Solve sqrt{144}*2^2+3^4div15^0-3^1*sqrt{49}

$$\sqrt{ 144 } \times { 2 }^{ 2 } + { 3 }^{ 4 } \div { 15 }^{ 0 } - { 3 }^{ 1 } \times \sqrt{ 49 }$$

Evaluate

$$108$$

Solution Steps

Calculate the square root of 144 and get 12.

$$12\times 2^{2}+\frac{3^{4}}{15^{0}}-3^{1}\sqrt{49}$$

Calculate 2 to the power of 2 and get 4.

$$12\times 4+\frac{3^{4}}{15^{0}}-3^{1}\sqrt{49}$$

Multiply 12 and 4 to get 48.

$$48+\frac{3^{4}}{15^{0}}-3^{1}\sqrt{49}$$

Calculate 3 to the power of 4 and get 81.

$$48+\frac{81}{15^{0}}-3^{1}\sqrt{49}$$

Calculate 15 to the power of 0 and get 1.

$$48+\frac{81}{1}-3^{1}\sqrt{49}$$

Anything divided by one gives itself.

$$48+81-3^{1}\sqrt{49}$$

Add 48 and 81 to get 129.

$$129-3^{1}\sqrt{49}$$

Calculate 3 to the power of 1 and get 3.

$$129-3\sqrt{49}$$

Calculate the square root of 49 and get 7.

$$129-3\times 7$$

Multiply 3 and 7 to get 21.

$$129-21$$

Subtract 21 from 129 to get 108.

$$108$$

Factor

$$2^{2}\times 3^{3}$$