Answer:
Step-by-step explanation:
height of pyramid = √(10²-(5/2)²) = √93.75
volume of pyramid = ⅓b²h = ⅓·5²√93.75 ≅ 80.69 in³
lateral area = 2×5×10 = 100 in²
base area = 5² = 25 in²
surface area = 100+25 = 125 in²
Because you can have a negative inside the absolute value so even if it is inside the absolute value of -3 will come out as 3.
Answer:
Step-by-step explanation:
hello :
p²+30p = p(p+30)
Chicken sandwich with my baby and my grandma is a little too much to do and I wanna was a way better then you got a baby and stuff and I wanna just like a
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).