Question

y varies jointly as a and b, and inversely as the square root of c. y = 24 when a = 6, b = 2,and c=4. Find y when a = 3, b=5, and c= 25.y=

Answer 1

We have an unknown quantity, which is y. We are told that it varies joint as a and b. This translates as follows

`y=k\cdot a\cdot\text{b }`

where k is a constant.

Now, we are also told that it varies inversely as the square root of c. So this look as follows

`y=\frac{k}{\sqrt[]{c}}`

where k is another constant. We have to combine both facts as follows

`y=\frac{k\cdot a\cdot b}{\sqrt[]{c}}`

in this case, k is a constant. Now, we are told that whenever c=4, b=2 and a=6 we get y=24. So we get the following equation

`24=\frac{k\cdot6\cdot2}{\sqrt[]{4}}=(k\cdot6\cdot2)/(2)=6\cdot k`

By dividing by 6 on both sides, we get

`k=(24)/(6)=4`

So the general expression of y looks like this

`y=\frac{4\cdot a\cdot b}{\sqrt[]{c}}`

Now, we want to calculate the value of y whenever a=3 and b=5 and c=25. So we get

`y=\frac{4\cdot3\cdot5}{\sqrt[]{25}}=(4\cdot3\cdot5)/(5)=4\cdot3=12`

so y=12 whenever a=3, b=5 and c=25.