Question 911963
For what value of k  are the graphs of 12y=-3x+8 and 6y=kx-5 perpendicular?
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Find the slopes::
y = -3/12 *x + 2/3
slope = -1/4
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y = (k/6)x - (5/6)
slope = k/6
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To be perpendicular, solve 
(-1/4)(k/6) = -1
Solve for "k"::
(1/4)(k/6) = 1
k = 24
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Cheers,
Stan H.
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