One way to answer this question is the following:
y = [(3-b)/5]x + b where b is greater than 3.
This is a family of lines with negative slope as required and passes through the point (5,3).
The area of the triangle is (b/2)[-5b/(3-b)].
Preston H.
asked 08/25/22A right triangle is formed in the first quadrant by the x- and y-axes, from (0,y) and (x,0) and a line through the point (5, 3). Write the area A of the triangle as a function of x, and determine the domain of the function. (Enter your answer for the domain using interval notation.)
One way to answer this question is the following:
y = [(3-b)/5]x + b where b is greater than 3.
This is a family of lines with negative slope as required and passes through the point (5,3).
The area of the triangle is (b/2)[-5b/(3-b)].
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