Question 222324
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The perimeter of a triangle is the sum of the measures of the three sides, so for your triangle,

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 4(x -3) + 3x + 5x + 4]

Simplify:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 12x - 8]

The perimeter can be no more than 100, so it is less than or equal to 100 and no less than 28, so it is greater than or equal to 28, so:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 28 \leq 12x - 8 \leq 100]

Add 8 to all three parts of the compound inequality:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 36 \leq 12x \leq 108]

Divide all three parts of the compound inequality by 12:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3 \leq x \leq 9]

However, we have to change the inequality sign on the low end of the interval.  That is because *[tex \Large x \neq 3].  If *[tex \Large x = 3] then the measure of the left side of the triangle would be 0, i.e. *[tex \Large 4(3 - 3) = 0], and there would be no triangle.

Hence, the final interval for *[tex \Large x] is:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3 < x \leq 9]

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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