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Doss [256]
3 years ago
11

Michael plays 1/5 of a song in 1/15 pf a minute. How many minutes will it take to play whole song

Mathematics
1 answer:
mr Goodwill [35]3 years ago
6 0
Well, a whole will be five fifths or 5/5.

now, he only plays 1/5 in 1/15, how many minutes will it take to play the whole 5/5 of the song?

\bf \begin{array}{ccll}
\stackrel{part}{song}&\stackrel{part}{minutes}\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
\frac{1}{5}&\frac{1}{15}\\\\
\frac{5}{5}&m
\end{array}\implies \cfrac{\frac{1}{5}}{\frac{5}{5}}=\cfrac{\frac{1}{15}}{m}\implies \cfrac{\frac{1}{5}}{\frac{5}{5}}=\cfrac{\frac{1}{15}}{\frac{m}{1}}\implies \cfrac{1}{5}\cdot \cfrac{5}{5}=\cfrac{1}{15}\cdot \cfrac{1}{m}

\bf \cfrac{1}{5}=\cfrac{1}{15m}\implies 15m=5\implies m=\cfrac{5}{15}\implies m=\stackrel{\textit{minutes}}{\cfrac{1}{3}}
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How do you solve this limit of a function math problem? ​
hram777 [196]

If you know that

e=\displaystyle\lim_{x\to\pm\infty}\left(1+\frac1x\right)^x

then it's possible to rewrite the given limit so that it resembles the one above. Then the limit itself would be some expression involving e.

For starters, we have

\dfrac{3x-1}{3x+3}=\dfrac{3x+3-4}{3x+3}=1-\dfrac4{3x+3}=1-\dfrac1{\frac34(x+1)}

Let y=\dfrac34(x+1). Then as x\to\infty, we also have y\to\infty, and

2x-1=2\left(\dfrac43y-1\right)=\dfrac83y-2

So in terms of y, the limit is equivalent to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^{\frac83y-2}

Now use some of the properties of limits: the above is the same as

\displaystyle\left(\lim_{y\to\infty}\left(1-\frac1y\right)^{-2}\right)\left(\lim_{y\to\infty}\left(1-\frac1y\right)^y\right)^{8/3}

The first limit is trivial; \dfrac1y\to0, so its value is 1. The second limit comes out to

\displaystyle\lim_{y\to\infty}\left(1-\frac1y\right)^y=e^{-1}

To see why this is the case, replace y=-z, so that z\to-\infty as y\to\infty, and

\displaystyle\lim_{z\to-\infty}\left(1+\frac1z\right)^{-z}=\frac1{\lim\limits_{z\to-\infty}\left(1+\frac1z\right)^z}=\frac1e

Then the limit we're talking about has a value of

\left(e^{-1}\right)^{8/3}=\boxed{e^{-8/3}}

# # #

Another way to do this without knowing the definition of e as given above is to take apply exponentials and logarithms, but you need to know about L'Hopital's rule. In particular, write

\left(\dfrac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\ln\left(\frac{3x-1}{3x+3}\right)^{2x-1}\right)=\exp\left((2x-1)\ln\frac{3x-1}{3x+3}\right)

(where the notation means \exp(x)=e^x, just to get everything on one line).

Recall that

\displaystyle\lim_{x\to c}f(g(x))=f\left(\lim_{x\to c}g(x)\right)

if f is continuous at x=c. \exp(x) is continuous everywhere, so we have

\displaystyle\lim_{x\to\infty}\left(\frac{3x-1}{3x+3}\right)^{2x-1}=\exp\left(\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}\right)

For the remaining limit, write

\displaystyle\lim_{x\to\infty}(2x-1)\ln\frac{3x-1}{3x+3}=\lim_{x\to\infty}\frac{\ln\frac{3x-1}{3x+3}}{\frac1{2x-1}}

Now as x\to\infty, both the numerator and denominator approach 0, so we can try L'Hopital's rule. If the limit exists, it's equal to

\displaystyle\lim_{x\to\infty}\frac{\frac{\mathrm d}{\mathrm dx}\left[\ln\frac{3x-1}{3x+3}\right]}{\frac{\mathrm d}{\mathrm dx}\left[\frac1{2x-1}\right]}=\lim_{x\to\infty}\frac{\frac4{(x+1)(3x-1)}}{-\frac2{(2x-1)^2}}=-2\lim_{x\to\infty}\frac{(2x-1)^2}{(x+1)(3x-1)}=-\frac83

and our original limit comes out to the same value as before, \exp\left(-\frac83\right)=\boxed{e^{-8/3}}.

3 0
3 years ago
PLZ HELP ME LOL
Serga [27]
40/7.50=5.33
So Diego can buy 5 pizzas with a little money left over.
3 0
2 years ago
Read 2 more answers
A term for a geometric figure that is enclosed by a circle.
Verizon [17]
A geometric figure enclosed by a circle is called an inscribed figure.
6 0
3 years ago
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A soccer ball has an interior diameter of 22 cm. How many cubic centimeters of air does a soccer ball
Delvig [45]

Answer:

Volume of soccer ball = 5,572.45 Cm³ (Approx)

Step-by-step explanation:

Given:

Diameter of soccer ball = 22 Cm

Radius of soccer ball = 22 Cm / 2 = 11 Cm

Find:

Volume of soccer ball = ?

Computation:

Volume\ of\ soccer\ ball = \frac{4}{3} \pi r^3\\\\ Volume\ of\ soccer\ ball = \frac{4}{3} \times\frac{22}{7} \times 11^3\\\\ Volume\ of\ soccer\ ball = 5,572.45332

Volume of soccer ball = 5,572.45 Cm³ (Approx)

Therefore, Soccer ball contain 5,572.5 Cm³ of air.

5 0
3 years ago
Quadrilateral GHIJ is similar to quadrilateral KLMN. Find the measure of side NK.
Leokris [45]

Answer:
NK = 14.4

Step-by-step explanation:

Given the following question:

We know that the two sides are similar
JI = 3, NM = 7.2

<u>The scale factor is 2.4 since....</u>

3\times2.4=7.2


Knowing the two sides are similar and if we multiply JI by 2.4, we get NM we can easily do the same for JG and NK.

JG = 6
6\times2.4=14.4
=14.4

NK = 14.4

Hope this helps.

6 0
2 years ago
Read 2 more answers
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