L4 Detormine two positive integers between which \( \sqrt{33} \) lies. 5 Simplity completely \( 1.5 .13 x(2 x-4 x y) \) \( 1.52(x-3)^{2} \) \( 1.5 .3(2 r-p)\left(3 r^{2}-4 r p+p^{2}\right) \)

To find two positive integers between which \( \sqrt{33} \) lies, let's look at the perfect squares. We know \( 5^2 = 25 \) and \( 6^2 = 36 \). Therefore, since \( 25 < 33 < 36 \), we can conclude that \( \sqrt{33} \) is between 5 and 6. So, the answer is 5 and 6! Now, let’s simplify the expressions. For \( 1.5 \cdot 0.13 x(2x - 4xy) \), you can factor out \( 2x \) from within the parentheses giving \( 1.5 \cdot 0.13 \cdot 2x(x - 2y) \). With \( 1.52(x - 3)^2 \), you just keep it as is unless asked for further action. Lastly, for \( 1.5 \cdot 0.3(2r - p)(3r^2 - 4rp + p^2) \), you can simplify the constant, \( 1.5 \cdot 0.3 = 0.45 \), and keep the product as \( 0.45(2r - p)(3r^2 - 4rp + p^2) \). Happy simplifying!