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Question Number 82034    Answers: 0   Comments: 1

Prove by maths induction tbat n^5 − n^3 is divisible by 24.

$$\boldsymbol{{P}}{rove}\:\:{by}\:\:{maths}\:\:{induction}\:\:{tbat} \\ $$$$\boldsymbol{{n}}^{\mathrm{5}} \:−\:\boldsymbol{{n}}^{\mathrm{3}} \:\:\boldsymbol{{is}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{24}. \\ $$

Question Number 81971    Answers: 1   Comments: 0

Question Number 81843    Answers: 2   Comments: 5

Question Number 81851    Answers: 0   Comments: 0

1)find ∫ (dx/((x+1)^3 (x^2 +1)^2 )) 2) calculate ∫_0 ^∞ (dx/((x+1)^3 (x^2 +1)^2 ))

$$\left.\mathrm{1}\right){find}\:\int\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 81797    Answers: 0   Comments: 1

Question Number 81704    Answers: 0   Comments: 0

find Γ((1/3)) and Γ((2/3))

$${find}\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:{and}\:\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$

Question Number 81657    Answers: 0   Comments: 2

∫_0 ^3 ((x+1)/((x^2 +2x)^(15) ))=....

$$\:\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\frac{\mathrm{x}+\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\right)^{\mathrm{15}} }=.... \\ $$

Question Number 81545    Answers: 0   Comments: 3

Question Number 81518    Answers: 0   Comments: 0

Question Number 81514    Answers: 0   Comments: 0

Hello sirs ... what are the graphic maker Apps can you suggest me for my android phone ...please.

$${Hello}\:{sirs}\:...\:{what}\:{are}\:{the}\:{graphic} \\ $$$${maker}\:{Apps}\:{can}\:{you}\:{suggest}\:{me}\: \\ $$$${for}\:{my}\:{android}\:{phone}\:...{please}. \\ $$

Question Number 81458    Answers: 0   Comments: 2

if a_1 = 3 ,a_2 =2 a_(n+2) = a_(n+1) +(a_1 /2) find a_6 =? mister W method a_n =A(((1+(√3))/2))^n +B(((1−(√3))/2))^n a_1 = A(((1+(√3))/2))+B(((1−(√3))/2))=3 a_2 = A(((1+(√3))/2))^2 +B(((1−(√3))/2))^2 =2 ⇒A+B+(A−B)(√3) =6 ⇒2(A+B)+(A−B)(√3) =4 A= ((4−(√3))/(√3)) , B = ((−4−(√3))/3) a_n = (((4−(√3))/(√3)))(((1+(√3))/2))^n −(((4+(√3))/(√3)))(((1−(√3))/2))^n

$$\mathrm{if}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{3}\:,\mathrm{a}_{\mathrm{2}} =\mathrm{2} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{2}} \:=\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} +\frac{\mathrm{a}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{a}_{\mathrm{6}} \:=? \\ $$$$\mathrm{mister}\:\mathrm{W}\:\mathrm{method} \\ $$$$\mathrm{a}_{\mathrm{n}} \:=\mathrm{A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} +\mathrm{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$$$\mathrm{a}_{\mathrm{1}} =\:\mathrm{A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\mathrm{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{3} \\ $$$$\mathrm{a}_{\mathrm{2}} \:=\:\mathrm{A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow\mathrm{A}+\mathrm{B}+\left(\mathrm{A}−\mathrm{B}\right)\sqrt{\mathrm{3}}\:=\mathrm{6} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{A}+\mathrm{B}\right)+\left(\mathrm{A}−\mathrm{B}\right)\sqrt{\mathrm{3}}\:=\mathrm{4} \\ $$$$\mathrm{A}=\:\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}\:,\:\mathrm{B}\:=\:\frac{−\mathrm{4}−\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\mathrm{a}_{\mathrm{n}} \:=\:\left(\frac{\mathrm{4}−\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}\right)\left(\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} −\left(\frac{\mathrm{4}+\sqrt{\mathrm{3}}}{\sqrt{\mathrm{3}}}\right)\left(\frac{\mathrm{1}−\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$

Question Number 81124    Answers: 1   Comments: 0

Question Number 81115    Answers: 0   Comments: 0

e^x ∫((2csec^2 θ dθ)/([4+(2tanθ)^2 ]^(3/2) )) = e^x ∫((2csec^2 θ dθ)/([4+(2tanθ)^2 ]^(3/2) )) =

$$\mathrm{e}^{\mathrm{x}} \int\frac{\mathrm{2}{csec}^{\mathrm{2}} \theta\:{d}\theta}{\left[\mathrm{4}+\left(\mathrm{2}{tan}\theta\right)^{\mathrm{2}} \right]^{\mathrm{3}/\mathrm{2}} }\:\:= \\ $$$$\mathrm{e}^{\mathrm{x}} \int\frac{\mathrm{2}{csec}^{\mathrm{2}} \theta\:{d}\theta}{\left[\mathrm{4}+\left(\mathrm{2}{tan}\theta\right)^{\mathrm{2}} \right]^{\mathrm{3}/\mathrm{2}} }\:\:= \\ $$

Question Number 80997    Answers: 0   Comments: 4

give a rational fraction example : cancelling in -1 and 2 having as a set defnition R

$${give}\:{a}\:{rational}\:{fraction}\:{example}\:: \\ $$$${cancelling}\:{in}\:-\mathrm{1}\:{and}\:\mathrm{2}\:{having}\:{as}\:{a}\:{set}\:{defnition}\:\mathbb{R} \\ $$

