A) 5000 m²
b) A(x) = x(200 -2x)
c) 0 < x < 100
Step-by-step explanation:
b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:
A(x) = x(200 -2x)
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a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²
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c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.
100-35 = 65
65% de 2300 = 1495 (mes 1)
2300-1495 = 805 (mes 2)
65% de 805 = 523,25 (mes 3)
65% de 523,25 = 340,1125 ~ 340,10 (mes 4)
M<UWV =180- 99 -36= 45
45 = 1/2(94+20- arcUV)
45=57 - arcUV/2
arcUV / 2 = 57-45
arcUV / 2 =12
arcUV = 12(2)
arcUV = 24
answer is B. 24
Answer:
4 1/2
Step-by-step explanation:
Just do the math work it out on paper and check it!