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Sholpan [36]
3 years ago
11

What is the distance between the points (-4, 2) and (1, -3) on the coordinate plane?

Mathematics
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

A. 7.07 units

Step-by-step explanation:

Use the distance formula for coordinates:

d=\sqrt{(x_{2}-x_{1})^2 +(y_{2}-y_{1})^2   }

Fill in the coordinates accordingly:

d=\sqrt{(1-(-4))^2+(-3-2)^2}

which simplifies down to

d=\sqrt{25+25}

which is of course

d=\sqrt{50}

Plug that into your calculator and you'll get the answer

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2.
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Answer: the system has no solution.

Step-by-step explanation:

\displaystyle\\

\left \{ {{x^2y=16\ \ \ \ \ (1)} \atop {x^2+4y+16=0\ \ \ \ \ (2)}} \right. .\\

Multiply\  both\  sides\  of\  the\  equation\ (2)\   by\  y\  (y\neq 0):\\

x^2y+4y^2+16y=0\\

We\  substitute\  equation\  (1)\  into\  equation\  (2):\\

16+4y^2+16y=0\\

4y^2+16y+16=0\\

4*(y^2+4y+4)=0\\

4*(y^2+2*y*2+2^2)=0\\

4*(y+2)^2=0\\

Divide\  both\  sides\  of\  the \ equation\  by\  4:\\

(y+2)^2=0\\

(y+2)*(y+2)=0\\

So,\ y+2=0\\

y=-2.\\

\displaystyle\\We\  substitute\  the\  value\  of\  y\  into\  equation\  (1):\\x^2*(-2)=16\\Divide\  both\  sides\  of\  the \ equation\  by\  -2:\\x^2=-8\\x^2\geq 0\\ Hence,\ the\ system\ has\ no\ solution.

7 0
1 year ago
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