(i) √?+ 2 (ii) t2 + 5t – 1 (iii) √3×2 – 2x (iv) 8 (v) 1 – √2x (vi)

Question

(i) √?+ 2 (ii) t2 + 5t – 1 (iii) √3×2 – 2x (iv) 8 (v) 1 – √2x (vi) (vii) 0 (viii) y½ + 3y + 2​

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Allison 1 month 2021-09-17T10:14:43+00:00 1 Answer 0 views 0

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    2021-09-17T10:16:09+00:00

    Step-by-step explanation:

    Given that:

    (i) √x+ 2 (ii) t2 + 5t – 1 (iii) √3×2 – 2x (iv) 8 (v) 1 – √2x (vi) 0 (vii) y½ + 3y + 2

    To prove:

    Identify Polynomial

    Explanation: The expression is said to be polynomial if degree /power of variable is +ve integer.

    (i)  \sqrt{x}  + 2 \\  \\ Ans:No,\:it\:is \: not \: a \: polynomial \: because \\  \: power \: of \: x \: is \: rational \: ( \frac{1}{2} ) \\

    (ii) {t}^{2}  + 5t  - 1 \\  \\ Ans:Yes, \: it \: is \: a \: polynomial \:  \\

     (iii)\sqrt{3}  {x}^{2}  - 3x \\ Ans: \: Yes ,\: it \: is \: a \: polynomial

    (iv)8 \\ Ans: \: Yes, \: it \: is \: a \: polynomial \: of \: degree \: 0 \\

    (v)1 -  \sqrt{2} x \\ Ans: \: Yes, \: it \: is \: a \: polynomial

    (vi)0 \\ Ans: \: Yes, \: it \: is \: a \: polynomial \: of \: zero \: degree

    (vii) {y}^{ \frac{1}{2} }  + 3y + 2 \\ Ans: \: No, \: it \: is \: not \: a \: polynomial \\ because \: y \: has \: rational \: power

    Thus,(ii),(iii),(iv),(v) and (vi) are polynomial.

    Hope it helps you.

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