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Sergeeva-Olga [200]
2 years ago
12

If d = m/v, what is the equation when solved for v? **please explain**

Mathematics
1 answer:
inn [45]2 years ago
8 0
V= m/d.  When you get questions asking about density/mass/volume draw a triangle just like in the picture i have attached and with your finger cover up the one you are looking for. So to find the equation for volume you cover up volume and you are left with mass divided by density. 

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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:

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

The value of the function at x = 36 is:

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

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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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