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3 years ago

Which of the following terms means “a period of one thousand years”?

Mathematics

2 answers:

juin [17]3 years ago

6 0

A millennium is a period of one thousand years

weeeeeb [17]3 years ago

6 0

I think its a decade

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Enguun earns $17 per hour tutoring student-athletes at Brooklyn University.

Juliette [100K]

Answer:

a. Earnings of Enguun for 12 hours this month (G)

G = 204 dollars

b. Earnings of Enguun for 19.5 hours last month (G)

G = 331.5 dollars

Step-by-step explanation:

Nomenclature

G: Total earnings of Enguun in dollars

E: Enguun's earnings for 1 hour of tutoring (dollars/hour)

n: number of hours that Enguun tutored per month (hour/month)

Formula

G = E*n

E = $17 per hour = 17 dollars/hour

Problem development

a. Earnings of Enguun for 12 hours this month

n = 12 hours

G = E*n

$G= 17\frac{dollars}{hour} * 12 hour$ : We eliminate hours

G = 204 dollars

b. Earnings of Enguun for 19.5 hours last month

G = E*n

G = 17 * 19.5

G = 331.5 dollars

7 0

3 years ago

Read 2 more answers

Find the value of x (7x - 12) 114

Harrizon [31]

Answer:

(7x-12)=-5

-5x÷-5 114÷-5=

x=29

5 0

1 year ago

Read 2 more answers

Alana is 4 years older than her brother Ernie Ernie is two years older than her sister Amelia Amelia is 10 years younger than th

OverLord2011 [107]

Answer:

Alana is 13 years old.

Step-by-step explanation:

Assuming you meant Mazo = Maisel,

You start from the bottom:

Amelia is 10 years younger than 17-year-old Maisel.

Amelia is 7 years old. (17-10)

Ernie is 2 years older than 7-year-old Amelia.

Ernie is 9. (2+7)

Alana is 4 years older than 9-year-old Ernie.

Alana is 13 years old. (4+9)

Hope this helps!

-EmV

4 0

3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$