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Maru [420]
3 years ago
8

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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Answer:

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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:

d = 240 - 44t

Step-by-step explanation:

Distance of the raft to the waterfall

The raft is heading for the waterfall, therefore, the distance between the raft and the waterfall is diminishing.

Waterfall=240 ft from the tunnel

Speed of the raft =44 ft per second

Therefore the equation of a function rule that represent the distance of the raft to the waterfall is:

d = 240 - 44t

Where t=time in seconds

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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]

Answer:

14.1 cm

Step-by-step explanation:

first find the base diagonal :

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so,

diagonal = Square root (10^2 + 10^2) = 14.1 cm (ANS)

3 0
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]

"The graph shows the distribution of time (in hours) that teenagers spend playing video games per week. ", The statement that best describes the distribution is  "The distribution is approximately Normal, with a mean of 14 hours and a standard deviation of 4" Option B.

This is further explained below.

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

Generally, Data closer to the mean tends to occur more often than data further from the mean, as shown by the normal distribution, also known as the Gaussian distribution. The normal distribution may be represented graphically as a "bell curve."

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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4 0
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 <em>µ(x, y)</em> 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 <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

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}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
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