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Given the axis of symmetry and two points, how do you find the equation of a parabola?

Jim H
Featured 2 months ago

Explanation:

Given that the axis is $x = - 3$, we know the vertex is at $x = - 3$, so the equation can be written in the form

$y = a {\left(x + 3\right)}^{2} + k$

Now use the two points given to get:

From $\left(- 1 , 3\right)$, we have $3 = a {\left(2\right)}^{2} + k$ so $4 a + k = 3$ and $k = 3 - 4 a$

From $\left(2 , - 5\right)$, we get $- 5 = a {\left(5\right)}^{2} + k$ so $25 a + k = - 5$

Replace $k$ in the last equation with $3 - 4 a$ to get

$25 a + \left(3 - 4 a\right) = - 5$, so

$21 a = - 8$ and $a = - \frac{8}{21}$

Using $k = 3 - 4 a$ again we get $k = 3 - \left(- \frac{32}{21}\right) = \frac{63}{21} + \frac{32}{21} = \frac{95}{21}$.

The equation is $y = - \frac{8}{21} {\left(x + 3\right)}^{2} + \frac{95}{21}$

If course, there are other ways to write the answer.

graph{(y+8/21(x+3)^2-95/21)(x+3)((x+1)^2+(y-3)^2-0.02)((x-2)^2+(y+5)^2-0.02)=0 [-12, 12, -6, 6]}

How do you solve the system 5x + 5y - 6z = -21, -2x + y - 3z = -18, -x +6y - 6z = -25?

Featured 2 months ago

The solution is $\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}1 \\ 2 \\ 6\end{matrix}\right)$

Explanation:

We perform the Gauss Jordan elimination with the augmented matrix

$\left(\begin{matrix}5 & 5 & - 6 & : & - 21 \\ - 2 & 1 & - 3 & : & - 18 \\ - 1 & 6 & - 6 & : & - 25\end{matrix}\right)$

$R 1 \leftarrow - R 3$ and $R 3 \leftarrow R 1$

$\left(\begin{matrix}1 & - 6 & 6 & : & 25 \\ - 2 & 1 & - 3 & : & - 18 \\ 5 & 5 & - 6 & : & - 21\end{matrix}\right)$

$R 3 \leftarrow 2 R 3 + 5 R 2$

$\left(\begin{matrix}1 & - 6 & 6 & : & 25 \\ - 2 & 1 & - 3 & : & - 18 \\ 0 & 15 & - 27 & : & - 132\end{matrix}\right)$

$R 2 \leftarrow R 2 + 2 R 1$

$\left(\begin{matrix}1 & - 6 & 6 & : & 25 \\ 0 & - 11 & 9 & : & 32 \\ 0 & 15 & - 27 & : & - 132\end{matrix}\right)$

$R 3 \leftarrow 11 R 3 - 15 R 2$

$\left(\begin{matrix}1 & - 6 & 6 & : & 25 \\ 0 & - 11 & 9 & : & 32 \\ 0 & 0 & - 162 & : & - 972\end{matrix}\right)$

$R 3 \leftarrow \frac{R 3}{- 162}$

$\left(\begin{matrix}1 & - 6 & 6 & : & 25 \\ 0 & - 11 & 9 & : & 32 \\ 0 & 0 & 1 & : & 6\end{matrix}\right)$

$R 2 \leftarrow R 2 - 9 R 3$

$\left(\begin{matrix}1 & - 6 & 6 & : & 25 \\ 0 & - 11 & 0 & : & - 22 \\ 0 & 0 & 1 & : & 6\end{matrix}\right)$

$R 2 \leftarrow \frac{R 2}{- 11}$

$\left(\begin{matrix}1 & - 6 & 6 & : & 25 \\ 0 & 1 & 0 & : & 2 \\ 0 & 0 & 1 & : & 6\end{matrix}\right)$

$R 1 \leftarrow R 1 + 6 R 2 - 6 R 3$

$\left(\begin{matrix}1 & 0 & 0 & : & 1 \\ 0 & 1 & 0 & : & 2 \\ 0 & 0 & 1 & : & 6\end{matrix}\right)$

Find all the 8^(th) roots of 3i-3?

Steve M
Featured 1 month ago

Explanation:

Let $\omega = - 3 + 3 i$, and let ${z}^{8} = \omega$

First, we will put the complex number into polar form:

$| \omega | = \sqrt{{\left(- 3\right)}^{2} + {3}^{2}} = 3 \sqrt{2}$
$\theta = \arctan \left(\frac{3}{- 3}\right) = \arctan \left(- 1\right) = - \frac{\pi}{4}$
$\implies a r g \setminus \omega = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3 \pi}{4}$

So then in polar form we have:

$\omega = 3 \sqrt{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

We now want to solve the equation ${z}^{8} = \omega$ for $z$ (to gain $8$ solutions):

${z}^{8} = 3 \sqrt{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of $2 \pi$, so we can equivalently write (incorporating the periodicity):

