Cameron invested $7,400 in an account paying an interest rate of 4\tfrac{3}{8}4 8 3 ​ % compounded continuously. Savannah invested $7,400 in an account paying an interest rate of 4\tfrac{3}{4}4 4 3 ​ % compounded daily. After 7 years, how much more money would Savannah have in her account than Cameron, to the nearest dollar?

$266

Step-by-step explanation:

Cameron :

r = 4 3/8% = 0.04375

P = 7400

For account compounded continuously :

A = Pe^it

i = interest ; t number of years ;

A = 7400 * e^0.04375*7

A = 7400 * e^0.30625

A = 10051.581

Savanna:

r = 4 3/4% = 0.0475

P = 7400

A = P(1 + r/n)^nt

n = number of compounding times per period

A = final amount ; P = initial investment

Compounded daily :

A = 7400(1 + 0.0475/365)^365*7

A = 7400(1 + 0.0001301)^2555

A = 7400(1.0001301)^2555

A = 7400 * 1.3942879

A = 10317.731

Difference :

$(10317.731 - 10051.581)

= $266.15

= $266 (nearest dollar)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.