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AlgebraQuestion and Answers: Page 239

Question Number 115325 Answers: 0 Comments: 5

Question Number 115218 Answers: 0 Comments: 0

Show that ∀n∈N, ∀u_0 ,u_1 ,...,u_n ,v_0 ,v_1 ,...v_n ∈C ∀k≤n; u_k =Σ_(i=0) ^k ((k),(i) )v_i ⇔∀k≤n; v_k =Σ_(i=0) ^k (−1)^(k−1) ((k),(i) )u_i

$$\mathrm{Show}\:\mathrm{that}\:\forall\mathrm{n}\in\mathbb{N},\:\forall\mathrm{u}_{\mathrm{0}} ,\mathrm{u}_{\mathrm{1}} ,...,\mathrm{u}_{\mathrm{n}} ,\mathrm{v}_{\mathrm{0}} ,\mathrm{v}_{\mathrm{1}} ,...\mathrm{v}_{\mathrm{n}} \in\mathbb{C} \\ $$$$\forall\mathrm{k}\leqslant\mathrm{n};\:\mathrm{u}_{\mathrm{k}} =\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{k}} \begin{pmatrix}{\mathrm{k}}\\{\mathrm{i}}\end{pmatrix}\mathrm{v}_{\mathrm{i}} \Leftrightarrow\forall\mathrm{k}\leqslant\mathrm{n};\:\mathrm{v}_{\mathrm{k}} =\sum_{\mathrm{i}=\mathrm{0}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}−\mathrm{1}} \begin{pmatrix}{\mathrm{k}}\\{\mathrm{i}}\end{pmatrix}\mathrm{u}_{\mathrm{i}} \\ $$

Question Number 115136 Answers: 1 Comments: 0

Question Number 115135 Answers: 1 Comments: 1

Question Number 115023 Answers: 2 Comments: 0

solve { ((7x−5y+3z=6)),((2x+4y−5z=−5)),((9x−8y+2z=−1)) :}. Find x+y+z

$${solve}\:\begin{cases}{\mathrm{7}{x}−\mathrm{5}{y}+\mathrm{3}{z}=\mathrm{6}}\\{\mathrm{2}{x}+\mathrm{4}{y}−\mathrm{5}{z}=−\mathrm{5}}\\{\mathrm{9}{x}−\mathrm{8}{y}+\mathrm{2}{z}=−\mathrm{1}}\end{cases}.\:{Find}\:{x}+{y}+{z}\: \\ $$

Question Number 115018 Answers: 2 Comments: 0

If 9^x +9^(−x) = 3^(2+x) +3^(2−x) −20, then 27^x +27^(−x) =?

$${If}\:\mathrm{9}^{{x}} +\mathrm{9}^{−{x}} \:=\:\mathrm{3}^{\mathrm{2}+{x}} +\mathrm{3}^{\mathrm{2}−{x}} \:−\mathrm{20},\:{then}\: \\ $$$$\mathrm{27}^{{x}} +\mathrm{27}^{−{x}} \:=? \\ $$

Question Number 114956 Answers: 0 Comments: 5

Question Number 114919 Answers: 0 Comments: 0

Question Number 114912 Answers: 0 Comments: 0

Question Number 114917 Answers: 0 Comments: 3

Question Number 114809 Answers: 1 Comments: 1

{ ((1−((12)/(y+3x))=(2/( (√x))))),((1+((12)/(y+3x))=(6/( (√y))))) :}

$$\begin{cases}{\mathrm{1}−\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}=\frac{\mathrm{2}}{\:\sqrt{{x}}}}\\{\mathrm{1}+\frac{\mathrm{12}}{{y}+\mathrm{3}{x}}=\frac{\mathrm{6}}{\:\sqrt{{y}}}}\end{cases} \\ $$

Question Number 114806 Answers: 1 Comments: 0

Question Number 114799 Answers: 1 Comments: 0

Question Number 114796 Answers: 2 Comments: 0

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Question Number 114758 Answers: 1 Comments: 0

Π_(n=1) ^∞ ((4n^2 (10n−6)(10n−4))/((2n−1)^2 (10n−1)(10n+1))) =?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\mathrm{4}{n}^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{6}\right)\left(\mathrm{10}{n}−\mathrm{4}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{1}\right)\left(\mathrm{10}{n}+\mathrm{1}\right)}\:=? \\ $$

Question Number 114739 Answers: 1 Comments: 0

find the largest and smallest coefficient in (4+3x)^(−5) .

