How do you evaluate #7+(-4x^2)# for #x=0#?

Question

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Answer 1 · 2017-06-19T14:27:58

7

#7+(-4x^2)# for #x=0#

Solution
Since #x=0# you have to substitute #0# to anywhere you see #x#

#:.# #7+(-4(0^2))#

#7+(-4(0))#

#7+(-0)#

#7-0#

#Answer# #-># #7#

Answer 2 · 2017-06-19T14:31:54

The answer is 7.

All you have to do is simply substitute 0 in for x:
#7+(-4(0^2))#
#7+(-4(0))#
#7+0=7#