Solve 11*1-3*3/(1+0.2)^3*30

$$\frac { 11 \cdot 1 - 3 \cdot 3 } { ( 1 + 0.2 ) ^ { 3 } } \times 30$$

Evaluate

$$\frac{625}{18}\approx 34.722222222$$

Solution Steps

Multiply 11 and 1 to get 11.

$$\frac{11-3\times 3}{\left(1+0.2\right)^{3}}\times 30$$

Multiply 3 and 3 to get 9.

$$\frac{11-9}{\left(1+0.2\right)^{3}}\times 30$$

Subtract 9 from 11 to get 2.

$$\frac{2}{\left(1+0.2\right)^{3}}\times 30$$

Add 1 and 0.2 to get 1.2.

$$\frac{2}{1.2^{3}}\times 30$$

Calculate 1.2 to the power of 3 and get 1.728.

$$\frac{2}{1.728}\times 30$$

Expand \frac{2}{1.728} by multiplying both numerator and the denominator by 1000.

$$\frac{2000}{1728}\times 30$$

Reduce the fraction \frac{2000}{1728} to lowest terms by extracting and canceling out 16.

$$\frac{125}{108}\times 30$$

Multiply \frac{125}{108} and 30 to get \frac{625}{18}.

$$\frac{625}{18}$$

Factor

$$\frac{5 ^ {4}}{2 \cdot 3 ^ {2}} = 34\frac{13}{18} = 34.72222222222222$$