You add and subtract what's in the parentheses and they that will give u your awnser for x I hope I helped u
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
I’m pretty sure it’s c
Explanation
Answer:
It would be the third answer
Step-by-step explanation:
Answer:
D. y - 2 = 2(x + 1)
Step-by-step explanation:
<h3>y - y₁ = m (x - x₁)</h3>
y - 2 = 2 (x - (-1))
y - 2 = 2 (x + 1)
<h3>#CMIIW</h3>
