I think the answer is 19.46
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.
Answer:
D. d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
C. d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
B. d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
Step-by-step explanation:
Given two points (x1, y1) and (x2,y2) we can find the distance using
d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
The order of the terms inside the square doesn't matter
d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
When we are squaring are term, we can take the absolute value before we square and it does not change the value
d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
Answer:
...........the answer is 7?!
Answer:
Yes it has infinity of solutions.
Step-by-step explanation:
If you plug in certain numbers it will make the statements true.