A right triangle has one leg with length 7 and
2 answers:
C because of the length of the number
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This is a hard question but I think that u can answer this 83;2233
Is cbad similar to yxwz ?
<h2>✒️MATHEMATICS</h2>
![\Large \bold{SOLUTION\ 1:}](https://tex.z-dn.net/?f=%20%5CLarge%20%5Cbold%7BSOLUTION%5C%201%3A%7D%20)
![\small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array}](https://tex.z-dn.net/?f=%20%5Csmall%20%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctext%7BFirst%2C%20we%20need%20to%20check%20if%20the%20given%20differential%7D%20%5C%5C%20%5Ctext%7Bequation%20is%20exact.%7D%20%5C%5C%20%5C%5C%20%281-xy%29%5E%7B-2%7D%20dx%20%20%2B%20%5Cbig%5By%5E2%20%2B%20x%5E2%20%281-xy%29%5E%7B-2%7D%5Cbig%5Ddy%20%3D%200%20%5C%5C%20%5C%5C%20%5Cdfrac%7Bdx%7D%7B%281-xy%29%5E2%7D%20%2B%20%5Cleft%5By%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%5Cright%5Ddy%20%3D%200%20%5C%5C%20%5C%5C%20%5Cquad%20M%28x%2C%20y%29%20dx%20%2B%20N%28x%2C%20y%29%20dy%20%3D%200%20%5Cend%7Barray%7D%20)
![\small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array}](https://tex.z-dn.net/?f=%20%5Csmall%20%5Cbegin%7Barray%7D%7Bl%20l%7D%5Ctt%5C%3A%20M%28x%2Cy%29%20%3D%20%5Cdfrac%7B1%7D%7B%281%20-%20xy%29%5E2%7D%2C%20%20%26%20N%28x%2Cy%29%20%3D%20y%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%5C%5C%20%5C%5C%5Ctt%20%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%20%3D%20%5Cdfrac%7B-2%28-x%29%7D%7B%281%20-%20xy%29%5E3%7D%2C%20%26%20%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D%20%3D%20%5Cdfrac%7B2x%7D%7B%281%20-%20xy%29%5E2%7D%20%2B%20%5Cdfrac%7B-2%28-y%29x%5E2%7D%7B%281%20-%20xy%29%5E3%7D%20%5C%5C%20%5C%5C%5Ctt%20%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%20%3D%20%5Cdfrac%7B2x%7D%7B%281%20-%20xy%29%5E3%7D%2C%20%26%20%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D%20%3D%20%5Cdfrac%7B2x%281%20-%20xy%29%2B2x%5E2y%7D%7B%281%20-%20xy%29%5E3%7D%20%5C%5C%20%5C%5C%5Ctt%20%5C%3A%20%26%20%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D%20%3D%20%5Cdfrac%7B2x%7D%7B%281%20-%20xy%29%5E3%7D%20%5Cend%7Barray%7D%20)
![\small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array}](https://tex.z-dn.net/?f=%20%5Csmall%20%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctt%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%20%3D%20%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D%20%5Cimplies%20%5Ctext%7BDifferential%20equation%20is%20exact.%7D%20%5C%5C%20%5C%5C%5Ctt%20%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%20%3D%20M%28x%2C%20y%29%20%3D%20%5Cdfrac%7B1%7D%7B%281%20-%20xy%29%5E2%7D%20%5C%5C%20%5Ctt%5Cdisplaystyle%20F%28x%2C%20y%29%20%3D%20%5Cint%20%5Cdfrac%7B1%7D%7B%281%20-%20xy%29%5E2%7D%20%5Cpartial%20x%20%3D%20-%5Cdfrac%7B1%7D%7By%7D%20%5Cint%20%5Cdfrac%7B1%7D%7B%281%20-%20xy%29%5E2%7D%28-y%29%5Cpartial%20x%20%5C%5C%20%5C%5C%20%5Ctt%5C%3AF%28x%2C%20y%29%20%3D%20%5Cdfrac%7B1%7D%7By%281%20-%20xy%29%7D%20%2B%20h%28y%29%20%5C%5C%20%5C%5C%20%5Ctt%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%20%3D%20N%28x%2C%20y%29%20%3D%20y%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%20%5C%5C%20%5C%5C%5Ctt%20%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Cleft%5B%5Cdfrac%7B1%7D%7By%281%20-%20xy%29%7D%20%2B%20h%28y%29%5Cright%5D%20%3D%20y%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%20%5C%5C%20%5C%5C%20%5Ctt-%5Cdfrac%7B1%20-%20xy%20%2B%20y%28-x%29%7D%7By%5E2%281%20-%20xy%29%5E2%7D%20%2B%20h%27%28y%29%20%3D%20y%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%20%5C%5C%20%5C%5C%20%5Ctt-%5Cdfrac%7B1%20-%202xy%7D%7By%5E2%281%20-%20xy%29%5E2%7D%20%2