Let’s call the number of gallons of 30% solution “x”. If we mix together 40 gallons and x gallons of solution, we’re going to come out with a total of 40 + x gallons of that 20% solution. Mathematically, this is
40(12%) + x(30%) = (40 + x)(20%)
With the percents as decimals:
40(0.12) + x(0.3) = (40 + x)(0.2)
Simplifying, we have:
4.8 + 0.3x = 8 + 0.2x
Collecting the x terms on one side and the constants on the other:
0.1x = 3.2
And multiplying both sides by 10 to find x:
x = 32
So, you’ll need 32 gallons of 30% solution to yield 40 + 32 = 72 gallons of 20% solution.
Solve for x:x/5 - 2 = x/2 + 3
Put each term in x/5 - 2 over the common denominator 5: x/5 - 2 = x/5 - (10)/5:x/5 - (10)/5 = x/2 + 3
x/5 - (10)/5 = (x - 10)/5:(x - 10)/5 = x/2 + 3
Put each term in x/2 + 3 over the common denominator 2: x/2 + 3 = x/2 + 6/2:(x - 10)/5 = x/2 + 6/2
x/2 + 6/2 = (x + 6)/2:(x - 10)/5 = (x + 6)/2
Multiply both sides by 10:(10 (x - 10))/5 = (10 (x + 6))/2
10/5 = (5×2)/5 = 2:2 (x - 10) = (10 (x + 6))/2
10/2 = (2×5)/2 = 5:2 (x - 10) = 5 (x + 6)
Expand out terms of the left hand side:2 x - 20 = 5 (x + 6)
Expand out terms of the right hand side:2 x - 20 = 5 x + 30
Subtract 5 x from both sides:(2 x - 5 x) - 20 = (5 x - 5 x) + 30
2 x - 5 x = -3 x:-3 x - 20 = (5 x - 5 x) + 30
5 x - 5 x = 0:-3 x - 20 = 30
Add 20 to both sides:(20 - 20) - 3 x = 20 + 30
20 - 20 = 0:-3 x = 30 + 20
30 + 20 = 50:-3 x = 50
Divide both sides of -3 x = 50 by -3:(-3 x)/(-3) = 50/(-3)
(-3)/(-3) = 1:x = 50/(-3)
Multiply numerator and denominator of 50/(-3) by -1:Answer: x = (-50)/3
Answer:
1 / We have the area of a rectangle ABCD = 8 x 16 = 128 ft²
2 / Find the area of an isosceles Δ DEF:
The base of the isosceles ΔDEF => DF = DC- FC = 16 -12 = 4 ft
So the area of Δ DEF = 1/2 (DF x EH) = 1/2 (4 x 6) = 12 ft²
3 / Area of the irregular shape = area ABCB + Area DEF = 128 + 12 = 140 ft²
Step-by-step explanation:
Answer:
He can add through calculater or by doing addition in his mind
I suspect 4/2 should actually be 4/3, since 4/2 = 2, while 4/3 would make V the volume of a sphere with radius r. But I'll stick with what's given:
In Mathematica, you can check this result via
D[4/2*Pi*r^3, r]
