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:
5
Step-by-step explanation:
f(-4)= -2(-4)-3
=8-3
=5
The answer is C. 50 in, 48 in, 14 in
In a right triangle, a² + b² = c², where c is the longest side (hypotenuse), a and b are the other two sides.
14² + 48² = 196 + 2304 = 2500
50² = 2500
Answer: About 3.06
Step-by-step explanation:
We can use trigonometry functions to solve for AC. Let the ?, representing AC, be "x" in our mathematical work.
Since we have the hypotenuse and x is adjacent to the angle given, I am going to use cosine.
cos(θ) = 
cos(40) = 
0.766 ≈
3.06 ≈ x
x ≈ 3.06
Angle between 16 and 18 = 90 + 55 = 145°
