How many L of a 10% alcohol solution must be mixed with 9l of a 80% alcohol solution to make a 55% solution

Step-by-step explanation:

The best thing to do for these solution/mixture problems is to make a table:

                          #L           *           %alcohol         =            L alcohol

    10% sol.

+   80% sol.                                                                                            

    55% sol.  

In this way, we can keep track of our information AND figure out the equation we need to use to solve it. Notice first, the equation along the top of the table:

#L * %alcohol = L alcohol  This tells us that we are multiplying the #L column times the %alcohol column to get the L alcohol column. Notice second, that there is a + sign to the far left, indicating that we are ADDING rows 1 and 2 together to get the mix. Let's start filling this in. The easy part is the % alcohol column. 10% alcohol has .10 alcohol as a decimal, likewise for the 80% and 55%:

                          #L             *            %alcohol          =         L alcohol

   10% sol.                                             .10

+  80% sol.                                            .80               =                          

   55% sol.                                            .55

Now to fill in the first column. We know that our unknown, from the problem, is the number of Liters, #L in the 10% solution, so that is x, and we also know that we have 9 L of the 80% alcohol. Filling that in:

                             #L            *             %alcohol           =         L alcohol

  10% sol.                x            *                .10

+ 80% sol.               9           *                 .80                                                

  55%                                                      .55

Now look back at the + sign. We are told that we are mixing the 10% with the 80%, so we are adding them together. So let's do that. We will also follow the rule for the table and multiply the first column times the second column to fill in the last column to complete the table:

                          #L           *             %alcohol            =              L alcohol

   10% sol.           x            *                 .10                  =                 .10x

+  80% sol.          9           *                 .80                  =                  7.2          

  55% sol.      (9 + x)         *                 .55                  =         .55x + 4.95

If we add the 2 solutions together to get the new solution in the first column, we will also add the L alcohol in the last column to get our equation:

.10x + 7.2 = .55x + 4.95 and

- .45x = - 2.25 so

x = 5L

We need 5 Liters of the 10% solution if we want to mix that with 9L of 80% solution to get 14 L of 55% solution.

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.