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2 years ago

Frances gets 6 paychecks in 12 weeks. How many paychecks does she get in 52 weeks?

Mathematics

2 answers:

MA_775_DIABLO [31]2 years ago

5 0

She would get 26

52/2=26

Xelga [282]2 years ago

3 0

Answer:

Twenty six, 26.

Step-by-step explanation:

12 divided by 6= 2- that's the unit rate.

52 divided by 2=<u>26</u>

<u />

If you need pictures to see this, I have included it in a picture:

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Michelle has read 0.05625 percent of a book. If she has read 18 pages, how many total pages are in the book?

faust18 [17]

You read 2/5 of the remainder on Sunday. That means 3/5 of the remainder are unread (1- 2/5). It is given that 36 pages are unread. Therefore 3/5 of the remainder is equal to 36. Set up equation and solve for x which will be the remainder

3/5x=36
x=36(5/3)
x=60

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3 years ago

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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

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

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Lady bird [3.3K]

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3 years ago

3 kilograms of red hots and I

MrRa [10]

Answer:

Step-by-step explanation:

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3 years ago

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sophia borrowed $12,109 at 4.5% simple interest for 6 months. How much will the interest amount to? What is the total amount tha

Yuri [45]

The applicable formula is
I = Prt
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I = ($12,109)(.0.45)(1/2)
I = $272.45

b) The amount Sophia will have to pay back is the sum of the principal amount and the interest owed.
$12,109 + 272.45 = $12,381.45

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3 years ago