Name 5 that are near, but not equivalent to, 4/12
1 answer:
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Answer: The length is 13cm and the width is 7cm
Step-by-step explanation:
For a rectangle of length L and width W, the area is:
A = W*L
In this case we have:
L = (2*x + 9) cm
W=(3*x + 1) cm
Then the area of the rectangle is:
A = (2*x + 9)*(3*x + 1) cm^2
A = (6*x^2 + 2*x + 27*x + 9) cm^2
A = (6*x^2 + 29*x + 9) cm^2
now we remove two squares with sides of x cm
The area of each one of these squares is (x cm)*(x cm) = x^2 cm^2
Then the area of the figure will be:
area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2
area = (4*x^2 + 29*x + 9) cm^2
Now we know that the area of this shape is 83 cm^2, then we need to solve:
83 cm^2 = (4*x^2 + 29*x + 9) cm^2
0 = (4*x^2 + 29*x + 9) cm^2 - 83 cm^2
0 = (4*x^2 + 29*x - 74) cm^2
Then we need to solve:
0 = 4*x^2 + 29*x - 74
Here we can use Bhaskara's equation, the solutions of this equation are given by:
Then the two solutions are:
x = (-29 - 45)/8 = -9.25 (for how the length and width are defined, we can not have x as a negative number, then this solution can be discarded).
The other solution is:
x = (-29 + 45)/8 = 2
x = 2
Then the length and width of the rectangle are:
Length = (2*2 + 9)cm = 13 cm
Width = (3*2 + 1)cm = 7cm
Try this solution:
distance=
answer: 7
distance=
answer: 7