The slope of the line is given as -3 and it passes through the point (2, -6).
Thus we can write the equation of the line in point-slope form;
![\begin{gathered} \frac{y-y_1}{x-x_1}=-3 \\ \frac{y-(-6)}{x-2}=-3 \\ \frac{y+6}{x-2}=-3 \\ y+6=-3(x-2) \\ \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7By-y_1%7D%7Bx-x_1%7D%3D-3%20%5C%5C%20%5Cfrac%7By-%28-6%29%7D%7Bx-2%7D%3D-3%20%5C%5C%20%5Cfrac%7By%2B6%7D%7Bx-2%7D%3D-3%20%5C%5C%20y%2B6%3D-3%28x-2%29%20%5C%5C%20%20%5Cend%7Bgathered%7D)
Therefore, the correct option is D
I think its 18.24 because if you multiply it by 3.14 then subtract u should get 18.24
![\bf \cfrac{d}{dx}\left[ ln \left[\sqrt{\cfrac{(x+1)^5}{(x+2)^{20}}} \right]\right]\\\ -----------------------------\\\\ \sqrt{\cfrac{(x+1)^5}{(x+2)^{20}}}\implies \left[ \cfrac{(x+1)^5}{(x+2)^{20}} \right]^{\frac{1}{2}}\implies \cfrac{(x+1)^{\frac{5}{2}}}{(x+2)^{10}} \\\\\\ \textit{using the quotient rule} \\\\\\ \cfrac{\frac{5}{2}(x+1)^{\frac{3}{2}}\cdot 1\cdot (x+2)^{10}-(x+1)\frac{5}{2}\cdot 10(x+2)^9\cdot 1}{\left[ (x+2)^{10} \right]^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%20ln%20%5Cleft%5B%5Csqrt%7B%5Ccfrac%7B%28x%2B1%29%5E5%7D%7B%28x%2B2%29%5E%7B20%7D%7D%7D%20%5Cright%5D%5Cright%5D%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Csqrt%7B%5Ccfrac%7B%28x%2B1%29%5E5%7D%7B%28x%2B2%29%5E%7B20%7D%7D%7D%5Cimplies%20%5Cleft%5B%20%5Ccfrac%7B%28x%2B1%29%5E5%7D%7B%28x%2B2%29%5E%7B20%7D%7D%20%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B%28x%2B1%29%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7D%7B%28x%2B2%29%5E%7B10%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Busing%20the%20quotient%20rule%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Cfrac%7B5%7D%7B2%7D%28x%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Ccdot%201%5Ccdot%20%28x%2B2%29%5E%7B10%7D-%28x%2B1%29%5Cfrac%7B5%7D%7B2%7D%5Ccdot%2010%28x%2B2%29%5E9%5Ccdot%201%7D%7B%5Cleft%5B%20%28x%2B2%29%5E%7B10%7D%20%5Cright%5D%5E2%7D)
![\bf \cfrac{\frac{5(x+1)\sqrt{x+1}(x+2)^{10}-2(x+1)^2\sqrt{x+1}\cdot 10(x+2)^9}{2}}{(x+2)^{20}} \\\\\\ \cfrac{5(x+1)\sqrt{x+1}(x+2)^{10}-20(x+1)^2\sqrt{x+1}(x+2)^9}{2(x+2)^{20}} \\\\\\ \textit{common factor of }(x+2)^{9}\textit{ on top and bottom}\\ \\\\\\ \cfrac{5(x+1)\sqrt{x+1}(x+2)-20(x+1)^2\sqrt{x+1}}{2(x+2)^{11}} \\\\\\ \textit{common factor atop of }5(x+1)\sqrt{x+1} \\\\\\ \cfrac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Cfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%28x%2B2%29%5E%7B10%7D-2%28x%2B1%29%5E2%5Csqrt%7Bx%2B1%7D%5Ccdot%2010%28x%2B2%29%5E9%7D%7B2%7D%7D%7B%28x%2B2%29%5E%7B20%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%28x%2B2%29%5E%7B10%7D-20%28x%2B1%29%5E2%5Csqrt%7Bx%2B1%7D%28x%2B2%29%5E9%7D%7B2%28x%2B2%29%5E%7B20%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bcommon%20factor%20of%20%7D%28x%2B2%29%5E%7B9%7D%5Ctextit%7B%20on%20top%20and%20bottom%7D%5C%5C%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%28x%2B2%29-20%28x%2B1%29%5E2%5Csqrt%7Bx%2B1%7D%7D%7B2%28x%2B2%29%5E%7B11%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bcommon%20factor%20atop%20of%20%7D5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%5Cleft%5B%20%28x%2B2%29-4%28x%2B1%29%20%5Cright%5D%7D%7B2%28x%2B2%29%5E%7B11%7D%7D)
