(-3-2)^2 + (5--4)^2

=

(-5)^2 + (9)^2

=25 + 81

= 106

So d=Square Root of 106

=

(-5)^2 + (9)^2

=25 + 81

= 106

So d=Square Root of 106

(-3-2)^2 + (5--4)^2

=

(-5)^2 + (9)^2

=25 + 81

= 106

So d=Square Root of 106

=

(-5)^2 + (9)^2

=25 + 81

= 106

So d=Square Root of 106

The formula for distance is d^{2} = (x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}

let x1 = 2

x2 = -3

y1 = -4

y2 = 5

putting in the formula we get

d^{2}= (-3-2)^{2} + (5+4)^{2}d^{2}= 106

d= 10.29563014098700031579

let x1 = 2

x2 = -3

y1 = -4

y2 = 5

putting in the formula we get

d

d= 10.29563014098700031579

(-3-2)^2+ (5--4)^2

If EF=2x, FG=5x, and EG=8x+1 what is the value of x?

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