Question

find p(0) ,p(1),p(2) for p(y)=y^2-y+1​

Answer 1

Answer:

Perfect questions for guys interested in maths..

2. The value of x + x(xx) when x = 2 is:

(a) 10, (b) 16, (c) 18, (d) 36, (e) 64

Solution:

x + x(xx)

Put the value of x = 2 in the above expression we get,

2 + 2(22)

= 2 + 2(2 × 2)

= 2 + 2(4)

= 2 + 8

= 10

Answer: (a)

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Answer 2

Answer:

p(y)=y²-y+1

=>p(0)=0²-0+1 = 1

=> p(1)=1²-1+1 = 1

=> p(2)=2²-2+1 = 3

Step-by-step explanation:

p(0)=1

p(1)=1

p(2)=3