05 22 –27<span> 08 40 37 –03 22 –22 –23 –02 –20 48 53 18 31 –41 –38 –29 –47 30 – 13 46 –07 29 –06 –40 – 19 –20 02 22 20 –06 17 35 31 –01 –37 –34 – 18 –30 ... </span>27 28<span> 10 —ll 16 10 –50 –52 – 10 — 13 35 40 </span>27 28<span> –54 –70 –36 –39 –48 –05 03 03 – </span>12<span> –24 – 13 – 14 – 14 –46 –29 –40 39 </span>60<span> 38 48 –</span>59<span> –</span>59–43<span> –</span>27<span> –</span>44<span> –01 ...ik this didnt help but ill never know if it acually did </span>
Answer: Last Option
![4x^5\sqrt[3]{3x}](https://tex.z-dn.net/?f=4x%5E5%5Csqrt%5B3%5D%7B3x%7D)
Step-by-step explanation:
To make the product of these expressions you must use the property of multiplication of roots:
![\sqrt[n]{x^m}*\sqrt[n]{x^b} = \sqrt[n]{x^{m+b}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%2A%5Csqrt%5Bn%5D%7Bx%5Eb%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%2Bb%7D%7D)
we also know that:
![\sqrt[3]{x^3} = x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%20%3D%20x)
So
![\sqrt[3]{16x^7}*\sqrt[3]{12x^9}\\\\\sqrt[3]{16x^3x^3x}*\sqrt[3]{12(x^3)^3}\\\\x^2\sqrt[3]{16x}*x^3\sqrt[3]{12}\\\\x^5\sqrt[3]{16x*12}\\\\x^5\sqrt[3]{2^4x*2^2*3}\\\\x^5\sqrt[3]{2^6x*3}\\\\4x^5\sqrt[3]{3x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B16x%5E7%7D%2A%5Csqrt%5B3%5D%7B12x%5E9%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B16x%5E3x%5E3x%7D%2A%5Csqrt%5B3%5D%7B12%28x%5E3%29%5E3%7D%5C%5C%5C%5Cx%5E2%5Csqrt%5B3%5D%7B16x%7D%2Ax%5E3%5Csqrt%5B3%5D%7B12%7D%5C%5C%5C%5Cx%5E5%5Csqrt%5B3%5D%7B16x%2A12%7D%5C%5C%5C%5Cx%5E5%5Csqrt%5B3%5D%7B2%5E4x%2A2%5E2%2A3%7D%5C%5C%5C%5Cx%5E5%5Csqrt%5B3%5D%7B2%5E6x%2A3%7D%5C%5C%5C%5C4x%5E5%5Csqrt%5B3%5D%7B3x%7D)
Answer:
I think p= -3 and q= -2. Or the other way around. I'm not completely sure.
Answer:
bd=cc=oc2
Step-by-step explanation: