Let h(x)=(x^(2)-49)/(x+7) when x!=-7. h is continuous for all real numbers. Find h(-7). Choose 1 answer: (A) -14 (B) 7 (C) 14 (D) -7

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Let  h(x)=(x^(2)-49)/(x+7) when  x!=-7.  h is continuous for all real numbers. Find  h(-7). Choose 1 answer: (A) -14 (B) 7 (C) 14 (D) -7

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Definition of h(x): The function h(x) is defined as h(x) = (x^2 - 49) / (x + 7) for x ≠ -7. To find h(-7), we need to evaluate the limit of h(x) as x approaches -7, since the function is continuous for all real numbers.Factoring the numerator: First, we factor the numerator of h(x): x^2 - 49 can be factored as (x + 7)(x - 7).Rewriting h(x) using factored form: Now, we rewrite h(x) using the factored form: h(x) = [(x + 7)(x - 7)] / (x + 7).Canceling out common terms: We notice that (x + 7) appears in both the numerator and the denominator. Since we are looking for the limit as x approaches -7, and x is not exactly -7, we can simplify the expression by canceling out the (x + 7) terms.Simplified expression for h(x): After canceling, we get h(x) = x - 7 for all x ≠ -7.Substituting x with -7: Now, we can find h(-7) by substituting x with -7: h(-7) = -7 - 7.Calculating h(-7): Calculating the value, we get h(-7) = -14.

Let $h(x)=\frac{x^{2}-49}{x+7}$ when $x \neq-7$ . $\newline$ $h$ is continuous for all real numbers. $\newline$ Find $h(-7)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-14$ $\newline$ (B) $7$ $\newline$ (C) $14$ $\newline$ (D) $-7$

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Let h(x)=x2−49x+7 h(x)=\frac{x^{2}-49}{x+7} h(x)=x+7x2−49​ when x≠−7 x \neq-7 x=−7.\newlineh h h is continuous for all real numbers.\newlineFind h(−7) h(-7) h(−7).\newlineChoose 111 answer:\newline(A) −14-14−14\newline(B) 777\newline(C) 141414\newline(D) −7-7−7

Let $h(x)=\frac{x^{2}-49}{x+7}$ when $x \neq-7$ . $\newline$ $h$ is continuous for all real numbers. $\newline$ Find $h(-7)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-14$ $\newline$ (B) $7$ $\newline$ (C) $14$ $\newline$ (D) $-7$