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3 years ago

Write an equation to represent the following description. The difference of seven times a number and 15 is 41.

Mathematics

2 answers:

chubhunter [2.5K]3 years ago

7 0

Answer: 3

7x + 15 = 41 or 7x - 15 = 41

egoroff_w [7]3 years ago

6 0

Answer:

7x - 15 = 41

Step-by-step explanation:

The key word is "difference" so keep that in mind. 7 times a number and 15 is 41. The "number" is unknown so a variable can be put in place until the number is known. The whole equation equals 41. The word "and" is where the "difference" or subtraction sign should go. Now we have a translated description! I hope this helps and if I'm wrong, please let me know!

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Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=

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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:

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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