Simplify 8√6m+6√8mn/2√2mn
1 answer:
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Answer:
a) Each corral should be 33⅓ ft long and 25 ft wide
b) The total enclosed area is 1666⅔ ft²
Step-by-step explanation:
I assume that the corrals have identical dimensions and are to be fenced as in the diagram below
Let x = one dimension of a corral
and y = the other dimension
(a) Dimensions to maximize the area
The total length of fencing used is:
4x + 3y = 200
4x = 200 – 3y
x = 50 - ¾y
The area of one corral is A = xy, so the area of the two corrals is
A = 2xy
Substitute the value of x
A = 2(50 - ¾y)y
A = 100 y – (³/₂)y²
This is the equation for a downward-pointing parabola:
A = (-³/₂)y² + 100y
a = -³/₂; b = 100; c = 0
The vertex (maximum) occurs at
y = -b/(2a) = 100 ÷ (2׳/₂) = 100 ÷ 3 = 33⅓ ft
4x + 3y = 100
Substitute the value of y
4x + 3(33⅓) = 200
4x + 100 = 200
4x = 100
x = 25 ft
Each corral should measure 33⅓ ft long and 25 ft wide.
Step 2. Calculate the total enclosed area
The enclosed area is 50 ft long and 33⅓ ft wide.
A = lw = 50 × 100/3 = 5000/3 = 1666⅔ ft²
Answer:
Zeros : 1 , -1, 3
Degree : 4
End Behaviour : At x-> ∞ f(x) -> ∞ and x->-∞ f(x) -> ∞
Y - intercept : -3
Extra Points: (0,-3), (2,-3)
Step-by-step explanation:
f(x) = 0 to find the zeros
Clearly x = -1,1,3
Here 1 is a repeating root as it is (x-1)²
Degree is highest power of x in f(x)
Clearly it is x*x²*x = x⁴ is the maximum power of x
Thus degree is 4
Looking at end behavior we substitute x->∞ and x-> -∞
Clearly f(x)>0 as all terms are positive and f(x)->∞
Similarly when x->-∞
f(x)>0 as 2 terms are -ve and their product is positive thus f(x)-> ∞
Y-Intercept is f(0)
f(0) = (0+1)(0-1)²(0-3) = 1*1*-3 = -3
Thus Y-Intercept is -3
Substitute x = 0 , 2 for extra points
Thus f(0) = -3
and f(2) = -3
Thus points on the graph (0,-3), (0,2)
We can use all this information to draw a graph remember that 1 is a repeating root so that will be a point of minima. The graph is a parabola that passes through x-axis at x = -1, 3.