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Sveta_85 [38]
3 years ago
10

Mrs. Hernanadez is covering the top of a kitchen shelf that is26 3/8 inches long and 15 1/2 inches wide. She incorrectly measure

s the length to be 25 inches long. How will this error affect her calculations?
Mathematics
1 answer:
miss Akunina [59]3 years ago
5 0

Answer:

The  calculation of Mrs. Hernanadez will affect by 1.375 inches lesser.

Step-by-step explanation:

Given:

Length of the Kitchen shelf = 26 3/8 inches

Width of the Kitchen Shelf =  15 1/2 inches

Incorrect measurement of the length of the kitchen shelf = 25 inches

Solution:

The Actual length of the kitchen shelf is 26\frac{3}{8}

But she measured as 25 inches

So the error in the measurement will be

=>Correct length of the shelf - Incorrect length of the shelf

=> 26\frac{3}{8} - 25

Coverting the mixed fraction to fraction,

=> \frac{211}{8} - 25

=> 26.375 - 25

=>1.375

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Answer:

The function \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}} is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}

The function is continuous at point x=36 if:

  1. The function \\ f(x) exists at x=36.
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  3. The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

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\\ f(36) = \frac{36*6}{900} = \frac{6}{25}

The limit of the \\ f(x) is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}

\\ \lim_{x \to 36} f(x) = \frac{6}{25}

Since

\\ f(36) = \frac{6}{25}

And

\\ \lim_{x \to 36} f(x) = \frac{6}{25}

Then, the function \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}} is continuous at x = 36.

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