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Marat540 [252]
3 years ago
7

Terrance buys a car for 23,000. The car depricates in value 3% per year. Write an exponential funtuon to model the value of the

car. What would be the value of the car after 6 years
Mathematics
2 answers:
alukav5142 [94]3 years ago
8 0

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ              

tatyana61 [14]3 years ago
7 0

Answer:Terrence buys a new car for $20,000. The value of the car depreciates by 15% each year. t is time in years. Therefore the the value of the car after x years is represented by f(x) = 20,000(0.85)x .

Step-by-step explanation:

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Because I've gone ahead with trying to parameterize S directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over S straight away, let's close off the hemisphere with the disk D of radius 9 centered at the origin and coincident with the plane y=0. Then by the divergence theorem, since the region S\cup D is closed, we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV

where R is the interior of S\cup D. \vec F has divergence

\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z

so the flux over the closed region is

\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0

\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}

Parameterize D by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k

with 0\le u\le9 and 0\le v\le2\pi. Take the normal vector to D to be

\vec s_u\times\vec s_v=-u\,\vec\jmath

Then the flux of \vec F across S is

\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0

\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}

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AD= 36 cm. Points C, B∈AD, such that AB:BC:CD=2:3:4. Find the distance of midpoints of the segments AB and CD.
brilliants [131]

Answer:

The distance of the midpoints of AB and CD is 24 cm

Step-by-step explanation:

The given information are;

The length of AD = 36 cm.

C and B are points on AD

The ratio of AB:BC:CD 2:3:4

The distance of the midpoints of segments AB and CD required

Therefore, we have the following proportion of the total length, 36

Segment AB = 2/(2 + 3 + 4) = 2/9×36 = 8 cm

Segment BC = 3/(2 + 3 + 4) = 3/9×36 = 12 cm

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Which gives;

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The distance of midpoint of segment CD  from A = 28 cm

And the distance of the midpoints of AB and CD = 24 - 4 = 24 cm.

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