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Question Number 78440 Answers: 0 Comments: 0

ab = p (a−2b)(a+b)(a+2b)=−24q Find ((a+b)/3) in terms of p, q.

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}\:=\:{p} \\ $$$$\:\:\left({a}−\mathrm{2}{b}\right)\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)=−\mathrm{24}{q} \\ $$$${Find}\:\:\:\frac{{a}+{b}}{\mathrm{3}}\:\:\:\:{in}\:{terms}\:{of}\:{p},\:{q}. \\ $$

Question Number 78427 Answers: 2 Comments: 0

please i need it urgently show that the midpoint of the hypotenuse of a right triangle is equidistant from its vertices

$${please}\:{i}\:{need}\:{it}\:{urgently} \\ $$$$ \\ $$$${show}\:{that}\:{the}\:{midpoint}\:{of}\:{the}\:{hypotenuse} \\ $$$${of}\:{a}\:{right}\:{triangle}\:{is}\:{equidistant}\:{from}\:{its}\:{vertices} \\ $$

Question Number 78425 Answers: 3 Comments: 0

how to solve find inf and sup of A 1. A={(m/n)+((4n)/m) m,n ∈N} 2. A={((mn)/(4m^2 + n^2 )) m∈Z,n ∈N} 3. A={(m/(∣m∣ + n)) m∈Z,n ∈N}

$${how}\:{to}\:{solve} \\ $$$${find}\:{inf}\:{and}\:{sup}\:{of}\:{A} \\ $$$$ \\ $$$$\mathrm{1}.\:{A}=\left\{\frac{{m}}{{n}}+\frac{\mathrm{4}{n}}{{m}}\:\:\:{m},{n}\:\in{N}\right\} \\ $$$$\mathrm{2}.\:{A}=\left\{\frac{{mn}}{\mathrm{4}{m}^{\mathrm{2}} \:+\:{n}^{\mathrm{2}} }\:\:{m}\in{Z},{n}\:\in{N}\right\} \\ $$$$\mathrm{3}.\:{A}=\left\{\frac{{m}}{\mid{m}\mid\:+\:{n}}\:\:{m}\in{Z},{n}\:\in{N}\right\} \\ $$$$ \\ $$$$ \\ $$

Question Number 78400 Answers: 0 Comments: 0

the convolute function of both f and g is marked f∗g And define by (f∗g)(x)=∫_0 ^x f(x−t)g(t)dt Let noted E=the set of function define on R_+ 0) Prove that there exist a function f_0 ∈E such as for all x>0 , ∫_0 ^∞ f_0 (t)e^(−xt) dt=1 1)Prove that (E,∗) is a semigroup 2)Prove that (E,+,∗) is an integrity domain.Is it a hull? 3) Prove that the sub−set J(x)={f∈E , ∫_0 ^∞ f(t)e^(−xt) dt=0} is a maximal ideal of E for all x>0. 4) let U(E) be the set of units of E Prove that H(x)={f∈U(E), f≡f_0 modJ(x)} is an invariant sub−group of U(E). 5) let be I_n the ideal formed by g_n : x→x^n witb n≥1, let mark I_n =(g_n )=g_n E Prove that , I_n ={∫∫..∫fdx_1 ...dx_n , f∈E} 6) By using the fondamental analysis theorem , can we say that E is a principal ring??

