For this case we have an equation in which we must apply operations to rewrite it in an equivalent way. We must start by raising both sides of the equation to the square. Thus, we eliminate the square root of the left side of equality and finally solve the binomial of the right side of equality.

So we have:

$\sqrt {x ^ 2 + 81} = x + 10\\(\sqrt {x ^ 2 + 81}) ^ 2 = (x + 10) ^ 2\\x ^ 2 + 81 = (x + 10) ^ 2$

By definition: $(a + b) ^ 2 = a ^ 2 + 2ab + b ^ 2$

$x ^ 2 + 81 = x ^ 2 + 2 (x) (10) + (10) ^ 2\\x ^ 2 + 81 = x ^ 2 + 20x +100$

Thus, $x ^ 2 + 81 = x ^ 2 + 20x +100$ is equivalent to $\sqrt {x ^ 2 + 81} = x + 10$

Answer:

$x ^ 2 + 81 = x ^ 2 + 20x +100$

Option D

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3 years ago

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G es to prove that if a, b, and c are integers, where a ≠ 0 and c ≠ 0, such that ac | bc, then a |

Alex Ar [27]

$ac\mid bc$ means there exists an integer $n$ such that $ac=bcn$ . We can divide both sides by $c$ because $c\neq0$ . This implies $a=bn$ for some integer, which is identically $a\mid b$ .

3 0

3 years ago