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suter [353]
3 years ago
6

Suppose a new car that costs $25,000 depreciates by 12% each year. In about how many years will the car be worth $15,000? Use th

e equation 15,000 = (25,000)(0.88)x and round the value of x to the nearest year.
Mathematics
2 answers:
VLD [36.1K]3 years ago
8 0
Because the equation is already set up, we get a jump start on this problem!

15000 = (25000)*(.88)*(x) 

To solve for x, we just need to get x alone on its side! Our first step should be to divide both sides by 25000, which turns the equation into:

.06 = (.88)*(x)

Next, we need to divide both sides by .88 which should get x all alone, giving us our answer!

.68181... = x!

Because the question says to round the answer to the nearest year, it would take 1 year for the car's value to depreciate to $15000.<span />
lapo4ka [179]3 years ago
5 0
0.12 x 3 years = 0.36 

<span>0.36 X $25,000 = $9,000 </span>

<span>25,000 - 9000 = $16,000 </span>

<span>so its 3 1/2 years</span>
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Answer:

Step-by-step explanation:

I think you wish is:

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Charlene is knitting a baby blanket. She wants its width, w, to be at least half its length, l. She estimates that she has enoug
zepelin [54]

Answer:

W > 0.5L

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÷3       ÷3

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A car takes 54.9 seconds to travel 1 mile. How long does it take the car to travel 4.3 miles? First round to the nearest whole n
fiasKO [112]

Answer: Rounded: 236 seconds

Not rounded: 236.07 seconds

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5 0
3 years ago
Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

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\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
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