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
Answer:
<em>m∠EBC = 34° </em>
Step-by-step explanation:
m∠DBC = m∠DBE + m∠ EBC
m∠DBC - m∠DBE = m∠EBC
(12x - 3)° - (5x + 12)° = (3x + 13)°
12x - 5x - 3 - 12 = 3x + 13
4x = 28
x = 7
<em>m∠EBC </em>= (3(7) + 13)° <em>= 34° </em>
The x-coordinate after the rotation is -10
Answer:
C
Step-by-step explanation:
I explained in my last answer but someone deleted it
