

$a_{2} = - \frac{40}{24} = - \frac{5}{3} = - 1 \frac{2}{3}$ $a_2 = -40/24 = -5/3 = -1 2/3$

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If a varies jointly as b and c, and a = -10, when b = 12 and c = 2, how do you find a when b = 8 and c = ½?

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