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
|-9| means the absolute value of -9 which is 9
so we have 9 - 9 = 0 (answer)
Answer:
1 or 3 sorry if wrong
Step-by-step explanation:
Answer:
The approximate volume of the ball is, 24,416.64 cubic cm
Step-by-step explanation:
A ball has a radius of 18 cm.What is the approximate volume of the ball?Use 3.14 to approximate pi. Round to the nearest hundredth if necessary.
Answer:
points (2,2) and (-2,-4):
y-intercept= -1, the equation is y= 3/2x -1
points (-2,0) and (0,-5):
y-intercept = -5, the equation is y= -5/2- 5
points (0,3) and (i can’t read the other one)
y-intercept= 3, the equation is y=4x+ 3!
ask if you need further assistance ^_^