B%20h%27%28y%29%20%3D%20y%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%20%5C%5C%20%5C%5C%20h%27%28y%29%20%3D%20y%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%20%2B%20%5Cdfrac%7B1%20-%202xy%7D%7By%5E2%281%20-%20xy%29%5E2%7D%20%5C%5C%20%5C%5C%20%5Ctt%5C%3Ah%27%28y%29%20%3D%20y%5E2%20%2B%20%5Cdfrac%7Bx%5E2y%5E2%20-%202xy%20%2B%201%7D%7By%5E2%281-xy%29%5E2%7D%20%3D%20y%5E2%20%2B%20%5Cdfrac%7B1%7D%7By%5E2%7D%20%5C%5C%20%5C%5C%20h%28y%29%20%3D%20%5Cdfrac%7By%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7By%7D%20%2B%20C%20%5C%5C%20%5C%5C%20%5Ctt%5Ctext%7BSubstituting%20to%20%7DF%28x%2Cy%29%2C%5Ctext%7Bwe%20get%7D%20%5C%5C%20%5C%5C%20%5Cdfrac%7B1%7D%7By%281%20-%20xy%29%7D%20%2B%20%5Cdfrac%7By%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7By%7D%20%3D%20C%20%5C%5C%20%5C%5C%20%5Cquad%20%5Cquad%20%5Ctext%7Bor%7D%20%5C%5C%20%5C%5C%20%5Ctt%5Cred%7B%5Cboxed%7B%5Cdfrac%7Bx%7D%7B1%20-%20xy%7D%20%2B%20%5Cdfrac%7By%5E3%7D%7B3%7D%20%3D%20C%7D%20%5CLongleftarrow%20%5Ctextit%7BAnswer%7D%7D%20%5Cend%7Barray%7D%20)
![\Large \bold{SOLUTION\ 2:}](https://tex.z-dn.net/?f=%20%5CLarge%20%5Cbold%7BSOLUTION%5C%202%3A%7D%20)
![\small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array}](https://tex.z-dn.net/?f=%20%5Csmall%20%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctt%5Ctext%7BSince%20we%20already%20know%20that%20the%20equation%20is%20exact%2C%7D%20%5C%5C%20%5Ctext%7Bwe%20can%20then%20continue%20solving%20for%20the%20solution%20by%7D%20%5C%5C%20%5Ctext%7Binspection%20method%20or%20by%20algebraic%20manipulation.%7D%20%5C%5C%20%5C%5C%20%5Ctt%281-xy%29%5E%7B-2%7D%20dx%20%20%2B%20%5Cbig%5By%5E2%20%2B%20x%5E2%20%281-xy%29%5E%7B-2%7D%5Cbig%5Ddy%20%3D%200%20%5C%5C%20%5C%5C%20%5Ctt%5Cdfrac%7Bdx%7D%7B%281-xy%29%5E2%7D%20%2B%20%5Cleft%5By%5E2%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%5Cright%5Ddy%20%3D%200%20%5C%5C%20%5C%5C%20%5Ctt%5Cdfrac%7Bdx%7D%7B%281-xy%29%5E2%7D%20%2B%20y%5E2%20dy%20%2B%20%5Cdfrac%7Bx%5E2%7D%7B%281-xy%29%5E2%7D%20dy%20%3D%200%20%5C%5C%20%5C%5C%20%5Ctt%5Cdfrac%7Bdx%20%2B%20x%5E2dy%7D%7B%281-xy%29%5E2%7D%20%2B%20y%5E2%20dy%20%20%3D%200%20%5C%5C%20%5C%5C%20%5Ctt%5Ctext%7BDivide%20both%20numerator%20and%20denominator%20of%20the%7D%20%5C%5C%20%5Ctt%5Ctext%7Bfraction%20by%20%7Dx%5E2.%20%5Cend%7Barray%7D%20)
![\small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array}](https://tex.z-dn.net/?f=%20%5Csmall%20%5Cbegin%7Barray%7D%7Bc%7D%5Ctt%20%5Cdfrac%7B%5Cdfrac%7B1%7D%7Bx%5E2%7Ddx%20%2B%20dy%7D%7B%5Cdfrac%7B%281-xy%29%5E2%7D%7Bx%5E2%7D%7D%20%2B%20y%5E2%20dy%20%3D%200%20%5C%5C%20%5Ctt%5C%5C%20%5Ctt%5Cdfrac%7B%5Cdfrac%7B1%7D%7Bx%5E2%7Ddx%20%2B%20dy%7D%7B%5Cleft%28%5Cdfrac%7B1%7D%7Bx%7D-y%5Cright%29%5E2%7D%20%2B%20y%5E2%20dy%20%3D%200%20%5C%5C%20%5C%5C%20%5Ctt-%5Cdfrac%7B%5Cleft%28-%5Cdfrac%7B1%7D%7Bx%5E2%7Ddx%20-%20dy%5Cright%29%7D%7B%5Cleft%28%5Cdfrac%7B1%7D%7Bx%7D-y%5Cright%29%5E2%7D%20%2B%20y%5E2%20dy%20%3D%200%20%5C%5C%20%5C%5C%20%5Ctt%5Cdisplaystyle%20%7B%5Clarge%7B%5Cint%7D%7D%20-%5Cfrac%7Bd%5Cleft%28%5Cdfrac%7B1%7D%7Bx%7D-y%5Cright%29%7D%7B%5Cleft%28%5Cdfrac%7B1%7D%7Bx%7D-y%5Cright%29%5E2%7D%20%2B%20%5Cint%20y%5E2%20dy%20%3D%20%5Cint%200%20%5C%5C%20%5C%5C%20%5Ctt%5Cimplies%5Ctt%20%5Cdfrac%7B1%7D%7B%5Cdfrac%7B1%7D%7Bx%7D-y%7D%20%2B%20%5Cdfrac%7By%5E3%7D%7B3%7D%20%3D%20C%20%5C%5C%20%5C%5C%5Ctext%7Bor%7D%20%5C%5C%20%5C%5C%20%20%5Ctt%5Cred%7B%5Cboxed%7B%5Cdfrac%7Bx%7D%7B1%20-%20xy%7D%20%2B%20%5Cdfrac%7By%5E3%7D%7B3%7D%20%3D%20C%7D%20%5CLongleftarrow%20%5Ctextit%7BAnswer%7D%7D%20%5Cend%7Barray%7D%20)
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40.00 for premium - 15.00 more than standard
40-15=35.00 for standard