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Category: Geometry

2sinx-3cosx-3sinx-4cosx-dx-

Question Number 23796 by tapan das last updated on 06/Nov/17 $$\int\frac{\mathrm{2sinx}+\mathrm{3cosx}}{\mathrm{3sinx}+\mathrm{4cosx}}\:\mathrm{dx} \\ $$ Answered by ajfour last updated on 06/Nov/17 $$\mathrm{2sin}\:{x}+\mathrm{3cos}\:{x}={A}\left(\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{B}\left(\mathrm{3cos}\:{x}−\mathrm{4sin}\:{x}\right) \\ $$$$\Rightarrow\:\mathrm{3}{A}−\mathrm{4}{B}=\mathrm{2}\:\:{and}…

guys-how-was-kvpy-SA-tinkutara-physicslover-etc-i-screwd-in-bio-completely-how-much-you-guys-are-expecting-and-do-you-have-any-idea-of-cutoff-

Question Number 23769 by math solver last updated on 05/Nov/17 $$\mathrm{guys}\:,\:\mathrm{how}\:\mathrm{was}\:\mathrm{kvpy}\:\left(\:\mathrm{SA}\right)?? \\ $$$$:\:\mathrm{tinkutara}\:,\:\mathrm{physicslover},\mathrm{etc}……. \\ $$$$\mathrm{i}\:\mathrm{screwd}\:\mathrm{in}\:\mathrm{bio}\:\mathrm{completely}. \\ $$$$\mathrm{how}\:\mathrm{much}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{are}\:\mathrm{expecting} \\ $$$$\mathrm{and}\:\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{any}\:\mathrm{idea}\:\mathrm{of}\: \\ $$$$\mathrm{cutoff}\:? \\ $$ Commented by…

1-2-x-3-1-

Question Number 23752 by pombekali last updated on 05/Nov/17 $$\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{1}=? \\ $$ Answered by Joel577 last updated on 05/Nov/17 $$\left[\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} \:+\:{x}\right]_{\mathrm{1}} ^{\mathrm{2}}…

Let-ABC-be-a-triangle-with-AB-AC-and-BAC-30-Let-A-be-the-reflection-of-A-in-the-line-BC-B-be-the-reflection-of-B-in-the-line-CA-C-be-the-reflection-of-C-in-the-line-AB-Show-that-A-B-C

Question Number 23592 by Tinkutara last updated on 02/Nov/17 $$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:{AB}\:=\:{AC} \\ $$$$\mathrm{and}\:\angle{BAC}\:=\:\mathrm{30}°.\:\mathrm{Let}\:{A}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{A}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{BC};\:{B}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{B}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{CA};\:{C}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{C}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{AB}.\:\mathrm{Show}\:\mathrm{that}\:{A}',\:{B}',\:{C}' \\ $$$$\mathrm{form}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$ Answered…

Question-89097

Question Number 89097 by TawaTawa1 last updated on 15/Apr/20 Commented by mr W last updated on 15/Apr/20 $$\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{10}+\mathrm{5}\right)^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{10}} \\ $$$${s}=\mathrm{5}\sqrt{\mathrm{10}}−\frac{\mathrm{5}}{\mathrm{5}\sqrt{\mathrm{10}}}\left(\mathrm{15}+\mathrm{5}\right)=\mathrm{3}\sqrt{\mathrm{10}} \\ $$$${area}\:={s}^{\mathrm{2}} =\mathrm{9}×\mathrm{10}=\mathrm{90}…