A chemical company makes 2 brands of antifreeze.the first brand is 30%pure antifreeze and the second brand is 55%pure antifreeze.in order to obtain 170 gallons of a mixture that contains 35% pure antifreeze how many gallons of each brand antifreeze must be used
To solve this problem, we can set up a system of equations.
Let x be the number of gallons of brand 1 antifreeze (30% pure).
Let y be the number of gallons of brand 2 antifreeze (55% pure).
We are given the following information:
1. We need to obtain 170 gallons of a mixture that contains 35% pure antifreeze.
This information can be expressed in the following equation:
x + y = 170 (equation 1)
2. Brand 1 antifreeze is 30% pure and brand 2 antifreeze is 55% pure.
To obtain a mixture that contains 35% pure antifreeze, we can use the weighted average formula:
(30% * x + 55% * y) / (x + y) = 35% (equation 2)
Now, we can solve the system of equations.
Substitute equation 1 into equation 2:
(30% * x + 55% * y) / 170 = 35%
Multiply both sides by 170 to eliminate the fraction:
30% * x + 55% * y = 35% * 170
Simplify:
0.3x + 0.55y = 0.35 * 170
Multiply 0.35 by 170:
0.3x + 0.55y = 59.5 (equation 3)
Now, we have a system of two equations:
x + y = 170 (equation 1)
0.3x + 0.55y = 59.5 (equation 3)
We can solve this system of equations through substitution, elimination, or other methods.
Let's solve by substitution:
Solve equation 1 for x:
x = 170 - y
Substitute this value of x into equation 3:
0.3(170 - y) + 0.55y = 59.5
Distribute:
51 - 0.3y + 0.55y = 59.5
Combine like terms:
0.25y = 8.5
Divide by 0.25:
y = 34
Now, substitute this value of y back into equation 1 to solve for x:
x + 34 = 170
x = 136
Therefore, we need 136 gallons of brand 1 antifreeze and 34 gallons of brand 2 antifreeze to obtain a mixture of 170 gallons that contains 35% pure antifreeze.
Divide
9ay/2y^4 /3a/4y
Simply as much as possible
To divide the expression (9ay/2y^4) by (3a/4y), we can simplify by following these steps:
First, let's write down the expression:
(9ay/2y^4) / (3a/4y)
To divide fractions, we keep the first fraction as it is and change the division sign to multiplication. Then, we take the reciprocal of the second fraction. Therefore, the expression becomes:
(9ay/2y^4) * (4y/3a)
Now, let's simplify by canceling out common factors.
The 2 in the numerator cancels out with the 2 in the denominator:
(9ay/(1y^4)) * (4y/3a)
Next, we can simplify the variables:
9ay * 4y / 1y^4 / 3a
Combining like terms, we have:
36ay^2 / 1y^4 / 3a
To divide fractions in the denominator, we multiply by the reciprocal:
36ay^2 * 3a / 1y^4
Multiply the numerators and denominators:
(36ay^2 * 3a) / (1 * 1y^4)
Simplify further:
108a^2y^2 / y^4
Using the laws of exponents, we can simplify y^2 in the numerator and denominator:
108a^2 / y^(4-2)
Simplifying further, we get:
108a^2 / y^2
Therefore, the simplified expression is 108a^2 / y^2.
To find out how many gallons of each brand of antifreeze must be used, we can set up a system of equations based on the given information. Let's say x represents the amount of the first brand of antifreeze (30% pure) and y represents the amount of the second brand of antifreeze (55% pure) in gallons.
1. We know that the total amount of antifreeze required is 170 gallons:
x + y = 170
2. We also know that the resulting mixture should be 35% pure antifreeze:
(0.30x + 0.55y)/(x + y) = 0.35
Now, let's solve this system of equations. First, let's rearrange equation 1 to solve for x:
x = 170 - y
Substitute this value of x into equation 2:
(0.30(170 - y) + 0.55y)/170 = 0.35
Now, simplify and solve for y:
(51 - 0.30y + 0.55y)/170 = 0.35
(51 + 0.25y)/170 = 0.35
51 + 0.25y = 0.35 * 170
51 + 0.25y = 59.5
0.25y = 59.5 - 51
0.25y = 8.5
y = 8.5 / 0.25
y = 34
Now substitute the value of y back into equation 1 to find x:
x + 34 = 170
x = 170 - 34
x = 136
Therefore, in order to obtain 170 gallons of a mixture that contains 35% pure antifreeze, you will need 136 gallons of the first brand (30% pure) and 34 gallons of the second brand (55% pure).