${z}^{8} = 3 \sqrt{2} \left(\cos \left(\frac{3 \pi}{4} + 2 n \pi\right) + i \sin \left(\frac{3 \pi}{4} + 2 n \pi\right)\right) \setminus \setminus \setminus n \in \mathbb{Z}$

By De Moivre's Theorem we can write this as:

$z = 3 \sqrt{2} {\left(\cos \left(\frac{3 \pi}{4} + 2 n \pi\right) + i \sin \left(\frac{3 \pi}{4} + 2 n \pi\right)\right)}^{\frac{1}{8}}$
$\setminus \setminus = {\left(3 \sqrt{2}\right)}^{\frac{1}{8}} {\left(\cos \left(\frac{3 \pi}{4} + 2 n \pi\right) + i \sin \left(\frac{3 \pi}{4} + 2 n \pi\right)\right)}^{\frac{1}{8}}$
$\setminus \setminus = {3}^{\frac{1}{8}} {2}^{\frac{1}{16}} \left(\cos \left(\frac{\frac{3 \pi}{4} + 2 n \pi}{8}\right) + i \sin \left(\frac{\frac{3 \pi}{4} + 2 n \pi}{8}\right)\right)$
$\setminus \setminus = {3}^{\frac{1}{8}} {2}^{\frac{1}{16}} \left(\cos \theta + i \sin \theta\right)$

Where:

$\theta = \frac{\frac{3 \pi}{4} + 2 n \pi}{8} = \frac{\left(3 + 8 n\right) \pi}{32}$

And we will get $8$ unique solutions by choosing appropriate values of $n$. Working to 3dp, and using excel to assist, we get:

After which the pattern continues (due the above mentioned periodicity).

We can plot these solutions on the Argand Diagram:

How is an inner product defined ?

Aritra G.
Featured 4 weeks ago

In linear algebra, one studies the notion of linear vector spaces over different scalar fields.

So assuming that you are familiar with the notion of a linear vector space, let us consider an LVS $V \left(F\right)$.

Then for two elements $u$ and $v$ in $V$, we can associate a scalar (in the field $F$) such that the following properties hold,

1) Complex conjugate of $\left(u , v\right) = \left(v , u\right)$

2) $\left(u , u\right) \ge 0$ and the equality holds only for $u = 0$

3) For four vectors $u , v , w , z$ in $V$ and two arbitrary scalars $a$ and $b$ in $F$, $\left(a \left(u + v\right) , b \left(w + z\right)\right) = \overline{a} b \left[\left(u , w\right) + \left(u , z\right) + \left(v , w\right) + \left(v , z\right)\right]$

Where $\overline{a}$ denotes complex conjugation.

If these properties hold, we call $\left(u , v\right)$ an inner product of $u$ and $v$ and $V$ is then an inner product space.

However, there is no general recipe for defining the inner product of two vectors (or functions). We just have to make sure that the properties 1-3 are satisfied.

For an example, an inner product of two wave functions $\psi$ and $\phi$ in Quantum mechanics is defined as,

$< \psi | \phi > = \int \overline{\psi} \phi d \tau$ where integration is over all volume of the space and $\overline{\psi}$ denotes complex conjugate of $\psi$.

One can easily check if the properties 1-3 hold.

X^4-2x-4x^2+2x+3 is divided by x^2+2x+1 Using in long division, find the quotients And the remainder when: can u please calculate for me ?.thank u

Geoff K.
Featured 4 weeks ago

$\left({x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3\right) \div \left({x}^{2} + 2 x + 1\right) = {x}^{2} - 4 x + 3$, with no remainder.

Explanation:

Set up the long division like this:

$\textcolor{m a \ge n t a}{{x}^{2}} + 2 x + 1 \overline{| \text{ } \textcolor{m a \ge n t a}{{x}^{4}} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$

Divide $\textcolor{m a \ge n t a}{{x}^{4} / {x}^{2}}$, giving $\textcolor{red}{{x}^{2}}$; put this quotient above the ${x}^{4}$:

color(white)(x^2+2x+1bar(|"  "color(red)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$

Multiply $\textcolor{red}{{x}^{2}} \times \left({x}^{2} + 2 x + 1\right)$, giving $\textcolor{b l u e}{{x}^{4} + 2 {x}^{3} + {x}^{2}}$; put this product below the ${x}^{4} - 2 {x}^{3} - 4 {x}^{2}$:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(blue)(x^4+2x^3+color(white)(1)x^2))

Subtract $\left({x}^{4} - 2 {x}^{3} - 4 {x}^{2}\right) - \left(\textcolor{b l u e}{{x}^{4} + 2 {x}^{3} + {x}^{2}}\right)$, giving color(orange)(–4x^3-5x^2); draw a line under $\textcolor{b l u e}{{x}^{4} + 2 {x}^{3} + {x}^{2}}$ and write this difference below the line:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ "color(white)(x^4" ")color(orange)(-4x^3-5x^2)" }}$

