Can some one give me the answers to the slope questions please ?!
1 answer:
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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:
A. The graph is not proportional, although it has a consistent slope, it does now go through the point of origin or (0,0). The equation of a proportional line is y=kx, so the y-intercept must be at (0,0).
B. Slope is found by , from 1 point to the next, the y-value (rise) increases by 2 and the x-value (run) increases by 1,
giving us a slope of 4.
C. The y-intercept is the lines point of intersection with the y-axis. In the graph, this point is (0,10).
D. y = mx + b is the slope-intercept form where m is the slope and b is the y-intercept, the slope (m) is 4 and the y-intercept (b) is 10. y = 4x + 10
E. The equation is not proportional because the b value is 10 rather than 0.