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4 years ago

-4|x+5y| for x=7 and y=-3

Mathematics

2 answers:

yuradex [85]4 years ago

7 0

The answer is negatif 32

Likurg_2 [28]4 years ago

3 0

First of all, let's plug the values. You simply have to substitute every occurrence of "x" with "7" and every occurrence of "y" with "-3", which leads to

$-4|x+5y| \to -4|7+5\cdot (-3)| = -4|7-15| = -4|-8|$

Now, the absolute value of a number $x$ , i.e. $|x|$ , returns the positive versione of a number. So, if x is positive, you have $|x| = x$ whereas if x is negative you have $|x| = -x$ .

In this case, the absolute value of -8 is 8, because -8 is negative. So, the expression becomes

$-4|-8| = -4 \cdot 8 = -32$

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3 years ago

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3 years ago

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irinina [24]

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

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

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

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

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

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

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

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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shepuryov [24]

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3 years ago

Read 2 more answers