Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1957

Question Number 1942    Answers: 1   Comments: 1

Let N be a positive integer with prime factorisation N=p_1 ^m_1 p_2 ^m_2 p_3 ^m_3 ×...×p_(n−1) ^m_(n−1) p_n ^m_n where n,m_i ∈Z^+ and p_r is prime. How many proper factors does N have? Investigate cases where n=1,n=2, n=3 and n=4. What is the smallest positive integer with 12 proper factors? What is the smallest positive integer with at least 12 proper factors? (A proper factor of a positive number N is positive nteger M such that M≠1 and M≠N.)

$${Let}\:{N}\:{be}\:{a}\:{positive}\:{integer}\:{with}\:{prime} \\ $$$${factorisation}\: \\ $$$$\:\:{N}={p}_{\mathrm{1}} ^{{m}_{\mathrm{1}} } {p}_{\mathrm{2}} ^{{m}_{\mathrm{2}} } {p}_{\mathrm{3}} ^{{m}_{\mathrm{3}} } ×...×{p}_{{n}−\mathrm{1}} ^{{m}_{{n}−\mathrm{1}} } {p}_{{n}} ^{{m}_{{n}} } \\ $$$${where}\:\:{n},{m}_{{i}} \in\mathbb{Z}^{+} \:{and}\:{p}_{{r}} \:{is}\:{prime}. \\ $$$${How}\:{many}\:{proper}\:{factors}\:{does}\:{N}\:{have}? \\ $$$${Investigate}\:{cases}\:{where}\:{n}=\mathrm{1},{n}=\mathrm{2},\:{n}=\mathrm{3} \\ $$$${and}\:{n}=\mathrm{4}.\:{What}\:{is}\:{the}\:{smallest}\:{positive} \\ $$$${integer}\:{with}\:\mathrm{12}\:{proper}\:{factors}? \\ $$$${What}\:{is}\:{the}\:{smallest}\:{positive}\:{integer} \\ $$$${with}\:{at}\:{least}\:\mathrm{12}\:{proper}\:{factors}? \\ $$$$\left({A}\:{proper}\:{factor}\:{of}\:{a}\:{positive}\:{number}\:{N}\right. \\ $$$$\left.{is}\:{positive}\:{nteger}\:{M}\:{such}\:{that}\:{M}\neq\mathrm{1}\:{and}\:{M}\neq{N}.\right) \\ $$

Question Number 1937    Answers: 1   Comments: 0

•Is ′⇔′ necessary and suficient for two inequalities to be equivalent? •If a>b : Are A>B and A+a > B+b equivalent?

$$\bullet{Is}\:\:\:'\Leftrightarrow'\:\:{necessary}\:{and}\:{suficient}\:{for}\:{two} \\ $$$${inequalities}\:{to}\:{be}\:{equivalent}? \\ $$$$\bullet{If}\:\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}\:: \\ $$$${Are}\:\:\boldsymbol{\mathrm{A}}>\boldsymbol{\mathrm{B}}\:{and}\:\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{a}}\:>\:\boldsymbol{\mathrm{B}}+\boldsymbol{\mathrm{b}}\:{equivalent}? \\ $$

Question Number 1936    Answers: 0   Comments: 2

f^2 (x)−f(x^2 )=a [G.Q1902] f(x)=?

$${f}^{\mathrm{2}} \left({x}\right)−{f}\left({x}^{\mathrm{2}} \right)={a}\:\left[\mathrm{G}.\mathrm{Q1902}\right] \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 1930    Answers: 2   Comments: 0

f ′(x)−g(x)=0 f(x)+g′(x)=0 f(x)=? g(x)=?

$${f}\:'\left({x}\right)−{g}\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)+{g}'\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$

Question Number 1928    Answers: 2   Comments: 10

Prove that, if p>q>0 and x≥0, then (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1).