$${find}\:{the}\:{largest}\:{and}\:{smallest} \\ $$$${coefficient}\:{in}\:\left(\mathrm{4}+\mathrm{3}{x}\right)^{−\mathrm{5}} . \\ $$

Question Number 114720 Answers: 2 Comments: 0

Question Number 114653 Answers: 2 Comments: 0

solve 6x^4 −25x^3 +12x^2 −25x+6=0

$${solve}\:\mathrm{6}{x}^{\mathrm{4}} −\mathrm{25}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{6}=\mathrm{0} \\ $$

Question Number 114625 Answers: 1 Comments: 0

Question Number 114593 Answers: 1 Comments: 0

((∣x−1∣)/(∣x∣−1)) ≤ 1

$$\frac{\mid{x}−\mathrm{1}\mid}{\mid{x}\mid−\mathrm{1}}\:\leqslant\:\mathrm{1} \\ $$

Question Number 114592 Answers: 3 Comments: 0

Solve : x^2 +2x−1≡0(mod 3)

$$\mathrm{Solve}\::\:{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$

Question Number 114551 Answers: 1 Comments: 3

The value of x in the given equation 4^x −3^(x−(1/2)) = 3^(x+(1/2)) −2^(2x−1) , is _,

$${The}\:{value}\:{of}\:{x}\:{in}\:{the}\:{given}\:{equation} \\ $$$$\mathrm{4}^{{x}} −\mathrm{3}^{{x}−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{3}^{{x}+\frac{\mathrm{1}}{\mathrm{2}}} \:−\mathrm{2}^{\mathrm{2}{x}−\mathrm{1}} \:,\:{is}\:\_, \\ $$

Question Number 114538 Answers: 1 Comments: 4

how many trees can be planted in one square meter area if the distance between each tow trees is 3cm?

$${how}\:{many}\:{trees}\:{can}\:{be}\:{planted}\:{in}\:{one} \\ $$$${square}\:{meter}\:{area}\:{if}\:{the}\:{distance}\: \\ $$$${between}\:{each}\:{tow}\:{trees}\:{is}\:\mathrm{3}{cm}? \\ $$

Question Number 114511 Answers: 2 Comments: 3

Find numbers which are common terms of the two following arithmetic progression: 3,7,11,...,407 and 2,9,16,...,709

$$\mathrm{Find}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{common} \\ $$$$\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{following}\:\mathrm{arithmetic} \\ $$$$\mathrm{progression}: \\ $$$$\mathrm{3},\mathrm{7},\mathrm{11},...,\mathrm{407}\:\mathrm{and}\:\mathrm{2},\mathrm{9},\mathrm{16},...,\mathrm{709} \\ $$

Question Number 114388 Answers: 0 Comments: 2

Developed the formula for the cubic equation when it has three real roots and Cardano is just not suitable as discriminant gets negative.. Given x^3 −bx−c=0 with b>0 , c>0 (t/b)=1+((15)/2)((c^2 /b^3 ))+(√(((c^2 /b^3 ))[49−((75)/4)((c^2 /b^3 ))])) x=(√(t+((4c^2 )/b^2 ))) −((3c)/b) ★ mrW Sir, MjS Sir please check!

$${Developed}\:{the}\:{formula}\:{for}\:{the}\:{cubic} \\ $$$${equation}\:{when}\:{it}\:{has}\:{three}\:{real}\:{roots} \\ $$$${and}\:{Cardano}\:{is}\:{just}\:{not}\:{suitable}\:{as} \\ $$$${discriminant}\:{gets}\:{negative}.. \\ $$$$\:\:\:{Given}\:\:\:\:\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{bx}}−\boldsymbol{{c}}=\mathrm{0} \\ $$$$\:\:{with}\:\:\boldsymbol{{b}}>\mathrm{0}\:,\:\boldsymbol{{c}}>\mathrm{0} \\ $$$$\:\frac{\boldsymbol{{t}}}{\boldsymbol{{b}}}=\mathrm{1}+\frac{\mathrm{15}}{\mathrm{2}}\left(\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{3}} }\right)+\sqrt{\left(\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{3}} }\right)\left[\mathrm{49}−\frac{\mathrm{75}}{\mathrm{4}}\left(\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{3}} }\right)\right]} \\ $$$$\:\:\:\:\:\boldsymbol{{x}}=\sqrt{\boldsymbol{{t}}+\frac{\mathrm{4}\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} }}\:−\frac{\mathrm{3}\boldsymbol{{c}}}{\boldsymbol{{b}}}\:\:\bigstar \\ $$$${mrW}\:{Sir},\:{MjS}\:{Sir}\:\:{please}\:{check}! \\ $$

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