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KatRina [158]
2 years ago
15

The vertex of a function is (-3, 0) and is a minimum. What is the axis of symmetry and how is it determined?

Mathematics
1 answer:
nalin [4]2 years ago
5 0
The axis of symmetry would be -3 because it is always the (x) of the vertex
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Select the equation in which the graph of the line has a positive slope, and the y-intercept equals -8.
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The second one. 10x - 5y = 40
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3 years ago
I WILL GIVE A CROWN JUST NEED HELP ASAP
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Answer:

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3 0
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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Please help another question
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Answer is D) Y. It's matching with the results given

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On the day a coffee shop first opened, it had 140 customers. 65% of the customers ordered a large coffee.
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3 years ago
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