A person invested \$5,600$5,600 in an account growing at a rate allowing the money to double every 6 years. How long, to the nearest tenth of a year would it take for the value of the account to reach \$72,500$72,500?

22.2 years

Step-by-step explanation:

Doubling formula is given as:

P(t) = Po × (2)^t/k

Where

Po = Initial amount invested = $5600

P(t) = Amount after time t = $72500

k = Time it takes to double = 6 years

t = Time in years = ??

Hence:

72500 = 5600 × (2)^t/6

Divide both sides by 5600

72500/5600 =( 5600 × (2)^t/6)5600

12.946428571 = (2)^t/6

Take the In of both sides

In 12.946428571 = In (2)^t/6

In 12.946428571 = t/6 In (2)

Divide both sides by In 2

In 12.946428571/In 2 = t/6

3.6944822628 = t/6

Cross Multiply

t = 6 × 3.6944822628

t = 22.166893577 years

Approximately

t = 22.2 years

Therefore, it would take 22.2 years

-4

step-by-step explanation:

(-4+8)=-(4-8)=-8+4=-4

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic