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kupik [55]
3 years ago
10

A car is traveling at a constant speed of 55 miles per hour. Using m for miles and t for time, write 2 equations to represent th

is situation. One equation to solve for miles and the other equation to solve for hours.
Mathematics
1 answer:
Viefleur [7K]3 years ago
3 0

Answer:

m = 55t

t = m/55

Step-by-step explanation:

According to the formula for calculating speed expressed as;

Speed = distance/time

Given;

Distance = 55mph

required:

distance as a function of t

time as a function of distance m

If distance is m and time is t

From the formula;

55 = m/t

m = 55t

Hence the equation to solve for miles m is m = 55t

To get t, we will cross multiply from 55 = m/t;

55t = m

Divide both sides by 55

55t/55 = m/55

t = m/55

Hence the equation to solve for time in hours is t = m/55

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Berry Boulevard is 4 4/5 Miles long.There are two traffic lights on Berry Boulevard every 2/5 mile after the street begins .Ther
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Answer: There are 24 traffic lights on Berry Boulevard.

Step-by-step explanation:

Since we have given that

Total length of road = 4\dfrac{4}{5}=\dfrac{24}{5}\ miles

Number of roads on every \dfrac{2}{5}\ mile = 2

So, According to unitary method,

In every \dfrac{2}{5}\ mile, number of roads = 2

In every 1 mile, number of roads = \dfrac{5}{2}\times 2=5

So, in every \dfrac{24}{5}\ miles,  the number of roads = \dfrac{24}{5}\times 5=24

Hence, there are 24 traffic lights on Berry Boulevard.

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Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of
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Close off the hemisphere S by attaching to it the disk D of radius 3 centered at the origin in the plane z=0. By the divergence theorem, we have

\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV

where R is the interior of the joined surfaces S\cup D.

Compute the divergence of \vec F:

\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2

Compute the integral of the divergence over R. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}

So the volume integral is

\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5

From this we need to subtract the contribution of

\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S

that is, the integral of \vec F over the disk, oriented downward. Since z=0 in D, we have

\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k

Parameterize D by

\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

where 0\le u\le 3 and 0\le v\le2\pi. Take the normal vector to be

\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k

Then taking the dot product of \vec F with the normal vector gives

\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v

So the contribution of integrating \vec F over D is

\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==>  486π/5 - (-81π/4) = 2349π/20

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