Answer: 13.375% per year
Explanation:
1) Depreciation is the loss of value: $ 20,000.00 - $ 14,650.00 = $ 5,350
2) The percent of depreciation is amount of the depreciation divided by the value of the car when purchased, times 100.
That is (5,350 / $ 20,000) * 100 = 26.75 %
2) The rate is percent of depreciation per year:
depreciation rate = % of depreciation / number of years = 26.75% / 2 = 13.375% per year.
Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t
Answer:
The 3rd set
Step-by-step explanation:
The first set has a range of 6, not 5
The second set has a range of 5, but not a mean of 5
The third set has both a mean and a range of 5
X and y vary inversely means y is inversely proportional to x, this means:
y=k/x, where k is a constant.
We are given that when x=-4, y=7/2, we replace these values to find k.
7/2=k/-4
k= (-4*7)/2=-14
So we have y=-14/x
Therefore, when x=4, y=-14/4= -7/2
