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There are four species of hyaena: the spotted hyaenaZoo Reflection - Elephantsbrown hyaena or strandwolf (this one is a bit wind-swept)Brown Hyena at dawn - Stock Image - C034/3743striped hyaenaStriped Hyena (With images) | Striped hyena, Hyena, Rare albino animalsand aardwolf (it means earth-wolf, because they burrow).Aardwolf (With images) | Animals beautiful, Weird animals, Rare animalsNone could breed with a wolf or a dog. They’re actually much more closely related to cats, although they couldn’t breed with them either. Some of the ancient sabre-toothed and scimitar-toothed cats had the same kind of dropped rump.Homotherium serum

Here is the State Quota cutoffs (last admitted student after the 2nd round of State Counselling) for BJMC, Ahmedabad:UR;AIR- 9300, GR- 160, Marks- 558SEBC;AIR- 44504, GR- 782, Marks- 458SC;AIR- 57246, GR- 992, Marks- 434ST;AIR- 179330, GR- 3743, Marks- 296Marks= NEET Score (out of 720)GR= Gujarat RankHope you got it! ;-)Edit: As the medical counselling for Gujarat state is over, you can find the final cutoffs for BJ Medical College, Ahmedabad here - BJMC, Ahmedabad Cutoffs .

As other answers have said it already, the rational root theorem doesn’t give us any solution.Let’s solve it using something else for the sake of training…Here is how we solve this equation[math]2x^3–4x^2–5x-550 = 0[/math]We make a first variable change to get rid of the [math]x^2[/math] term and obtain the depressed form of the equation which is[math]x= y +\dfrac{2}{3}[/math]Replacing this into the equation, I get this:[math]2(y +\dfrac{2}{3})^3-4(y +\dfrac{2}{3})^2-5(y +\dfrac{2}{3})-550 = 0[/math]which is[math]2(y^3 [/math][math][/math][math]+ 2y^2+\dfrac{8}{3}y+\dfrac{8}{27})-4(y^2+\dfrac{4}{3}y+\dfrac{4}{9})-5(y +\dfrac{2}{3})-550 = 0[/math]which gets simplified to[math]2y^3 - \dfrac{23}{3}y-\dfrac{14972}{27} = 0[/math]and even better like this[math]y^3 - \dfrac{23}{6}y-\dfrac{7486}{27} = 0[/math]by dividing the equation by [math]2[/math].From this depressed cubic equation, there are two equivalent methods I recently learned which can take us to the solutions.Tartaglia’s methodTo find the solution from to the depressed cubic equation, we do this variable change:[math]y = w+\dfrac{23}{18w}[/math]Replacing it into the depressed equation, we obtain this[math](w+\dfrac{23}{18w})^3 - \dfrac{23}{6}(w+\dfrac{23}{18w})-\dfrac{7486}{27} = 0[/math]Expanding this, we have[math]w^3+\dfrac{23^3}{18^3w^3}+\dfrac{3\cdot 23^2}{18^2w}+\dfrac{3\cdot 23}{18}w-\dfrac{23}{6}w-\dfrac{23^2}{6\cdot 18w}-\dfrac{7486}{27} = 0[/math]We then multiply the whole equation with [math]w^3[/math] and that yields[math]w^6+\dfrac{23^3}{18^3}+\dfrac{3\cdot 23^2}{18^2}w^2+\dfrac{3\cdot 23}{18}w^4-\dfrac{23}{6}w^4-\dfrac{23^2}{6\cdot 18}w^2-\dfrac{7486}{27}w^3 = 0[/math]and simplifying it yields[math]w^6 -\dfrac{7486}{27}w^3+\dfrac{23^3}{18^3} = 0[/math]Thank you God, we are near the solution!Look at this equation, it can be easily turned into a quadratic equation by setting [math]z = w^3[/math]. The equation thus gives[math]z^2 -\dfrac{7486}{27}z+\dfrac{23^3}{18^3} = 0[/math]which we solve like this[math]\left(z-\dfrac{3743}{27}\right)^2-\dfrac{3743^2}{27^2}+\dfrac{23^3}{18^3} = 0[/math][math]\left(z-\dfrac{3743}{27}\right)^2-\dfrac{14010049}{729}+\dfrac{23^3}{18^3} = 0[/math][math]\left(z-\dfrac{3743}{27}\right)^2-\dfrac{112080392}{5832}+\dfrac{23^3}{5832} = 0[/math][math]\left(z-\dfrac{3743}{27}\right)^2=\dfrac{112092559}{5832}[/math]So[math]z-\dfrac{3743}{27} = \pm\dfrac{\sqrt{112092559}}{54\cdot\sqrt{2}}[/math]Thus [math]z =\dfrac{7486\sqrt{2}\pm\sqrt{112092559}}{54\cdot\sqrt{2}}[/math]Ouf! Now, let’s say something a bit magic. If you set [math]z_1 =\left(\dfrac{7486\sqrt{2}+\sqrt{112092559}}{54\cdot \sqrt{2}}\right)^{\frac{1}{3}}[/math] and [math]z_2 =\left(\dfrac{7486\sqrt{2}-\sqrt{112092559}}{54\cdot\sqrt{2}}\right)^{\frac{1}{3}}[/math]. Then let [math]\alpha[/math] and [math]\beta [/math][math][/math][math][/math] be [math]\alpha = \dfrac{-1+\sqrt{-3}}{2}[/math] and [math]\beta = \dfrac{-1-\sqrt{-3}}{2}.[/math]For those who don’t understand why I wrote [math]\sqrt{-3}[/math], it's just a placeholder for any one of the complex numbers [math]t[/math] such that [math]t^2 = -3[/math].Fine. So from now be assured that[math]z_1+z_2+\dfrac{2}{3}[/math], [math]\alpha z_1+\beta z_2+\dfrac{2}{3}[/math] and [math]\beta z_1+\alpha z_2+\dfrac{2}{3}[/math] are the three solutions to your original cubic equation.I say magic because I can’t find any simple way to prove it to myself. I read it a few days ago in a book. If you can explain it elegantly why this does work, please don’t hesitate to write a comment.We’re done!del Ferro-Cardano’s methodWell, I’m tired of writing and honestly I’m not a big fan of such calculations with tons of numbers so I’ll give you the outline (like you asked anything, haha!).You have the following compressed equation[math]y^3 - \dfrac{23}{6}y-\dfrac{7486}{27} = 0[/math]Let’s do the following variable change: [math]y = u+v.[/math]After doing the algebra, you’ll find that[math]u^3 [/math][math][/math][math]+ v^3 +(u+v)\left(3uv - \dfrac{23}{6}\right) -\dfrac{7486}{27}= 0[/math]Next we set [math]3uv - \dfrac{23}{6} = 0[/math], and the above equation becomes [math]u^3 [/math][math][/math][math]+ v^3 = \dfrac{7486}{27}[/math]And we are left with these two equations:[math]u^3 [/math][math][/math][math]+ v^3 = -\dfrac{7486}{27}[/math][math]u^3v^3 = \dfrac{(\frac{23}{18})^3}{27}[/math]Solve for [math]u[/math] and [math]v[/math]. Then add the solutions.You’ve just found a solution to the cubic. The others will thus be easy to find.