A variable P varies partly as x and partly varies inversely as the square root of x. Given P=17, x=9, P=42, x=16. Write a formula to illustrate it. Find P when x=81.

For condition 1:

$$17 = A(9) + \frac{B}{\sqrt{9}}$$ $$17 = 9A + \frac{B}{3} \quad (*)$$

For condition 2:

$$42 = A(16) + \frac{B}{\sqrt{16}}$$ $$42 = 16A + \frac{B}{4} \quad (**)$$

From $(*)$, multiply by 3:

$$51 = 27A + B \implies B = 51 - 27A$$

Substitute this expression for B into $(**)$:

$$42 = 16A + \frac{51 - 27A}{4}$$

Multiply the entire equation by 4 to eliminate the denominator:

$$4 \times 42 = 4 \times 16A + (51 - 27A)$$ $$168 = 64A + 51 - 27A$$ $$168 - 51 = 64A - 27A$$ $$117 = 37A$$ $$A = \frac{117}{37}$$

Now substitute the value of A back into the expression for B:

$$B = 51 - 27\left(\frac{117}{37}\right)$$ $$B = \frac{51 \times 37 - 27 \times 117}{37}$$ $$B = \frac{1887 - 3159}{37}$$ $$B = \frac{-1272}{37}$$

a) Write a formula to illustrate it.
Substitute the values of A and B into the general variation formula:

$$P = \frac{117}{37}x - \frac{1272}{37\sqrt{x}}$$

The formula is $\boxed{P = \frac{117}{37}x - \frac{1272}{37\sqrt{x}}}$.

b) Find P when x = 81.

$$P = \frac{117}{37}(81) - \frac{1272}{37\sqrt{81}}$$ $$P = \frac{117 \times 81}{37} - \frac{1272}{37 \times 9}$$ $$P = \frac{9477}{37} - \frac{1272}{333}$$

To combine the fractions, find a common denominator, which is $37 \times 9 = 333$.

$$P = \frac{9477 \times 9}{333} - \frac{1272}{333}$$ $$P = \frac{85293 - 1272}{333}$$ $$P = \frac{84021}{333}$$

This fraction can be simplified by dividing the numerator and denominator by 3:

$$P = \frac{28007}{111}$$

When $x = 81$, $P = \boxed{\frac{28007}{111}}$.

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: P varies directly as the cube root of V.
This can be written as:

$$P = k \sqrt[3]{V}$$

Given: $P = 24$ when $V = 64$.

$$24 = k \sqrt[3]{64}$$ $$24 = k \times 4$$ $$k = \frac{24}{4}$$ $$k = 6$$

So the formula is $P = 6\sqrt[3]{V}$.

a) Find P when V = 125.

$$P = 6 \sqrt[3]{125}$$ $$P = 6 \times 5$$ $$P = 30$$

When $V = 125$, $P = \boxed{30}$.

b) Find V when P = 36.

$$\frac{36}{6} = \sqrt[3]{V}$$ $$6 = \sqrt[3]{V}$$

To find V, cube both sides:

$$V = 6^3$$ $$V = 216$$

When $P = 36$, $V = \boxed{216}$.

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: T varies directly as the square root of m.
This can be written as:

$$T = k \sqrt{m}$$

Given: $T = 18$ when $m = 9$.

$$18 = k \sqrt{9}$$ $$18 = k \times 3$$ $$k = \frac{18}{3}$$ $$k = 6$$

So the formula is $T = 6\sqrt{m}$.

a) Find T when m = 36.

$$T = 6 \sqrt{36}$$ $$T = 6 \times 6$$ $$T = 36$$

When $m = 36$, $T = \boxed{36}$.

b) Find m when T = 30.

$$\frac{30}{6} = \sqrt{m}$$ $$5 = \sqrt{m}$$

To find m, square both sides:

$$m = 5^2$$ $$m = 25$$

When $T = 30$, $m = \boxed{25}$.