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2 years ago

5

Ethan and three friends kept track of their average points scored in each basketball game over the season. The results are shown

below. Which player is the most consistent?

2 answers:

Likurg_2 [28]2 years ago

6 0

Answer: Joaquin

He’s the most consistent player since he has the lowest Inter Quartile Range.

I remember seeing one of these questions earlier.

vodka [1.7K]2 years ago

3 0

Answer:

JOAQUIN

Step-by-step explanation:

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Which of the following is the complete list of roots for the polynomial function f (x) = (x2 + 6x + 8) (x2 + 6x + 13)?

Natali5045456 [20]

Answer:

-2, -4, -3 + 2i, -3-2i

Step-by-step explanation:

Equaling the function to zero we have:

(x ^ 2 + 6x + 8) (x ^ 2 + 6x + 13) = 0

For the first parenthesis we have:

(x ^ 2 + 6x + 8) = 0\\(x + 4) (x + 2) = 0

Therefore the roots are:

x = - 4\\x = - 2

For the second parenthesis we have:

(x ^ 2 + 6x + 13) = 0

By completing squares we have:

x ^ 2 + 6x = -13

x ^ 2 + 6x + (\frac{6}{2}) ^ 2 = -13 + (\frac{6}{2}) ^ 2\\x ^ 2 + 6x + 9 = -13 + 9\\(x + 3) ^ 2 = - 4\\x + 3 = +/- \sqrt{-4}

Therefore the roots are:

x = -3 + 2i\\x = -3 - 2i

Hope this was helpful

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3 years ago

Find the solution to the system 4x - y = -12 and -x - y = 3

rjkz [21]

Answer:

Equation Form : x=-3 y =0

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2 years ago

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What are two solutions to x>29

Mamont248 [21]

Answer:

43 and 56/12

Step-by-step explanation:

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2 years ago

Simplify the expression. 10 (n^2+ n) - 6 (n^2 - 2)

Gnom [1K]

<h3>10(n²+n)-6(n²+2)</h3><h3>10n²+10n-6n²-12</h3><h3>10n²-6n²+10n-12</h3><h3>4n²+10n-12</h3>

please mark this answer as brainlist

3 0

2 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

7 0

3 years ago