$${Prove}\:{that},\:{if}\:{p}>{q}>\mathrm{0}\:{and}\:{x}\geqslant\mathrm{0},\:{then} \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{{p}}\left(\frac{{x}^{{p}} }{{p}+\mathrm{1}}−\mathrm{1}\right)\geqslant\frac{\mathrm{1}}{{q}}\left(\frac{{x}^{{q}} }{{q}+\mathrm{1}}−\mathrm{1}\right).\: \\ $$

Question Number 1925    Answers: 0   Comments: 1

Evaluate the following integral. I=∫_5 ^6 ((ln(11x+12))/(x^2 +42))dx

$${Evaluate}\:{the}\:{following}\:{integral}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\int_{\mathrm{5}} ^{\mathrm{6}} \frac{{ln}\left(\mathrm{11}{x}+\mathrm{12}\right)}{{x}^{\mathrm{2}} +\mathrm{42}}{dx} \\ $$

Question Number 1952    Answers: 0   Comments: 2

Inequality relation starting a new thread (x^p /(p(p+1)))−(1/p)≥(x^q /(q(q+1)))−(1/q) p=2, q=1, x=1 (x^p /(p(p+1)))=(1/6) (x^q /(q(q+1)))=(1/2) (x^p /(p(p+1)))−(1/p) = (1/6)−(1/2)=−(1/3) (x^p /(p(p+1))) = (1/2) − 1 = −(1/2) A≥B and c>d ⇏A−c≥B−d Since you are subtracting a larger quantity c from A, so if (c−d)>(A−B) then A−c<B−d

$$\mathrm{Inequality}\:\mathrm{relation}\:\mathrm{starting}\:\mathrm{a}\:\mathrm{new}\:\mathrm{thread} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{p}}\geqslant\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{q}} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{1},\:{x}=\mathrm{1} \\ $$$$\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\frac{{x}^{{q}} }{{q}\left({q}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{p}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\frac{{x}^{{p}} }{{p}\left({p}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:−\:\mathrm{1}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${A}\geqslant{B}\:\mathrm{and}\:{c}>{d}\:\nRightarrow{A}−{c}\geqslant{B}−{d} \\ $$$$\mathrm{Since}\:\mathrm{you}\:\mathrm{are}\:\mathrm{subtracting}\:\mathrm{a}\:\mathrm{larger}\:\mathrm{quantity} \\ $$$${c}\:\mathrm{from}\:{A},\:\mathrm{so}\:\mathrm{if}\:\left({c}−{d}\right)>\left({A}−{B}\right)\:\mathrm{then}\:{A}−{c}<{B}−{d} \\ $$

Question Number 1902    Answers: 1   Comments: 4

f^( 2) (x)−f(x^2 )=2 , f^( 2) (x) stands for [f(x)]^2 f(x)=? (If possible solve stepwise)

$${f}^{\:\mathrm{2}} \left({x}\right)−{f}\left({x}^{\mathrm{2}} \right)=\mathrm{2}\:,\:{f}^{\:\mathrm{2}} \left({x}\right)\:{stands}\:{for}\:\left[{f}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({If}\:{possible}\:{solve}\:{stepwise}\right) \\ $$

Question Number 1899    Answers: 1   Comments: 1

Consider the system of equations 2yz+zx−5xy=2 yz−zx+2xy=1 yz−2zx+6xy=3. Show that xyz=±6 and find the possible values of x,y and z.

$${Consider}\:{the}\:{system}\:{of}\:{equations} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{yz}+{zx}−\mathrm{5}{xy}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−{zx}+\mathrm{2}{xy}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{yz}−\mathrm{2}{zx}+\mathrm{6}{xy}=\mathrm{3}. \\ $$$${Show}\:{that}\:{xyz}=\pm\mathrm{6}\: \\ $$$${and}\:{find}\:{the}\:{possible}\:{values} \\ $$$${of}\:{x},{y}\:{and}\:{z}. \\ $$

Question Number 1898    Answers: 1   Comments: 0

(df/dt)=αf+βt+γ f(t)=??

