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4 years ago

14

To divide 572 divide 4, Stanley estimated to place the first digit of the quotient. In which place is the first digit of the quo

tient?

1 answer:

andreyandreev [35.5K]4 years ago

6 0

It would go over the five, so im guessing the hundreds

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What quadratic inequality does the graph below represent?

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The quadratic graph crosses the y-axis at -3, so the equation is $y= x^{2} -3$

The shaded region is on 'outer' part of the graph, hence the inequality is

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Find the equation of ellipse passing throgh (1,4) and (-3,2)

irinina [24]

Answer:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

Step-by-step explanation:

we want to figure out the ellipse equation which passes through <u>(</u><u>1</u><u>,</u><u>4</u><u>)</u><u> </u>and <u>(</u><u>-</u><u>3</u><u>,</u><u>2</u><u>)</u>

the standard form of ellipse equation is given by:

$\displaystyle \frac{(x - h {)}^{2} }{ {a}^{2} } + \frac{(y - k {)}^{2} }{ {b}^{2} } = 1$

where:

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b is the vertical radius

since the centre of the equation is not mentioned, we'd assume it (0,0) therefore our equation will be:

$\displaystyle \frac{ {x}^{2} }{ {a}^{2} } + \frac{{y}^{2} }{ {b}^{2} } = 1$

substituting the value of x and y from the point (1,4),we'd acquire:

$\displaystyle \frac{ 1}{ {a}^{2} } + \frac{16}{ {b}^{2} } = 1$

similarly using the point (-3,2), we'd obtain:

$\displaystyle \frac{ 9}{ {a}^{2} } + \frac{4 }{ {b}^{2} } = 1$

let 1/a² and 1/b² be q and p respectively and transform the equation:

$\displaystyle \begin{cases} q + 16p = 1 \\ 9q + 4p = 1 \end{cases}$

solving the system of linear equation will yield:

$\displaystyle \begin{cases} q = \dfrac{3}{35} \\ \\ p = \dfrac{2}{35} \end{cases}$

substitute back:

$\displaystyle \begin{cases} \dfrac{1}{ {a}^{2} } = \dfrac{3}{35} \\ \\ \dfrac{1}{ {b}^{2} } = \dfrac{2}{35} \end{cases}$

divide both equation by 1 which yields:

$\displaystyle \begin{cases} {a}^{2} = \dfrac{35}{ 3} \\ \\ {b}^{2} = \dfrac{35}{2} \end{cases}$

substitute the value of a² and b² in the ellipse equation , thus:

$\displaystyle \frac{ {x}^{2} }{ \dfrac{35}{3} } + \frac{{y}^{2} }{ \dfrac{35}{2} } = 1$

simplify complex fraction:

$\displaystyle \frac{ {3x}^{2} }{ 35 } + \frac{{2y}^{2} }{ 35 } = 1$

and we're done!

(refer the attachment as well)

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