“b” varies inversely as the square root of “c”. If b = 1 when c = 16, find b when c = 6.

b = k/x
1= k/SQRT(16)
k/SQRT(16) = 1
k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
k = 1 * SQRT(16)
k = 4

Now that I have the variation constant, I can plug in the x-value they gave me, and find the value of y:

y = 4/6

y = 2/3

2. Originally Posted by bobbyboy1111
“b” varies inversely as the square root of “c”. If b = 1 when c = 16, find b when c = 6.

b = k/x Mr F says: This makes no sense. The denominator is meant to be ${\color{red}\sqrt{c}}$.
1= k/SQRT(16)
k/SQRT(16) = 1
k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
k = 1 * SQRT(16)
k = 4

Now that I have the variation constant, I can plug in the x-value they gave me, and find the value of y: Mr F says: There are no x's or y's in the question. The pronumerals are b and c.

y = 4/6

y = 2/3
Given: $b = \frac{k}{\sqrt{c}}$.

$k = 4$.

Therefore: $b = \frac{4}{\sqrt{c}}$. Now substitute $c = 6$.

3. So....????

b = k/SQRT(c)
1= k/SQRT(16)
k/SQRT(16) = 1
k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
k = 1 * SQRT(16)
k = 4

....

b = 4/SQRT(c)

b = 4/SQRT(6)

b = 4/2.44

b = 1.63 or 163/100

Thanks again, Mr. Fantastic

4. Originally Posted by bobbyboy1111
So....????

b = k/SQRT(c)
1= k/SQRT(16)
k/SQRT(16) = 1
k/SQRT(16) * SQRT(16) = 1 * SQRT(16)
k = 1 * SQRT(16)
k = 4

....

b = 4/SQRT(c)

b = 4/SQRT(6)

b = 4/2.44

b = 1.63 or 163/100

Thanks again, Mr. Fantastic
You should leave the final answer in exact form unless instructed otherwise by the question: $b = \frac{4}{\sqrt{6}} = \frac{2 \sqrt{6}}{3}$.