Solve 3+5*3-4*3+3^2+6/3-2^43/8

$$3+5 \cdot 3-4 \cdot 3+ { 3 }^{ 2 } + \frac{ 6 }{ 3 } - \frac{ { 2 }^{ 4 } 3 }{ 8 }$$

Evaluate

$$11$$

Solution Steps

Multiply 5 and 3 to get 15.

$$3+15-4\times 3+3^{2}+\frac{6}{3}-\frac{2^{4}\times 3}{8}$$

Add 3 and 15 to get 18.

$$18-4\times 3+3^{2}+\frac{6}{3}-\frac{2^{4}\times 3}{8}$$

Multiply 4 and 3 to get 12.

$$18-12+3^{2}+\frac{6}{3}-\frac{2^{4}\times 3}{8}$$

Subtract 12 from 18 to get 6.

$$6+3^{2}+\frac{6}{3}-\frac{2^{4}\times 3}{8}$$

Calculate 3 to the power of 2 and get 9.

$$6+9+\frac{6}{3}-\frac{2^{4}\times 3}{8}$$

Add 6 and 9 to get 15.

$$15+\frac{6}{3}-\frac{2^{4}\times 3}{8}$$

Divide 6 by 3 to get 2.

$$15+2-\frac{2^{4}\times 3}{8}$$

Add 15 and 2 to get 17.

$$17-\frac{2^{4}\times 3}{8}$$

Calculate 2 to the power of 4 and get 16.

$$17-\frac{16\times 3}{8}$$

Multiply 16 and 3 to get 48.

$$17-\frac{48}{8}$$

Divide 48 by 8 to get 6.

$$17-6$$

Subtract 6 from 17 to get 11.

$$11$$

Factor

$$11$$