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polet [3.4K]
3 years ago
5

2m = 1 + m what do I need to do to solve

Mathematics
2 answers:
devlian [24]3 years ago
4 0

So all you need to do is subtract m from both sides.

Then you have your answer

m = 1

I hope this helps

-ayden

Fantom [35]3 years ago
4 0

2m=1+m


first, 2m=1+m

-m -m

Second, m=1

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Answer:

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Step-by-step explanation:

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Twenty-one fish are swimming in an aquarium. Eleven are removed, leaving 10 fish. Which number is the additive inverse of the nu
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4 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}

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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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