Solve for P if P varies as the square of Q, and P=4 when Q=16. Find R if R varies inversely as the square of its radius, and R=0.80 when r=0.4.

Here are the solutions to the problems you sent. Problem 1: Step 1: Write the variation equation. P varies as the square of Q, so we can write this relationship as: P = kQ^2 where k is the constant of proportionality. Step 2: Find the constant k using the given values. We are given P = 4 when Q = 16. Substitute these values into the equation: 4 = k(16)^2 4 = k(256) Divide by 256 to find k: k = (4)/(256) k = (1)/(64) Step 3: Write the specific variation equation. Substitute the value of k back into the general equation: P = (1)/(64)Q^2 Step 4: Calculate P for the new value of Q. We need to find P when Q = 64. Substitute Q = 64 into the specific variation equation: P = (1)/(64)(64)^2 P = (1)/(64) × 64 × 64 P = 64 The value of P is 64. Problem 4: Step 1: Write the variation equation. The electrical resistance (R) of a wire varies inversely as the square of its radius (r). This can be written as: R = (k)/(r^2) where k is the constant of proportionality. Step 2: Find the constant k using the given values. We are given that the resistance is 0.80 \, when the radius is 0.4 \, cm. Substitute these values into the equation: 0.80 = (k)/((0.4)^2) 0.80 = (k)/(0.16) Multiply both sides by 0.16 to find k: k = 0.80 × 0.16 k = 0.128 Step 3: Write the specific variation equation. Substitute the value of k back into the general equation: R = (0.128)/(r^2) Step 4: Calculate the resistance for the new radius. We need to find the resistance when the radius is 0.7 \, cm. Substitute r = 0.7 into the specific variation equation: R = (0.128)/((0.7)^2) R = (0.128)/(0.49) R ≈ 0.26122 Rounding to two significant figures, the resistance is approximately: R ≈ 0.26 \, The resistance is approximately 0.26 . That's 2 down. 3 left today — send the next one.