Solve left|6*2+(-8)*2-5right|/sqrt{6^2+8^2}

$$\frac{ \left| 6 \times 2+(-8) \times 2-5 \right| }{ \sqrt{ { 6 }^{ 2 } + { 8 }^{ 2 } } }$$

Evaluate

$$\frac{9}{10}=0.9$$

Solution Steps

Multiply 6 and 2 to get 12. Multiply -8 and 2 to get -16.

$$\frac{|12-16-5|}{\sqrt{6^{2}+8^{2}}}$$

Subtract 16 from 12 to get -4.

$$\frac{|-4-5|}{\sqrt{6^{2}+8^{2}}}$$

Subtract 5 from -4 to get -9.

$$\frac{|-9|}{\sqrt{6^{2}+8^{2}}}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -9 is 9.

$$\frac{9}{\sqrt{6^{2}+8^{2}}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{9}{\sqrt{36+8^{2}}}$$

Calculate 8 to the power of 2 and get 64.

$$\frac{9}{\sqrt{36+64}}$$

Add 36 and 64 to get 100.

$$\frac{9}{\sqrt{100}}$$

Calculate the square root of 100 and get 10.

$$\frac{9}{10}$$

Factor

$$\frac{3 ^ {2}}{2 \cdot 5} = 0.9$$