I think its camp. would be the answer
Answer:
a. A(x) = (1/2)x(9 -x^2)
b. x > 0 . . . or . . . 0 < x < 3 (see below)
c. A(2) = 5
d. x = √3; A(√3) = 3√3
Step-by-step explanation:
a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...
A(x) = (1/2)(x)(y)
A(x) = (1/2)(x)(9-x^2)
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b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).
On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).
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c. A(2) = (1/2)(2)(9 -2^2) = 5
The area is 5 when x=2.
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d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).
Answer:
50%
Step-by-step explanation:
No matter how many times you flip a coin, there is always a 50% probability that you will flip heads because there are only two outcomes.
Answer:
x=17 p=20
Step-by-step explanation:
Answer:
9 in
Step-by-step explanation:
For an n-sided polygon, the length of the apothem is ...
a = r·cos(180°/n)
We assume your problem statement is saying the radius is 10 inches. For a hexagon, n=6 and we have ...
a = (10 in)cos(30°) ≈ 8.66 in
Rounded to the nearest inch, the apothem is 9 in.