$$\frac{{df}}{{dt}}=\alpha{f}+\beta{t}+\gamma \\ $$$${f}\left({t}\right)=?? \\ $$

Question Number 1895    Answers: 2   Comments: 5

Let us generalise the result of taking the inverse tangent of a complex number to the form tan^(−1) (c+id)=a+ib where a,b,c,d∈R and i=(√(−1)). Determine a and b respectively in terms of c and d.

$${Let}\:{us}\:{generalise}\:{the}\:{result}\:{of}\:{taking}\:{the}\:{inverse}\:{tangent}\:{of}\:{a}\:{complex}\:{number} \\ $$$${to}\:{the}\:{form}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}^{−\mathrm{1}} \left({c}+{id}\right)={a}+{ib} \\ $$$${where}\:{a},{b},{c},{d}\in\mathbb{R}\:{and}\:{i}=\sqrt{−\mathrm{1}}.\:{Determine}\:{a}\:{and}\:{b}\:{respectively}\:{in}\:{terms} \\ $$$${of}\:{c}\:{and}\:{d}.\: \\ $$

Question Number 1890    Answers: 2   Comments: 3

Solve for x tan x +tan 2x −tan 3x =0 Some one had posted this question and I had answered it but then thread was deleted! I think that the question is not importanceless , so I hav reposted it.

$${Solve}\:{for}\:{x} \\ $$$$\:\:\:\:\:\:{tan}\:{x}\:+{tan}\:\mathrm{2}{x}\:−{tan}\:\mathrm{3}{x}\:=\mathrm{0} \\ $$$${Some}\:{one}\:{had}\:{posted}\:{this}\:{question}\:{and}\:{I}\:{had}\:{answered}\:{it} \\ $$$${but}\:{then}\:{thread}\:{was}\:{deleted}! \\ $$$${I}\:{think}\:{that}\:{the}\:{question}\:{is}\:{not}\:{importanceless}\:,\:{so}\:{I}\:{hav} \\ $$$${reposted}\:{it}. \\ $$

Question Number 1882    Answers: 0   Comments: 0

∫_(−∞) ^∞ ∫_(−∞) ^∞ ∣ψ(r,t)∣∈∣φ(r,t)∣⇒λ(x) in quantum mechanics would this be true, if so how?

$$\underset{−\infty} {\overset{\infty} {\int}}\underset{−\infty} {\overset{\infty} {\int}}\mid\psi\left({r},{t}\right)\mid\in\mid\phi\left({r},{t}\right)\mid\Rightarrow\lambda\left({x}\right)\: \\ $$$${in}\:{quantum}\:{mechanics}\:{would}\:{this}\:{be} \\ $$$${true},\:{if}\:{so}\:{how}? \\ $$$$ \\ $$$$ \\ $$

Question Number 1880    Answers: 1   Comments: 0

Solve 1) (sin 2x+ (√)3 cos 2x)^2 =2 −2 cos (((2π)/3)−x) 2) cos x − cos 2x = sin 3x 3) sin (15 °+x) + cos (45°+x)+ (1/2)=0 4) tan (70°+x) + tan (20°−x) = 2

$${Solve} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\:\left({sin}\:\mathrm{2}{x}+\:\sqrt{}\mathrm{3}\:{cos}\:\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{2}\:−\mathrm{2}\:{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−{x}\right) \\ $$$$\left.\mathrm{2}\right)\:{cos}\:{x}\:−\:{cos}\:\mathrm{2}{x}\:=\:{sin}\:\mathrm{3}{x} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:{sin}\:\left(\mathrm{15} °+{x}\right)\:+\:{cos}\:\left(\mathrm{45}°+{x}\right)+\:\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left.\mathrm{4}\right)\:{tan}\:\left(\mathrm{70}°+{x}\right)\:+\:{tan}\:\left(\mathrm{20}°−{x}\right)\:=\:\mathrm{2} \\ $$$$ \\ $$

