p(y)=5y2-3y+7 at y=1,-1
find value
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Answered by
28
Hey!!
y = 1, - 1
Given polynomial
P( y ) = 5y^2 - 3y + 7
Put y = 1
P ( 1 ) = 5 ( 1 ) ^2 - 3 ( 1 ) + 7
=> 5 - 3 + 7
=> 12 - 3
=> 9
Put y = - 1
P ( - 1 ) = 5 ( - 1 )^2 - 3 ( - 1 ) + 7
=> 5 + 3 + 7
=> 12 + 3
=> 15
Answers :- 9, 15
Hope it helps!
y = 1, - 1
Given polynomial
P( y ) = 5y^2 - 3y + 7
Put y = 1
P ( 1 ) = 5 ( 1 ) ^2 - 3 ( 1 ) + 7
=> 5 - 3 + 7
=> 12 - 3
=> 9
Put y = - 1
P ( - 1 ) = 5 ( - 1 )^2 - 3 ( - 1 ) + 7
=> 5 + 3 + 7
=> 12 + 3
=> 15
Answers :- 9, 15
Hope it helps!
swapnesh16:
hey @naina what's ur age
please reply
hey
Answered by
12
1. P(y) = 5y+2-3y+7
p(1)= 5(1)^2 - 3(1) + 7
p(1)= 5- 3+7
p(1)= 2+7
p(1)= 9
2.
p(-1) = 5(-1)^2 - 3(-1) + 7
p(-1) = 5 + 3 + 7
P(-1)= 8+ 7
p(-1) = 15
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