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nadya68 [22]
3 years ago
13

The value of m in the following system of equations is:

Mathematics
1 answer:
Andrews [41]3 years ago
7 0
N=m-2

m-2n=8, using n from above in this equation yields:

m-2(m-2)=8  perform indicated multiplication

m-2m+4=8 combine like terms on left side

-m+4=8 subtract 4 from both sides

-m=4  divide both sides by -1

m=-4

So the correct answer is A) m=-4
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Find the limit (enter 'DNE' if the limit does not exist)
vaieri [72.5K]

Answer:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)=2

Step-by-step explanation:

We need to first simplify the expression using rationalization(i.e. if a square root term exists in the denominator, then multiply and divide the whole expression by the denominator(but the change the sign of its middle term))

here, we need to find:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\right)

first we'll rationalize our expression:

\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\left(\dfrac{\sqrt{-2x^2-6y^2+1}+1}{\sqrt{-2x^2-6y^2+1}+1}\right)

\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{(\sqrt{-2x^2-6y^2+1}+1)^2-(1)^2}

\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-2x^2-6y^2+1-1}

\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-(2x^2+6y^2)}

\sqrt{-2x^2-6y^2+1}+1

this is our simplified expression, now we can apply our limits:

\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)

\sqrt{-2(0)^2-6(0)^2+1}+1

1+1

2

the limit does exists and it is 2.

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Answer:

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Step-by-step explanation:

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3 years ago
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aksik [14]
Slot method
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pashok25 [27]

Answer:

Step-by-step explanation:

<h2><u>Part A</u></h2>

in interval ( 0 ; 2)

<h2><u>Part B</u></h2>

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<h2><u>Part C</u></h2>

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