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Elza [17]
2 years ago
6

Rachel is a stunt driver. One time, during a gig where she escaped from a building about to explode(!), she drove to get to the

safe zone at 24 meters per second. After 4 seconds of driving, she was 70 meters away from the safe zone. Let y represent the distance (in meters) from the safe zone after x seconds. Complete the equation for the relationship between the distance and number of seconds.
Mathematics
2 answers:
NNADVOKAT [17]2 years ago
7 0

Answer:

24x=y or 24x-y=0 or y-24x=0

Step-by-step explanation:

To Solve , here we use the formula for velocity:

Velocity = \frac{Distance}{Time}

As given in the statement, the distance is represented by Y and Time by X

So,

Distance = y meters

Time = x seconds

Velocity = \frac{y}{x}

The given velocity is 24 m/s

24=\frac{y}{x}

Cross multiplying:

24(x) = y

Moving x to the left hand side:

or 24x-y =0

Multiplying both sides by -1

or y-24x = 0

<u>Note: The mentioned time 4 seconds and distance travelled 70 meters is not needed to calculate the relation between distance and number of seconds.</u>

marishachu [46]2 years ago
5 0

y=−24x+166

Step-by-step explanation:

hope this helps <3

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Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
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y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
6 0
2 years ago
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