Asum of $4200 was invested, part at 8% and the remainder at 11%. if $426.00 was earned in interest after one year, how much was invested at 11%?

$3000 was invested at 11%.

Step-by-step explanation:

The total interest was $426.  This was comprised of interest earned at 8% (represented by e) and (separately) interest earned at 11% (represented by v).

Then e + v = $4200 total investment, and  

i = $426 = e(0.08)(1 year) + v(0.11)(1 year)

We eliminate the variable e as follows:  since e + v = $4200, e = $4200 - v.  Thus,

i = $426 = e(0.08)(1 year) + v(0.11)(1 year) becomes:

i = $426 = ($4200 - v)(0.08)(1 year) + v(0.11)(1 year)  

This is one equation in one unknown, the amount of $ invested at 11%.

Performing the indicated multiplications:

426 = 4200(0.08) - 0.08v + 0.11v.  Simplifying this further, we get:

426 = 336 + 0.03v.

Then 90 = 0.03v, and v = 90 / 0.03 = $3000.

$3000 was invested at 11%.

three-fourths of an hour

36÷48=0.75

0.75=

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

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