Soybean meal is 18% protein; cornmeal is 9% protein. how many pounds of each should be mixed together in order to get 360lb mixture that is 11% protein?
How many pounds of the cornmeal should be in the mixture?
How many pounds of the soybeans meal should be in the mixture?
The perimeter of a rectangle is 150 inches. The length exceeds the with by 3 inches. Find the length and the width.
The length is _______-inches
The width is __________inches
The perimeter of a rectangle is 136 m. The length is 2m more thatn twice the width. Find the dimensions.
What is the length?________m
What is the width?__________m
You will have better luck getting a reply if you restrict yourself to one question at a time.
I will do the first one....
let the amount of cornmeal be x pounds
then the amount of soybean is 360-x pounds
so 18/100(360-x) + (9/100)x = (11/100)360
multiply each term by 100, expand and solve
... I got x = 280
To solve the first problem, let's assign variables to the unknown values:
Let x = pounds of soybean meal to be mixed
Let y = pounds of cornmeal to be mixed
We are given:
Soybean meal protein content = 18%
Cornmeal protein content = 9%
Total mixture weight = 360 lbs
Target protein content = 11%
To find the amount of cornmeal needed, we can start by setting up a system of equations based on the protein content and weight:
Equation 1: x + y = 360 (since the total weight of the mixture is 360 lbs)
Equation 2: (0.18 * x) + (0.09 * y) = 0.11 * 360 (since the total protein content of the mixture is 11% of the total weight)
To solve this system of equations:
1. Rearrange Equation 1 to express y in terms of x: y = 360 - x
2. Substitute this expression for y in Equation 2: (0.18 * x) + (0.09 * (360 - x)) = 0.11 * 360
3. Simplify and solve for x:
0.18x + 32.4 - 0.09x = 39.6
0.09x = 7.2
x = 80
So, you would need 80 pounds of soybean meal.
To find the amount of cornmeal needed, substitute this value of x into Equation 1: 80 + y = 360
Solving for y:
y = 360 - 80
y = 280
Therefore, you would need 280 pounds of cornmeal.
For the second problem, let's assign variables as follows:
Let w = width of the rectangle
Length = w + 3 (since the length exceeds the width by 3 inches)
We are given:
Perimeter = 150 inches
To find the length and width, we can set up the equation based on the formula for the perimeter of a rectangle:
Perimeter = 2 * (Length + Width)
Plug in the given values and solve for w:
150 = 2 * (w + (w + 3))
Simplify and solve for w:
150 = 2 * (2w + 3)
150 = 4w + 6
4w = 144
w = 36
The width of the rectangle is 36 inches.
Now, substitute the width value into the equation for the length:
Length = w + 3
Length = 36 + 3
Length = 39
So, the length of the rectangle is 39 inches.
For the third problem, let's assign variables as follows:
Let w = width of the rectangle
Length = 2w + 2 (since the length is 2m more than twice the width)
We are given:
Perimeter = 136 m
To find the length and width, we can set up the equation based on the formula for the perimeter of a rectangle:
Perimeter = 2 * (Length + Width)
Plug in the given values and solve for w:
136 = 2 * ((2w + 2) + w)
Simplify and solve for w:
136 = 2 * (3w + 2)
136 = 6w + 4
6w = 132
w = 22
The width of the rectangle is 22 m.
Now, substitute the width value into the equation for the length:
Length = 2w + 2
Length = 2 * 22 + 2
Length = 44 + 2
Length = 46
So, the length of the rectangle is 46 m.