$$\mathrm{the}\:\mathrm{convolute}\:\mathrm{function}\:\mathrm{of}\:\mathrm{both}\:\mathrm{f}\:\mathrm{and}\:\mathrm{g}\:\mathrm{is}\:\mathrm{marked}\:\mathrm{f}\ast\mathrm{g} \\ $$$$\mathrm{And}\:\mathrm{define}\:\mathrm{by}\:\:\left(\mathrm{f}\ast\mathrm{g}\right)\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{f}\left(\mathrm{x}−\mathrm{t}\right)\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{Let}\:\mathrm{noted}\:\mathrm{E}=\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{function}\:\mathrm{define}\:\mathrm{on}\:\mathbb{R}_{+} \\ $$$$\left.\mathrm{0}\right)\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{a}\:\mathrm{function}\:\mathrm{f}_{\mathrm{0}} \in\mathrm{E}\:\mathrm{such}\:\mathrm{as}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}>\mathrm{0}\:,\:\int_{\mathrm{0}} ^{\infty} \mathrm{f}_{\mathrm{0}} \left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{xt}} \mathrm{dt}=\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\mathrm{Prove}\:\mathrm{that}\:\left(\mathrm{E},\ast\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{semigroup} \\ $$$$\left.\mathrm{2}\right)\mathrm{Prove}\:\mathrm{that}\:\left(\mathrm{E},+,\ast\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{integrity}\:\mathrm{domain}.\mathrm{Is}\:\mathrm{it}\:\mathrm{a}\:\mathrm{hull}?\: \\ $$$$\left.\mathrm{3}\right)\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sub}−\mathrm{set}\:\mathrm{J}\left(\mathrm{x}\right)=\left\{\mathrm{f}\in\mathrm{E}\:,\:\int_{\mathrm{0}} ^{\infty} \mathrm{f}\left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{xt}} \mathrm{dt}=\mathrm{0}\right\}\:\mathrm{is}\:\mathrm{a}\:\mathrm{maximal}\:\mathrm{ideal}\:\mathrm{of}\:\mathrm{E}\:\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}>\mathrm{0}. \\ $$$$\left.\mathrm{4}\right)\:\mathrm{let}\:\mathrm{U}\left(\mathrm{E}\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{units}\:\mathrm{of}\:\mathrm{E} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\mathrm{H}\left(\mathrm{x}\right)=\left\{\mathrm{f}\in\mathrm{U}\left(\mathrm{E}\right),\:\mathrm{f}\equiv\mathrm{f}_{\mathrm{0}} \mathrm{modJ}\left(\mathrm{x}\right)\right\}\:\mathrm{is}\:\mathrm{an}\:\mathrm{invariant}\:\mathrm{sub}−\mathrm{group}\:\mathrm{of}\:\mathrm{U}\left(\mathrm{E}\right). \\ $$$$\left.\mathrm{5}\right)\:\mathrm{let}\:\mathrm{be}\:\mathrm{I}_{\mathrm{n}} \:\mathrm{the}\:\mathrm{ideal}\:\mathrm{formed}\:\mathrm{by}\:\mathrm{g}_{\mathrm{n}} :\:\mathrm{x}\rightarrow\mathrm{x}^{\mathrm{n}} \:\mathrm{witb}\:\mathrm{n}\geqslant\mathrm{1},\:\mathrm{let}\:\mathrm{mark}\:\:\mathrm{I}_{\mathrm{n}} =\left(\mathrm{g}_{\mathrm{n}} \:\right)=\mathrm{g}_{\mathrm{n}} \mathrm{E} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:,\:\mathrm{I}_{\mathrm{n}} =\left\{\int\int..\int\mathrm{fdx}_{\mathrm{1}} ...\mathrm{dx}_{\mathrm{n}} \:,\:\:\:\:\:\mathrm{f}\in\mathrm{E}\right\} \\ $$$$\left.\mathrm{6}\right)\:\mathrm{By}\:\mathrm{using}\:\mathrm{the}\:\mathrm{fondamental}\:\mathrm{analysis}\:\mathrm{theorem}\:,\:\mathrm{can}\:\mathrm{we}\:\mathrm{say}\:\mathrm{that}\:\mathrm{E}\:\mathrm{is}\:\mathrm{a}\:\mathrm{principal}\:\mathrm{ring}?? \\ $$

Question Number 78399 Answers: 2 Comments: 14

dear sir W, Mjs the set {1,4,n} have the condition that if two different elements are selected and 2112 is added to the result , then the result is a perfect square if n is a positif number . then the number of possible values of n is (A) 8 (B) 7 (C) 6 (D) 5 (E) 4

$${dear}\:{sir}\:{W},\:{Mjs}\: \\ $$$${the}\:{set}\:\left\{\mathrm{1},\mathrm{4},{n}\right\}\:{have}\:{the}\:{condition}\:{that}\: \\ $$$${if}\:{two}\:{different}\:{elements}\:{are} \\ $$$${selected}\:{and}\:\mathrm{2112}\:{is}\:{added}\:{to} \\ $$$${the}\:{result}\:,\:{then}\:{the}\:{result}\: \\ $$$${is}\:{a}\:{perfect}\:{square}\:{if}\:{n}\:{is}\:{a}\: \\ $$$${positif}\:{number}\:.\:{then}\:{the}\:{number}\: \\ $$$${of}\:{possible}\:{values}\:{of}\:{n}\:{is}\: \\ $$$$\left({A}\right)\:\mathrm{8}\:\:\:\:\left({B}\right)\:\mathrm{7}\:\:\:\:\:\left({C}\right)\:\mathrm{6}\:\:\:\:\:\left({D}\right)\:\mathrm{5} \\ $$$$\left({E}\right)\:\mathrm{4} \\ $$

