Aretiree receives $5120 a year interest from $40,000 placed in two bonds, one paying 14% and the other 12%. how much is invested in each bond? 3)

Investment in first bond is $ 16,000

In second bond is $ 24,000

Step-by-step explanation:

Let x be the amount invested in first bond,

Since, the total invested amount = $ 40,000,

So, the total amount invested in second bond = ( 40,000 - x ) dollars,

Given,

In first bond,

Interest rate = 14%,

While in second bond,

Interest rate = 12%,

Thus, the total interest from both bonds in a year,

I = 14% of x + 12% of ( 40,000 - x )

According to the question,

I = $ 5120,

Hence, the amount invested in first bond = $ 16,000,

The amount invested in second bond = $ 40,000 - $ 16,000 = $ 24,000

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.

they charge the same amount at 20 minutes.

step-by-step explanation: