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4 years ago

First gets brainliest! Please and thanks you :)

Mathematics

2 answers:

horrorfan [7]4 years ago

6 0

Answer:

3.72 × 10 to the power of 4

Step-by-step explanation:

firstly convert 2/5 to percentage form, which is 40%

then find 10% of earths distance from the sun so divide 93000000 by 10 to get 9300000

multiply that by 4 to get 37200000

standard form = 3.72 × 10 to the power of 7

erma4kov [3.2K]4 years ago

3 0

Answer:

Its B

Step-by-step explanation:

93000000 times 2/5 is 37200000 aka 3.72 times 10 to the 7th power

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False, should be 54.03

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3 years ago

What is the solution to log 25x = 3?

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3 years ago

A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner

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

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