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4 years ago

A copy machine can print 480 copies every 4 minutes. How many copies can it print in 10 minutes?

Mathematics

2 answers:

Irina-Kira [14]4 years ago

6 0

Step-by-step explanation:

So you need to figure out how much it can print in 1 minute. You can find that out by dividing 480 by 4. You get 120 which you then multiple by 10.

So the answer is 1,200

labwork [276]4 years ago

3 0

Answer:

10x120=1,200 That will be your answer.

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Sally read that a recent survey found that 56% of all American adults own a smartphone. She asked her friend Alexandra what the

Crank

Answer:

This survey means that about 14 in every 25 adults in the survey had a smartphone.

Step-by-step explanation:

3 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:

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

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

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

How do you do solve this?

solniwko [45]

The relative frequency is the frequency of the item divided by the total number of all items. The total number of all items here is: 50 + 40 + 90 +20 = 200.

$Burgers = \frac{50}{200} = 25%$
$Pasta = \frac{40}{200} = 20%$
$Pizza = \frac{90}{200} = 45%$
$Salad = \frac{20}{200} = 10%$

Hope this helps!

3 0

3 years ago

Solve the system of equations 2x-2y=-14 and 3x-y=-1 by combining the equations.

Veseljchak [2.6K]

The solution to the system of equations is x = 3 and y = 10

<h3>How to solve the equations?</h3>

The system is given as:

2x-2y=-14

3x-y=-1

Multiply (1) by 1 and (2) by 2

So, we have:

1(2x-2y=-14)

2(3x-y=-1 )

This gives

2x - 2y=-14

6x - 2y=-2

Subtract the equations

-4x = -12

Divide by -4

x = 3

Substitute x = 3 in 3x-y=-1

3(3)-y=-1

Evaluate

9 - y = -1

Solve for y

y = 10

Hence, the solution to the system of equations is x = 3 and y = 10

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7 0

2 years ago

Given r(x) = startfraction 11 over (x minus 4) squared endfraction , which represents a domain restriction on r(x) and the corre

Dmitry_Shevchenko [17]

The domain and inverse of the function

$r(x) = \frac{11}{(x - 4)^2}$ is

Domain = $\mathbb{R} - \{4\}$ ,

$r^{-1}(x) = \pm \sqrt{\frac{11}{x}} + 4$

What is a function?

A function from A to B is a rule that assigns to each element of A a unique element of B. A is called the domain of the function and B is called the codomain of the function.

There are different operations on functions like addition, subtraction, multiplication, division and composition of functions.

The given function is

$r(x) = \frac{11}{(x - 4)^2}$

r(x) is not defined if x - 4 = 0

r(x) is not defined for x = 4

Domain = $\mathbb{R} - \{4\}$ ,

Where $\mathbb{R}$ is the set of all real number

Let r(x) = y

$\frac{11}{(x-4)^2} = y\\(x - 4)^2 = \frac{11}{y}\\x - 4 = \pm \sqrt{\frac{11}{y}}\\x = \pm\sqrt{\frac{11}{y}} +4$

<em /> $r^{-1}(x) = \pm \sqrt{\frac{11}{x}} + 4$

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3 0

1 year ago