A person invests 4000 dollars in a bank. The bank pays 6.25% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money
in the bank until it reaches 9600 dollars?

14.4

Step-by-step explanation:

Given that :

Principal = 4000

Interest (r) = 6.25% compounded annually

Calculate time, t, if final amount A = 9600

Using the compound interest formula :

A = P(1 + r/n)^n*t

A = final amount

n = number of times interest is applied per period

9600 = 4000(1 + 0.0625)^t

9600 = 4000(1.0625)^t

9600/4000 = 1.0625^t

2.4 = 1.0625^t

Take the log of both sides

0.3802112 = 0.0263289t

t = 0.3802112/0.0263289

t = 14.440811

t = 14.4 ( nearest tenth)

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)