The amount invested at 4% is $4,200. This problem can be solved by using two linear equation with two different variables which represent the investment's partition and the investment's return. The first equation is x+y=$7,600 and the second equation is 0.04x + 0.05y=$338, then you must substitute the y variable in the second question with the first equation (0.04x + 0.05($7,600-x)=$338). Therefore, you will get a new equation to solve the problem which is 0.04x+380-0.05x=338 resulting x=$4,200.
The scientific notation of 8888800 is 8.8888 X 106.
Let the necessary amount of water be x. Then (1.00)(1 gallon antifreeze) = 0.40(x+1 gallon), or 1 = .40x + .40. Combining like terms: 0.60 = 0.40x.
Then x = 0.60 / 0.40 = 1.5.
Adding 1.5 gallons of water to that 1 gallon of pure antifreeze results in 2.5 gallons of 40% antifreeze.
![\bf \begin{cases} f(x)=\cfrac{2}{x}\\[1em] g(x)=x^2+9 \end{cases}~\hspace{5em}f(~~g(x)~~)=\cfrac{2}{g(x)}\implies f(~~g(x)~~)=\cfrac{2}{x^2+9}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20f%28x%29%3D%5Ccfrac%7B2%7D%7Bx%7D%5C%5C%5B1em%5D%20g%28x%29%3Dx%5E2%2B9%20%5Cend%7Bcases%7D~%5Chspace%7B5em%7Df%28~~g%28x%29~~%29%3D%5Ccfrac%7B2%7D%7Bg%28x%29%7D%5Cimplies%20f%28~~g%28x%29~~%29%3D%5Ccfrac%7B2%7D%7Bx%5E2%2B9%7D)
that's one combination for f(x) and g(x), off many combinations.
They spend $360 on clothing and entertainment.
