Hank invested a total of $20,000, part at 7% and part at 10%. How much did he invest at each rate if the total interest earned in one year was $1640?

The amount invested at 7% is $12,000 while the amount invested at 10% is $8,000

Step-by-step explanation:

The given parameters are;

The total amount Hank invested = $20,000

The interest earned in one year = $1640

Hank invested part of the $20,000 at 7% interest and part of the $20,000 at 10%

Let the part invested at 7% = X and the part invested at 10% = Y, we have;

7% of X + 10% of Y = $1640 which gives;

0.07×X + 0.1×Y = $1640(1)

X + Y = $20,000(2)

Multiply equation (1) by 10 and subtract equation (1) from equation (2) to get;

X + Y - 10×(0.07×X + 0.1×Y) = $20,000 - 10×$1640

X - 0.7·X - Y - Y = $20,000 - $16,400 = $3,600

0.3·X = $3,600

X = $3,600/0.3 = $12,000

X = $12,000

Y = $20,000 - X = $20,000 - $12,000 = $8,000

Y = $8,000

Therefore, the amount invested at 7% = $12,000 and the amount invested at 10% = $8,000.

$8,000 at 10%

$12,000 at 7%

Step-by-step explanation:

Let the part invested at 10% = x

Let the part invested at 7% = $20,000 - x

7% = 0.07

10% = 0.10

Hank invested a total of $20,000, part at 7%

0.07(20000 - x)

And part at 10%

0.10(x)

Hence, we have:

0.10x + 0.07(20000 - x) = 1640

0.10x + 1400 - 0.07x = 1640

0.10x - 0.07x = 1640 - 1400

0.03x = 240

x = 240/0.03

x = 8000

Hence, the amount he invested at 10% = $8,000

The amount he invested at 7% =

$20,000 - x

= $20,000 - $8,000

= $12,000

throw the computer

step-by-step explanation:

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.