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Nutka1998 [239]
3 years ago
5

Solve.

Mathematics
1 answer:
LuckyWell [14K]3 years ago
3 0
-\frac{s}{3} > 6\ \ \ |change\ signs\\\\\frac{s}{3} < -6\ \ \ \ |multiply\ both\ sides\ by\ 3\\\\\boxed{s < -18\Rightarrow s\in(-\infty;-18)}\\\\Answer:\boxed{\bxoed{?}}\\\\I\ think,\ correct\ equation\ is:-\frac{s}{3} \geq6\\\\therefore\ answer\ is\ \boxed{\boxed{D.\ s\leq-18}}
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Answer:

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where d = diameter

Step-by-step explanation:

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The expression 5,430,900 is written in standard notation.
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frozen [14]

Answer:

I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner I_{corner} is:

I_{corner} = \int\limits \int\limits_R (x^2+y^2)  \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx

where :

(a and b are the length and the breath of the rectangle respectively )

I_{corner} =  \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx

I_{corner} =  \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx

I_{corner} =  \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}

I_{corner} =  \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]

I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

Thus; the moment of inertia of the lamina about one corner is I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

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3 years ago
45 centimeters is equivalent to how many inches?
Fiesta28 [93]

It is equivalent to 17.7165 inches.

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