Let h(x)=(x-2)/(sqrt(x+7)-3) when x!=2. h is continuous for all x -7. Find h(2). Choose 1 answer: (A) 6 (B) 2 (c) 4 (D) -3

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Let  h(x)=(x-2)/(sqrt(x+7)-3) when  x!=2.  h is continuous for all  x > -7. Find  h(2). Choose 1 answer: (A) 6 (B) 2 (c) 4 (D) -3

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Understand the Problem: Understand the problem. We need to find the value of the function h(x) at x = 2. However, the function is not defined at x = 2 because of the denominator becoming zero. We need to find a way to evaluate the limit of h(x) as x approaches 2.Simplify the Function: Simplify the function. To find the limit as x approaches 2, we can try to simplify the function by rationalizing the denominator. This means we will multiply the numerator and the denominator by the conjugate of the denominator.Multiply by the Conjugate: Multiply by the conjugate. The conjugate of sqrt(x+7)-3 is sqrt(x+7)+3. We multiply both the numerator and the denominator by this conjugate to rationalize the denominator.Perform the Multiplication: Perform the multiplication. h(x) = (x-2)/(sqrt(x+7)-3) * (sqrt(x+7)+3)/(sqrt(x+7)+3)Apply the Difference of Squares: Apply the difference of squares. When we multiply the denominators, we get (sqrt(x+7)-3)(sqrt(x+7)+3) which simplifies to (x+7) - 9 because (a-b)(a+b) = a^2 - b^2.Simplify the Expression: Simplify the expression. The denominator simplifies to x - 2. The numerator, when multiplied out, is (x-2)(sqrt(x+7)+3).Cancel out Common Terms: Cancel out the common terms. The term (x-2) is present in both the numerator and the denominator, so they cancel each other out.Evaluate the Limit: Evaluate the limit. Now that the (x-2) terms are canceled, we are left with the limit of (sqrt(x+7)+3) as x approaches 2. We can now directly substitute x = 2 into this expression.Substitute x = 2: Substitute x = 2. h(2) = sqrt(2+7) + 3 = sqrt(9) + 3 = 3 + 3 = 6Choose the Correct Answer: Choose the correct answer. The value of h(2) is 6, which corresponds to answer choice (A).

Let $h(x)=\frac{x-2}{\sqrt{x+7}-3}$ when $x \neq 2$ . $\newline$ $h$ is continuous for all x>-7 . $\newline$ Find $h(2)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $6$ $\newline$ (B) $2$ $\newline$ (C) $4$ $\newline$ (D) $-3$

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Let h(x)=x−2x+7−3 h(x)=\frac{x-2}{\sqrt{x+7}-3} h(x)=x+7​−3x−2​ when x≠2 x \neq 2 x=2.\newlineh h h is continuous for all x>-7 .\newlineFind h(2) h(2) h(2).\newlineChoose 111 answer:\newline(A) 666\newline(B) 222\newline(C) 444\newline(D) −3-3−3

Let $h(x)=\frac{x-2}{\sqrt{x+7}-3}$ when $x \neq 2$ . $\newline$ $h$ is continuous for all x>-7 . $\newline$ Find $h(2)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $6$ $\newline$ (B) $2$ $\newline$ (C) $4$ $\newline$ (D) $-3$