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stellarik [79]
3 years ago
8

The velocity of a roller coaster as it moves down a hill is v=√v²+64h. where v is the initial velocity and h is the vertical dro

p in feet. the designer wants the coaster to have a velocity of 200 feet per second when it reaches the bottom of the hill. if the initial velocity of the coaster is 30 feet per second, how high should the designer make the hill?
Mathematics
1 answer:
brilliants [131]3 years ago
4 0

Answer: So required height is 124.22feets.

Step-by-step explanation:

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They will all flash their lights together at 5:00 pm
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Doug's teacher told him that the standardized score (z-score) for his mathematics
BigorU [14]

Answer:

Doug’s test score is 1.20 standard deviations above the average test score of the students in the course.

I just took the test and got it right :)

Step-by-step explanation:

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You are in talks to settle a potential lawsuit. The defendant has offered to make annual payments of $15,000, $25,000, $40,000,
Sphinxa [80]

Answer:

All payments will be made at the end of the year by using the present value of inflows

Step-by-step explanation:

Present Value Of Inflows = Cash Inflow × Present Value Of Discounting Factor (Rate%,Time Period)

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8 0
4 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
Solve this and show work​
just olya [345]

Answer:

1.71

Step-by-step explanation:

x/4 = 3/7

4 * 3 = 12

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Rounded to the nearest hundreths, the answer is  1.71

6 0
3 years ago
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