Identify the zero that has a multiplicity greater than one: f(x) = (x - 2)(x + 2)(x – 5)
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Let the amount of the first brand be x, and let the amount of the second brand be y.
0.09x + 0.14y = 240 * 0.13 .................(1)
x + y = 240 ..............(2)
y = 240 - x .......................(3)
Plugging the value for y from equation (3) into equation (1), we get:
...............(4)
Equation (4) simplifies to:
-0.05x = -2.4
giving the value for the required amount of 9% vinegar as 48 ml and the required amount of 14% vinegar as 240 - 48 = 192 ml.
0.09x + 0.14y = 240 * 0.13 .................(1)
x + y = 240 ..............(2)
y = 240 - x .......................(3)
Plugging the value for y from equation (3) into equation (1), we get:
Equation (4) simplifies to:
-0.05x = -2.4
giving the value for the required amount of 9% vinegar as 48 ml and the required amount of 14% vinegar as 240 - 48 = 192 ml.
Use this systems of equations to solve:
x = first antifreeze
y = second antifreeze
Isolate y.
x + y = 15
Subtract x from both sides.
y = -x + 15
Substitute y into the other equation.
.2x + .12(-x + 15) = .18(15)
Simplify.
.2x - .12x + 1.8 = 2.7
Subtract 1.8 from both sides.
.08x = .9
Divide both sides by .08
x = 11.25
Substitute x in the equation that we isolated y in.
y = -11.25 + 15
y = 3.75
11.25 L of the first antifreeze and 3.75 L of the second.
x = first antifreeze
y = second antifreeze
Isolate y.
x + y = 15
Subtract x from both sides.
y = -x + 15
Substitute y into the other equation.
.2x + .12(-x + 15) = .18(15)
Simplify.
.2x - .12x + 1.8 = 2.7
Subtract 1.8 from both sides.
.08x = .9
Divide both sides by .08
x = 11.25
Substitute x in the equation that we isolated y in.
y = -11.25 + 15
y = 3.75
11.25 L of the first antifreeze and 3.75 L of the second.