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`x=3t+5 , y=7-2t , -1

The formula of arc length of a parametric equation on the interval `alt=tlt=b` is:


`L = int_a^b sqrt((dx/dt)^2+(dy/dt)^2) dt`


The given parametric equation is:


`x=3t + 5`


`y=7 - 2t`


The derivative of x and y are:


`dx/dt = 3`


`dy/dt = -2`


So the integral needed to compute the arc length of the given parametric equation on the interval `-1lt=tlt=3` is:


`L = int_(-1)^3 sqrt(3^2+(-2)^2) dt`


The simplified form of the integral is:


`L = int_(-1)^3 sqrt13 dt`


Evaluating this yields:


`L = sqrt13t `  `|_(-1)^3`


`L = sqrt(13)*3 - sqrt13*(-1)`


`L=3sqrt13 + sqrt13`


`L=4sqrt13`


Therefore, the arc length of the curve is `4sqrt13` units.

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