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AllQuestion and Answers: Page 72

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

Question Number 209228    Answers: 1   Comments: 0

Question Number 209223    Answers: 2   Comments: 0

Question Number 209221    Answers: 0   Comments: 0

Question Number 209220    Answers: 1   Comments: 0

Question Number 209211    Answers: 0   Comments: 1

Question Number 209206    Answers: 0   Comments: 2

2 YouTube channels I think you might find useful.

$$\mathrm{2}\:\mathrm{YouTube}\:\mathrm{channels}\:\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{might} \\ $$$$\mathrm{find}\:\mathrm{useful}. \\ $$

Question Number 209193    Answers: 0   Comments: 1

A pin 6cm high is placed in front of a diverging lens of focal length 15cm, Calculate the position of the image formed

A pin 6cm high is placed in front of a diverging lens of focal length 15cm, Calculate the position of the image formed

Question Number 209187    Answers: 3   Comments: 0

:: α , β and γ are roots of the following equation . Find the value of ” F ” : Equation : x^( 3) −2x −1=0 F := α^( 5) + β^( 5) + γ^( 5)

$$ \\ $$$$\:\:\:::\:\:\:\alpha\:,\:\beta\:\:{and}\:\:\gamma\:\:{are}\:{roots}\:{of}\:{the} \\ $$$$\:\:\:\:\:{following}\:\:{equation}\:.\:{Find}\:{the} \\ $$$$\:\:\:\:\:{value}\:\:{of}\:\:\:''\:\:\mathrm{F}\:\:''\::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{E}{quation}\::\:\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{2}{x}\:\:−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{F}\::=\:\alpha^{\:\mathrm{5}} \:+\:\beta^{\:\mathrm{5}} \:+\:\gamma^{\:\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Question Number 209185    Answers: 0   Comments: 0

Question Number 209217    Answers: 2   Comments: 0

calculate : I= ∫_(0 ) ^( ∞) (( tan^( −1) (x))/((1 + x^( 2) )^( 2) )) dx = ?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{calculate}}\:: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\:\int_{\mathrm{0}\:} ^{\:\infty} \frac{\:{tan}^{\:−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}\:+\:{x}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\: \\ $$$$ \\ $$

Question Number 209167    Answers: 2   Comments: 0

please convert 2531_((5000) ) to base 5002. thanks.

$${please}\:{convert}\:\:\mathrm{2531}_{\left(\mathrm{5000}\right)\:} {to}\:\:{base}\:\mathrm{5002}.\:\:{thanks}.\:\: \\ $$

Question Number 209166    Answers: 1   Comments: 0

Question Number 209162    Answers: 2   Comments: 0

Cyclic quadrilateral ABCD is inscribed in circle. Point S is intersection of diagonals AC and BD (S is not center of the circle). If AB=BC=6 and BS=4, what is length of BD?

$$ \\ $$Cyclic quadrilateral ABCD is inscribed in circle. Point S is intersection of diagonals AC and BD (S is not center of the circle). If AB=BC=6 and BS=4, what is length of BD?

Question Number 209161    Answers: 3   Comments: 2

Question Number 209137    Answers: 2   Comments: 0

Question Number 209131    Answers: 1   Comments: 0

prove : curve { ((x(t)=((a+r.cos(t))/(a^2 +r^2 +2ar.cos(t))))),((y(t)=((r.sin(t))/(a^2 +r^2 +2ar.cos(t))))) :} 0≤t≤2π is circle , find center & radius

$${prove}\:: \\ $$$${curve}\:\begin{cases}{{x}\left({t}\right)=\frac{{a}+{r}.{cos}\left({t}\right)}{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}.{cos}\left({t}\right)}}\\{{y}\left({t}\right)=\frac{{r}.{sin}\left({t}\right)}{{a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}.{cos}\left({t}\right)}}\end{cases}\:\:\:\:\mathrm{0}\leqslant{t}\leqslant\mathrm{2}\pi \\ $$$${is}\:{circle}\:,\:{find}\:{center}\:\&\:{radius} \\ $$

Question Number 209129    Answers: 1   Comments: 0

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