How old is Emily if Kelcey is 45 and she is also three more than three times Emily's age.

Iyana Jackson
2022-09-14
Answered

How old is Emily if Kelcey is 45 and she is also three more than three times Emily's age.

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mercuross8

Answered 2022-09-15
Author has **16** answers

K is for Kelcey's age and E is for Emily's age.

$K=45=(3\times E)+3$

$45-3=(3\times E)$

$42=3\times E$

$\frac{42}{3}=14=E$

Emily is 14 years old.

$K=45=(3\times E)+3$

$45-3=(3\times E)$

$42=3\times E$

$\frac{42}{3}=14=E$

Emily is 14 years old.

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I came across the following question:

Let $V$ be the real vector space of polynomials with degree $\le 2$. For $\alpha ,\beta \in \mathbb{R}$ let ${W}_{\alpha}=\{f\in V|f(\alpha )=0\}$ and ${D}_{\beta}=\{f\in V|{f}^{\prime}(\beta )=0\}$. ${f}^{\prime}$ describes the derivative of $f$.

Find a basis for ${W}_{\alpha}$ and ${D}_{\beta}$.

Unfortunatly I am not quite getting my head wrapped around vector spaces with polynomials yet. So after failing the question I had a look at the solution and there it says:

By solving the system of linear equations:

${W}_{\alpha}:\{X-\alpha ,{X}^{2}-{\alpha}^{2}\}$

${D}_{\beta}:\{1,{X}^{2}-2\beta X\}$

More explanation is not given.

I know that the elements in ${W}_{\alpha}$ in the form of $f(x)={c}_{2}{X}^{2}+{c}_{1}X+{c}_{0}$ are zero at $\alpha $, so $f(\alpha )={c}_{2}{\alpha}^{2}+{c}_{1}\alpha +{c}_{0}=0$.

Respectively for the elements in ${D}_{\beta}$:${f}^{\prime}(\beta )=2{c}_{2}\beta +{c}_{1}=0$.

However, it is not clear to me how I get from that information to the basis of ${W}_{\alpha}$ and ${D}_{\beta}$ using systems of linear equations.

Let $V$ be the real vector space of polynomials with degree $\le 2$. For $\alpha ,\beta \in \mathbb{R}$ let ${W}_{\alpha}=\{f\in V|f(\alpha )=0\}$ and ${D}_{\beta}=\{f\in V|{f}^{\prime}(\beta )=0\}$. ${f}^{\prime}$ describes the derivative of $f$.

Find a basis for ${W}_{\alpha}$ and ${D}_{\beta}$.

Unfortunatly I am not quite getting my head wrapped around vector spaces with polynomials yet. So after failing the question I had a look at the solution and there it says:

By solving the system of linear equations:

${W}_{\alpha}:\{X-\alpha ,{X}^{2}-{\alpha}^{2}\}$

${D}_{\beta}:\{1,{X}^{2}-2\beta X\}$

More explanation is not given.

I know that the elements in ${W}_{\alpha}$ in the form of $f(x)={c}_{2}{X}^{2}+{c}_{1}X+{c}_{0}$ are zero at $\alpha $, so $f(\alpha )={c}_{2}{\alpha}^{2}+{c}_{1}\alpha +{c}_{0}=0$.

Respectively for the elements in ${D}_{\beta}$:${f}^{\prime}(\beta )=2{c}_{2}\beta +{c}_{1}=0$.

However, it is not clear to me how I get from that information to the basis of ${W}_{\alpha}$ and ${D}_{\beta}$ using systems of linear equations.

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