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Question Number 229184 Answers: 0 Comments: 0

h^2 +k^2 =r^2 (((25)/2)−h)^2 +k^2 =((5/2)+r)^2 (13−h)^2 +(6−k)^2 =r^2 solve for h,k &r

$${h}^{\mathrm{2}} +{k}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{25}}{\mathrm{2}}−{h}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} =\left(\frac{\mathrm{5}}{\mathrm{2}}+{r}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{13}−{h}\right)^{\mathrm{2}} +\left(\mathrm{6}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${solve}\:{for}\:{h},{k}\:\&{r} \\ $$

Question Number 229154 Answers: 1 Comments: 6

If you need to post hyperlinks please add a plain text message and paste link there. Code should automatically create a clickable link

$$\mathrm{If}\:\mathrm{you}\:\mathrm{need}\:\mathrm{to}\:\mathrm{post}\:\mathrm{hyperlinks}\:\mathrm{please} \\ $$$$\mathrm{add}\:\mathrm{a}\:\mathrm{plain}\:\mathrm{text}\:\mathrm{message}\:\mathrm{and}\:\mathrm{paste} \\ $$$$\mathrm{link}\:\mathrm{there}.\:\mathrm{Code}\:\mathrm{should}\: \\ $$$$\mathrm{automatically}\:\mathrm{create}\:\mathrm{a}\:\mathrm{clickable}\:\mathrm{link} \\ $$

Question Number 229144 Answers: 0 Comments: 11

Question Number 229137 Answers: 1 Comments: 1

Question Number 229128 Answers: 4 Comments: 1

Question Number 229105 Answers: 2 Comments: 0

∫_0 ^1 ((ln (x+1))/(x^2 +1))dx=?

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=? \\ $$

Question Number 229085 Answers: 2 Comments: 0

Question Number 229067 Answers: 1 Comments: 13

Question Number 229037 Answers: 1 Comments: 4

Question Number 229010 Answers: 1 Comments: 11

Question Number 229003 Answers: 1 Comments: 7

Question Number 229012 Answers: 0 Comments: 1

Question Number 228981 Answers: 0 Comments: 4

2∫_0 ^(2tan^(−1) (R/((a−R)(√2)))) (√((1(sin α+(√2))tan α)^2 +(1+(√(1^2 −((√2)tan α(5−2)−1)^2 ))cot (((((3π)/4)−sin^(−1) ((√(1^2 −((√2)tan α(5−1)−1)^2 ))/((5−1)(√2))))/2))−(1−(√(1^2 −((√2)tan α(5−1)−1)^2 )))tan ((π/4)−sin^(−1) ((√(1^2 −((√2)tan α(5−1)−1)^2 ))/((5−1)(√2)))))^2 ))×((5(√2))/((cos α)^2 ))dα

$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2tan}^{−\mathrm{1}} \frac{{R}}{\left({a}−{R}\right)\sqrt{\mathrm{2}}}} \sqrt{\left(\mathrm{1}\left(\mathrm{sin}\:\alpha+\sqrt{\mathrm{2}}\right)\mathrm{tan}\:\alpha\right)^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha\left(\mathrm{5}−\mathrm{2}\right)−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{cot}\:\left(\frac{\frac{\mathrm{3}\pi}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha\left(\mathrm{5}−\mathrm{1}\right)−\mathrm{1}\right)^{\mathrm{2}} }}{\left(\mathrm{5}−\mathrm{1}\right)\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)−\left(\mathrm{1}−\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha\left(\mathrm{5}−\mathrm{1}\right)−\mathrm{1}\right)^{\mathrm{2}} }\right)\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha\left(\mathrm{5}−\mathrm{1}\right)−\mathrm{1}\right)^{\mathrm{2}} }}{\left(\mathrm{5}−\mathrm{1}\right)\sqrt{\mathrm{2}}}\right)\right)^{\mathrm{2}} }×\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\left(\mathrm{cos}\:\alpha\right)^{\mathrm{2}} }{d}\alpha \\ $$

Question Number 229070 Answers: 0 Comments: 12

Question Number 228975 Answers: 1 Comments: 2

Question Number 228988 Answers: 0 Comments: 3

Question Number 229082 Answers: 1 Comments: 2

Question Number 228908 Answers: 1 Comments: 0

∫_0 ^(tan^(−1) (((2×1(5(√2)−1))/( 5(√2)(5(√2)−2))))) 2(1+(√(1^2 −((5−1)(√2)tan α−1)^2 ))cot (((((3π)/4)−sin^(−1) (((√(1^2 −((5−1)(√2)tan α−1)^2 ))/( (√2)(5−1)))))/2))−(1−(√(1^2 −((5−1)(√2)tan α−1)^2 )))tan ((π/4)−sin^(−1) (((√(1^2 −((5−1)(√2)tan α−1)^2 ))/( (√2)(5−1))))))((5(√2))/(cos^2 α))dα

$$\int_{\mathrm{0}} ^{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}×\mathrm{1}\left(\mathrm{5}\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\:\mathrm{5}\sqrt{\mathrm{2}}\left(\mathrm{5}\sqrt{\mathrm{2}}−\mathrm{2}\right)}\right)} \mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\left(\mathrm{5}−\mathrm{1}\right)\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{cot}\:\left(\frac{\frac{\mathrm{3}\pi}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\left(\mathrm{5}−\mathrm{1}\right)\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha−\mathrm{1}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}−\mathrm{1}\right)}\right)}{\mathrm{2}}\right)−\left(\mathrm{1}−\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\left(\mathrm{5}−\mathrm{1}\right)\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha−\mathrm{1}\right)^{\mathrm{2}} }\right)\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\left(\mathrm{5}−\mathrm{1}\right)\sqrt{\mathrm{2}}\mathrm{tan}\:\alpha−\mathrm{1}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}−\mathrm{1}\right)}\right)\right)\right)\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{cos}\:^{\mathrm{2}} \alpha}{d}\alpha \\ $$

Question Number 228907 Answers: 1 Comments: 1