$13,957 is invested, part at 7% and the rest at 6%. if the interest earned from the amount invested at 7% exceeds the interest earned from the amount invested at 6% by $833.73, how much is invested at each rate?

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102  

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% =  = $A

And The money invested at 6% = = $13957 - $A

Let The interest earn at 7% =

And The interest earn at 6% =

-   = $833.73

Let The time period = 1 year

Now, From Simple Interest method

Simple Interest =

Or,   =

Or,   =

And

=

Or,   =

∵   -   = $833.73

So, -   = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A =

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102   . Answer

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i

.

step-by-step explanation: