Equation of direct variation involving a common proportion. Then identify the constant of variation

Mathematics

1 answer:

inessss [21]3 years ago

8 0

The equation for direct variation is y=kx where k is the constant of variation.

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Evaluate the spherical coordinate integral

expeople1 [14]

Rewrite the equations of the given boundary lines:

y = -x + 1 ==> x + y = 1

y = -x + 4 ==> x + y = 4

y = 2x + 2 ==> -2x + y = 2

y = 2x + 5 ==> -2x + y = 5

This tells us the parallelogram in the x-y plane corresponds to the rectangle in the u-v plane with 1 ≤ u ≤ 4 and 2 ≤ v ≤ 5.

Compute the Jacobian determinant for this change of coordinates:

$J=\begin{bmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}1&1\\-2&1\end{bmatrix}\implies|\det J|=3$

Rewrite the integrand:

$-3x+4y=-3\cdot\dfrac{u-v}3+4\cdot\dfrac{2u+v}3=\dfrac{5u+7v}3$

The integral is then

$\displaystyle\iint_R(-3x+4y)\,\mathrm dx\,\mathrm dy=3\iint_{R'}\frac{5u+7v}3\,\mathrm du\,\mathrm dv=\int_2^5\int_1^45u+7v\,\mathrm du\,\mathrm dv=\boxed{333}$

5 0

3 years ago

What is the area of trapezoid ABCD ? (Enter your answer as a decimal or whole number)

MrRa [10]

Check the picture below.

bear in mind that, the "bases" are the two parallel sides, and the height is the distance between them.

$\bf \textit{area of this trapezoid}\\\\
A=\cfrac{AB(BC+AD)}{2}\\\\
-------------------------------\\\\
\textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -1}}\quad ,&{{ 5}})\quad 
% (c,d
B&({{ 3}}\quad ,&{{ 2}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
AB=\sqrt{[3-(-1)]^2+[2-5]^2}\implies AB=\sqrt{(3+1)^2+(2-5)^2}
\\\\\\
AB=\sqrt{16+9}\implies AB=\sqrt{25}\implies \boxed{AB=5}$

$\bf -------------------------------\\\\
\textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 3}}\quad ,&{{ 2}})\quad 
% (c,d
C&({{ 0}}\quad ,&{{ -2}})
\end{array}
\\\\\\
BC=\sqrt{(0-3)^2+(-2-2)^2}\implies BC=\sqrt{9+16}
\\\\\\
BC=\sqrt{25}\implies \boxed{BC=5}$

$\bf -------------------------------\\\\
\textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ -1}}\quad ,&{{ 5}})\quad 
% (c,d
D&({{ -13}}\quad ,&{{ -11}})
\end{array}
\\\\\\
AD=\sqrt{[-13-(-1)]^2+[-11-5]^2}
\\\\\\
AD=\sqrt{(-13+1)^2+(-16)^2}\implies AD=\sqrt{144+256}
\\\\\\
AD=\sqrt{400}\implies \boxed{AD=\sqrt{20}}$

so, the area for this trapezoid is then

$\bf A=\cfrac{5(5+20)}{2}\implies A=\cfrac{125}{2}\implies A=62\frac{1}{2}$