P varies directly as V and inversely as the square root of W. If V is decreased by 10% and W is increased by 21%, what is the percentage change in P?

Here are the solutions to your problems.

:

P varies directly as V and inversely as the square root of W.

Step 1: Write the initial relationship.

$$P = k \frac{V}{\sqrt{W}}$$

where $k$ is the constant of proportionality.

Step 2: Express the new values of V and W.
V is decreased by 10%, so $V_{new} = V - 0.10V = 0.90V$.
W is increased by 21%, so $W_{new} = W + 0.21W = 1.21W$.

Step 3: Calculate the new value of P ($P_{new}$).

$$P_{new} = k \frac{V_{new}}{\sqrt{W_{new}}} = k \frac{0.90V}{\sqrt{1.21W}}$$ $$P_{new} = k \frac{0.90V}{1.1\sqrt{W}}$$ $$P_{new} = \frac{0.90}{1.1} \left( k \frac{V}{\sqrt{W}} \right)$$

Since $P = k \frac{V}{\sqrt{W}}$, we have:

$$P_{new} = \frac{0.90}{1.1} P$$ $$P_{new} = \frac{9}{11} P$$

Step 4: Calculate the percentage change in P.
Percentage change $= \frac{P_{new} - P}{P} \times 100\%$
Percentage change = \left( \frac{\frac{9}{11}P - P}{P} \right) \times 100\%$$ Percentage change = \left( \frac{9}{11} - 1 \right) \times 100% Percentage change $= \left( \frac{9 - 11}{11} \right) \times 100\% Percentage change = \left( -\frac{2}{11} \right) \times 100\%$$ Percentage change \approx -18.18%$

The percentage change in P is a decrease of approximately 18.18%.

$$\boxed{\text{Decrease of } 18.18\%}$$

:

R varies directly as S and inversely as the square root of T.

a) Write an equation connecting R, S and T.

Step 1: Write the general variation equation.

$$R = k \frac{S}{\sqrt{T}}$$

where $k$ is the constant of proportionality.

b) Find the value of R when S = 360 and T = 1.5.

Step 1: Find the constant of proportionality ($k$) using the given values $R = 480$, $S = 150$, and $T = 5$.

$$480 = k \frac{150}{\sqrt{5}}$$ $$k = \frac{480 \sqrt{5}}{150}$$ $$k = \frac{48 \sqrt{5}}{15}$$ $$k = \frac{16 \sqrt{5}}{5}$$

Step 2: Write the specific equation for R.

$$R = \frac{16 \sqrt{5}}{5} \frac{S}{\sqrt{T}}$$

Step 3: Substitute the new values $S = 360$ and $T = 1.5$ into the equation.

$$R = \frac{16 \sqrt{5}}{5} \frac{360}{\sqrt{1.5}}$$ $$R = \frac{16 \sqrt{5}}{5} \frac{360}{\sqrt{\frac{3}{2}}}$$ $$R = \frac{16 \sqrt{5}}{5} \frac{360}{\frac{\sqrt{3}}{\sqrt{2}}}$$ $$R = \frac{16 \sqrt{5}}{5} \cdot \frac{360 \sqrt{2}}{\sqrt{3}}$$ $$R = \frac{16 \cdot 360 \sqrt{10}}{5 \sqrt{3}}$$ $$R = \frac{16 \cdot 72 \sqrt{10}}{\sqrt{3}}$$ $$R = \frac{1152 \sqrt{10}}{\sqrt{3}}$$

To rationalize the denominator:

$$R = \frac{1152 \sqrt{10} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}$$ $$R = \frac{1152 \sqrt{30}}{3}$$ $$R = 384 \sqrt{30}$$

The value of R is approximately $384 \times 5.477 = 2103.17$.

$$\boxed{R = 384 \sqrt{30} \approx 2103.17}$$

c) Find the percentage change in R if S is increased by 5% and T decreased by 20%.

Step 1: Express the new values of S and T.
S is increased by 5%, so $S_{new} = S + 0.05S = 1.05S$.
T is decreased by 20%, so $T_{new} = T - 0.20T = 0.80T$.

Step 2: Calculate the new value of R ($R_{new}$).

$$R_{new} = k \frac{S_{new}}{\sqrt{T_{new}}} = k \frac{1.05S}{\sqrt{0.80T}}$$ $$R_{new} = \frac{1.05}{\sqrt{0.80}} \left( k \frac{S}{\sqrt{T}} \right)$$

Since $R = k \frac{S}{\sqrt{T}}$, we have:

$$R_{new} = \frac{1.05}{\sqrt{0.80}} R$$

Convert decimals to fractions for precision: $1.05 = \frac{21}{20}$ and $0.80 = \frac{4}{5}$.

$$\sqrt{0.80} = \sqrt{\frac{4}{5}} = \frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}}$$ $$R_{new} = \frac{\frac{21}{20}}{\frac{2}{\sqrt{5}}} R$$ $$R_{new} = \frac{21}{20} \cdot \frac{\sqrt{5}}{2} R$$ $$$$