Answer:
a) see below
b) 40x20 meters
Step-by-step explanation:
Write down what you know:
- The area of the enclosure is length*width, so
- The length of the fencing is 80 meters, so
Now we have to combine these two equations above, and get rid of y in the process.
First rewrite the second as:
Then substitute for y in the first:
b) To maximize A, find the zero of the first derivative:
So y = (80-40)/2 = 20 meters.
Answer:
the height of the flagpole is three fourths the height of the school. the difference in there heights is 4.5 m. what is the height of the school?
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Equation:
f = (3/4)s
s-f = 4.5
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Substitute for "f" and solve for "s":
s-(3/4)s = 4.5
(1/4)s = 4.5
s = 4*4.5
s = 18 meters (height of the school)
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Since f = (3/4)s, f = (3/4)(18)
= (3/2)(9)= 13.5 meters (height of the flag pole)
===============================-step explanation: HOPE THIS HELPS
Answer:
(14,2)
Step-by-step explanation:
43/100 - 223/100
(43x100)-(23x100)/100x100
-18000/100000
(-18000/2000)/(100000/2000)
= -9/5 or -1 4/5
Answer:
Abc is the whole explanation.