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Nesterboy [21]
2 years ago
7

The pressure P (in pounds per square foot), in a pipe varies over time. Ten times an hour, the pressure oscillates from a low of

40 to a high of 280 and then back to a low of 40. The pressure at time t = 0 is 40. Let the function P = f(t) denote the pressure in pipe at time t minutes. Find the formula for the function P=f(t),
Mathematics
1 answer:
Vedmedyk [2.9K]2 years ago
6 0
<span>Pressure oscillating ten times every hour. So period n = 6 min. So the negative cos is represented in the graph and since it datrt at time 0, P = f(t) = Acos(Bt) + D Amplitude A = (Ph - Pl) / 2 = (280 - 40) / 2 = 240 / 2 = 120 Period B = 2xpi / 6 = pi /3 D = (Ph + Pl) / 2 = (280 + 40) / 2 = 320 / 2 = 160 Subtituting the equation we get f(t) = 120cos (pi x t ) / 3 + 160.</span>
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Answer:

Option a.

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Step-by-step explanation:

You have the following limit:

\lim_{x \to \frac{\pi}{2}{(3e)^{xcosx}

The method of direct substitution consists of substituting the value of \frac{\pi}{2} in the function and simplifying the expression obtained.

We then use this method to solve the limit by doing x=\frac{\pi}{2}

Therefore:

\lim_{x \to \frac{\pi}{2}}{(3e)^{xcosx} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})}

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By definition, any number raised to exponent 0 is equal to 1

So

\lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}cos(\frac{\pi}{2})} = \lim_{x\to \frac{\pi}{2}}{(3e)^{\frac{\pi}{2}(0)}\\\\

\lim_{x\to \frac{\pi}{2}}{(3e)^{0}} = 1

Finally

\lim_{x \to \frac{\pi}{2}}(3e)^{xcosx}=1

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