Solve |D|=sqrt{(-3-3)^2+(-4-4)^2}

$$| D | = \sqrt { ( - 3 - 3 ) ^ { 2 } + ( - 4 - 4 ) ^ { 2 } }$$

Solve for D

$$D=-10$$

$$D=10$$

Steps Using Definition of Absolute Value

Subtract 3 from -3 to get -6.

$$|D|=\sqrt{\left(-6\right)^{2}+\left(-4-4\right)^{2}}$$

Calculate -6 to the power of 2 and get 36.

$$|D|=\sqrt{36+\left(-4-4\right)^{2}}$$

Subtract 4 from -4 to get -8.

$$|D|=\sqrt{36+\left(-8\right)^{2}}$$

Calculate -8 to the power of 2 and get 64.

$$|D|=\sqrt{36+64}$$

Add 36 and 64 to get 100.

$$|D|=\sqrt{100}$$

Calculate the square root of 100 and get 10.

$$|D|=10$$

Use the definition of absolute value.

$$D=10 D=-10$$