I will give BRAINLIEST! HELP PLEASE.!!! 2. Find each measure. m<1 m<2 m<3
1 answer:
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Assuming a diagram similar to the one I've attached, ∠<em>YVZ</em> is a vertical angle to ∠<em>WVX</em>, which means they have an equal measure. Additionally, ∠<em>WVZ</em> and ∠<em>WVX</em> form a linear pair, which means they are supplementary (sum to 180°). That means we start out with the equation
We combine our like terms (the <em>x</em>'s get combined, then the constants get combined) and have:
Cancel the 9 first by subtraction:
Cancel the 19 by division:
Since we know that our angle we're looking for, ∠<em>YVZ</em>, is the same measure as ∠<em>WVX</em>, we substitute 9 in for <em>x</em>:
8(9)+28=72+28=100°
We combine our like terms (the <em>x</em>'s get combined, then the constants get combined) and have:
Cancel the 9 first by subtraction:
Cancel the 19 by division:
Since we know that our angle we're looking for, ∠<em>YVZ</em>, is the same measure as ∠<em>WVX</em>, we substitute 9 in for <em>x</em>:
8(9)+28=72+28=100°
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).
