What is the value of the expression #(cd - d^2) / 4# when #c = 0# and #d = -6#?

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Answer 1 · 2015-12-02T19:59:42

#-9#

You simply need to substitute the values:

#c*d \to 0*(-6)=0#

#d^2 = (-6)^2 = 36#

So, #\frac{cd-d^2}{4} = \frac{0-36}{4} = -36/4=-9#

Answer 2 · 2015-12-03T02:36:46

#-9#

Plug in the corresponding values at each instance.

#(cd-d^2)/4=>(0xx-6-(-6)^2)/4#

#=>(0-(-6)^2)/4#

#=>(0-36)/4#

#=>(-36)/4#

#=>-9#