Solve (15*154+10*143+16*079+4*078)div45

$$(15 \cdot 154+10 \cdot 143+16 \cdot 079+4 \cdot 078) \div 45$$

Evaluate

$$\frac{748}{9}\approx 83.111111111$$

Solution Steps

Multiply 15 and 154 to get 2310. Multiply 10 and 143 to get 1430. Multiply 16 and 0 to get 0. Multiply 4 and 0 to get 0.

$$\frac{2310+1430+0\times 79+0\times 78}{45}$$

Add 2310 and 1430 to get 3740.

$$\frac{3740+0\times 79+0\times 78}{45}$$

Multiply 0 and 79 to get 0.

$$\frac{3740+0+0\times 78}{45}$$

Add 3740 and 0 to get 3740.

$$\frac{3740+0\times 78}{45}$$

Multiply 0 and 78 to get 0.

$$\frac{3740+0}{45}$$

Add 3740 and 0 to get 3740.

$$\frac{3740}{45}$$

Reduce the fraction \frac{3740}{45} to lowest terms by extracting and canceling out 5.

$$\frac{748}{9}$$

Factor

$$\frac{2 ^ {2} \cdot 11 \cdot 17}{3 ^ {2}} = 83\frac{1}{9} = 83.11111111111111$$