1. x^2 + y^2 - 4x + 12y - 20 = 0 ---- (x-2)^2 + (y+6)^2 = 60
3. 3x^2+ 3y^2 + 12x +18y - 15 = 0 --- (x+2)^2 + (y+3)^2 = 18
5. 2x^2 + 2y^2 - 24x - 16y - 8 = 0 --- (x-6)^2 + (y-4)^2 = 56
I don't know the other two, hope I helped though.
Answer:12
Step-by-step explanation:
k
Answer: ![\sqrt[3]{125x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125x%5E3%7D)
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Step-by-step explanation: Given radical expression
![\sqrt[3]{5x} \times \sqrt[3]{25x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5x%7D%20%5Ctimes%20%5Csqrt%5B3%5D%7B25x%5E2%7D)
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According to the product property of roots.
![\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%5Ctimes%20%5Csqrt%5Bn%5D%7Bb%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%20%5Ctimes%20b%7D)
On applying above rule, we get
![\sqrt[3]{5x} \times \sqrt[3]{25x^2} = \sqrt[3]{5x \times 25x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5x%7D%20%5Ctimes%20%5Csqrt%5B3%5D%7B25x%5E2%7D%20%3D%20%5Csqrt%5B3%5D%7B5x%20%5Ctimes%2025x%5E2%7D)
5 × 25 = 125 and
Therefore,
![\sqrt[3]{5x \times 25x^2}= \sqrt[3]{125x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5x%20%5Ctimes%2025x%5E2%7D%3D%20%5Csqrt%5B3%5D%7B125x%5E3%7D)
<h3>So, the correct option would be second option
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9945/ 39= 255
If you were to try to divide 99 by thirty five it would not be a whole. You would be able to multiply 39 by 2 to get 78. 99 subtracted by 78 = 21 you would then carry down the 4. to equal 214 you would then figure out 39 multiplied by 5 is the highest number to figure a 195 you would have 19 left over. Carry down your 5 and would have 195. You would have 195 again and then multiply 39 by 5 and it would equal out to 195. Your answer would then be 255.
The intersection is at (2,-1)
