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3 years ago

How many real solutions does x2+6x+13=0 have?

Mathematics

1 answer:

Ainat [17]3 years ago

3 0

Answer:

There are NO real solutions to this equation as per the study of its discriminant.

Step-by-step explanation:

In order to answer the question, one needs to examine the "discriminant" associated with this quadratic equation.

Recall that the discriminant of a quadratic equation of the form:

$a\,x^2+b\,x+c=0$

is given by: $b^2-4\, (a)\,(c)$

in our case : $a=1\, ,\,b=6\,, and\,\,\,c=13$

Then the discriminant becomes:

$b^2-4\, (a)\,(c)=6^2-4\,(1)\,(13) =36-52=-16$

The discriminant is then a negative number, which means that there are NO real solutions (its solutions must involve imaginary numbers)

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HELP ME NOWWWWWWWWWWWWWWWWWWWWWW

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Answer:105.00

A = $105.00

I = A - P = $5.00

Equation:

A = P(1 + rt)

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r = R/100 = 0.5%/100 = 0.005 per year.

Solving our equation:

A = 100(1 + (0.005 × 10)) = 105

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3 years ago

Let’s play Pick-A-Ball with replacement! There are 10 colored balls: 3 red, 4 white, and 3 blue. The balls have been placed into

Anettt [7]

Answer:

0.48

Step-by-step explanation:

There are 10 colored balls: 3 red, 4 white, and 3 blue.

You selected 2 balls at random. They may be

RR, WW, BB, RW, RB, WB, WR, BR, BW.

To find the probability of selecting the color of ball that you just selected, find this probability in each of previous cases:

RR: (One red ball left and 8 balls left in total)

$P_{RR}=\dfrac{3}{10}\cdot \dfrac{2}{9}\cdot \dfrac{1}{8}=\dfrac{1}{120}$

WW: (Two white balls left and 8 balls left in total)

$P_{WW}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{2}{8}=\dfrac{1}{30}$

BB: (One blue ball left and 8 balls left in total)

$P_{BB}=\dfrac{3}{10}\cdot \dfrac{2}{9}\cdot \dfrac{1}{8}=\dfrac{1}{120}$

RW: (Two red and three white balls left and 8 balls left in total)

$P_{RW}=\dfrac{3}{10}\cdot \dfrac{4}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

RB: (Two red and two blue balls left and 8 balls left in total)

$P_{RB}=\dfrac{3}{10}\cdot \dfrac{3}{9}\cdot \dfrac{4}{8}=\dfrac{1}{20}$

WB: (Two blue and three white balls left and 8 balls left in total)

$P_{WB}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

WR: (Two red and three white balls left and 8 balls left in total)

$P_{WR}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

BR: (Two red and two blue balls left and 8 balls left in total)

$P_{BR}=\dfrac{3}{10}\cdot \dfrac{3}{9}\cdot \dfrac{4}{8}=\dfrac{1}{20}$

BW: (Two blue and three white balls left and 8 balls left in total)

$P_{BW}=\dfrac{4}{10}\cdot \dfrac{3}{9}\cdot \dfrac{5}{8}=\dfrac{1}{12}$

In total, the probability of selecting the color of ball that you just selected is

$\dfrac{1}{120}+\dfrac{1}{30}+\dfrac{1}{120}+2\cdot\dfrac{1}{12}+2\cdot \dfrac{1}{20}+2\cdot \dfrac{1}{12}=\\ \\=\dfrac{1}{120}+\dfrac{4}{120}+\dfrac{1}{120}+\dfrac{20}{120}+\dfrac{12}{120}+\dfrac{20}{120}=\dfrac{58}{120}=\dfrac{29}{60}\approx 0.48$

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3 years ago