A few days ago
M

Math help, might be easy, but i might be stressing over it. Please click here.?

When, and if u decide to answer, please show all of your work. Thanks in advance.

**Niagara Falls has an average of 212,000 cubic feet of water passing over it each second. Is this rate more or less than 10 cubic miles per year? Explain your calculations. (Hint: 1 cubic mile= 5283 ^3 cubic feet.)** I PUT LESS THAN**

**At noon car A is traveling north at 30 miles per hour and is located 20 miles north of car B. Car B is traveling west at 50 miles per hour. Approximate the distance between the cars at 12:45 P.M. to the nearest miles**

**Let f(x)= 0.5 x^2 + 50 compute the outside temperature in degrees Farenheit at X p.m. where 1 < or equal to X < or equal to 5. a) Calculate the average rate of change of F from 1 P.M. to 4 P.M. b) Interpret this average rate of change. **Please help me. I have a test tomorrow** If you want to explain to me in very specific detail, you can INSTANT MESSAGE me. I am online. Click on my icon and then click "IM this contact." Thanks in advance.

Top 1 Answers
A few days ago
supensa

Favorite Answer

1. You must convert twice, once to convert ft^3/sec to ft^3/year, and then again to convert ft^3/year to mile^3/year.

How many seconds in a year? 60 * 60 * 24 * 365 = 31536000. Multiply by 212000 to get 6,685,632,000,000. Then divide that by 5280^3 to get 45.4 mi^3/year. So your answer is incorrect.

2. This one isn’t as difficult as it seems! It’s just a right triangle. After 3/4 of an hour, car A has travelled 20 + (3/4)(30), or 42.5 miles. Car B has travelled (3/4)(50), or 37.5. Solve for C by Pythagorean theorem: 42.5^2 + 37.5^2 = C^2. Solving this gets an answer of 57 miles.

3. a)

f(1) = 0.5(1)^2 + 50 = 50.5

f(4) = 0.5(4)^2 + 50 = 58

The average rate of change is (58 – 50.5)/2, or 3.75.

b) The temperature is gradually going up by an average rate of 3.75 degrees per hour.

Hope this helps!

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