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AlgebraQuestion and Answers: Page 67

Question Number 187233 Answers: 1 Comments: 1

Find x if : x^4 +a=x ∀ (0<a<(2/9))

$${Find}\:{x}\:{if}\::\:\:\:{x}^{\mathrm{4}} +{a}={x}\:\:\:\:\forall\:\left(\mathrm{0}<{a}<\frac{\mathrm{2}}{\mathrm{9}}\right)\:\:\: \\ $$

Question Number 187209 Answers: 0 Comments: 1

Question Number 187166 Answers: 0 Comments: 0

x^3 =x+c let x=((mt)/(1−t)) m^3 t^3 =mt(1−t)^2 +c(1−t)^3 ⇒ (m^3 −m−c)t^3 +(2m−3c)t^2 +(3c−m)t−c=0 t^3 +At^2 +Bt+C=0 let AB=C ⇒ (2m−3c)(m−3c)=c(m^3 −m−c) ⇒ m^3 −(2/c)m^2 −8m+10c=0 ...

$${x}^{\mathrm{3}} ={x}+{c} \\ $$$${let}\:\:{x}=\frac{{mt}}{\mathrm{1}−{t}} \\ $$$${m}^{\mathrm{3}} {t}^{\mathrm{3}} ={mt}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} +{c}\left(\mathrm{1}−{t}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left({m}^{\mathrm{3}} −{m}−{c}\right){t}^{\mathrm{3}} +\left(\mathrm{2}{m}−\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\left(\mathrm{3}{c}−{m}\right){t}−{c}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +{At}^{\mathrm{2}} +{Bt}+{C}=\mathrm{0} \\ $$$${let}\:\:{AB}={C} \\ $$$$\Rightarrow\:\left(\mathrm{2}{m}−\mathrm{3}{c}\right)\left({m}−\mathrm{3}{c}\right)={c}\left({m}^{\mathrm{3}} −{m}−{c}\right) \\ $$$$\Rightarrow\:{m}^{\mathrm{3}} −\frac{\mathrm{2}}{{c}}{m}^{\mathrm{2}} −\mathrm{8}{m}+\mathrm{10}{c}=\mathrm{0} \\ $$$$... \\ $$

Question Number 187124 Answers: 2 Comments: 0

(a/x)=(b/y)=(c/z)=(1/3) , a−2b+c=2 and −2y+z=1 x=? An altered form of q#187020 (In this case solveable)

$$\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{\mathrm{1}}{\mathrm{3}}\:, \\ $$$${a}−\mathrm{2}{b}+{c}=\mathrm{2}\:\:{and}\:\:−\mathrm{2}{y}+{z}=\mathrm{1}\:\:\:\: \\ $$$${x}=? \\ $$$${An}\:{altered}\:{form}\:{of}\:\:{q}#\mathrm{187020} \\ $$$$\left({In}\:{this}\:{case}\:{solveable}\right) \\ $$

Question Number 187008 Answers: 2 Comments: 0

(a/3)=(b/4)=(c/5) 3a+c=42 b=? how is solution

$$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{a}+{c}=\mathrm{42}\:\:\:\:\:\:\:\:{b}=? \\ $$$${how}\:{is}\:{solution} \\ $$

Question Number 186941 Answers: 0 Comments: 1

((sinx)/x^b )=((Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))x^(2n+1) )/x^b ) prove that

$$\frac{{sinx}}{{x}^{{b}} }=\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}^{{b}} } \\ $$$${prove}\:{that} \\ $$

Question Number 186929 Answers: 3 Comments: 0

If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?

$$\:\:{If}\:\begin{cases}{{B}+{R}+{P}=−\mathrm{1}}\\{{B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\mathrm{17}}\\{{B}^{\mathrm{3}} +{R}^{\mathrm{3}} +{P}^{\mathrm{3}} =\mathrm{11}}\end{cases} \\ $$$$\:{then}\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} \:=? \\ $$

Question Number 186867 Answers: 2 Comments: 0

Question Number 186864 Answers: 2 Comments: 0

if AB,BA and BB three tow digits natural numbers if (((AB+BA))/(BB))=4 then find the maximum volue of (A+B)=?

$${if}\:\:{AB},{BA}\:{and}\:{BB}\:\:{three}\:\:{tow}\:{digits}\:{natural} \\ $$$${numbers}\:{if}\:\:\frac{\left({AB}+{BA}\right)}{{BB}}=\mathrm{4} \\ $$$${then}\:{find}\:{the}\:{maximum}\:{volue}\:{of} \\ $$$$\left({A}+{B}\right)=? \\ $$

Question Number 186772 Answers: 0 Comments: 0

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Question Number 186701 Answers: 2 Comments: 2

Question Number 186690 Answers: 1 Comments: 0

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 = ?

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

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