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AlgebraQuestion and Answers: Page 251

Question Number 109063 Answers: 0 Comments: 1

Question Number 108309 Answers: 2 Comments: 0

((△BeMath△)/∴) Given (√(5+(√(9+2(√(15)))))) +(√(5−(√(9+2(√(15)))))) = x find the value of (x−(1/x))^2

$$\:\:\:\:\:\frac{\bigtriangleup\mathcal{B}{e}\mathcal{M}{ath}\bigtriangleup}{\therefore} \\ $$$${Given}\:\sqrt{\mathrm{5}+\sqrt{\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}}}\:+\sqrt{\mathrm{5}−\sqrt{\mathrm{9}+\mathrm{2}\sqrt{\mathrm{15}}}}\:=\:{x} \\ $$$${find}\:{the}\:{value}\:{of}\:\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \\ $$

Question Number 108295 Answers: 3 Comments: 0

((BobHans)/(βo♭)) (1) { (((x+y)(x^2 −y^2 ) = 9)),(((x−y)(x^2 +y^2 ) = 5)) :} find the solution (2) x (dy/dx) = x^2 +y^2 when x=1 give y = 2

$$\:\:\:\:\frac{\mathbb{B}\mathrm{ob}\mathbb{H}\mathrm{ans}}{\beta\mathrm{o}\flat} \\ $$$$\:\left(\mathrm{1}\right)\begin{cases}{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{9}}\\{\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\:=\:\mathrm{5}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\mathrm{when}\:\mathrm{x}=\mathrm{1}\:\mathrm{give}\:\mathrm{y}\:=\:\mathrm{2}\: \\ $$

Question Number 108291 Answers: 2 Comments: 0

((∥ BeMath ∥)/(°∫ dx°)) (1) Given (x+(√(1+x^2 )))(y+(√(1+y^2 )))=1 find (x+y)^2

$$\:\:\:\frac{\parallel\:\mathcal{B}{e}\mathcal{M}{ath}\:\parallel}{°\int\:{dx}°} \\ $$$$\left(\mathrm{1}\right)\:{Given}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$${find}\:\left({x}+{y}\right)^{\mathrm{2}} \: \\ $$

Question Number 108316 Answers: 3 Comments: 0

((▽BeMath▽)/△) If x^4 +x^2 = ((11)/5) , find the value of Ω = (((x+1)/(x−1)))^(1/3) + (((x−1)/(x+1)))^(1/3)

$$\:\:\:\:\frac{\bigtriangledown\mathcal{B}{e}\mathcal{M}{ath}\bigtriangledown}{\bigtriangleup} \\ $$$${If}\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} \:=\:\frac{\mathrm{11}}{\mathrm{5}}\:,\:{find}\:{the}\:{value}\:{of} \\ $$$$\Omega\:=\:\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}\:+\:\sqrt[{\mathrm{3}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\: \\ $$

Question Number 108270 Answers: 0 Comments: 1

If x, y, z > −1, show that ((1 + x^2 )/(1 + y + z^2 )) + ((1 + y^2 )/(1 + z + x^2 )) + ((1 + z^2 )/(1 + x + y^2 )) ≥ 2

$$\mathrm{If}\:{x},\:{y},\:{z}\:>\:−\mathrm{1},\:\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{1}\:+\:{x}^{\mathrm{2}} }{\mathrm{1}\:+\:{y}\:+\:{z}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}\:+\:{y}^{\mathrm{2}} }{\mathrm{1}\:+\:{z}\:+\:{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}\:+\:{z}^{\mathrm{2}} }{\mathrm{1}\:+\:{x}\:+\:{y}^{\mathrm{2}} }\:\geqslant\:\mathrm{2} \\ $$

Question Number 108238 Answers: 4 Comments: 0

y=e^x ln(sin2x) (dy/dx)=??

$${y}={e}^{{x}} {ln}\left({sin}\mathrm{2}{x}\right)\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=?? \\ $$