Question Number 1875    Answers: 0   Comments: 0

Given that: Z={0, 1, 2, ...} all integers ≥0 R={0, 0.01, ..., 1, 1.01, ...} all reals ≥0 Prove that ∣R∣>∣Z∣

$$\mathrm{Given}\:\mathrm{that}: \\ $$$${Z}=\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:...\right\}\:\mathrm{all}\:\mathrm{integers}\:\geqslant\mathrm{0} \\ $$$${R}=\left\{\mathrm{0},\:\mathrm{0}.\mathrm{01},\:...,\:\mathrm{1},\:\mathrm{1}.\mathrm{01},\:...\right\}\:\mathrm{all}\:\mathrm{reals}\:\geqslant\mathrm{0} \\ $$$$\:\mathrm{Prove}\:\mathrm{that}\:\mid{R}\mid>\mid{Z}\mid \\ $$

Question Number 1873    Answers: 1   Comments: 0

f_(n+1) (x)=f_n (x)(x−n)(x+n) f_0 (x)=1 f_5 (x)=? f_4 (2x)=0,x=?

$${f}_{{n}+\mathrm{1}} \left({x}\right)={f}_{{n}} \left({x}\right)\left({x}−{n}\right)\left({x}+{n}\right) \\ $$$${f}_{\mathrm{0}} \left({x}\right)=\mathrm{1} \\ $$$${f}_{\mathrm{5}} \left({x}\right)=? \\ $$$${f}_{\mathrm{4}} \left(\mathrm{2}{x}\right)=\mathrm{0},{x}=? \\ $$

Question Number 1870    Answers: 1   Comments: 0

Solve 2sin 2x− 3(sin x + cos x) + 2 =0

$${Solve}\: \\ $$$$ \\ $$$$\mathrm{2}{sin}\:\mathrm{2}{x}−\:\mathrm{3}\left({sin}\:{x}\:+\:{cos}\:{x}\right)\:+\:\mathrm{2}\:=\mathrm{0} \\ $$

Question Number 1865    Answers: 1   Comments: 0

proof that (√2) is an irrational number

$${proof}\:{that}\:\sqrt{\mathrm{2}}\:{is}\:{an}\:{irrational}\:{number} \\ $$$$ \\ $$

Question Number 1862    Answers: 2   Comments: 0

f[x−f(y)]=f(x)−f(y),∀(x,y)∈R f:R→R f(x)=?

$${f}\left[{x}−{f}\left({y}\right)\right]={f}\left({x}\right)−{f}\left({y}\right),\forall\left({x},{y}\right)\in\mathbb{R} \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 1861    Answers: 1   Comments: 0

V(ξ)=πρ^2 ∫_(−∞) ^ξ e^(2z) dz S(ξ)=2πρ∫_(−∞) ^ξ e^z (√(1+ρ^2 e^(2z) ))dz V(ξ)−S(ξ)=?

$$\mathrm{V}\left(\xi\right)=\pi\rho^{\mathrm{2}} \underset{−\infty} {\overset{\xi} {\int}}{e}^{\mathrm{2}{z}} {dz} \\ $$$$\mathrm{S}\left(\xi\right)=\mathrm{2}\pi\rho\underset{−\infty} {\overset{\xi} {\int}}{e}^{{z}} \sqrt{\mathrm{1}+\rho^{\mathrm{2}} {e}^{\mathrm{2}{z}} }{dz} \\ $$$$\mathrm{V}\left(\xi\right)−\mathrm{S}\left(\xi\right)=? \\ $$

Question Number 1858    Answers: 1   Comments: 0

[((x(ρ,θ,ξ))),((y(ρ,θ,ξ))),((z(ρ,θ,ξ))) ]= [((ρe^ξ cos θ)),((ρe^ξ sin θ)),(ξ) ] { ((ρ∈[0,+∞))),((θ∈[0,2π))),((ξ∈R)) :} r(ρ,θ,ξ)=x(ρ,θ,ξ)i+y(ρ,θ,ξ)j+z(ρ,θ,ξ)k a_ρ =(∂r/∂ρ) a_θ =(∂r/∂θ) a_ξ =(∂r/∂ξ) a_ρ ∙a_ρ +a_θ ∙a_θ +a_ξ ∙a_ξ =??? a_ρ ∙a_θ +a_ρ ∙a_ξ +a_θ ∙a_ξ =??? a_ρ ×a_θ +a_ρ ×a_ξ +a_θ ×a_ξ =???

