Consider a circle whose equation is x^2+y^2-2x-8=0. Which statements are true? Select three options.

Mathematics

1 answer:

lidiya [134]2 years ago

5 0

Answer:

The radius of the circle is 3 units

The center of the circle lies on the x-axis.

The radius of this circle is the same as the radius of the circle whose equation is x^2 + y^2 = 9

Step-by-step explanation:

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2 years ago

Simplify the expression.<br> -5+i/2i

Lostsunrise [7]

Answer:

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Step-by-step explanation:

$\rm Simplify \: the \: following: \\ \rm \longrightarrow - 5 + \dfrac{i}{2} i \\ \\ \rm Combine \: powers. \\ \rm \dfrac{i \times i}{2} = \dfrac{i^{1 + 1}}{2}: \\ \rm \longrightarrow - 5 + \dfrac{i^{1 + 1}}{2} \\ \\ \rm 1 + 1 = 2: \\ \rm \longrightarrow - 5 + \dfrac{i^2}{2} \\ \\ \rm i^2 = -1: \\ \rm \longrightarrow - 5 + \dfrac{( - 1)}{2} \\ \\ \rm \longrightarrow - 5 - \dfrac{1}{2} \\ \\ \rm Put \: - 5 - \dfrac{1}{2} \: over \: the \: common \: denominator \: 2. \\ \rm - 5 - \dfrac{1}{2} = \dfrac{2( - 5)}{2} - \dfrac{1}{2} : \\ \rm \longrightarrow \dfrac{ - 5 \times 2}{2} - \dfrac{1}{2} \\ \\ \rm 2 (-5) = -10: \\ \rm \longrightarrow \dfrac{ - 10}{2} - \dfrac{1}{2} \\ \\ \rm \dfrac{ - 10}{2} - \dfrac{1}{2} = \dfrac{ - 10 - 1}{2} : \\ \rm \longrightarrow \dfrac{ - 10 - 1}{2} \\ \\ \rm -10 - 1 = -11: \\ \rm \longrightarrow - \dfrac{11}{2}$