The time it takes to travel a fixed distance varies inversely with the speed traveled. It takes

1 answer:

6 0

first off, let's notice that Purple's time is in minutes, whilst the rate is in miles per hour, the units of both must correspond, so, we can either change the time from minutes to hours or the rate from hours to minutes, hmmm let's change the time to hours.

so 40 minutes, we know there are 60 minutes in 1 hour, so 40 minutes will be 40/60 of an hr, or namely 2/3.

$\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill$

$\stackrel{\begin{array}{llll} \textit{\tiny "t"ime varies}\\ \textit{\tiny inversely with "s"peed} \end{array}}{t = \cfrac{k}{s}}\qquad \textit{we know that} \begin{cases} t=\stackrel{minutes}{40}\to \stackrel{hrs}{\frac{2}{3}}\\ s=\stackrel{m/h}{9} \end{cases} \implies \cfrac{2}{3}~~ = ~~\cfrac{k}{9} \\\\\\ 18=3k\implies \cfrac{18}{3}=k\implies 6=k~\hfill \boxed{t=\cfrac{6}{s}} \\\\\\ \textit{when s = }\stackrel{m/h}{12}\textit{ what is "t"?}\qquad t=\cfrac{6}{12}\implies t=\cfrac{1}{2}$

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Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

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Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

$\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\$

So, this integral is convergent.

$\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\$

So, this integral is divergent.

$\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\$