Solve (-10)^2+2*10^0-3(1/10)^-2-10^-2

$$( - 10 ) ^ { 2 } + 2 \times 10 ^ { 0 } - 3 ( \frac { 1 } { 10 } ) ^ { - 2 } - 10 ^ { - 2 }$$

Evaluate

$$-\frac{19801}{100}=-198.01$$

Solution Steps

Calculate -10 to the power of 2 and get 100.

$$100+2\times 10^{0}-3\times \left(\frac{1}{10}\right)^{-2}-10^{-2}$$

Calculate 10 to the power of 0 and get 1.

$$100+2\times 1-3\times \left(\frac{1}{10}\right)^{-2}-10^{-2}$$

Multiply 2 and 1 to get 2.

$$100+2-3\times \left(\frac{1}{10}\right)^{-2}-10^{-2}$$

Add 100 and 2 to get 102.

$$102-3\times \left(\frac{1}{10}\right)^{-2}-10^{-2}$$

Calculate \frac{1}{10} to the power of -2 and get 100.

$$102-3\times 100-10^{-2}$$

Multiply 3 and 100 to get 300.

$$102-300-10^{-2}$$

Subtract 300 from 102 to get -198.

$$-198-10^{-2}$$

Calculate 10 to the power of -2 and get \frac{1}{100}.

$$-198-\frac{1}{100}$$

Subtract \frac{1}{100} from -198 to get -\frac{19801}{100}.

$$-\frac{19801}{100}$$

Factor

$$-\frac{19801}{100} = -198\frac{1}{100} = -198.01$$