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steposvetlana [31]
4 years ago
5

A set of equations is given below:

Mathematics
2 answers:
Gemiola [76]4 years ago
5 0
2d+1=3d+5.
If both equations equal c, then both equations can be equal to each other.
katovenus [111]4 years ago
4 0

Answer:

The correct option is 1.

Step-by-step explanation:

The given set of equations is

Equation A: c = 2d + 1

Equation B: c = 3d + 5

Using equation A and equation B, equate the value of c.

2d+1=3d+5

1-5=3d-2d

-4=d

The value of d is -4. So the value of c is

c=2d+1=2(-4)+1=-7

The equation 2d+1=3d+5 is used to find the solution to the set of equations, therefore the correct option is 1.

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Step-by-step explanation:

Starting with x^2+y^2+6x-2y+3, group like terms, first x terms and then y terms:  x^2 + 6x            + y^2 -2y             = 3.  Please note that there has to be an " = " sign in this equation, and that I have taken the liberty of replacing " +3" with " = 3 ."  

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Now re-write the perfect square x^2 + 6x + 9 by (x + 3)^2.  Then we have x^2 + 6x + 9 - 9; also y^2 - 1y + 1 - 1.  Making these replacements:

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Then the original equation now looks like (x + 3)^2 + (y - 1)^2 = 13, and this 13 is the square of the radius, r:  r^2 = 13, so that the radius is r = √13.


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