Jeremiah invested $x at a 6% simple annual interest rate account. Six times that amount was invested at an 9% simple annual interest rate account. How much was invested at the 6% account if the total annual return was $800.00?

Step-by-step explanation:

We will make a table with the values for both a 6% account and a 9% account.

The formula for this problem is Prt = I, where P is the amount invested in each account, r is the interest rate each carries in decimal form, t is the time in years, and I is the interest earned from the multiplication of the 3 previous values. We don't know how much is invested in either account, but we do know that no matter how much is invested in the 6% account, there is 6 times that in the 9% account. We know that the 6% account has a decimal rate of .06 and that the 9% account has a decimal rate of .09. "Annual" means 1 year, so the time is 1 year. Filling in the table, then:

                     P  *    r     *     t     =     I

Acct 6%         x     .06         1

Acct 9%       6x    .09          1

What we do with those number is multiply them straight across each row to get the amount of interest earned from each:

                    P     *     r     *     t     =     I

Acct 6%       x      *  .06    *    1      =   .06x

Acct 9%      6x     *  .09    *    1      =   .54x

The amount of Interest for both ADDS UP to 800; therefore:

.06x + .54x = 800 and

.6x = 800 so

x = 1333.33

That's how much was invested in the account that earned 6% interest annually.  

the answer is b.

64

step-by-step explanation:

c=2πr=2·π·8≈64

the answer is a.

606.0 sq. in.

i got it down.