Question Number 78395 Answers: 0 Comments: 6

what minimum value of y = sin x+cosec x+2

$${what}\:{minimum}\:{value}\:{of}\: \\ $$$${y}\:=\:\mathrm{sin}\:{x}+\mathrm{cosec}\:{x}+\mathrm{2}\: \\ $$

Question Number 78689 Answers: 0 Comments: 0

Question Number 78392 Answers: 1 Comments: 0

Question Number 78391 Answers: 0 Comments: 0

Question Number 78390 Answers: 1 Comments: 0

solve for different digits a,b,c,d such that abcd=(ab+cd)^2 .

$${solve}\:{for}\:\boldsymbol{{different}}\:{digits}\:{a},{b},{c},{d}\: \\ $$$${such}\:{that}\:\boldsymbol{{abcd}}=\left(\boldsymbol{{ab}}+\boldsymbol{{cd}}\right)^{\mathrm{2}} . \\ $$

Question Number 78358 Answers: 0 Comments: 1

Q. solve (2^(sin^2 (x)) /(sin^2 (x) )) + (3^(cos^2 (x)) /(cos^2 (x))) = 6

$${Q}.\:{solve} \\ $$$$ \\ $$$$\frac{\mathrm{2}^{{sin}^{\mathrm{2}} \left({x}\right)} }{{sin}^{\mathrm{2}} \left({x}\right)\:}\:+\:\frac{\mathrm{3}^{{cos}^{\mathrm{2}} \left({x}\right)} }{{cos}^{\mathrm{2}} \left({x}\right)}\:=\:\mathrm{6} \\ $$

Question Number 78357 Answers: 2 Comments: 7

Show that: (((√(1 + 6x)) − (1/(√(1 − 6x))))/((√(1 + 3x)) − (1/(√(1 − 3x))))) = 4 + 6x Ignoring higher power of x in the expansion

$$\mathrm{Show}\:\mathrm{that}:\:\:\:\frac{\sqrt{\mathrm{1}\:+\:\mathrm{6x}}\:\:−\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}\:−\:\mathrm{6x}}}}{\sqrt{\mathrm{1}\:+\:\mathrm{3x}}\:\:−\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}\:−\:\mathrm{3x}}}}\:\:\:\:\:=\:\:\:\mathrm{4}\:+\:\mathrm{6x} \\ $$$$\mathrm{Ignoring}\:\mathrm{higher}\:\mathrm{power}\:\mathrm{of}\:\:\mathrm{x}\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$

Question Number 78352 Answers: 1 Comments: 1

Find ∫(1/(cos^2 x(1−tanx)^2 )) dx

$${Find}\:\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}\left(\mathrm{1}−{tanx}\right)^{\mathrm{2}} }\:{dx} \\ $$

Question Number 78351 Answers: 2 Comments: 1

Let a,b,c ∈ R^+ and (a+b)(b+c) = 1 , where 0 <b≤ 1 . Prove that ∣a−b∣∣b−c∣ ≥ ((∣(√a)−(√b)∣∣(√b)−(√c)∣)/2) .

$$\mathrm{Let}\:\:{a},{b},{c}\:\in\:\mathrm{R}^{+} \:\:\mathrm{and}\:\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{1}\:,\:\mathrm{where}\:\:\mathrm{0}\:<{b}\leqslant\:\mathrm{1}\:. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\mid{a}−{b}\mid\mid{b}−{c}\mid\:\geqslant\:\frac{\mid\sqrt{{a}}−\sqrt{{b}}\mid\mid\sqrt{{b}}−\sqrt{{c}}\mid}{\mathrm{2}}\:. \\ $$

Question Number 78353 Answers: 0 Comments: 0

Question Number 78343 Answers: 0 Comments: 1

Question Number 78342 Answers: 0 Comments: 0

sppose that (R,+,.)be aring and we have the ring (R×Z,+^(′ ) ,.^′ ) prove that (R×0,+^′ ,.′) it was ideal in (R×Z,+^′ ,.^′ ) and prove (0×Z,+,.)be isomorphic in (Z,+,.) and if a∈R identity element (a^2 =a)prove that (−a,1)be identity element in the ring (R×Z,+^′ ,.^′ ) pleas sir help me am neding this pleas?

