Triangle abc is such that ab=4 and ac=8 if m is the midpoint of bc and am=3, what is the length of bc?
1 answer:
This is an interesting question. I chose to tackle it using the Law of Cosines.
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
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Step-by-step explanation:
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The graphs of inverse functions are reflections of each other across the line y = x.
In the first diagram, the graph of ƒ(x) (blue) is the reflection of g(x) (red) about the line y = x (black)
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(a) Rename h(x) as y
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(b) Solve for x:
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(c) Switch x and y
y = x + 2
(e) Rename y as the inverse function
h⁻¹(x) = x + 2
(f) Compare with your function
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The graph of h(x) (blue) reflects ƒ(x) (red) across the line y = x (black).
