Solve 3^{0}+(-3)^2*3^-2+(-2)^2+(-1)^2=

$$3 ^ { 0 } + ( - 3 ) ^ { 2 } \cdot 3 ^ { - 2 } + ( - 2 ) ^ { 2 } + ( - 1 ) ^ { 2 } =$$

Evaluate

$$7$$

Solution Steps

Calculate 3 to the power of 0 and get 1.

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

Calculate -3 to the power of 2 and get 9.

$$1+9\times 3^{-2}+\left(-2\right)^{2}+\left(-1\right)^{2}$$

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$1+9\times \frac{1}{9}+\left(-2\right)^{2}+\left(-1\right)^{2}$$

Multiply 9 and \frac{1}{9} to get 1.

$$1+1+\left(-2\right)^{2}+\left(-1\right)^{2}$$

Add 1 and 1 to get 2.

$$2+\left(-2\right)^{2}+\left(-1\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$2+4+\left(-1\right)^{2}$$

Add 2 and 4 to get 6.

$$6+\left(-1\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$6+1$$

Add 6 and 1 to get 7.

$$7$$

Factor

$$7$$