Question

If x=1, y=2 and z=5 , find the value of (3) 2x^2-3y^2+z^2​

Answer 1

Explanation:

2*1*1-3*2*2+5*5

2-12+25

27-12

15

Answer 2

Answer:

15

To find : 2x^2 - 3y^2 + z^2

solution :

x^2 = 1x1 = 1

y^2 = 2x2 = 4

z^2= 5x5 = 25

=> 2x^2 - 3y^2 + z^2 = 2×1 - 3×4 + 25 = 15