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Question Number 209314 Answers: 0 Comments: 11

Question Number 209309 Answers: 0 Comments: 0

m , n ∈ N m ≥ 2 and n ≥ 2 p > 0 and q > 0 p + q = 1 Prove that: (1−q^n )^m + (1−p^m )^n ≥ 1

$$\mathrm{m}\:,\:\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{m}\:\geqslant\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{n}\:\geqslant\:\mathrm{2} \\ $$$$\mathrm{p}\:>\:\mathrm{0}\:\:\:\mathrm{and}\:\:\:\mathrm{q}\:>\:\mathrm{0} \\ $$$$\mathrm{p}\:+\:\mathrm{q}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\left(\mathrm{1}−\mathrm{q}^{\boldsymbol{\mathrm{n}}} \right)^{\boldsymbol{\mathrm{m}}} \:+\:\left(\mathrm{1}−\mathrm{p}^{\boldsymbol{\mathrm{m}}} \right)^{\boldsymbol{\mathrm{n}}} \:\geqslant\:\mathrm{1} \\ $$

Question Number 209308 Answers: 1 Comments: 0

Donner l′e^ quivalence simple de I_n =∫^( 1) _( 0) (t^n /(t^n −t+1))dt

$$\mathrm{Donner}\:\mathrm{l}'\acute {\mathrm{e}quivalence}\:\mathrm{simple} \\ $$$$\mathrm{de}\:\mathrm{I}_{\mathrm{n}} =\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{{t}^{{n}} }{{t}^{{n}} −{t}+\mathrm{1}}{dt} \\ $$

Question Number 209307 Answers: 2 Comments: 3

Question Number 209304 Answers: 1 Comments: 0

Question Number 209301 Answers: 0 Comments: 0

Question Number 209290 Answers: 0 Comments: 1

a^2 −a−^(1000) (√((1+8000a)))=1000 find a

$$\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}−^{\mathrm{1000}} \sqrt{\left(\mathrm{1}+\mathrm{8000}\boldsymbol{\mathrm{a}}\right)}=\mathrm{1000} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{a}} \\ $$

Question Number 209289 Answers: 1 Comments: 0

Question Number 209288 Answers: 1 Comments: 0

Question Number 209281 Answers: 2 Comments: 2

Find the value of r, if ^(10) C_r = ^(10) C_(2r + 1)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{r},\:\mathrm{if}\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{r}} \:\:=\:\:\overset{\mathrm{10}} {\:}\mathrm{C}_{\mathrm{2r}\:\:+\:\:\mathrm{1}} \\ $$

Question Number 209272 Answers: 2 Comments: 1

Question Number 209263 Answers: 0 Comments: 5

6 different letters were written to 6 different people and 6 different envelopes were prepared with the addresses of these people written on them. In how many different ways can you put a letter in each envelope without putting a letter written to this person in the envelope with the name of any person?

$$ \\ $$6 different letters were written to 6 different people and 6 different envelopes were prepared with the addresses of these people written on them. In how many different ways can you put a letter in each envelope without putting a letter written to this person in the envelope with the name of any person?

Question Number 210090 Answers: 0 Comments: 9

Question Number 209246 Answers: 1 Comments: 0

Find f(x)=∫^( x) _( 0) (dt/(t+e^(f(t)) ))

$$\mathrm{Find}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{x}} \frac{\mathrm{dt}}{\mathrm{t}+\mathrm{e}^{\mathrm{f}\left(\mathrm{t}\right)} } \\ $$

Question Number 209241 Answers: 2 Comments: 0

Question Number 209240 Answers: 1 Comments: 0

If x + ((49)/(x + 48)) = − 34 find (2x + 83)^3 + (1/((2x + 83)^3 ))

$${If}\:\:{x}\:\:+\:\:\frac{\mathrm{49}}{{x}\:+\:\mathrm{48}}\:\:=\:\:−\:\mathrm{34} \\ $$$${find}\:\:\left(\mathrm{2}{x}\:+\:\mathrm{83}\right)^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\left(\mathrm{2}{x}\:+\:\mathrm{83}\right)^{\mathrm{3}} } \\ $$

Question Number 209234 Answers: 2 Comments: 0

Arrange in descending order: (√5) − (√2), (√7) − (√5) , (√(13)) − (√(11)) , (√(19)) − (√(17))

$$\mathrm{Arrange}\:\mathrm{in}\:\mathrm{descending}\:\mathrm{order}: \\ $$$$\:\:\:\:\sqrt{\mathrm{5}}\:\:−\:\:\sqrt{\mathrm{2}},\:\:\:\:\:\sqrt{\mathrm{7}}\:\:−\:\:\sqrt{\mathrm{5}}\:,\:\:\:\sqrt{\mathrm{13}}\:\:−\:\:\sqrt{\mathrm{11}}\:,\:\:\:\sqrt{\mathrm{19}}\:\:−\:\:\sqrt{\mathrm{17}} \\ $$

Question Number 209232 Answers: 1 Comments: 0

u_0 = a, u_(n+1) = (√(u_n v_n )) v_0 = b ∈ ]0,1[ , v_(n+1) = (1/(2(u_n +v_n ))) • show that a≤u_n ≤u_(n+1) ≤v_n ≤v_(n+1) ≤b • show that v_n − u_n ≤ ((a+b)/2^n )

$${u}_{\mathrm{0}} \:=\:{a},\:{u}_{{n}+\mathrm{1}} \:=\:\sqrt{{u}_{{n}} {v}_{{n}} } \\ $$$$\left.{v}_{\mathrm{0}} \:=\:{b}\:\in\:\right]\mathrm{0},\mathrm{1}\left[\:,\:{v}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\left({u}_{{n}} +{v}_{{n}} \right)}\right. \\ $$$$\bullet\:{show}\:{that}\:{a}\leqslant{u}_{{n}} \leqslant{u}_{{n}+\mathrm{1}} \leqslant{v}_{{n}} \leqslant{v}_{{n}+\mathrm{1}} \leqslant{b} \\ $$$$\bullet\:{show}\:{that}\:{v}_{{n}} \:−\:{u}_{{n}} \:\leqslant\:\frac{{a}+{b}}{\mathrm{2}^{{n}} } \\ $$

Question Number 209229 Answers: 6 Comments: 2