Question Number 108237 Answers: 3 Comments: 0

y=(√(x^2 +1))−ln((1/x)+(√(1+(1/x^2 )))) (dy/dx)=?

$${y}=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{ln}\left(\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\right) \\ $$$$\frac{{dy}}{{dx}}=? \\ $$

Question Number 108219 Answers: 0 Comments: 3

Question Number 107999 Answers: 3 Comments: 5

Question Number 107976 Answers: 1 Comments: 0

Rewrite cos6xcos 4x as a sum or difference

$${Rewrite}\:\mathrm{cos6}{x}\mathrm{cos}\:\mathrm{4}{x}\:{as}\:{a}\:{sum}\:{or} \\ $$$${difference} \\ $$

Question Number 107947 Answers: 1 Comments: 0

Question Number 107945 Answers: 3 Comments: 0

((○BeMath○)/(∧⌣∧)) { ((x^4 +(1/x^4 ) = 23)),((x^3 −(1/x^3 ) = ?)) :}

$$\:\:\:\:\:\:\frac{\circ\mathbb{B}{e}\mathbb{M}{ath}\circ}{\wedge\smile\wedge} \\ $$$$\:\:\:\begin{cases}{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:=\:\mathrm{23}}\\{{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:?}\end{cases} \\ $$

Question Number 107867 Answers: 0 Comments: 0

Question Number 107794 Answers: 2 Comments: 0

Question Number 107779 Answers: 1 Comments: 3

Question Number 107690 Answers: 1 Comments: 0

“BeMath“ Let the complex number z satisfies the equation 3(z−1)= i(z+1) (1) find z in the form a+bi where a,b ∈R (2) find the value of ∣z∣ and ∣z−z^∗ ∣

$$\:\:\:\:\:\:\:\:``\mathcal{B}{e}\mathcal{M}{ath}`` \\ $$$${Let}\:{the}\:{complex}\:{number}\:{z}\:{satisfies}\:{the} \\ $$$${equation}\:\mathrm{3}\left({z}−\mathrm{1}\right)=\:{i}\left({z}+\mathrm{1}\right)\: \\ $$$$\left(\mathrm{1}\right)\:{find}\:{z}\:{in}\:{the}\:{form}\:{a}+{bi}\:{where}\:{a},{b}\:\in\mathbb{R}\: \\ $$$$\left(\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\mid{z}\mid\:{and}\:\mid{z}−{z}^{\ast} \mid\: \\ $$$$ \\ $$

Question Number 107674 Answers: 2 Comments: 0

f(x)=(√x)(√x) D_f =???

$${f}\left({x}\right)=\sqrt{{x}}\sqrt{{x}}\:\:\:\:\:\:\:{D}_{{f}} =??? \\ $$

Question Number 107673 Answers: 3 Comments: 0

Question Number 107486 Answers: 3 Comments: 1

∦BeMath∦ (2/5)+(5/(25))+(8/(125))+((11)/(625))+((14)/(3125))+... = ?

$$\:\:\:\:\:\:\nparallel\mathcal{B}{e}\mathcal{M}{ath}\nparallel \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{5}}{\mathrm{25}}+\frac{\mathrm{8}}{\mathrm{125}}+\frac{\mathrm{11}}{\mathrm{625}}+\frac{\mathrm{14}}{\mathrm{3125}}+...\:=\:? \\ $$

Question Number 107484 Answers: 2 Comments: 0

∦BeMath∦ (√(x+(√(x+(√(x+(√(x+...)))))))) = (√(4(√(4(√(4(√(4...)))))))) x=?

$$\:\:\:\:\:\nparallel\mathcal{B}{e}\mathcal{M}{ath}\nparallel \\ $$$$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}}}\:=\:\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}\sqrt{\mathrm{4}...}}}} \\ $$$${x}=?\: \\ $$

