if it is friday then letica family will have pizza for dinner. if letica family has pizza for dinner, letica will song at the di
1 answer:
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The correct answer is: [D]: " 
" .
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Explanation:
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;
Simplify the "denominator" ; → (b + 2b = 1b + 2b = 3b) ;
and rewrite:
;
Cancel out the "b" in the "numerator" to a "1" ;
and cancel the "3b" in the "denominator to a "3" ;
{since: "b ÷ b = 1 "; and since: "3b ÷ b = 3" } ;
and rewrite as:
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;
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→ which is: Answer choice: [D]: " " .
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Explanation:
__________________________________________________
Simplify the "denominator" ; → (b + 2b = 1b + 2b = 3b) ;
and rewrite:
Cancel out the "b" in the "numerator" to a "1" ;
and cancel the "3b" in the "denominator to a "3" ;
{since: "b ÷ b = 1 "; and since: "3b ÷ b = 3" } ;
and rewrite as:
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→ which is: Answer choice: [D]: "
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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).
