Solve 1^2+4(1)+4/1^3+41^2

$$\frac{ { 1 }^{ 2 } +4(1)+4 }{ { 1 }^{ 3 } +4 { 1 }^{ 2 } }$$

Evaluate

$$\frac{9}{5}=1.8$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

$$\frac{1+4\times 1+4}{1^{3}+4\times 1^{2}}$$

Multiply 4 and 1 to get 4.

$$\frac{1+4+4}{1^{3}+4\times 1^{2}}$$

Add 1 and 4 to get 5.

$$\frac{5+4}{1^{3}+4\times 1^{2}}$$

Add 5 and 4 to get 9.

$$\frac{9}{1^{3}+4\times 1^{2}}$$

Calculate 1 to the power of 3 and get 1.

$$\frac{9}{1+4\times 1^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{9}{1+4\times 1}$$

Multiply 4 and 1 to get 4.

$$\frac{9}{1+4}$$

Add 1 and 4 to get 5.

$$\frac{9}{5}$$

Factor

$$\frac{3 ^ {2}}{5} = 1\frac{4}{5} = 1.8$$