Question Number 107483 Answers: 2 Comments: 0

⋇JS⋇ ∣1+(1/x) ∣−∣x−3∣ > 2 find solution set. (A) 3−(√(10)) < x < 2−(√3) ; x≠0 (B) 3−(√(10)) < x < 3+(√(10)) ; x≠0 (C) 3−(√(10)) < x < 2+(√(10)) ; x≠0 (D) 2+(√(10)) < x < 3+(√(10)) ; x≠0 (E) none of these

$$\:\:\:\:\:\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$$\:\:\:\:\:\mid\mathrm{1}+\frac{\mathrm{1}}{{x}}\:\mid−\mid{x}−\mathrm{3}\mid\:>\:\mathrm{2}\: \\ $$$${find}\:{solution}\:{set}. \\ $$$$\left({A}\right)\:\mathrm{3}−\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{2}−\sqrt{\mathrm{3}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({B}\right)\:\mathrm{3}−\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{3}+\sqrt{\mathrm{10}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({C}\right)\:\mathrm{3}−\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{2}+\sqrt{\mathrm{10}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({D}\right)\:\mathrm{2}+\sqrt{\mathrm{10}}\:<\:{x}\:<\:\mathrm{3}+\sqrt{\mathrm{10}}\:;\:{x}\neq\mathrm{0} \\ $$$$\left({E}\right)\:{none}\:{of}\:{these}\: \\ $$

Question Number 107454 Answers: 2 Comments: 0

Given the function f(x) = ((x + 3)/(x−2)) and g(x) = (1/2)xe^x (1) Find the centre of symmetry of f. (2) Define the monotony of g and if possible draw a variation table for g(x). (3) Sketch the function g(x) (4) determine if f and g intersect.

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:=\:\frac{{x}\:+\:\mathrm{3}}{{x}−\mathrm{2}}\:\mathrm{and}\:\mathrm{g}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{xe}^{{x}} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{symmetry}\:\mathrm{of}\:{f}. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Define}\:\mathrm{the}\:\mathrm{monotony}\:\mathrm{of}\:\mathrm{g}\:\mathrm{and}\:\mathrm{if}\:\mathrm{possible}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{variation} \\ $$$$\mathrm{table}\:\mathrm{for}\:\mathrm{g}\left({x}\right). \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Sketch}\:\mathrm{the}\:\mathrm{function}\:\mathrm{g}\left({x}\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{determine}\:\mathrm{if}\:{f}\:\mathrm{and}\:\mathrm{g}\:\mathrm{intersect}. \\ $$

Question Number 107438 Answers: 1 Comments: 0

Question Number 107761 Answers: 0 Comments: 2

Question Number 107385 Answers: 0 Comments: 7

⋰BeMath⋰ Given 6x^2 −6px+14p−2=0 has the roots are u & v where u,v ∉Z If u,v ≥ 1 , then the value of ∣u−v∣ . (a)14 (b)15 (c)16 (d)17 (e) 18

$$\:\:\:\:\:\iddots\mathcal{B}{e}\mathcal{M}{ath}\iddots \\ $$$${Given}\:\mathrm{6}{x}^{\mathrm{2}} −\mathrm{6}{px}+\mathrm{14}{p}−\mathrm{2}=\mathrm{0} \\ $$$${has}\:{the}\:{roots}\:{are}\:\:{u}\:\&\:{v}\:{where}\:{u},{v}\:\notin\mathbb{Z} \\ $$$${If}\:{u},{v}\:\geqslant\:\mathrm{1}\:,\:{then}\:{the}\:{value}\:{of}\:\mid{u}−{v}\mid\:. \\ $$$$\left({a}\right)\mathrm{14}\:\:\:\:\:\left({b}\right)\mathrm{15}\:\:\:\:\:\left({c}\right)\mathrm{16}\:\:\:\:\:\left({d}\right)\mathrm{17}\:\:\:\left({e}\right)\:\mathrm{18} \\ $$

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