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3 years ago

The monthly fees for the local pool are $8 per month

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

3 0

Given:

Monthly fees for the local pool are $8 per month and $2 per visit.

Hector pays $34 in pool fees total for the month.

To find:

The number of times he visit the pool.

Solution:

We have,

Monthly fee of pool = $8

Additional fee = $2 per visit

Let Hector visit x times.

Additional fee for x times = $2x

Total fee = Monthly fee + Additional fee

$34=8+2x$

$34-8=2x$

$26=2x$

Divide both sides by 2.

$\dfrac{26}{2}=x$

$13=x$

Therefore, Hector visit the pool 13 times.

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Differential Equations - Reduction of order

iren [92.7K]

With reduction of order, we assume a solution of the form $y_2=zy_1=ze^{2x}$ , with $z=z(x)$ . Then

${y_2}'=(z'+2z)e^{2x}$
${y_2}''=(z''+4z'+4z)e^{2x}$

and substituting into the ODE gives

$x(z''+4z'+4z)e^{2x}-(2x+1)(z'+2z)e^{2x}+2ze^{2x}=0$
$x(z''+4z'+4z)-(2x+1)(z'+2z)+2z=0$
$xz''+(2x-1)z'=0$

Let $\xi(x)=z'(x)$ , so that $\xi'=z''$ . This gives the linear ODE

$x\xi'+(2x-1)\xi=0$

This equation is also separable, so you can write

$\dfrac{\xi'}{\xi}=\dfrac{1-2x}x$

Integrating both sides with respect to $x$ gives

$\ln|\xi|=-2x+\ln x+C_1$
$\xi=C_1xe^{-2x}$

Next, solve $z'=\xi$ for $z$ by integrating both sides again with respect to $x$ .

$z'=\xi$
$\implies z=\displaystyle\int C_1xe^{-2x}\,\mathrm dx$
$\implies z=C_1e^{-2x}(2x+1)+C_2$

And finally, solve for $y_2$ .

$y_2=zy_1=C_1(2x+1)+C_2e^{2x}$

and note that $y_1$ is already taken into account as part of $y_2$ , so this is the general solution to the ODE.

6 0

3 years ago

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atroni [7]

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6 0

3 years ago

One light-year equals 5.9 x 1012 miles. How many light-years are in 6.79

inna [77]

Option B

The number of light years in $6.79 \times 10^{16}$ miles is 11508 light years

<em><u>Solution:</u></em>

Given that,

One light-year equals 5.9 x 10^12 miles

Therefore,

$1 \text{ light year } = 5.9 \times 10^{12} \text{ miles }$

To find: Number of light years in $6.79 \times 10^{16}$ miles

Let "x" be the number of light years in $6.79 \times 10^{16}$ miles

Then number of light years in $6.79 \times 10^{16}$ miles can be found by dividing $6.79 \times 10^{16}$ miles by miles in 1 light year

$\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79 \times 10^{16}}{5.9 \times 10^{12}}\\\\\text{Use the law of exponent }\\\\\frac{a^m}{a^n} = a^{m-n}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = \frac{6.79}{5.9} \times 10^{16-12}\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 1.1508 \times 10^4\\\\\text{Number of light years in } 6.79 \times 10^{16} miles = 11508$