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.
2x+4-3x>-1
-x+4>-1
-x>-1-4
-x>-5
X<-5/-1
Hence x<5
Answer:
14 × 23 = 322 this is you answer
Answer: D is correct, 52.9%
Step-by-step explanation:
% decrease from 340 to 160, Where 340 is the old number and 160 is the new number
Percent change = (New number - Old number) / Old number x 100%
Percent change
= (160 - 340) / 340 x 100
= (-180) / 340x 100 = -52.94%
I hope this is clear, please mark as brainliest answer.
