Solve:9⋅32+23⋅4 \sqrt{9}\cdot3^2+2^3\cdot\sqrt{4} 9⋅32+23⋅4
First, we need to evaluate the square roots: 9=3 \sqrt{9} = 3 9=3 and 4=2 \sqrt{4} = 2 4=2.
Substitute these values back into the expression:
3⋅32+23⋅2 3\cdot3^2+2^3\cdot2 3⋅32+23⋅2
Calculate each term separately:
32=9 3^2 = 9 32=9, so 3⋅9=27 3\cdot9 = 27 3⋅9=27
23=8 2^3 = 8 23=8, so 8⋅2=16 8\cdot2 = 16 8⋅2=16
Add these values together:
27+16=33 27 + 16 = 33 27+16=33
33