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Alenkinab [10]
3 years ago
11

The length of a rectangle is 4 inches more than its width, and its perimeter is 34 inches. Find the length and width of the

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer:

Width = 6.5 inches

Length = 10.5 inches

Step-by-step explanation:

Let

"l" be length of rectangle

and

"w" be width of rectangle

Since length of rectangle is 4 inches MORE THAN WIDTH, we can write:

l = 4 + w

The perimeter is sum of all sides, so

Perimeter = l + l + w + w = 2l + 2w

The perimeter is 34, so we can write:

2l + 2w = 34

Putting 1st equation in 2nd, we get equation in "w" and solve for this first:

2l + 2w = 34\\2(4+w) + 2w = 34\\8+2w+2w=34\\4w=26\\w=\frac{26}{4}=6.5

Width is 6.5 in

Now, l = 4 + w, so

l = 4 + 6.5 = 10.5

Length is 10.5 in

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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

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The inner integral is evaluated using integration by parts.

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

\implies du=xdy

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We substitute the limits of integration and evaluate to get:

I_0=xe^x

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

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

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

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3 0
3 years ago
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Answer:

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