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Liula [17]
3 years ago
5

Emanuel was charged $32 for a 14 2/9 km taxi ride. What was the cost per kilometer?

Mathematics
2 answers:
Lady_Fox [76]3 years ago
7 0

Answer:

$2.25 per kilometer

Step-by-step explanation:

Here we are asked for the unit rate, dollar per kilometer, that is \$/km.

If Emanuel was charged $32 we need to divided that amount by the kilometers traveled, so:

x=\frac{\$32}{(14+\frac{2}{9})km}\\x=\frac{\$32}{\frac{128}{9}km}\\\\x= \frac{\$288}{128km}\\\\x=2.25\frac{\$}{km}

So the cost per kilometer traveled is $2.25.

padilas [110]3 years ago
4 0
32 / (14 2/9) = 
32 / (128/9) =
32 * 9 / 128 =
288 / 128 =
$ 2.25 per km
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Answer:

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Step-by-step explanation:

Greatest common factor of 66 and 36 = 6

66+36 = 6*11 + 6*6 = 6(11+6)

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natima [27]

Answer:

The result of the integral is 81π

Step-by-step explanation:

We can use Stoke's Theorem to evaluate the given integral, thus we can write first the theorem:

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S

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Given F(x,y,z) = < yz, 4xz, e^{xy} > we have:

curl \vec F =\left|\begin{array}{ccc} \hat i &\hat j&\hat k\\ \cfrac{\partial}{\partial x}& \cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\yz&4xz&e^{xy}\end{array}\right|

Working with the determinant we get

curl \vec F = \left( \cfrac{\partial}{\partial y}e^{xy}-\cfrac{\partial}{\partial z}4xz\right) \hat i -\left(\cfrac{\partial}{\partial x}e^{xy}-\cfrac{\partial}{\partial z}yz \right) \hat j + \left(\cfrac{\partial}{\partial x} 4xz-\cfrac{\partial}{\partial y}yz \right) \hat k

Working with the partial derivatives

curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(4z-z\right) \hat k\\curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k

Integrating using Stokes' Theorem

Now that we have the curl we can proceed integrating

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot \hat n dS

where the normal to the circle is just \hat n= \hat k since the normal is perpendicular to it, so we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S \left(\left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k\right) \cdot \hat k dS

Only the z-component will not be 0 after that dot product we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S 3z dS

Since the circle is at z = 3 we can just write

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Thus the integral represents the area of a circle, the given circle x^2+y^2 = 9 has a radius r = 3, so its area is A = \pi r^2 = 9\pi, so we get

\displaystyle \int\limits_C \vec F \cdot d\vec r = 9(9\pi)\\\displaystyle \int\limits_C \vec F \cdot d\vec r = 81 \pi

Thus the result of the integral is 81π

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