$$\begin{bmatrix}{{x}\left(\rho,\theta,\xi\right)}\\{{y}\left(\rho,\theta,\xi\right)}\\{{z}\left(\rho,\theta,\xi\right)}\end{bmatrix}=\begin{bmatrix}{\rho{e}^{\xi} \mathrm{cos}\:\theta}\\{\rho{e}^{\xi} \mathrm{sin}\:\theta}\\{\xi}\end{bmatrix}\begin{cases}{\rho\in\left[\mathrm{0},+\infty\right)}\\{\theta\in\left[\mathrm{0},\mathrm{2}\pi\right)}\\{\xi\in\mathbb{R}}\end{cases} \\ $$$$\boldsymbol{{r}}\left(\rho,\theta,\xi\right)={x}\left(\rho,\theta,\xi\right)\boldsymbol{{i}}+{y}\left(\rho,\theta,\xi\right)\boldsymbol{{j}}+{z}\left(\rho,\theta,\xi\right)\boldsymbol{{k}} \\ $$$$\boldsymbol{{a}}_{\rho} =\frac{\partial\boldsymbol{{r}}}{\partial\rho} \\ $$$$\boldsymbol{{a}}_{\theta} =\frac{\partial\boldsymbol{{r}}}{\partial\theta} \\ $$$$\boldsymbol{{a}}_{\xi} =\frac{\partial\boldsymbol{{r}}}{\partial\xi} \\ $$$$\boldsymbol{{a}}_{\rho} \centerdot\boldsymbol{{a}}_{\rho} +\boldsymbol{{a}}_{\theta} \centerdot\boldsymbol{{a}}_{\theta} +\boldsymbol{{a}}_{\xi} \centerdot\boldsymbol{{a}}_{\xi} =??? \\ $$$$\boldsymbol{{a}}_{\rho} \centerdot\boldsymbol{{a}}_{\theta} +\boldsymbol{{a}}_{\rho} \centerdot\boldsymbol{{a}}_{\xi} +\boldsymbol{{a}}_{\theta} \centerdot\boldsymbol{{a}}_{\xi} =??? \\ $$$$\boldsymbol{{a}}_{\rho} ×\boldsymbol{{a}}_{\theta} +\boldsymbol{{a}}_{\rho} ×\boldsymbol{{a}}_{\xi} +\boldsymbol{{a}}_{\theta} ×\boldsymbol{{a}}_{\xi} =??? \\ $$

Question Number 1851    Answers: 1   Comments: 0

A plane has equation x−z=4(√3). The line l has vector equation r=λ[(cosθ+(√3))i+((√2)sinθ)j+(cosθ−(√3))k] where λ is a scalar parameter. If l meets the plane at P, show that, as θ varies, P describes a circle.

$${A}\:{plane}\:{has}\:{equation}\:{x}−{z}=\mathrm{4}\sqrt{\mathrm{3}}. \\ $$$${The}\:{line}\:{l}\:{has}\:{vector}\:{equation} \\ $$$$\boldsymbol{{r}}=\lambda\left[\left({cos}\theta+\sqrt{\mathrm{3}}\right)\boldsymbol{{i}}+\left(\sqrt{\mathrm{2}}{sin}\theta\right)\boldsymbol{{j}}+\left({cos}\theta−\sqrt{\mathrm{3}}\right)\boldsymbol{{k}}\right] \\ $$$${where}\:\lambda\:{is}\:{a}\:{scalar}\:{parameter}. \\ $$$${If}\:{l}\:{meets}\:{the}\:{plane}\:{at}\:{P},\:{show}\:{that}, \\ $$$${as}\:\theta\:{varies},\:{P}\:\:{describes}\:{a}\:{circle}.\: \\ $$

