I don’t know but I hope you have a good day
You have to add the same kind of terms, in this case the ones that have a variable (5.4q and -4.5q) and the ones that don't (2.4 and 3.6), like this:
(5.4q - 4.5q) + (2.4 + 3.6) = 0.9q + 6
The answer is c. 0.9 +6
Here are the answers:
1.) DC=D’C’
2.) 3.) ED=E’D’
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.
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Answer:
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Step-by-step explanation:
Look at the photo below for the details.
:)