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2 years ago

10

The length of a rectangle is 2x + 6. Its width is 52. Find the value of ac if the perimeter is 138m².

1 answer:

Sergeeva-Olga [200]2 years ago

8 0

Answer:

ac = 18.5

Step-by-step explanation:

perimeter of a rectangle = $2(l+b)$

= 138m^2= 2(2x + 6 ) + 52

= 4x + 12 + 52 = 138

= 4x + 64 = 138

= 4x = 138 - 64

= 4x = 74

= x = 74/4

= x = 18.5

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Write an equation that is parallel to the line y = -5x + 2 and passes through the point (0, 3).

nikklg [1K]

---------------------------------------
Slope :
---------------------------------------
y = -5x + 2
Slope = -5

---------------------------------------
y - intercept :
---------------------------------------
At (0. 3), y-intercept = 3

---------------------------------------
Equation :
---------------------------------------
y = mx + b
y = -5x + 3

---------------------------------------
Answer : y = -5 + 3
---------------------------------------

7 0

3 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

Alona [7]

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.

3 0

3 years ago

What is the nearest tenth P(4,-2) Q (1,3)

irinina [24]

I hope this helps you

4 0

3 years ago

Please explain step by step

Alja [10]

Let's find the value of x

Since the sum of internal angles of a triangle is 180°

therefore,

$\hookrightarrow \: (2x - 2) \degree + (x + 5) \degree + 90 \degree = 180 \degree$

$\hookrightarrow \: 2x - 2 \degree + x + 5\degree = 180\degree - 90\degree$

$\hookrightarrow \: 3x + 3\degree = 90\degree$

$\hookrightarrow \: 3(x + 1) = 3(30)$

$\hookrightarrow \: x + 1\degree = 30\degree$

$\hookrightarrow \: x = 29\degree$

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3 years ago

Consider this sphere inside the cylinder. A sphere is inside of a cylinder with a same radius and height. Which statements are t

sashaice [31]

Answer:

A,C,E

Step-by-step explanation:

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3 years ago

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