To Find :
How many gallons of a 80% antifreeze solution must be mixed with 100 gallons of 20% antifreeze to get a mixture that is 70% antifreeze.
Solution :
Let, x gallons of 80% antifreeze solution is required :
So,
![(x \times 80) + ( 100 \times 20 ) = (x + 100 )\times 70\\\\8x + 200 = 7( x + 100)\\\\x = 500](https://tex.z-dn.net/?f=%28x%20%5Ctimes%2080%29%20%2B%20%28%20100%20%5Ctimes%2020%20%29%20%3D%20%28x%20%2B%20100%20%29%5Ctimes%2070%5C%5C%5C%5C8x%20%2B%20200%20%3D%207%28%20x%20%2B%20100%29%5C%5C%5C%5Cx%20%3D%20500)
Therefore, 500 gallons of 80% antifreeze solution is required.
Answer:
Step-by-step explanation:
Rewrite this as
![\sqrt{(-1)(100)}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-1%29%28100%29%7D)
Knowing that i-squared = -1:
![\sqrt{(i^2)(100)}](https://tex.z-dn.net/?f=%5Csqrt%7B%28i%5E2%29%28100%29%7D)
Both i-squared and 100 are perfect squares, so this simplifies to
±10i
The distributive property of 18(12), or simply 18 * 12, is 216. (This is your answer.)
Answer:
1 1/4 miles
Step-by-step explanation:
3/4 + 2/4 (1/2 = 2/4) = 5/4 = 1 1/4