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3 years ago

What is a number between 0.0006 and 0.00029

Mathematics

2 answers:

Kisachek [45]3 years ago

8 0

Answer:

0.00035

0.00055

0.00045

Step-by-step explanation.

Anything between 0.00029 and 0.00060 would be a correct answer.

faltersainse [42]3 years ago

5 0

0.00010 it can be that

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3 years ago

A fluid moves through a tube of length 1 meter and radius r=0.006±0.00025 meters under a pressure p=4⋅105±2000 pascals, at a rat

iris [78.8K]

Answer:

Maximum error for viscosity is 17.14%

Step-by-step explanation:

We know that everything is changing with respect to the time, "r" is changing with respect to the time, and also "p" just "v" will not change with the time according to the information given, so we can find the implicit derivative with respect to the time, and since

$n = (\frac{\pi}{8}) (\frac{pr^4}{v})\\$

The implicit derivative with respect to the time would be

$\frac{dn}{dt} = \frac{\pi}{8} ( \frac{r^4}{v} \frac{dp}{dt} + \frac{4pr^3}{v}\frac{dr}{dt} )$

If we multiply everything by dt we get

$dn = \frac{\pi}{8} ( \frac{r^4}{v} dp + \frac{4pr^3}{v} dr})$

Remember that the error is given by $\frac{dn}{n}$ therefore doing some algebra we get that

$\frac{dn}{n} = 4 \frac{dr}{r} + \frac{dp}{p}$

Since, r = 0.006 , dr = 0.00025 , p = 4*105 , dp = 2000 we get that

$\frac{dn}{n} = 0.1714$

Which means that the maximum error for viscosity is 17.14%.

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3 years ago

Kellie walks 1/4 mile in 1/14 hour. At what rate does kellie walk

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3 years ago

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[20 pts] evaluate the logarithm log(6)1/36, show work pls!

Delvig [45]

Answer:

$\large\boxed{\log_6\dfrac{1}{36}=-2}$

Step-by-step explanation:

$\text{We know:}\\\\\log_ab=c\iff a^c=b\\\\\log_6\dfrac{1}{36}\qquad\text{use}\ a^{-1}=\dfrac{1}{a}\\\\=\log_636^{-1}=\log_6(6^2)^{-1}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\log_66^{(2)(-1)}=\log_66^{-2}\qquad\text{use}\log_ab^n=n\log_ab\\\\=-2\log_66\qquad\text{use}\ \log_aa=1\\\\=-2(1)=-2$