A rancher wants to enclose a rectangular field with 220 ft of fencing.
One side is a river and will not require a fence.
What is the maximum area that can be enclosed?
:
The field requires only 3 sides of fence, therefore
L + 2W = 220
L = (220-2W); we can use this for substitution in the Area equation
:
Area
A = L*W
replace L with (220-2W)
A = W(220-2W)
A = -2W^2 + 220W
A quadratic equation, max A occurs at the axis of symmetry, x = -b/(2a)
In this equation
W = %28-220%29%2F%282%2A-2%29
W = 55 ft is the width for max area
:
Find the max area
A = -2(55^2) + 220(55)
A = -2(3025) + 12100
A = 6050 sq/ft is the max area
:
:
Confirm this with the dimensions calculated; 110 * 55 = 6050
Vertical angles are congruent (equal in measure) and they have the same vertex point.
Answer:200 = 100*2, so √200 = √100 * √2. √100 = 10, so I can replace it with 10 to get 10 * √2 as my final answer for the square root of 200.
Step-by-step explanation:
Uniform?
Pretty sure its not a unimodal/left skewed and right skewed as theres no sudden drop and the frequency are all similar
Answer:
no it is not
Step-by-step explanation:
