Question

in a tornado the wind velocity in meters per second can be described by the function v(p)=12.3 sqrt 1134-3p where 'p' is th air pressure in millibars. What is the air pressure of a tornado in which the wind velocity is 49.3 meters per second?

Answer 1

It would be B. 373 millibars

Explanation:

Radical Functions&Rational Exponents Practice

1. A

2. A = 5x^3

3. B =0.9

4. D

5. D

6. C

7. C =13

8. D

9. C = 50 seconds

10. B

11. A

12. B = 373 millibars

Answer 2

We are given with the function
v(p) = 12.3 √(1134 - 3p)
We are asked to find the air pressure when the velocity is 49.3 m/s
Substituting this to v(p)
49.3 = 12.3 √(1134 - 3p)
49.3/12.3 = √(1134 - 3p)
4.01 = √(1134 - 3p)
4.01² = √(1134 - 3p)²
16.07 = 1134 - 3p
3p = 1134 - 16.07
3p = 1117.92
p = 1117.92/3
p = 372.64 millibars