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3 years ago

Plz help giving brainlest

Mathematics

2 answers:

lilavasa [31]3 years ago

5 0

Answer:

190km

Step-by-step explanation:

If we're trying to find an area of a rectangle, we're going to use the following formula:

a=l×w

A being our answer, area, l being length, our 19km, and w, being our width, our 10km.

We multiply 19 and 10 and get the result 190km!

I hope that this helped out!

Have a good rest of the day/evening.

UkoKoshka [18]3 years ago

4 0

Answer:

190

Step-by-step explanation:

are of rectangle is length times width

19 times 10 = 190

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4 years ago

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A raft in an amusement park ride comes out of a tuner and heads straight toward a waterfall at a speed of 44 feet per second. Th

Ilia_Sergeevich [38]

Answer:

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Step-by-step explanation:

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4 years ago

Find the length of the diagonal of an 8cmx 6cmx 10cm rectangular prism. Round to the nearest tenth. (HELP PLEASE!!)

Hatshy [7]

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14.1 cm

Step-by-step explanation:

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4 years ago

he graph shows the distribution of time (in hours) that teenagers spend playing video games per week. A graph titled Video games

Goshia [24]

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This is further explained below.

<h3>What is Normal distribution?</h3>

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In conclusion, "the graph illustrates the distribution of the amount of time, in hours, that teens spend playing video games over the course of a week," The statement "The distribution is nearly normal, with a mean of 14 hours and a standard deviation of 4" is the one that gives the most accurate account of how the data was distributed.

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2 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 :

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