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k^(2)=m^(2)+n^(2)
For any right triangle, the given equation relates the length of the hypotenuse, 
k, to the lengths of the other two sides of the triangle, 
m and 
n. Which of the following equations correctly gives 
m in terms of 
k and 
n ?
Choose 1 answer:
(A) 
m=k-n
(B) 
m=sqrt(k^(2))-n^(2)
(C) 
m=sqrt(k^(2)-n^(2))
(D) 
m=sqrt(k^(2)+n^(2))
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k2=m2+n2 k^{2}=m^{2}+n^{2} \newlineFor any right triangle, the given equation relates the length of the hypotenuse, k k , to the lengths of the other two sides of the triangle, m m and n n . Which of the following equations correctly gives m m in terms of k k and n n ?\newlineChoose 11 answer:\newline(A) m=kn m=k-n \newline(B) m=k2n2 m=\sqrt{k^{2}}-n^{2} \newline(C) m=k2n2 m=\sqrt{k^{2}-n^{2}} \newline(D) m=k2+n2 m=\sqrt{k^{2}+n^{2}} \newlineShow calculator

Full solution

Q. k2=m2+n2 k^{2}=m^{2}+n^{2} \newlineFor any right triangle, the given equation relates the length of the hypotenuse, k k , to the lengths of the other two sides of the triangle, m m and n n . Which of the following equations correctly gives m m in terms of k k and n n ?\newlineChoose 11 answer:\newline(A) m=kn m=k-n \newline(B) m=k2n2 m=\sqrt{k^{2}}-n^{2} \newline(C) m=k2n2 m=\sqrt{k^{2}-n^{2}} \newline(D) m=k2+n2 m=\sqrt{k^{2}+n^{2}} \newlineShow calculator
  1. Isolate m in equation: We start with the given equation for a right triangle: k2=m2+n2 k^2 = m^2 + n^2 . To solve for m, we need to isolate m on one side of the equation.
  2. Subtract n^22: Subtract n2 n^2 from both sides of the equation to get m2 m^2 by itself: k2n2=m2 k^2 - n^2 = m^2 .
  3. Take square root: Take the square root of both sides of the equation to solve for m: m=k2n2 m = \sqrt{k^2 - n^2} . This will give us the value of m in terms of k and n.

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