Find the zeroes of
p (y)= 5y^2 + 7y
(koi faltu ke bakvas ans type naa karta karta ans report thase )
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Answered by
1
Answer:
there is not zero
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31
5y2 + 7y + 0
y( 5y + 7)
the value of y(5y+ 7) is zero when y = 0 and y = -7/5
therefore the zeroes are 0 and -7/5
hope it's help ✌✌
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