Over five weeks, you earned a total of $1,472.50. You were paid $7.75 per hour and worked the same number of hours each week. How many hours per week did you work?

Nigro6f
2022-10-20
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Pradellalo

Answered 2022-10-21
Author has **16** answers

$5\phantom{\rule{1ex}{0ex}}\text{weeks}=\$1472.50\mid :5}$

$1\phantom{\rule{1ex}{0ex}}\text{week}=\$294.5}$

Each week you earned $294.5.

$\text{hours}\phantom{\rule{1ex}{0ex}}\cdot \phantom{\rule{1ex}{0ex}}\text{hourly wage}=\text{total wage}$

$\text{hours}\phantom{\rule{1ex}{0ex}}\cdot \$7.75=\$294.5\mid :\$7.75}$

$\text{hours}\phantom{\rule{1ex}{0ex}}=38$

$1\phantom{\rule{1ex}{0ex}}\text{week}=\$294.5}$

Each week you earned $294.5.

$\text{hours}\phantom{\rule{1ex}{0ex}}\cdot \phantom{\rule{1ex}{0ex}}\text{hourly wage}=\text{total wage}$

$\text{hours}\phantom{\rule{1ex}{0ex}}\cdot \$7.75=\$294.5\mid :\$7.75}$

$\text{hours}\phantom{\rule{1ex}{0ex}}=38$

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I am coursing differential equations and recently encountered with the concept of integrating factors. I have seen them to solve two types of ODEs: inexact and linear. A linear equation is an ODE in the form:

$\frac{dy}{dx}+p(x)y=q(y)$

the integrating factor ends up being:

$u(x)={e}^{\int p(x)dx}$

so that the equation comes to:

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the equation can now be solved if $\int q(x)u(x)dx$ can be computed.

An inexact equation is an equation in the form

$A(x,y)dx+B(x,y)dy=0$

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(i.e. $Adx+Bdy$ is not an exact differential)

The integrating factor for these equations (I will call it $\mu $ for inexact equations) is a function such that

$(\mu A{)}_{y}=(\mu B{)}_{x}$

Expanding,

$\mu {A}_{y}+{\mu}_{y}A=\mu {B}_{x}+{\mu}_{x}A$

I have read the Wikipedia article, which says that to solve this equation where $\mu =\mu (x,y)$ requires partial differential equations, but if $\mu =\mu (x)$ or $\mu =\mu (y)$, then there is a straightforward formula for both, in terms of $A$ and $B$ (and their partial derivatives, respectively). But here is the important part: it says

"[...] in which case we only need to find $\mu $ with a first-order linear differential equation or a separable differential equation [...]"

Does this mean that this method can only be used for linear ODEs? In that case, I think the first method is way faster.

$\frac{dy}{dx}+p(x)y=q(y)$

the integrating factor ends up being:

$u(x)={e}^{\int p(x)dx}$

so that the equation comes to:

$\frac{d}{dx}(uy)=q(x)u(x)$

the equation can now be solved if $\int q(x)u(x)dx$ can be computed.

An inexact equation is an equation in the form

$A(x,y)dx+B(x,y)dy=0$

where

${A}_{y}\ne {B}_{x}$

(i.e. $Adx+Bdy$ is not an exact differential)

The integrating factor for these equations (I will call it $\mu $ for inexact equations) is a function such that

$(\mu A{)}_{y}=(\mu B{)}_{x}$

Expanding,

$\mu {A}_{y}+{\mu}_{y}A=\mu {B}_{x}+{\mu}_{x}A$

I have read the Wikipedia article, which says that to solve this equation where $\mu =\mu (x,y)$ requires partial differential equations, but if $\mu =\mu (x)$ or $\mu =\mu (y)$, then there is a straightforward formula for both, in terms of $A$ and $B$ (and their partial derivatives, respectively). But here is the important part: it says

"[...] in which case we only need to find $\mu $ with a first-order linear differential equation or a separable differential equation [...]"

Does this mean that this method can only be used for linear ODEs? In that case, I think the first method is way faster.

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I already know how the set of solutions of system of linear equations over real numbers infinite field $PP$ is expressed.

When there is only single solution then it is just a vector of scalars, where each scalar is a real number.

When there are more than 1 solution, and actually infinite solutions, then it is just a parametric linear vector space in the form:$\mathrm{\forall}t\in \mathbb{R}:\overrightarrow{p}+\overrightarrow{v}\cdot t$ where $\overrightarrow{p}\in {\mathbb{R}}^{n}l{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\overrightarrow{v}\in {\mathbb{R}}^{n}$ where n denotes the number of real variables in each linear equation thus $n\in \mathbb{N}$

But my question is how the set of solutions of system of linear equations over the finite field$\mathbb{Z}}_{2$ or galois field GF(2) is expressed?

I already know that when there is only single solution then it is also just a vector of scalars, but where each scalar is a binary number either zero or one in$\mathbb{Z}}_{2$ where $\mathbb{Z}}_{2}=\{0,1\$ , but when there are more than 1 solution, but always finite number of solutions, then how are they expressed?

Is this similar to how real solutions are expressed by parametric linear vector space by modulo 2? Or something else? I don't know. I am trying to google the answer for this question for days but I don't find the answer anywhere. It seems like nobody talks about this topic.

How?

When there is only single solution then it is just a vector of scalars, where each scalar is a real number.

When there are more than 1 solution, and actually infinite solutions, then it is just a parametric linear vector space in the form:

But my question is how the set of solutions of system of linear equations over the finite field

I already know that when there is only single solution then it is also just a vector of scalars, but where each scalar is a binary number either zero or one in

Is this similar to how real solutions are expressed by parametric linear vector space by modulo 2? Or something else? I don't know. I am trying to google the answer for this question for days but I don't find the answer anywhere. It seems like nobody talks about this topic.

How?