each box weighed 10 more pounds
and he lifted 12 boxes
so if you multiply 12 x the weight and then 12 times 10 more pounds each you would have 12c +12*10
which uses the equation 12(c+10)
The answer is 39.6 because when you add 36 plus 10% you get 39.6
From all this information we can come up with a few equations to work with...
first we have that the worth of each coin times the number of those coins all added together equals 4.65
![0.05n + 0.10d + 0.25q = 4.65](https://tex.z-dn.net/?f=%200.05n%20%2B%200.10d%20%2B%200.25q%20%3D%204.65%20)
here I have n = nickels, d = dimes, and q = quaters
then we are told that there are "three more quarters than dimes"
which translates to the equation
![q = d + 3](https://tex.z-dn.net/?f=%20q%20%3D%20d%20%2B%203%20)
or
![d = q - 3](https://tex.z-dn.net/?f=%20d%20%3D%20q%20-%203%20)
we are also told that there are "twice as many nickels as quarters"
this translates to
![n = 2q](https://tex.z-dn.net/?f=%20n%20%3D%202q%20)
to solve for quarters, we plug in d (in terms of q) and n (in terms of q) into the first equation we created.
![0.05(2q) + 0.10(q - 3) + 0.25q = 4.65](https://tex.z-dn.net/?f=%200.05%282q%29%20%2B%200.10%28q%20-%203%29%20%2B%200.25q%20%3D%204.65%20)
then we distribute and start trying to solve for
![q](https://tex.z-dn.net/?f=q)
![0.10q + 0.10q - 0.30 + 0.25 = 4.65](https://tex.z-dn.net/?f=%200.10q%20%2B%200.10q%20-%200.30%20%2B%200.25%20%3D%204.65%20)
then add like terms
![0.45q - 0.30 = 4.65](https://tex.z-dn.net/?f=%200.45q%20-%200.30%20%3D%204.65%20)
then add 0.30 to both sides
![0.45q = 4.95](https://tex.z-dn.net/?f=%200.45q%20%3D%204.95%20)
then divide both sides by 0.45
![q = \frac{4.95}{0.45} = 11](https://tex.z-dn.net/?f=%20q%20%3D%20%5Cfrac%7B4.95%7D%7B0.45%7D%20%3D%2011%20)
therefore our answer is B) 11 quarters