Answer:
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Algebra II
Step-by-step explanation:
Step 1: Define
Find points from graph.
Point A (-4, 2)
Point B (8, 6)
Step 2: Find distance d
Simply plug in the 2 coordinates into the distance formula to find distance d.
Answer: 0ft < X ≤ 5.73ft
Step-by-step explanation:
In this case, we have a rectangle.
For a rectangle of length L and width W, the area is:
A = L*W.
And we have:
L = X
W = X + 3ft.
Then the area will be:
A = X*(X + 3ft) = X^2 + 3ft*X.
And we want to have a maximum area of 50ft^2.
Then we can write:
A = X^2 + 3ft*X ≤ 50ft^2
Now let's solve this for X.
Now, the first thing we can see is that both coefficients in our quadratic equation are positive, so as the absolute value of X increases, also does the whole equation.
Then makes sense start for the upper limit of X, this is when:
X^2 + 3ft*X = 50ft^2.
Now we can solve the quadratic equation:
X^2 + 3ft*X - 50ft^2 = 0
Applying the Bhaskara formula, the solutions are:
Then we have two solutions:
X = (-3ft - 14.46ft)/2 = -8.73 ft.
X = (-3ft + 14.46ft)/2 = 5.73 ft
Because X represents a distance, it can only be positive, then we must select the option X = 5.73ft.
This is the maximum value of X, and we will have:
0ft < X ≤ 5.73ft
Where the lower limit is there because we can not have X = 0ft, as this does not have physical meaning.
0 –1 (0, –1)
1 4 (1, 4)
2 9 (2, 9)
A.
y = 5x – 1
B.
y = –3x + 3
C.
y = 3x – 4
D.
y = 3x + 3