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3 years ago

13

Can someone solve this pleaseit is solving equations in algebra: x-6=4

1 answer:

Sholpan [36]3 years ago

7 0

The answer to this equation is 10 = x

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Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

4 0

3 years ago