Question

1. a scuba diver is swimming 15 feet below the waters surface. the diver swims up slowly at 0.5 feet per second for 4 seconds and then stops to look at an eel. finally, the diver swims up for 3 more seconds at 0.4 feet per second. what is the divers new depth, in feet? 5. during 7 1/2 months of hibernation, a black bear lost 64.5 pounds. on average, what was the bears weight change per month?

Explanation:

Step1: Calculate the distance the diver swims up in the first part

The diver swims at 0.5 feet per second for 4 seconds. Using the formula distance = speed×time, the distance $d_1$ is $0.5\times4 = 2$ feet.

Step2: Calculate the distance the diver swims up in the second part

The diver swims at 0.4 feet per second for 3 seconds. Using the formula distance = speed×time, the distance $d_2$ is $0.4\times3=1.2$ feet.

Step3: Calculate the total distance the diver swims up

The total distance $D$ the diver swims up is $d_1 + d_2=2 + 1.2=3.2$ feet.

Step4: Calculate the new depth of the diver

The diver starts at 15 feet below the surface. The new depth $d$ is $15 - 3.2 = 11.8$ feet.

for question 5:

Step1: Convert the mixed - number of months to a decimal

$7\frac{1}{2}$ months is equal to 7.5 months.

Step2: Calculate the average weight change per month

The bear lost 64.5 pounds in 7.5 months. To find the average weight change per month, we use the formula $\text{average weight change}=\frac{\text{total weight change}}{\text{number of months}}$. So, $\frac{64.5}{7.5}=8.6$ pounds per month.

Answer:

11.8