Solve 7*4+(2^{6}:2^{4})^{0}-25^2:5^2+(7*3-5*4)*(5^3:5^2)=

$$7 \cdot 4 + ( 2 ^ { 6 } : 2 ^ { 4 } ) ^ { 0 } - 25 ^ { 2 } : 5 ^ { 2 } + ( 7 \cdot 3 - 5 \cdot 4 ) \cdot ( 5 ^ { 3 } : 5 ^ { 2 } ) =$$

Evaluate

$$9$$

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 4 from 6 to get 2.

$$7\times 4+\left(2^{2}\right)^{0}-\frac{25^{2}}{5^{2}}+\left(7\times 3-5\times 4\right)\times \frac{5^{3}}{5^{2}}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 0 to get 0.

$$7\times 4+2^{0}-\frac{25^{2}}{5^{2}}+\left(7\times 3-5\times 4\right)\times \frac{5^{3}}{5^{2}}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 2 from 3 to get 1.

$$7\times 4+2^{0}-\frac{25^{2}}{5^{2}}+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Multiply 7 and 4 to get 28.

$$28+2^{0}-\frac{25^{2}}{5^{2}}+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Calculate 2 to the power of 0 and get 1.

$$28+1-\frac{25^{2}}{5^{2}}+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Add 28 and 1 to get 29.

$$29-\frac{25^{2}}{5^{2}}+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Calculate 25 to the power of 2 and get 625.

$$29-\frac{625}{5^{2}}+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Calculate 5 to the power of 2 and get 25.

$$29-\frac{625}{25}+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Divide 625 by 25 to get 25.

$$29-25+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Subtract 25 from 29 to get 4.

$$4+\left(7\times 3-5\times 4\right)\times 5^{1}$$

Multiply 7 and 3 to get 21.

$$4+\left(21-5\times 4\right)\times 5^{1}$$

Multiply 5 and 4 to get 20.

$$4+\left(21-20\right)\times 5^{1}$$

Subtract 20 from 21 to get 1.

$$4+1\times 5^{1}$$

Calculate 5 to the power of 1 and get 5.

$$4+1\times 5$$

Multiply 1 and 5 to get 5.

$$4+5$$

Add 4 and 5 to get 9.

$$9$$

Factor

$$3^{2}$$