1/29 Warm Up

​The width of a rectangle is

 $3x^2y$ feet and the length of the rectangle is $10x^6y^3$  feet. 

If I need to find the area of the rectangle, which of the following would I do first?

• Add:  $3x^2y\ +10x^6y^3$  

• Multiply:  $\left(3x^2y\ \right)\left(10x^6y^3\right)$  

• Subtract:  $10x^6y^3-3x^2y\$  

• Divide:  $\frac{10x^6y^3}{3x^2y}$  

Simplify the following: $\sqrt{68}$  

•  $2\sqrt{17}$  

•  $17\sqrt{2}$  

•  $\sqrt{34\cdot2}$  

•  $4\sqrt{17}$  

Given the expression $\frac{4x^2y^5z^{-6}}{20x^4y^3z^{-3}}$ , which of the following numbers would be divided?

•  $4\div20$  

•  $20\div4$  

•  $2\div4$  

•  $4\div2$