What is the value of #6x^2+4x+8# when #x=7#?

1 Answer
Wayne
Mar 9, 2016

The value of the trinomial #6x^2+ 4x + 8# when #x=7# is #330#

Explanation:

The variable in this question is #x#, which in this case we are told has a value of #7# because we are given # x=7#. By substituting this value for #x# into the trinomial in place of #x# we get:
#6(7)^2+4(7)+8#
#6(49)+28+8#
#294+28+8 = 330#

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