Copy the $\textcolor{g r e e n}{2 x}$ from the dividend down below this line:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + \textcolor{g r e e n}{2 x} + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ "color(white)(x^4)-4x^3-5x^2" } \textcolor{g r e e n}{+ 2 x}}$

Repeat this process twice, dividing the latest leading term below your line by the leading ${x}^{2}$ from the divisor:

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-color(red)(4x))
$\textcolor{m a \ge n t a}{{x}^{2}} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - \textcolor{m a \ge n t a}{4 {x}^{3}} - 5 {x}^{2} + 2 x}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-color(red)(4x))
$\textcolor{red}{{x}^{2} + 2 x + 1} \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l u e}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + \textcolor{g r e e n}{3}}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} \textcolor{\mathmr{and} a n \ge}{3 {x}^{2} + 6 x} + \textcolor{g r e e n}{3}}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+color(red)3)
$\textcolor{m a \ge n t a}{{x}^{2}} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} \textcolor{m a \ge n t a}{3 {x}^{2}} + 6 x + 3}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+color(red)(3))
$\textcolor{red}{{x}^{2} + 2 x + 1} \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} 3 {x}^{2} + 6 x + 3}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ "bar(color(blue)(" } 3 {x}^{2} + 6 x + 3}$

...

color(white)(x^2+2x+1bar(|"  "color(black)(x^2-4x"  "+3)
${x}^{2} + 2 x + 1 \overline{| \text{ } {x}^{4} - 2 {x}^{3} - 4 {x}^{2} + 2 x + 3}$
color(white)(x^2+2x+1bar(|"  "color(black)(x^4+2x^3+color(white)(1)x^2))
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 |} \overline{\text{ } \textcolor{w h i t e}{{x}^{4}} - 4 {x}^{3} - 5 {x}^{2} + 2 x}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \overline{\text{ } \textcolor{b l a c k}{- 4 {x}^{3} - 8 {x}^{2} - 4 x}}}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ }} \overline{\textcolor{w h i t e}{- 4 {x}^{3} +} 3 {x}^{2} + 6 x + 3}$
$\textcolor{w h i t e}{{x}^{2} + 2 x + 1 | \text{ "bar(color(black)(" } 3 {x}^{2} + 6 x + 3}$
color(white)(x^2+2x+1|"                       ")bar(color(white)(3x^2+6x+color(orange)0)

What is the sum for 1+2+4+....+1024?

George C.
Featured 2 days ago

$2047$

Explanation:

Note that $1024 = {2}^{10}$

This is a geometric series, with common ratio $2$, initial term $1$ and $11$ terms.

The general term of a geometric series can be written:

${a}_{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ the common ratio.

Then we find:

$\left(1 - r\right) {\sum}_{n = 1}^{N} {a}_{n} = \left(1 - r\right) {\sum}_{n = 1}^{N} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} {a}_{n}} = {\sum}_{n = 1}^{N} a {r}^{n - 1} - r {\sum}_{n = 1}^{N} a {r}^{n - 1}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} {a}_{n}} = a + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N} a {r}^{n - 1}}}} - a {r}^{N}$

$\textcolor{w h i t e}{\left(1 - r\right) {\sum}_{n = 1}^{N} {a}_{n}} = a \left(1 - {r}^{N}\right)$

So dividing both ends by $\left(1 - r\right)$ we find:

${\sum}_{n = 1}^{N} {a}_{n} = \frac{a \left(1 - {r}^{N}\right)}{1 - r}$

In our example $N = 11$, $a = 1$, $r = 2$ and

${\sum}_{n = 1}^{N} {a}_{n} = \frac{\textcolor{b l u e}{1} \left(1 - {\textcolor{b l u e}{2}}^{\textcolor{b l u e}{11}}\right)}{1 - \textcolor{b l u e}{2}} = {2}^{11} - 1 = 2047$

Alternatively, in this particular example there is a simpler solution...

Note that:

$1 + \left(1 + 2 + 4 + 8 + \ldots + 1024\right) = 2 + 2 + 4 + 8 + \ldots + 1024$

$\textcolor{w h i t e}{1 + \left(1 + 2 + 4 + 8 + \ldots + 1024\right)} = 4 + 4 + 8 + 16 + \ldots + 1024$

$\textcolor{w h i t e}{1 + \left(1 + 2 + 4 + 8 + \ldots + 1024\right)} = 8 + 8 + 16 + 32 + \ldots + 1024$

$\textcolor{w h i t e}{1 + \left(1 + 2 + 4 + 8 + \ldots + 1024\right) =} \vdots$

$\textcolor{w h i t e}{1 + \left(1 + 2 + 4 + 8 + \ldots + 1024\right)} = 1024 + 1024$

$\textcolor{w h i t e}{1 + \left(1 + 2 + 4 + 8 + \ldots + 1024\right)} = 2048$

So:

$1 + 2 + 4 + 8 + \ldots + 1024 = 2048 - 1 = 2047$

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