Solve sqrt{25}+4*2^1+36:(3-2)^4+5-2^2=

$$\sqrt { 25 } + 4 \cdot 2 ^ { 1 } + 36 : ( 3 - 2 ) ^ { 4 } + 5 - 2 ^ { 2 } =$$

Evaluate

$$50$$

Solution Steps

Calculate the square root of 25 and get 5.

$$5+4\times 2^{1}+\frac{36}{\left(3-2\right)^{4}}+5-2^{2}$$

Calculate 2 to the power of 1 and get 2.

$$5+4\times 2+\frac{36}{\left(3-2\right)^{4}}+5-2^{2}$$

Multiply 4 and 2 to get 8.

$$5+8+\frac{36}{\left(3-2\right)^{4}}+5-2^{2}$$

Add 5 and 8 to get 13.

$$13+\frac{36}{\left(3-2\right)^{4}}+5-2^{2}$$

Subtract 2 from 3 to get 1.

$$13+\frac{36}{1^{4}}+5-2^{2}$$

Calculate 1 to the power of 4 and get 1.

$$13+\frac{36}{1}+5-2^{2}$$

Anything divided by one gives itself.

$$13+36+5-2^{2}$$

Add 13 and 36 to get 49.

$$49+5-2^{2}$$

Add 49 and 5 to get 54.

$$54-2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$54-4$$

Subtract 4 from 54 to get 50.

$$50$$

Factor

$$2\times 5^{2}$$