Question Number 1850    Answers: 1   Comments: 0

f_(n+1) (z+1)=[z−f_n (0)]f_n (z) f_1 (z)=z+1 f_3 (z)=???

$${f}_{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{{n}} \left(\mathrm{0}\right)\right]{f}_{{n}} \left({z}\right) \\ $$$${f}_{\mathrm{1}} \left({z}\right)={z}+\mathrm{1} \\ $$$${f}_{\mathrm{3}} \left({z}\right)=??? \\ $$

Question Number 1848    Answers: 2   Comments: 0

f_(n+1) (z+1)=[z−f_(n−1) (0)]f_n (z) f_0 (z)=0 f_1 (z)=z+1 f_3 (z)=????

$${f}_{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)=\left[{z}−{f}_{{n}−\mathrm{1}} \left(\mathrm{0}\right)\right]{f}_{{n}} \left({z}\right) \\ $$$${f}_{\mathrm{0}} \left({z}\right)=\mathrm{0} \\ $$$${f}_{\mathrm{1}} \left({z}\right)={z}+\mathrm{1} \\ $$$${f}_{\mathrm{3}} \left({z}\right)=???? \\ $$

Question Number 1844    Answers: 0   Comments: 5

lets two polynimies p_n ,q_n givwn by p_1 =q_1 =x p_(n+1) =p_n +q_n q_(n+1) =p_n q_n then (1,1)→(1,2)→(2,3) lets W(u,v)= determinant ((u,v),((u′),(v′))) is true that W(p_n ,q_n )≠0,∀n>1

$$\mathrm{lets}\:\mathrm{two}\:\mathrm{polynimies}\:{p}_{{n}} ,{q}_{{n}} \:\mathrm{givwn}\:\mathrm{by} \\ $$$${p}_{\mathrm{1}} ={q}_{\mathrm{1}} ={x} \\ $$$${p}_{{n}+\mathrm{1}} ={p}_{{n}} +{q}_{{n}} \\ $$$${q}_{{n}+\mathrm{1}} ={p}_{{n}} {q}_{{n}} \\ $$$$\mathrm{then}\:\left(\mathrm{1},\mathrm{1}\right)\rightarrow\left(\mathrm{1},\mathrm{2}\right)\rightarrow\left(\mathrm{2},\mathrm{3}\right) \\ $$$$\mathrm{lets}\:\mathrm{W}\left({u},{v}\right)=\begin{vmatrix}{{u}}&{{v}}\\{{u}'}&{{v}'}\end{vmatrix} \\ $$$$\mathrm{is}\:\mathrm{true}\:\mathrm{that} \\ $$$$\mathrm{W}\left({p}_{{n}} ,{q}_{{n}} \right)\neq\mathrm{0},\forall{n}>\mathrm{1} \\ $$

Question Number 1842    Answers: 1   Comments: 1

f(x,y)=f(x+y,xy) f(x,y)=x,−2≤y≤2 f(x,y)=y,∣x∣≥100∨∣y∣≥100 f(0,0)=? f(1,4)=?

$${f}\left({x},{y}\right)={f}\left({x}+{y},{xy}\right) \\ $$$${f}\left({x},{y}\right)={x},−\mathrm{2}\leqslant{y}\leqslant\mathrm{2} \\ $$$${f}\left({x},{y}\right)={y},\mid{x}\mid\geqslant\mathrm{100}\vee\mid{y}\mid\geqslant\mathrm{100} \\ $$$${f}\left(\mathrm{0},\mathrm{0}\right)=? \\ $$$${f}\left(\mathrm{1},\mathrm{4}\right)=? \\ $$

  Pg 1952      Pg 1953      Pg 1954      Pg 1955      Pg 1956      Pg 1957      Pg 1958      Pg 1959      Pg 1960      Pg 1961   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com