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AlgebraQuestion and Answers: Page 86

Question Number 186682 Answers: 0 Comments: 0

Prove the following set identities 1) A∪(B∪C)=(A∪B)∪C 2) A∪∅=A 3) A∩(B∪C)=(A∩B)∪(A∩C)

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{set}\:\mathrm{identities} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right)=\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{A}\cup\varnothing=\mathrm{A} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{A}\cap\left(\mathrm{B}\cup\mathrm{C}\right)=\left(\mathrm{A}\cap\mathrm{B}\right)\cup\left(\mathrm{A}\cap\mathrm{C}\right) \\ $$

Question Number 186732 Answers: 2 Comments: 0

Question Number 186657 Answers: 1 Comments: 0

Question Number 186616 Answers: 1 Comments: 0

Question Number 186627 Answers: 2 Comments: 1

Question Number 186582 Answers: 0 Comments: 0

(((10x^3 −c)/(4x^4 −x+1))+x^2 )^2 =x(x^3 +1)

$$\left(\frac{\mathrm{10}{x}^{\mathrm{3}} −{c}}{\mathrm{4}{x}^{\mathrm{4}} −{x}+\mathrm{1}}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}\left({x}^{\mathrm{3}} +\mathrm{1}\right) \\ $$

Question Number 186572 Answers: 1 Comments: 0

f={(1,2),(4,6),(5,8)} and g={(1,7),(2,4),(5,8)} the domain of f(g(x)) which number is not include? 1)1 2)2 3)5 4)all corect how is the solution?

$${f}=\left\{\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{4},\mathrm{6}\right),\left(\mathrm{5},\mathrm{8}\right)\right\}\:\:{and}\:\:\:{g}=\left\{\left(\mathrm{1},\mathrm{7}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{5},\mathrm{8}\right)\right\} \\ $$$${the}\:{domain}\:{of}\:\:\:{f}\left({g}\left({x}\right)\right)\:\:\:{which}\:{number}\:{is}\:{not}\:{include}? \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{1}\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\right){all}\:{corect} \\ $$$${how}\:{is}\:{the}\:{solution}? \\ $$

Question Number 186571 Answers: 2 Comments: 0

If x=−(1/((1+m)))((2/3)±(√((4/9)±((ia(√m))/(12))))) where (1+m)^2 =am and that 9a(a+16)^2 =(16)^3 Find real x.

$${If}\:\:\:{x}=−\frac{\mathrm{1}}{\left(\mathrm{1}+{m}\right)}\left(\frac{\mathrm{2}}{\mathrm{3}}\pm\sqrt{\frac{\mathrm{4}}{\mathrm{9}}\pm\frac{{ia}\sqrt{{m}}}{\mathrm{12}}}\right) \\ $$$${where}\:\:\:\left(\mathrm{1}+{m}\right)^{\mathrm{2}} ={am} \\ $$$$\:\:\:\:{and}\:{that}\:\:\mathrm{9}{a}\left({a}+\mathrm{16}\right)^{\mathrm{2}} =\left(\mathrm{16}\right)^{\mathrm{3}} \\ $$$${Find}\:{real}\:{x}. \\ $$

Question Number 186544 Answers: 0 Comments: 1

Question Number 186540 Answers: 0 Comments: 0

Complex Numbers Prove that 1. ∣Z_1 +Z_2 ∣≤∣Z_1 ∣+∣Z_2 ∣ 2. ∣Z_1 −Z_2 ∣≥∣Z_1 ∣−∣Z_2 ∣

$${Complex}\:{Numbers} \\ $$$${Prove}\:{that} \\ $$$$\:\mathrm{1}.\:\mid{Z}_{\mathrm{1}} +{Z}_{\mathrm{2}} \mid\leq\mid{Z}_{\mathrm{1}} \mid+\mid{Z}_{\mathrm{2}} \mid \\ $$$$\mathrm{2}.\:\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid\geq\mid{Z}_{\mathrm{1}} \mid−\mid{Z}_{\mathrm{2}} \mid \\ $$

Question Number 186503 Answers: 0 Comments: 1

function of , f (x) = ax + ∣ x ∣ is one to one .find ” a ” .

$$ \\ $$$$\:\:\:{function}\:{of}\:,\:{f}\:\left({x}\right)\:=\:{ax}\:\:+\:\mid\:{x}\:\mid\:{is}\:\:{one}\:{to}\:{one} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:.{find}\:\:\:\:''\:\:\:\:{a}\:\:\:\:''\:\:. \\ $$$$\:\: \\ $$

Question Number 186489 Answers: 1 Comments: 0

Question Number 186486 Answers: 0 Comments: 6

Question propose par Migma −−−−−−−−−−− calcul de X △ABC ∡ACB=40^° AB^2 =AC^2 +BC^2 −2AC×BCcos 40 AC=15 ; AB=X+4 ; BC=10 (X+4)^2 =15^2 +10^2 −300×cos 40 =325−229,81334 (X+4)^2 =95,18666706 posons Z=X+4 Z^2 =9,756^2 ⇒X+4=9,756 X=5,756

