Try this:
1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90.
2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4(x+18).
3) using the both parts: 0.5x+0.2*90=0.4(x+18) ⇒ x=54 gal. of <u>pure</u> weight.
4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal.
Answer: 108 gal.
Answer:
h−1(x)=−5x/3+5/6
Step-by-step explanation:
your answer is the second one
6x + 5x + (x + 16) + (3x - 1) = 360
11x + x + 16 + 3x - 1
15x + 15 = 360
15x = 345
x = 23
6(23) = 138
((23) + 16) = 39
(3(23) - 1) =68
5(23) = 115
138 + 39 + 68 + 115 = 360
First simplify the fractions: 1 1/2 + 3/10
Then add the fractions together: 0.8
Then you have 1.8.
Answer:
1/2x^2-2x
Step-by-step explanation:
x^2+2x+1/2x^4+2x^3-2x^2-2x
(x+1)(x+1)/2x(x+1)(x+1)(x+1)
