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Liula [17]
3 years ago
15

2. Consider the division operations below and choose whether the result can be rational,

Mathematics
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

All are rational.

Step-by-step explanation:

1)

=1/3/3/5

=1/3×5/3

=5/9

5/9 is non-terminating and recurring number so it is rational.

2)

4/3√9/5

=4/3*3/5

=4/5

4/5 is a terminating number it is rational.

3)

2*√16/√4

=2*4/2

=4

4 is a natural/whole number it is also rational.

Note<u>:</u><u>i</u><u>f</u><u> </u><u>y</u><u>o</u><u>u</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>a</u><u>s</u><u>k</u><u> </u><u>a</u><u>n</u><u>y</u><u> </u><u>questions</u><u> </u><u>please</u><u> </u><u>l</u><u>e</u><u>t</u><u> </u><u>m</u><u>e</u><u> </u><u>k</u><u>n</u><u>o</u><u>w</u><u>.</u>

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2 years ago
Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye
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Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let \rho(x,y) be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

m=\int\limits \int\limits_D \rho(x,y) \, dA.

From the question, the given density function is \rho (x,y)=xye^{x+y}.

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx.

Since D is a rectangular region, we can apply Fubini's Theorem to get:

I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx.

Let the inner integral be: I_0=\int\limits^1_0xye^{x+y}dy, then

I=\int\limits^1_0(I_0)dx.

The inner integral is evaluated using integration by parts.

Let u=xy, the partial derivative of u wrt y is

\implies du=xdy

and

dv=\int\limits e^{x+y} dy, integrating wrt y, we obtain

v=\int\limits e^{x+y}

Recall the integration by parts formula:\int\limits udv=uv- \int\limits vdu

This implies that:

\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy

\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}

I_0=\int\limits^1_0 xye^{x+y}dy

We substitute the limits of integration and evaluate to get:

I_0=xe^x

This implies that:

I=\int\limits^1_0(xe^x)dx.

Or

I=\int\limits^1_0xe^xdx.

We again apply integration by parts formula to get:

\int\limits xe^xdx=e^x(x-1).

I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1).

I=\int\limits^1_0xe^xdx=0-1(0-1).

I=\int\limits^1_0xe^xdx=0-1(-1)=1.

No unit is given, therefore the mass of the lamina is 1.

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