Question Number 80994    Answers: 0   Comments: 1

donnre un exenple de fraction rationnelle: 1)s′annulant en -1 et 2 ayant pour ensemble definition R

$$\mathrm{donnre}\:\mathrm{un}\:\mathrm{exenple}\:\mathrm{de}\:\mathrm{fraction}\:\mathrm{rationnelle}: \\ $$$$\left.\mathrm{1}\right)\mathrm{s}'\mathrm{annulant}\:\mathrm{en}\:-\mathrm{1}\:\mathrm{et}\:\mathrm{2}\:\mathrm{ayant}\:\mathrm{pour}\:\mathrm{ensemble}\:\mathrm{definition}\:\mathbb{R} \\ $$

Question Number 80974    Answers: 0   Comments: 2

Show that gcd (a , a + x) ∣ x hence show that any two consecutive integers are coprime

$$\:\mathrm{Show}\:\mathrm{that}\:\mathrm{gcd}\:\left({a}\:,\:{a}\:+\:{x}\right)\:\mid\:{x} \\ $$$${hence}\:{show}\:{that}\:{any}\:{two}\:{consecutive} \\ $$$${integers}\:{are}\:{coprime} \\ $$

Question Number 80973    Answers: 0   Comments: 0

Given that f(x) = { ((2x−7, 0 < x < 6)),((2^x , 7 < x < 8)) :} and f is periodic of period 4. find f(200)

$$\mathrm{Given}\:\mathrm{that}\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2}{x}−\mathrm{7},\:\:\mathrm{0}\:<\:{x}\:<\:\mathrm{6}}\\{\mathrm{2}^{{x}} ,\:\:\:\mathrm{7}\:<\:{x}\:<\:\mathrm{8}}\end{cases} \\ $$$$\mathrm{and}\:\mathrm{f}\:\mathrm{is}\:\mathrm{periodic}\:\mathrm{of}\:\mathrm{period}\:\mathrm{4}. \\ $$$$\mathrm{find}\:{f}\left(\mathrm{200}\right) \\ $$

Question Number 80890    Answers: 1   Comments: 0

(((1+i)/(2−i))+((2+i)/(1−i)))^3 =500x+500yi find x,y

$$\left(\frac{\mathrm{1}+\mathrm{i}}{\mathrm{2}−\mathrm{i}}+\frac{\mathrm{2}+\mathrm{i}}{\mathrm{1}−\mathrm{i}}\right)^{\mathrm{3}} =\mathrm{500}{x}+\mathrm{500}{y}\mathrm{i} \\ $$$${find}\:{x},{y} \\ $$

Question Number 80613    Answers: 1   Comments: 2

Q.find (d/dx)(x!)

$${Q}.{find}\:\:\:\frac{{d}}{{dx}}\left({x}!\right) \\ $$

Question Number 80550    Answers: 1   Comments: 1

Question Number 80540    Answers: 0   Comments: 1

Question Number 80515    Answers: 0   Comments: 1

∫(dx/((1+x^φ )^φ ))

$$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\phi} \right)^{\phi} } \\ $$

Question Number 80505    Answers: 0   Comments: 8

Given that 7^k ≡1 (mod 15) a) Write down three values of k. b) Find the general solution of the equation 7^k ≡ 1 (mod 15)

$$\mathrm{Given}\:\mathrm{that}\:\:\mathrm{7}^{{k}} \:\equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{15}\right) \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Write}\:\mathrm{down}\:\mathrm{three}\:\mathrm{values}\:\mathrm{of}\:{k}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\:\mathrm{7}^{{k}} \:\equiv\:\mathrm{1}\:\left({mod}\:\mathrm{15}\right) \\ $$

Question Number 80504    Answers: 0   Comments: 2

Solve the system of congruences x ≡ 2 (mod 3) x ≡ 5( mod 7)

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{congruences} \\ $$$${x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$${x}\:\equiv\:\mathrm{5}\left(\:\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\: \\ $$

Question Number 80341    Answers: 1   Comments: 0

A particle moves round the polar curve r = a(1 + cos θ) with constant angular velocity ω . Find the transverse component of the velocity.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{round}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{curve} \\ $$$${r}\:=\:{a}\left(\mathrm{1}\:+\:\mathrm{cos}\:\theta\right)\:\mathrm{with}\:\mathrm{constant}\:\mathrm{angular}\: \\ $$$$\mathrm{velocity}\:\omega\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{transverse}\:\mathrm{component} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{velocity}. \\ $$

Question Number 80340    Answers: 1   Comments: 0

If P = ((a,b,c,d),(c,d,a,b) ) , Q = ((a,b,c,d),(b,a,d,c) ) are permutations of the elements (a,b,c,d), then QP ≡

$$\mathrm{If}\:{P}\:=\:\begin{pmatrix}{{a}}&{{b}}&{{c}}&{{d}}\\{{c}}&{{d}}&{{a}}&{{b}}\end{pmatrix}\:\:,\:{Q}\:=\:\begin{pmatrix}{{a}}&{{b}}&{{c}}&{{d}}\\{{b}}&{{a}}&{{d}}&{{c}}\end{pmatrix}\:\mathrm{are} \\ $$$$\mathrm{permutations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{elements}\:\left({a},{b},{c},{d}\right),\:\mathrm{then}\: \\ $$$${QP}\:\equiv \\ $$$$\: \\ $$

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