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Sliva [168]
3 years ago
5

QUICK!!! PLEASE HELP!!! 50 points.. and BRAINLIEST FOR THE QUICK AND CORRECT ANSWER.

Mathematics
1 answer:
faust18 [17]3 years ago
5 0

Answer: e=-50|t-40|+2000

The cable car’s elevation will be 750 feet after 15 minutes or 65 minutes.

Step-by-step explanation:

Given: A cable car begins its trip by moving up a hill. As it moves up, it gains elevation at a constant rate of 50 feet/minute until it reaches the peak at 2,000 feet.

Then, total time taken to reach the peak = (Distance) ÷ (speed)

= (2,000 feet) ÷ ( 50 feet/minute)

= 40 minutes

Then, as the car moves down to the hill’s base, its elevation drops at the same rate.

The equation that models the cable car’s elevation, e, after t minutes is

e= (constant rate)|t- time to reach peak |+ Peak's height

e=-50|t-40|+2000

When the cable car’s elevation will be 750 feet after minutes, then we have

750=-50|t-40|+2000\\\\\Rightarrow\ -50|t-40|=750-2000\\\\\Rightarrow\ -50|t-40|=1250\\\\\Rightarrow|t-40|=-\dfrac{1250}{50}\\\\\Rightarrow|t-40|=-25\\\\\Rightarrow t-40=-25\text{ or }t-40=25\\\\\Rightarrow t=-25+40\text{ or }t=25+40\\\\\Rightarrow  t=15\text{ or }t=65

Time cannot be negative, so the cable car’s elevation will be 750 feet after 15 minutes or 65 minutes.

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Answer:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Step-by-step explanation:

GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...  \ \ \ (1)}

\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0}  \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!}  -\dfrac{x^6}{3!}+... \ \ \  (2)}

From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}

\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}

\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}

Multiplying equation (2) with (3); we have :

\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }

\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Finally , multiplying 5 with \mathtt{ e^{-x^2} cos (4x) } ; we have:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

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