If p (X)=2x^4-x^2+X, the p (-1/2)=?
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p (x)=2x^4-x^2+x
p (-1/2)=2 (-1/2)^4-(-1/2)^2+(-1/2)
=2 (1/16)-1/4-1/2
=1/8-2/8-4/8
=-5/8
p (-1/2)=2 (-1/2)^4-(-1/2)^2+(-1/2)
=2 (1/16)-1/4-1/2
=1/8-2/8-4/8
=-5/8
shrivastavavina:
Which answer is right I am confused
Answered by
4
Heya,
p(x) = 2x^4 - x^2 + x
p (-1/2) = 2 (-1/2)^4 - (-1/2)^2 + (-1/2)
p (-1/2) = 2/16 - 1/4 -1/2
[Take LCM of 16, 4 & 2]
p (-1/2) = 2/16 - 1/4 × 4/4 - 1/2 × 8/8
p (-1/2) = 2/16 - 4/16 - 8/16
p (-1/2) = 2/16 - 12/16
p (-1/2) = -10/16
p (-1/2) = -5/8
Hope my answer helps you :)
Regards,
Shobana
p(x) = 2x^4 - x^2 + x
p (-1/2) = 2 (-1/2)^4 - (-1/2)^2 + (-1/2)
p (-1/2) = 2/16 - 1/4 -1/2
[Take LCM of 16, 4 & 2]
p (-1/2) = 2/16 - 1/4 × 4/4 - 1/2 × 8/8
p (-1/2) = 2/16 - 4/16 - 8/16
p (-1/2) = 2/16 - 12/16
p (-1/2) = -10/16
p (-1/2) = -5/8
Hope my answer helps you :)
Regards,
Shobana
Which answer is right I am confused
sorry you are wrong
Wait a min I'm checking my answer
ya please
Yeah I edited. Sorry I forgot to change the symbol
it's ok
Sorry. :-) And thanks for rectifying my mistake
no need to say sorry and thanks
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