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tensa zangetsu [6.8K]
2 years ago
15

What is the solution to the system of equations?

Mathematics
1 answer:
elixir [45]2 years ago
6 0

Answer:

No solution

Step-by-step explanation:

-3x-4y-3z=-7\\2x-6y+2z=3\\5x-2y+5z=9\\\\

Lets consider the equation 2.

Here we multiply the LHS and RHS with (-1).

-3x-4y-3z=-7\\-(2x-6y+2z)=-3\\5x-2y+5z=9\\\\

Hence,

-3x-4y-3z=-7\\-2x+6y-2z=-3\\5x-2y+5z=9\\\\

When we add the equations we get,

0x+0y+0z=-10+9\\Hence,\\0=-1

As 0 doesn't equal to -1, answer is d) No solution

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valina [46]
<h2><u>Given:</u><u>-</u></h2>

  • Points C = (-7,2) → \sf{(X_1,Y_1)}
  • D = (3,12) → \sf{(X_2,Y_2)}

<h2><u>To </u><u>Find</u><u>:</u><u>-</u></h2>

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<h2><u>Required</u><u> </u><u>Response</u><u>:</u><u>-</u></h2>

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WKT,

\boxed{\sf{(x,y) = \frac{X_1+X_2}{2},\frac{Y_1+Y_2}{2}}}

→\;{\sf{\frac{-7+3}{2},\frac{2+12}{2}}}

→\;{\sf{\frac{-4}{2},\frac{14}{2}}}

→\;{\sf{-2,7}}

The Midpoint of CD ◕➜ \Large{\red{\mathfrak{(-2,7)}}}

Let,

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Radius = CO & OD

Here, C = (-7,2) → \sf{(X_1,Y_1)}

O = (-2,7) → \sf{(X_2,Y_2)}

\boxed{\sf{Distance = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}}}

→\;{\sf{\sqrt{(-2+7)²+(7-2)²}}}

→\;{\sf{\sqrt{5²+5²}}}

→\;{\sf{\sqrt{25+25}}}

→\;{\sf{\sqrt{50}}}

→\;{\sf{5√2 (or) 7.07}}

Radius of Circle ◕➜ \Large{\red{\mathfrak{7.07}}}

<h2>Option D.</h2>

Hope It Helps You ✌️

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