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murzikaleks [220]
3 years ago
6

5. A cube is a three-dimensional solid with six square faces.

Mathematics
1 answer:
Bond [772]3 years ago
6 0
Conditionallslslslod
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I dont even kno wat the BC is
Natasha2012 [34]

Answer:

<A

Step-by-step explanation:

The side BC is opposite an angle

The angle that is opposite BC is Angle A

4 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)

7 0
3 years ago
If p and q vary inversely and p is 10 when q is 19 , determine q when p is equal to 2
GenaCL600 [577]
You divide 10 by 5 is two so 19 decided by five is 3.8
6 0
3 years ago
Read 2 more answers
Helpppppppppp with this question???
irakobra [83]
Pretty sure it’s the second one
3 0
4 years ago
The axis of symmetry for the function f(x) = −x2 − 10x + 16 is x = −5. What are the coordinates of the vertex of the graph?
OLga [1]

Answer:

The coordinates of the vertex are (-5, 41)

Step-by-step explanation:

For a quadratic function of the form

f(x) = ax ^ 2 + bx + c

Where a, b and c are constants and represent the coefficients of the function, then the symmetry of the parabola always passes through its vertex.

In this case we have the following parabola

f (x) = -x2 - 10x + 16

And we know that its axis of symmetry is the line x = -5

Then we know that this axis of symmetry passes through the vertex of the parabola.

Therefore, the x coordinate of the vertex is -5.

To find the coordinate in y of the vertex, we substitute x = -5 in the function.

f (-5) = -(- 5) ^ 2 -10 (-5) +16\\\\f (-5) = 41

Finally, the vertices are in the point (-5, 41).

5 0
4 years ago
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