TIMEDFind the slope

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

4 0

Answer:

Step-by-step explanation:

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Solve the quadratic equation by using the quadratic formula. 2x^2-5x+1=0

ioda

Answer:

$x = \frac{5 + \sqrt{17} }{4}$

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3 years ago

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What decimal is represented by this expanded form?

Radda [10]

2*10000 is 20000.

Start your expanded form with 20000. Add 7000, add 400, add 9, add 0.3 and add 0.008:

27409.308 (answer)

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4 years ago

How do you write 0.045 as a reduced fraction

Anvisha [2.4K]

0.045 = 45/1000

Divide top and bottom by their greatest common factor.
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3 years ago

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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=

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:

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