A rubber ball dropped on a hard surface takes a sequence of bounces, each one 1/6 as high as the preceding one. If this ball is dropped from a height of 12 feet, how far will it have traveled when it hits the surface the fifth time?

Step-by-step explanation:

GIVEN: A rubber ball dropped on a hard surface takes a sequence of bounces, each one as high as the preceding one.

TO FIND: If this ball is dropped from a height of feet, how far will it have traveled when it hits the surface the fifth time.

SOLUTION:

once the ball is dropped on hard surface it bounces of preceding one and comes down the same distance.

When the ball is dropped from feet height

after first hit

new height

after second hit

new height

after third hit  

new height

after fourth hit

adding all distance

                                  feet

Hence the ball will travel   feet before it hits the surface fifth time.

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places