PLEASE HELP FAST!!!!!! How many triangles can be constructed using the following three side lengths: 4.8 cm, 2.1 cm, and 2.6 cm?
1 answer:
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Answer:
b = 12 and 28
Step-by-step explanation:
The absolute value equation |1/2b-8| = |1/4b-1| resolves to a piecewise linear function with three pieces. There are two solutions.
<h3>Domains</h3>The absolute value function on the left has a turning point where its value is zero:
1/2b -8 = 0
b -16 = 0
b = 16
The absolute value function on the right has a turning point where its value is zero:
1/4b -1 = 0
b -4 = 0
b = 4
For b > 16, both absolute value functions are identity functions. In this domain, the equation is ...
1/2b -8 = 1/4b -1
For 4 < b < 16, the function on the left negates its argument, so the equation in this domain is ...
-(1/2b -8) = 1/4b -1
For b < 4, both functions negate their arguments, so the equation in this domain is ...
-(1/2b -8) = -(1/4b -1)
For the purpose of finding the value of b, this is effectively identical to the equation for b > 16. (The value of b does not change if we multiply both sides of the equation by -1.)
<h3>Solutions</h3><u>Domain b < 4 ∪ 16 < b</u>
1/2b -8 = 1/4b -1
2b -32 = b -4 . . . . . . . . multiply by 4
b = 28 . . . . . . . . . . . . add 32-b to both sides
This solution is in the domain of the equation, so is one of the solutions to the original equation.
<u>Domain 4 < b < 16</u>
-(1/2b -8) = 1/4b -1 . . . . equation in this domain
-2b +32 = b -4 . . . . . . multiply by 4
36 = 3b . . . . . . . . . . . add 2b+4 to both sides
12 = b . . . . . . . . . . . . divide by 3
This solution is in the domain of the equation, so is the other solution to the original equation.
<h3>Graph</h3>For the purposes of the graph, we have define the function g(b) to be the difference of the two absolute value functions. The solutions are found where g(x) = 0, the x-intercepts. The graph shows those to be ...
b = 12 and b = 28
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<em>Additional comment</em>
Defining g(b) = |1/2b-8| -|1/4b-1|, we can rewrite it as ...
Then the solutions are the values of b that make g(b) = 0.
