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Ksju [112]
3 years ago
14

12. Carmen multiplied a non-zero rational number by its additive inverse. Which statement about the product must be true?

Mathematics
1 answer:
nignag [31]3 years ago
6 0

Answer:

<u>Option C. The product was negative.</u>

Step-by-step explanation:

Let the non-zero rational number is = \frac{x}{y}

So, its additive inverse = -\frac{x}{y}

The result of multiplication the non-zero rational number by its additive inverse = \frac{x}{y} *-\frac{x}{y} =-\frac{x^2}{y^2}

We will check the options:

A. The product was the multiplicative inverse of the original number.

<u>Wrong </u>because the product = \frac{x}{y} *-\frac{x}{y} =-\frac{x^2}{y^2}

B. The product was the additive inverse of the original number.

<u>Wrong </u>because the product = \frac{x}{y} *-\frac{x}{y} =-\frac{x^2}{y^2}

C. The product was negative.

<u>True</u>

D. The product was positive.

<u>Wrong </u>because the product was negative.

<u>So, the answer is option C.</u>

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barxatty [35]

Answer:

\large\boxed{-\dfrac{1}{3}-\dfrac{\sqrt{14}}{3}i,\ -\dfrac{1}{3}+\dfrac{\sqrt{14}}{3}i}

Step-by-step explanation:

Use

ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

3x^2+2x+5=0\\\\a=3,\ b=2,\ c=5\\\\b^2-4ac=2^2-4(3)(5)=4-60=-56\\\\\sqrt{b^2-4ac}=\sqrt{-56}=\sqrt{(4)(-14)}=\sqrt4\cdot\sqrt{-14}=2\sqrt{-14}

Use

i=\sqrt{-1}

\sqrt{-14}=\sqrt{(-1)(14)}=\sqrt{-1}\cdot\sqrt{14}=i\sqrt{14}

Therefore:

x=\dfrac{-2\pm 2i\sqrt{14}}{2(3)}=-\dfrac{2}{6}\pm\dfrac{2i\sqrt{14}}{6}=-\dfrac{1}{3}\pm \dfrac{\sqrt{14}}{3}i

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3 years ago
What is the factored expression of 63x^2-3x-6
Murljashka [212]

Answer:

3(3x-1)(7x +2)

Step-by-step explanation:

63x^2-3x-6

Suppose a generic quadratic equation

ax^2 + bx + c

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\frac{-b + \sqrt{b^2-4ac}}{2a}

and

\frac{-b + \sqrt{b^2-4ac}}{2a}

So, for the equation 63x ^ 2-3x-6 we have:

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and

\frac{3 - \sqrt{(-3)^2-4(63)(-6)}}{2(63)} = \frac{-2}{7}

So:

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Finally the polynomial is:

(3x-1)(7x +2)

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astraxan [27]
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hope helped 
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Answer:

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