$${sppose}\:{that}\:\left({R},+,.\right){be}\:{aring}\:{and}\:{we}\:{have}\:{the}\:{ring}\:\left({R}×{Z},+^{'\:} ,.^{'} \right)\:{prove}\:{that}\:\left({R}×\mathrm{0},+^{'} ,.'\right)\:{it}\:{was}\:{ideal}\:{in}\:\left({R}×{Z},+^{'} ,.^{'} \right) \\ $$$${and}\:{prove}\:\left(\mathrm{0}×{Z},+,.\right){be}\:{isomorphic}\:{in}\:\left({Z},+,.\right) \\ $$$${and}\:{if}\:{a}\in{R}\:{identity}\:{element}\:\left({a}^{\mathrm{2}} ={a}\right){prove}\:{that}\:\left(−{a},\mathrm{1}\right){be}\:{identity}\:{element}\:{in}\:{the}\:{ring}\:\left({R}×{Z},+^{'} ,.^{'} \right) \\ $$$${pleas}\:{sir}\:{help}\:{me}\:{am}\:{neding}\:{this}\:{pleas}? \\ $$

Question Number 78340 Answers: 2 Comments: 0

Let a,b,c > 0 and c^2 = ((ab+bc+ca)/3) . Prove that ((a^3 +b^3 −2c^3 )/(a^3 +b^3 +c^3 )) ≤ 3(((a^2 +b^2 −2c^2 )/(a^2 +b^2 +c^2 )))

$$\mathrm{Let}\:\:{a},{b},{c}\:>\:\mathrm{0}\:\:\mathrm{and}\:\:{c}^{\mathrm{2}} \:=\:\frac{{ab}+{bc}+{ca}}{\mathrm{3}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} −\mathrm{2}{c}^{\mathrm{3}} }{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }\:\leqslant\:\mathrm{3}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right) \\ $$

Question Number 78335 Answers: 1 Comments: 0

find the solution of (x/((x−2)^3 +(x−3)^3 −1)) ≥ 0

$${find}\:{the}\:{solution}\:{of} \\ $$$$\frac{{x}}{\left({x}−\mathrm{2}\right)^{\mathrm{3}} +\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{1}}\:\geqslant\:\mathrm{0} \\ $$

Question Number 78334 Answers: 0 Comments: 1

∫_0 ^(π/2) [cos^2 (cos x)+sin^2 (sin x) ] dx

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\left[\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{cos}\:{x}\right)+\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:\right]\:{dx} \\ $$

Question Number 78333 Answers: 1 Comments: 0

prove by contradiction that (√(2 )) is irrational.

$$\mathrm{prove}\:\mathrm{by}\:\mathrm{contradiction}\:\mathrm{that}\:\sqrt{\mathrm{2}\:}\:\mathrm{is}\:\mathrm{irrational}. \\ $$

Question Number 78332 Answers: 0 Comments: 2

Find the values of k and n for which x^3 and higher powers of x are negligeble given that (1+kx)^n =1+2x+6x^2 .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{k}\:\mathrm{and}\:\mathrm{n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{x}^{\mathrm{3}} \mathrm{and}\:\mathrm{higher}\:\mathrm{powers}\:\mathrm{of}\:\mathrm{x}\:\mathrm{are}\:\mathrm{negligeble} \\ $$$$\mathrm{given}\:\mathrm{that}\:\left(\mathrm{1}+\mathrm{kx}\right)^{\mathrm{n}} =\mathrm{1}+\mathrm{2x}+\mathrm{6x}^{\mathrm{2}} . \\ $$

Question Number 78330 Answers: 0 Comments: 0

given the regular pyramid T.ABCD with a square base . length AB = 8 , TC = 6. point P is mid BC. if x is the angle between TP and BD. determine the value of cos x.

$${given}\:{the}\:{regular}\:{pyramid} \\ $$$${T}.{ABCD}\:{with}\:{a}\:{square}\:{base} \\ $$$$.\:{length}\:{AB}\:=\:\mathrm{8}\:,\:{TC}\:=\:\mathrm{6}.\:{point} \\ $$$${P}\:{is}\:{mid}\:{BC}.\:{if}\:{x}\:{is}\:{the}\:{angle}\: \\ $$$${between}\:{TP}\:{and}\:{BD}.\:{determine} \\ $$$${the}\:{value}\:{of}\:\mathrm{cos}\:{x}. \\ $$

Question Number 78325 Answers: 0 Comments: 3

lim_(x→∞) (((5x^2 +3x)/(5x^2 −2x)))^(3x−1)

$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}}\right)^{\mathrm{3}{x}−\mathrm{1}} \\ $$

Question Number 78324 Answers: 0 Comments: 2

lim_(x→0) ((1+sin 4x−cos 2x)/((√(x+1))−(√(1−x))))=

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{2}{x}}{\sqrt{{x}+\mathrm{1}}−\sqrt{\mathrm{1}−{x}}}= \\ $$

Question Number 78319 Answers: 0 Comments: 1

57+6h=16h−33

$$\mathrm{57}+\mathrm{6}{h}=\mathrm{16}{h}−\mathrm{33} \\ $$

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