Question

Q23 The value of p(y) = y^2-y+1 for p(0) is *
​

Answer 1

Step-by-step explanation:

The value of p(y) = y^2-y+1 for p(0) is *ANSWER

Consider given the polynomial,

p(y)=y

2

−y+1 …….(1)

We have to find,p(0)=?

Put, y=0 in equation 1st ,

Now,

p(0)=0

2

−0+1

p(0)=1

Hence, this is the answer.

Answer 2

Answer:

1

Step-by-step explanation:

p(y) = y^2 - y + 1

p(0) = (0)^2 - 0 + 1

p(0)= 0 - 0 + 1

p(0) = 1

Hope it helps you !!!