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kobusy [5.1K]
3 years ago
9

How do you factor 6x^2+6x

Mathematics
1 answer:
Andreyy893 years ago
7 0
I think it might be 2 
 
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What is the median number of guest for each holiday
Ivan

You have not given the data for which you like to find the median.

I will however explain how you can find the median of a given set of data, and you can apply the same method to your problem.

Step-by-step explanation:

Median, just like it sounds, is simply the middle value of a given set of data.

Given a set of numbers:

a, b, c, d, e, ..., z.

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- Arrange the numbers a, b, c, d, e, ..., ..., z in an ascending or descending order.

- Count the whole numbers, if it is odd, you have a middle number, and that is the median.

If it is even, you have two middle numbers, then the median will be the addition of those two numbers divided by two.

Example: To find the median of

4, 5, 2, 1, 2, 6, 7, 3, 5.

First, we arrange in ascending:

1, 2, 2, 3, 4, 5, 5, 6, 7

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The MEDIAN is 4 .

Example 2: To find the median of

9, 4, 5, 2, 1, 2, 6, 7, 3, 5.

First, we arrange in ascending:

1, 2, 2, 3, 4, 5, 5, 6, 7, 9

Next, we locate the middle number(2), which are 4 and 5.

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3 0
3 years ago
Someone please be awesome and help me please :(
solong [7]

Answer:

(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0

(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

Step-by-step explanation:

x^2+\frac{b}{a}x+\frac{c}{a}=0

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared.  Whatever you add in, you must take out.

x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0

Now we are read to write that one part (the first three terms together) as a square:

(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0

I don't see this but what happens if we find a common denominator for those 2 terms after the square.  (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0

They put it in ( )

(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0

I'm going to go ahead and combine those fractions now:

(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

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

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

Now subtract b/(2a) on both sides:

x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}

Combine the fractions (they have the same denominator):

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

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