Ethanol fuel mixtures have "e" numbers that indicate the percentage of ethanol in the mixture by volume. for example, e10 is a mixture of 10% ethanol and 90% gasoline. how much e5 should be mixed with 6000 gal of e10 to make an e9 mixture?

According to the problem we start with 6000 gal of E10 which means there is 10% of ethanol  in this 6000, with ths information we can calculate the amount of ethanol already in the mixture:

The relationship between ethanol and total fluid can be represented like this:

Based on this relationship we can build the one that we are looking for:

the total solution will be:

where X is the E5 solution volume

and the TotalEthanol would be:

where E5ethanol (5% Ethanol 95% Gasoline) would be:

Now replacing in the relationship we found before:

Solving for X:

We should mix 1500 gal of E5 to make an E9 mixture

the answer is true.

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2