When taking square roots, you can't take square roots of negative roots of negative numbers. So, what will work for the domain of u(x) is what makes u(x) zero or more. We can make an inequality for that.
u(x) ≥ 0.
9x + 27 ≥ 0 by squaring both sides
9x ≥ -27
x ≥ -3
So the domain of the function is when x ≥ -3 is true.
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