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Integrate-dx-ax-2-bx-c-y-ax-2-bx-c-y-dy-dx-2ax-b-d-b-2-4ac-d-d-Case-1-d-2-lt-0-I-2-d-tan-1-y-d-C-tan-1-arctan-Case-2-d-2-0-I-2-y-C-Case

Question Number 66066 by Kunal12588 last updated on 08/Aug/19 $${Integrate}\:\int\frac{{dx}}{{ax}^{\mathrm{2}} +{bx}+{c}} \\ $$$${y}\:=\:{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${y}'=\frac{{dy}}{{dx}}=\mathrm{2}{ax}+{b} \\ $$$${d}\:=\:\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}} \\ $$$${d}'\:=\:\sqrt{−{d}} \\ $$$${Case}\:\mathrm{1}.\:{d}^{\mathrm{2}} \:<\:\mathrm{0} \\…

Question-131554

Question Number 131554 by Salman_Abir last updated on 06/Feb/21 Commented by EDWIN88 last updated on 06/Feb/21 $$\mathrm{A}−\mathrm{B}=\:\begin{bmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{6}}\end{bmatrix} \\ $$$$\mathrm{A}×\mathrm{B}^{\mathrm{T}} \:=\:\begin{bmatrix}{\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:\mathrm{7}}\end{bmatrix}\begin{bmatrix}{\mathrm{4}\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{4}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\begin{bmatrix}{\mathrm{8}+\mathrm{6}+\mathrm{12}\:\:\:\:\:\:\:−\mathrm{2}+\mathrm{12}+\mathrm{4}}\\{\mathrm{4}+\mathrm{10}+\mathrm{21}\:\:\:\:\:\:−\mathrm{1}+\mathrm{20}+\mathrm{7}}\end{bmatrix} \\ $$ Terms…

solve-for-x-1-2-x-6-11-

Question Number 456 by Karting7442 last updated on 25/Jan/15 $${solve}\:{for}\:{x}:\:\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{6}=\mathrm{11} \\ $$$$ \\ $$ Answered by prakash jain last updated on 09/Jan/15 $$\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{6}=\mathrm{11} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}=\mathrm{11}−\mathrm{6}…

x-1-3-1-1-4-2-1-5-3-1-1002-1000-x-1000-

Question Number 65992 by naka3546 last updated on 07/Aug/19 $${x}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{3}\centerdot\mathrm{1}!}\:+\:\frac{\mathrm{1}}{\mathrm{4}\centerdot\mathrm{2}!}\:+\:\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{3}!}\:+\:\ldots\:+\:\frac{\mathrm{1}}{\mathrm{1002}\centerdot\mathrm{1000}!} \\ $$$${x}\centerdot\mathrm{1000}!\:\:=\:\:? \\ $$ Commented by Prithwish sen last updated on 08/Aug/19 $$\mathrm{last}\:\mathrm{term}\:\frac{\mathrm{n}}{\left(\mathrm{n}+\mathrm{1}\right)!}\:\:\mathrm{where}\:\mathrm{n}=\:\mathrm{1001} \\ $$$$\therefore\:\left[\frac{\mathrm{n}+\mathrm{1}−\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\right]\:=\:\frac{\mathrm{1}}{\mathrm{n}!}\:−\:\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}…