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3 years ago

Estimate 7956 - 5853 by first rounding each number to the nearest hundred.

Mathematics

1 answer:

VARVARA [1.3K]3 years ago

8 0

Answer:

$2100$

Step-by-step explanation:

Given the following question:

$7956-5853$

To estimate by the nearest hundred, we have to find the hundreds place value and then look at the number next to it, to find out if that number is greater than or equal to five.

$7956-5853$

$7956$

$5=5$

$8000$

$5853$

$5=5$

$5900$

$8000-5900=2100$

$=2100$

Hope this helps.

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jasenka [17]

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:

The function $\\ f(x)$ exists at x=36.
The limit on both sides of 36 exists.
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:

$\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}$

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

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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