Question

x=3,y=5,z=7(2x²+2y²+3xy+4z)​

Answer 1

Answer:

141

Step-by-step explanation:

By substituting the Values we have

2(3²)+2(5²)+3(3)(5)+4(7)

=2(9)+2(25)+9(5)+4(7)

=18+50+45+28

=141

I hope it helps you

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Answer 2

Answer:

Given,

         x = 3

         y = 5

         z = 7

To solve → 2x²+2y²+3xy+4z

$2x^2+2y^2+3xy+4z = 2(3)^2+2(5)^2+3(3)(5)+4(7)$

                                  $= 2(9)+2(25)+45+28$

                                  $=18+50+45+28$

                                  $=141$

               Hope it helps you ^_^