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lord [1]
2 years ago
12

What is limit of startroot x squared minus 8 endroot as x approaches negative 3? –5 1 startroot 2 endroot startroot negative 14

endroot
Mathematics
1 answer:
DochEvi [55]2 years ago
3 0

The limit of \lim_{x \to -3}\sqrt{ x^2-8} as x approaches negative 3 is 1

<h3>Limit of a function </h3>

The limit of the function given in question can also be written as:

\lim_{x \to -3}\sqrt{ x^2-8}

To determine the required limit, we will simply substitute the value of x into the function as shown:

\lim_{x \to -3}\sqrt{ x^2-8}=\sqrt{ (-3)^2-8}\\\lim_{x \to -3}\sqrt{ x^2-8}=\sqrt{9-8}=1\\

Hence the limit of \lim_{x \to -3}\sqrt{ x^2-8} as x approaches negative 3 is 1

Learn more on limit here: brainly.com/question/1444047
#SPJ4

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3 years ago
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lana [24]

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Option C is correct. In a matrix equation, product Ax when defined, is a sum of products.

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3 years ago
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schepotkina [342]

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for new

if length=25℅0f 20+20=25m

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