$${Question}\:{propose}\:{par} \\ $$$${Migma} \\ $$$$−−−−−−−−−−− \\ $$$${calcul}\:{de}\:{X} \\ $$$$ \\ $$$$\bigtriangleup{ABC}\:\:\:\:\:\:\measuredangle{ACB}=\mathrm{40}^{°} \\ $$$$\mathrm{AB}^{\mathrm{2}} =\mathrm{AC}^{\mathrm{2}} +\mathrm{BC}^{\mathrm{2}} −\mathrm{2AC}×\mathrm{BCcos}\:\mathrm{40}\:\:\:\: \\ $$$$\mathrm{AC}=\mathrm{15}\:\:;\:\:\:\:\mathrm{AB}=\mathrm{X}+\mathrm{4}\:;\:\:\:\mathrm{BC}=\mathrm{10} \\ $$$$\left(\mathrm{X}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{300}×\mathrm{cos}\:\mathrm{40} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{325}−\mathrm{229},\mathrm{81334} \\ $$$$\left({X}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{95},\mathrm{18666706} \\ $$$${posons}\:\:\:{Z}={X}+\mathrm{4} \\ $$$$\:\:\:{Z}^{\mathrm{2}} =\mathrm{9},\mathrm{756}^{\mathrm{2}} \:\:\:\:\Rightarrow\mathrm{X}+\mathrm{4}=\mathrm{9},\mathrm{756} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{X}=\mathrm{5},\mathrm{756} \\ $$$$ \\ $$

Question Number 186419 Answers: 3 Comments: 0

If , (( 1 − lo^ g_( 2) (x)))^(1/3) + ((1+^ log_( 2) (x)))^(1/3) −1=0 ⇒ x = ?

$$ \\ $$$$\:\:\mathrm{I}{f}\:,\:\sqrt[{\mathrm{3}}]{\:\mathrm{1}\:−\:{l}\overset{} {{o}g}_{\:\mathrm{2}} \left({x}\right)}\:\:+\:\sqrt[{\mathrm{3}}]{\mathrm{1}\overset{} {+}{log}_{\:\mathrm{2}} \left({x}\right)}\:−\mathrm{1}=\mathrm{0}\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\:{x}\:=\:?\:\:\:\: \\ $$

Question Number 186330 Answers: 0 Comments: 1

Question Number 186329 Answers: 1 Comments: 0

Question Number 186305 Answers: 0 Comments: 3

log (((3.2^ )/(3.1^ ))) find Characteristic?

$$\mathrm{log}\:\left(\frac{\mathrm{3}.\bar {\mathrm{2}}}{\mathrm{3}.\bar {\mathrm{1}}}\right)\:\:\:\:\:\:\:{find}\:\mathrm{Characteristic}? \\ $$

Question Number 186300 Answers: 1 Comments: 0

solve in R ⌊ 2log_( 8) (x) + (1/3) ⌋ = log_( 4) (x )+ (1/2)

$$\:\:\: \\ $$$$\:\:\:\:\:\mathrm{solve}\:\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\lfloor\:\:\mathrm{2log}_{\:\mathrm{8}} \left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\rfloor\:=\:\mathrm{log}_{\:\mathrm{4}} \left({x}\:\right)+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 186292 Answers: 0 Comments: 0

(x−p)(x^3 −x−(1/3))=0 x^4 −px^3 −x^2 +(p−(1/3))x+(p/3)=0 (x^2 +ax+h)(x^2 +bx+k)=0 a+b=−p h+k+ab=−1 bh+ak=p−(1/3) hk=(p/3) −−−−−− say ab=t −−−−−− ah+bk+p−(1/3)=p(1+t) bh+ak−(p−(1/3))=0 ⇒ (a−b)(h−k)=pt−p+(2/3) squaring (p^2 −4t){(1+t)^2 −((4p)/3)}=(pt−p+(2/3))^2 say t+1=z (p^2 +4−4z)(z^2 −((4p)/3))=(pz−2p+(2/3))^2 ⇒ −4z^3 +(p^2 +4)z^2 +((16pz)/3)−((4p)/3)(p^2 +4) =p^2 z^2 −4p(p−(1/3))z+4(p−(1/3))^2 ⇒ z^3 −z^2 −p(p+1)z+(p−(1/3))^2 +(p/3)(p^2 +4)=0 .....

Question Number 186285 Answers: 0 Comments: 2

Prove that: R (m , n) ≤ C_(m+n) ^m Here R states the Ramsey theory

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathrm{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$\mathrm{Here}\:\:\mathrm{R}\:\:\mathrm{states}\:\mathrm{the}\:\:\mathrm{Ramsey}\:\:\mathrm{theory} \\ $$

Question Number 186265 Answers: 1 Comments: 0

f(x)= (√( x −a)) + (√(3a −x)) with ( a>0) is given .If , f_( max) . f_( min) = (√(32)) find , ” a ” = ?

$$ \\ $$$$\:\:\:{f}\left({x}\right)=\:\sqrt{\:{x}\:−{a}}\:\:+\:\sqrt{\mathrm{3}{a}\:−{x}}\:\:\:{with}\:\left(\:{a}>\mathrm{0}\right) \\ $$$$\:\:\:\:{is}\:\:{given}\:.{If}\:\:,\:\:{f}_{\:{max}} \:.\:{f}_{\:{min}} \:=\:\sqrt{\mathrm{32}} \\ $$$$\:\:\:\:\:{find}\:\:,\:\:\:\:\:\:\:''\:\:\:{a}\:\:''\:\:=\:? \\ $$

Question Number 186256 Answers: 0 Comments: 1

Prove that: R (m , n) ≤ C_(m+n) ^m

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathbb{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathbb{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$ \\ $$

Question Number 186253 Answers: 1 Comments: 1