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suppose that : E= (√( 9 + 4 ( Σ_(k=2) ^n a_( k) ^( 2) ))) is given let a_( k) = a_(k−1) . (a_( k−1) +1) and a_1 =1 , a_( 2) = 2 , a_( 3) = 6 , a_4 = 42 , a_( 5) = 1806 and etc find the value of E =?

$$ \\ $$$$\:\:\:\:\:{suppose}\:\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:{E}=\:\sqrt{\:\mathrm{9}\:+\:\mathrm{4}\:\left(\:\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\:\:{a}_{\:{k}} ^{\:\mathrm{2}} \:\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{is}\:\:{given} \\ $$$$\:\:\:\:\:{let}\:\:\:\:{a}_{\:{k}} \:=\:{a}_{{k}−\mathrm{1}} \:.\:\left({a}_{\:{k}−\mathrm{1}} \:+\mathrm{1}\right) \\ $$$$\:\:\:\:\:{and}\:\:\:{a}_{\mathrm{1}} \:=\mathrm{1}\:,\:\:\:{a}_{\:\mathrm{2}} \:=\:\mathrm{2}\:\:,\:{a}_{\:\mathrm{3}} =\:\mathrm{6}\:, \\ $$$$\:\:\:\:\:\:{a}_{\mathrm{4}} \:=\:\:\mathrm{42}\:\:\:,\:{a}_{\:\mathrm{5}} \:=\:\mathrm{1806}\:\:\:\:{and}\:\:{etc} \\ $$$$\:\:\:{find}\:{the}\:\:{value}\:\:{of}\:\:\:\:{E}\:=? \\ $$

Question Number 199092 Answers: 5 Comments: 0

Let the polynomial p(x)=5x^3 +3x^2 −10 have roots a,b and c. What is the value of (a/(b+c))+(b/(c+a))+(c/(a+b))?

$${Let}\:{the}\:{polynomial}\:{p}\left({x}\right)=\mathrm{5}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{10} \\ $$$${have}\:{roots}\:{a},{b}\:{and}\:{c}.\:{What}\:{is}\:{the}\:{value} \\ $$$${of}\:\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}? \\ $$

Question Number 198772 Answers: 0 Comments: 1

x^4 +ax^3 +bx^2 +cx+d=0

$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$

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Find a,b,c and d such that { ((abc = 1 (mod 11))),((abd = 2 (mod 11) )),((acd = 3 (mod 11) )),((bcd = 4 (mod 11) )) :}

$$\:\:\mathrm{Find}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{and}\:\mathrm{d}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{abc}\:=\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{11}\right)}\\{\mathrm{abd}\:=\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{acd}\:=\:\mathrm{3}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\\{\mathrm{bcd}\:=\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{11}\right)\:}\end{cases} \\ $$

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Question Number 198474 Answers: 2 Comments: 0

16000 = (x^3 /((1−x)^2 )) x=?

$$\mathrm{16000}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\: \\ $$$$\:\mathrm{x}=? \\ $$

Question Number 198460 Answers: 0 Comments: 0

In △ABC holds: Π (1 + (1/a) tan (A/2)) ≥ (1 + (1/(3R)))^3

$$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{holds}: \\ $$$$\Pi\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:\mathrm{tan}\:\frac{\mathrm{A}}{\mathrm{2}}\right)\:\geqslant\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{3R}}\right)^{\mathrm{3}} \\ $$

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Given function f(4567,321567)= 567+321=888. f(32156,12062)= 156+120=276 find the value of f(((20^(22) )/(21)) ).

$$\:\:\mathrm{Given}\:\mathrm{function}\: \\ $$$$\:\:\mathrm{f}\left(\mathrm{4567},\mathrm{321567}\right)=\:\mathrm{567}+\mathrm{321}=\mathrm{888}. \\ $$$$\:\:\mathrm{f}\left(\mathrm{32156},\mathrm{12062}\right)=\:\mathrm{156}+\mathrm{120}=\mathrm{276} \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\:\mathrm{f}\left(\frac{\mathrm{20}^{\mathrm{22}} }{\mathrm{21}}\:\right). \\ $$

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f : R → R f (3x − 1) = x + 5 Find: f(x) = ?

$$\mathrm{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{R} \\ $$$$\mathrm{f}\:\left(\mathrm{3x}\:−\:\mathrm{1}\right)\:=\:\mathrm{x}\:+\:\mathrm{5} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$

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(3/8)(3h−p)^2 +3ph=(3h^2 −1) and (((3h−p)^3 )/(16))+p(3h^2 −1)=h^3 −h−c Find p and h in terms of 0<c<(2/(3(√3)))∙

$$\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{3}{h}−{p}\right)^{\mathrm{2}} +\mathrm{3}{ph}=\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${and} \\ $$$$\frac{\left(\mathrm{3}{h}−{p}\right)^{\mathrm{3}} }{\mathrm{16}}+{p}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={h}^{\mathrm{3}} −{h}−{c} \\ $$$${Find}\:\:{p}\:{and}\:{h}\:\:{in}\:{terms}\:{of}\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\centerdot \\ $$

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