Solve 1=sqrt{left(3-3right)^2+left(7--1right)^2}

$$1 = \sqrt{ { \left(3-3 \right) }^{ 2 } + { \left(7--1 \right) }^{ 2 } }$$

Verify

Solution Steps

Subtract 3 from 3 to get 0.

$$1=\sqrt{0^{2}+\left(7-\left(-1\right)\right)^{2}}$$

Calculate 0 to the power of 2 and get 0.

$$1=\sqrt{0+\left(7-\left(-1\right)\right)^{2}}$$

The opposite of -1 is 1.

$$1=\sqrt{0+\left(7+1\right)^{2}}$$

Add 7 and 1 to get 8.

$$1=\sqrt{0+8^{2}}$$

Calculate 8 to the power of 2 and get 64.

$$1=\sqrt{0+64}$$

Add 0 and 64 to get 64.

$$1=\sqrt{64}$$

Calculate the square root of 64 and get 8.

$$1=8$$

Compare 1 and 8.

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