What two number multiply to negative 20 and add to 19

1 answer:

4 0

$x;\ y-the\ numbers\\\\ \left\{\begin{array}{ccc}xy=-20\\x+y=19&\Rightarrow&x=19-y\end{array}\right\\\\subtitute\ to\ first\ equation:\\\\(19-y)y=-20\\19y-y^2+20=0\\-y^2+19y+20=0\ \ \ |change\ signs\\y^2-19y-20=0\\y^2-20y+y-20=0\\y(y-20)+1(y-20)+0\\(y-20)(y+1)=0\iff y-20=0\ or\ y+1=0\\\\y=20\ or\ y=-1$

$x=19-y\\\\for\ y=20\Rightarrow x=19-20=-1\\\\for\ y=-1\Rightarrow x=19-(-1)=20\\\\Answer:\boxed{\boxed{20\ and\ -1}}$

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Which of the following expressions are equivalent to \left(9\dfrac{7}{8} + 2\dfrac{4}{5}\right)-\dfrac{1}{2}(9 87 +2 54 )− 21

lora16 [44]

Answer:

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Step-by-step explanation:

The given expression is

$\left(9\dfrac{7}{8} + 2\dfrac{4}{5}\right)-\dfrac{1}{2}$

We need to find the simplified form of the given expression.

It can be rewritten as

$\left(9+\dfrac{7}{8} + 2+\dfrac{4}{5}\right)-\dfrac{1}{2}$

Combine integers and fractions separately.

$(9+2)+(\dfrac{7}{8}+\dfrac{4}{5}-\dfrac{1}{2})$

Taking LCM we get

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In can be written as

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$11+1+\dfrac{7}{40}$

$12+\dfrac{7}{40}$

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Therefore, the expression $12\dfrac{7}{40}$ is equivalent to the given expression.

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Answer:

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Step-by-step explanation:

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