Question

find p(1)and p(2) of the polynomial
5y2-3y+7.​

Answer 1

$p(y) \: 5{y}^{2}-3y+7 \\ p(1) = 5( {1})^{2} - 3(1) + 7 \\ p(1) = 5(1) - 3(1) +7 \\ p(1) = 5 - 3 + 7 \\ p(1 ) = 9 \\ p(2) = 5 ({2)}^{2} - 3(2) + 7 \\ p(2) = 5(4) - 3(2) + 7 \\ p(2) = 20 - 6 + 7 \\ p(2) = 21$

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Answer 2

Solution :

⏭ Given:

$\star \: \boxed{ \tt{ \red{p(y) = 5 {y}^{2} - 3y + 7}}}$

⏭ To Find:

• p(1)
• p(2)

⏭ Calculation:

• p(1)

$\implies \sf \: p(1) = 5 {(1)}^{2} - 3(1) + 7 \\ \\ \implies \sf \: p(1) = 5 - 3 + 7 \\ \\ \implies \: \boxed{ \tt{ \blue{p(1) = 9}}}$

• p(2)

$\implies \sf \: p(2) = 5 {(2)}^{2} - 3(2) + 7 \\ \\ \implies \sf \: p(2) = (5 \times 4) - 6 + 7 \\ \\ \implies \sf \: p(2) = 20 - 6 + 7 \\ \\ \implies \: \boxed{ \tt{ \orange{p(2) = 21}}}$