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1 answer:
The problem statement tells you ∠MLK is 61°, so ∠LMK = 180° -68° -61° = 51°. Since a tangent is always perpendicular to a radius, triangles LJM and LJK are right triangles.
Trigonometry tells you ...
tangent = opposite / adjacent
so you can write two relations involving LJ.
tan(51°) = LJ/JM
tan(68°) = LJ/JK
The second equation can be used to write an expression for LJ that can be substituted into the first equation.
LJ = JK*tan(68°) = 3*tan(68°)
Substituting, we have
tan(51°) = 3*tan(68°)/JM
Multiplying by JM/tan(51°), we get
JM = 3*tan(68°)/tan(51°)
JM ≈ 6.01
The radius of circle M is about 6.01.
Trigonometry tells you ...
tangent = opposite / adjacent
so you can write two relations involving LJ.
tan(51°) = LJ/JM
tan(68°) = LJ/JK
The second equation can be used to write an expression for LJ that can be substituted into the first equation.
LJ = JK*tan(68°) = 3*tan(68°)
Substituting, we have
tan(51°) = 3*tan(68°)/JM
Multiplying by JM/tan(51°), we get
JM = 3*tan(68°)/tan(51°)
JM ≈ 6.01
The radius of circle M is about 6.01.
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Answer:
Step-by-step explanation:
We are given the second derivative:
And we want to find its inflection points.
To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.
Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.
So first, finding the zeros, we see that:
So, we can draw the following number-line:
<----(-4)--------------(3)----(6)---->
Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.
Testing for x < -4, we can use -5. So:
Since we acquired a positive result, g(x) is concave up for x < -4.
For -4 < x < 3, we can use 0. So:
Since we acquired a negative result, g(x) is concave down for -4 < x < 3.
And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.
For 3 < x < 6, we can use 4. So:
Since we acquired a negative result, g(x) is concave down for 3 < x < 6.
Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.
And finally, for x > 6, we can use 7. So:
So, g(x) is concave up for x > 6.
And since we changed signs before and after the inflection point at x = 6, x = 6 is indeed an inflection point.