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user100 [1]
3 years ago
15

Find the limit (enter 'DNE' if the limit does not exist)

Mathematics
1 answer:
vaieri [72.5K]3 years ago
5 0

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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Drag the tiles to the boxes to form correct pairs.
FrozenT [24]

Answer:

1 .4x2-9= 2x+3,2x-3

2 .16x2-1=4x-1,4x+1

3 .16x2-4=4(2x+1)(2x-1)

4 .4x2-1=(2x+1)(2x-1)

Step-by-step explanation:

16x² − 1  = (4x − 1)(4x + 1) ;  16x² − 4  = 4(2x + 1)(2x − 1); 4x² − 1  = (2x + 1)(2x − 1) ;

4x² − 9 = (2x + 3)(2x − 3)

16x² − 1  is the difference of squares.  This is because 16x² is a perfect square, as is 1.  To find the factors of the difference of squares, take the square root of each square; one factor will be the sum of these and the other will be the difference.

The square root of 16x² is 4x and the square root of 1 is 1; this gives us (4x-1)(4x+1).

16x² − 4 is also the difference of squares.  The difference of 16x² is 4x and the square root of 4 is 2; this gives us (4x-2)(4x+2).  However, we can also factor a 2 out of each of these binomials; this gives us

2(2x-1)(2)(2x+1) = 2(2)(2x-1)(2x+1) = 4(2x-1)(2x+1)

4x² − 1  is also the difference of squares.  The square root of 4x² is 2x and the square root of 1 is 1; this gives us (2x-1)(2x+1).

4x² − 9 is also the difference of squares.  The square root of 4x² is 2x and the square root of 9 is 3; this gives us (2x-3)(2x+3).

3 0
3 years ago
Evaluate the expression for x=-6 4more than the product of-4 and the number ​
son4ous [18]

Answer:

28

Step-by-step explanation:

The product of -4 and the unknown number n is -4n.

4 more than the product of -4 and the unknown number n is -4n + 4.

Letting n = -6 results in the value  -4(-6) + 4, or 24 + 4, or 28.

8 0
3 years ago
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saw5 [17]

Step-by-step explanation:

1st term =4×2(1-1)=0

2nd term=4×2(2-1)=8

3rd term=4×2(3-1)=16

4th term=4×2(4-1)=24

4 0
3 years ago
What is the solution of the equation 4p/6+27
anyanavicka [17]
The answer is 36 I'm pretty sure
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3 years ago
A parabola can be drawn given a focus of (-9, -7) and a directrix of x = 9. Write
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Check the picture below, so the parabola looks more or less like so, with a vertex at (0 , -7), let's recall the vertex is half-way between the focus point and the directrix.

so this horizontal parabola opens up to the left-hand-side, meaning that the "P" distance is a negative value.

\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

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4 0
2 years ago
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