Answer: See below
Explanation:
(x^2+10x+7)(2x-1) = 2x^3+19x^2+4x-7
2x^3-x^2+20x^2-10x+14x-7 = 2x^3+19x^2+4x-7
2x^3+19x^2+4x-7 = 2x^3+19x^2+4x-7
First one is reflection across x-axis.
Not visualizing 2nd one
Answer:ndbdnd ndbdnd 192
Step-by-step explanation:
BBB as haha
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.
The numbers will change and they can be increased or decreased
