Use the power rule and the power of a product or quotient rule to simplify the expression.2

A.
B.
C.
D.


Answer: A

Mathematics

You might also like to view...

Solve the problem.A pendulum bob swings through an arc 40 inches long on its first swing. Each swing thereafter, it swings only 60% as far as on the previous swing. How far will it swing altogether before coming to a complete stop?

A. 100 inches B. 50 inches C. 133 inches D. 67 inches

Mathematics

Use a calculator to approximate the logarithm to four decimal places.log 2.88

A. 1.0578 B. 0.4742 C. 0.4594 D. 0.4440

Mathematics

Solve the problem.A deposit of $8000 is made in an account that earns 6% interest compounded quarterly. The balance in the account after n quarters is given by the sequencean  = 8000 n  n = 1, 2, 3, ... Find the balance in the account after 7 years.

A. $11,975.78 B. $12,217.78 C. $12,219.78 D. $12,137.78

Mathematics

Solve the problem.The table below lists women's and men's total enrollments at all institutions of higher learning in Country X for various years. College Enrollments (Millions)YearWomenMen19886.35.919907.46.519948.16.820028.97.120069.27.4The enrollments (in millions) of women and men, respectively, can be modeled by the systemW = 0.14t + 5.69M = 0.07t + 5.62where t is the number of years since 1980.(i) Use "intersect" on a graphing calculator to estimate when women's and men's enrollments were approximately equal. What was that enrollment?(ii) Use the models to predict the total enrollment of women and men in 2011.

A. (i) Intersect at approximately (-2.20, 5.98). This means that in 1978, women's and men's enrollment was about 6.0 million students. (ii) Not enough information has been provided. B. (i) Intersect at approximately (-1.00, 5.55). This means that in 1979, women's and men's enrollment was about 5.6 million students. (ii) In 2011, W = 10.03 million students and M = 7.79 million students. C. (i) Intersect at approximately (-1.11, 6.32). This means that in 1979, women's and men's enrollment was about 6.3 million students. (ii) In 2011, so W = 10.31 million students and M = 7.93 million students. D. (i) The functions do not intersect. (ii) In 2011, W = 9.47 million students and M = 7.51 million students.

Mathematics