Step-by-step explanation:[tex]p {x}^{2} + 6xp + 9 {p} \\ ={( \sqrt{p} \times x)}^{2} + 2 \times( 3 \times \sqrt{p} ) \times \sqrt{p} + (3 \times \sqrt{p} )^{2} \\ = (\sqrt{p} x + 3 \sqrt{p} )^{2} [/tex][tex]r = \sqrt{ \frac{area}{\pi} } = \sqrt{ \frac{ {( \sqrt{p} x + 3 \sqrt{p} ) }^{2} }{\pi} } \\ = \frac{ \sqrt{p} x + 3 \sqrt{p} }{ \sqrt{\pi} } = \frac{ \sqrt{p} (x + 3)}{ \sqrt{\pi} } = (x + 3) \times \sqrt{ \frac{p}{\pi} } [/tex]well this is the final answer if p is a random variable. but if you wanted to refer to pi by p the the answer is simply x+3
Answer:
x+3
Step-by-step explanation:
Step-by-step explanation:[tex]p {x}^{2} + 6xp + 9 {p} \\ ={( \sqrt{p} \times x)}^{2} + 2 \times( 3 \times \sqrt{p} ) \times \sqrt{p} + (3 \times \sqrt{p} )^{2} \\ = (\sqrt{p} x + 3 \sqrt{p} )^{2} [/tex]
Step-by-step explanation:[tex]p {x}^{2} + 6xp + 9 {p} \\ ={( \sqrt{p} \times x)}^{2} + 2 \times( 3 \times \sqrt{p} ) \times \sqrt{p} + (3 \times \sqrt{p} )^{2} \\ = (\sqrt{p} x + 3 \sqrt{p} )^{2} [/tex][tex]r = \sqrt{ \frac{area}{\pi} } = \sqrt{ \frac{ {( \sqrt{p} x + 3 \sqrt{p} ) }^{2} }{\pi} } \\ = \frac{ \sqrt{p} x + 3 \sqrt{p} }{ \sqrt{\pi} } = \frac{ \sqrt{p} (x + 3)}{ \sqrt{\pi} } = (x + 3) \times \sqrt{ \frac{p}{\pi} } [/tex]
Step-by-step explanation:[tex]p {x}^{2} + 6xp + 9 {p} \\ ={( \sqrt{p} \times x)}^{2} + 2 \times( 3 \times \sqrt{p} ) \times \sqrt{p} + (3 \times \sqrt{p} )^{2} \\ = (\sqrt{p} x + 3 \sqrt{p} )^{2} [/tex][tex]r = \sqrt{ \frac{area}{\pi} } = \sqrt{ \frac{ {( \sqrt{p} x + 3 \sqrt{p} ) }^{2} }{\pi} } \\ = \frac{ \sqrt{p} x + 3 \sqrt{p} }{ \sqrt{\pi} } = \frac{ \sqrt{p} (x + 3)}{ \sqrt{\pi} } = (x + 3) \times \sqrt{ \frac{p}{\pi} } [/tex]well this is the final answer if p is a random variable. but if you wanted to refer to pi by p the the answer is simply x+3
Hope thus helps you. PLEASE MARK ME AS THE BRAINLIEST.