now, that's just the derivative of the funtion inside ln()
bear in mind that
![\bf \cfrac{d}{dx}ln[f(x)]\implies \cfrac{f'(x)}{f(x)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7Bd%7D%7Bdx%7Dln%5Bf%28x%29%5D%5Cimplies%20%5Ccfrac%7Bf%27%28x%29%7D%7Bf%28x%29%7D)
thus, let us give it the denominator to that
![\bf \cfrac{\frac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}}{\sqrt{\frac{(x+1)^5}{(x+2)^{20}}}} \\\\\\ \cfrac{\frac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}}{\frac{\sqrt{(x+1)^5}}{\sqrt{(x+2)^{20}}}} \\\\\\ \cfrac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]}{2(x+2)^{11}}\cdot \cfrac{\sqrt{(x+2)^{20}}}{\sqrt{(x+1)^5}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Cfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%5Cleft%5B%20%28x%2B2%29-4%28x%2B1%29%20%5Cright%5D%7D%7B2%28x%2B2%29%5E%7B11%7D%7D%7D%7B%5Csqrt%7B%5Cfrac%7B%28x%2B1%29%5E5%7D%7B%28x%2B2%29%5E%7B20%7D%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5Cfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%5Cleft%5B%20%28x%2B2%29-4%28x%2B1%29%20%5Cright%5D%7D%7B2%28x%2B2%29%5E%7B11%7D%7D%7D%7B%5Cfrac%7B%5Csqrt%7B%28x%2B1%29%5E5%7D%7D%7B%5Csqrt%7B%28x%2B2%29%5E%7B20%7D%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%5Cleft%5B%20%28x%2B2%29-4%28x%2B1%29%20%5Cright%5D%7D%7B2%28x%2B2%29%5E%7B11%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%7B%28x%2B2%29%5E%7B20%7D%7D%7D%7B%5Csqrt%7B%28x%2B1%29%5E5%7D%7D)
![\bf \cfrac{5(x+1)\sqrt{x+1}\left[ (x+2)-4(x+1) \right]\cdot (x+2)^{10}}{2(x+2)^{11}\cdot (x+1)^2\sqrt{x+1}} \\\\\\ \textit{now we take the common factors atop and bottom of}\\\\ (x+1)\sqrt{x+1}(x+2)^{10} \\\\\\ thus\implies \cfrac{5\left[ (x+2)-4(x+1) \right]}{2(x+2)(x+1)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B5%28x%2B1%29%5Csqrt%7Bx%2B1%7D%5Cleft%5B%20%28x%2B2%29-4%28x%2B1%29%20%5Cright%5D%5Ccdot%20%28x%2B2%29%5E%7B10%7D%7D%7B2%28x%2B2%29%5E%7B11%7D%5Ccdot%20%28x%2B1%29%5E2%5Csqrt%7Bx%2B1%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%20we%20take%20the%20common%20factors%20atop%20and%20bottom%20of%7D%5C%5C%5C%5C%0A%28x%2B1%29%5Csqrt%7Bx%2B1%7D%28x%2B2%29%5E%7B10%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Cimplies%20%5Ccfrac%7B5%5Cleft%5B%20%28x%2B2%29-4%28x%2B1%29%20%5Cright%5D%7D%7B2%28x%2B2%29%28x%2B1%29%7D)
yeah, we can simplify the numerator only, since the denominator is a cubic
so...
Answer:
D. 4
Step-by-step explanation:
We know that the square of (+4) is 16 and also for (—4) is 16. We normally takes interest in positive square value for a square root. Therefore, The principal square root of 16 is 4