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AlgebraQuestion and Answers: Page 222

Question Number 118936 Answers: 3 Comments: 0

show by recurrence that: for n ∈ N^∗ , 2^(6n−5) +3^(2n ) is divisible by 11.

$${show}\:{by}\:{recurrence}\:{that}: \\ $$$${for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}\:} \:{is}\:{divisible}\:{by} \\ $$$$\mathrm{11}. \\ $$

Question Number 118907 Answers: 0 Comments: 0

where can u fond a formula for denesting ramanujan type roots?

$${where}\:{can}\:{u}\:{fond}\:{a}\:{formula}\:{for} \\ $$$${denesting}\:{ramanujan}\:{type}\:{roots}? \\ $$

Question Number 118896 Answers: 1 Comments: 0

Question Number 118849 Answers: 1 Comments: 2

Question Number 118790 Answers: 0 Comments: 0

Show by recurence that (a+b)^n =Σ_(k=0 ) ^n C_n ^k ×a^k ×b^(n−k)

$$\mathrm{Show}\:\mathrm{by}\:\mathrm{recurence}\:\mathrm{that} \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}\:} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} ×\mathrm{a}^{\mathrm{k}} ×\mathrm{b}^{\mathrm{n}−\mathrm{k}} \\ $$

Question Number 118780 Answers: 2 Comments: 0

z and z′ ∈ C . show that: 1. zz′^(−) =z^− ×z′^(−) 2. ((z/(z′)))^(−) =(z^− /(z′^(−) ))

$$\mathrm{z}\:\mathrm{and}\:\mathrm{z}'\:\in\:\mathbb{C}\:. \\ $$$$\mathrm{show}\:\mathrm{that}: \\ $$$$\mathrm{1}.\:\:\:\:\:\:\overline {\mathrm{zz}'}=\overset{−} {\mathrm{z}}×\overline {\mathrm{z}'} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\:\overline {\left(\frac{\mathrm{z}}{\mathrm{z}'}\right)}=\frac{\overset{−} {\mathrm{z}}}{\overline {\mathrm{z}'}} \\ $$$$ \\ $$

Question Number 118777 Answers: 1 Comments: 0

lim_(n→∞) (((n!)/n^n ))^(1/n) =lim_(n→∞) ((((√(2nπ)) n^n )/(n^n ×e^n )))^(1/n) ⇒lim_(n→∞) (1/e)((√(2nπ)))^(1/n) =(1/e)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}{n}\pi}\:{n}^{{n}} }{{n}^{{n}} ×{e}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{e}}\left(\sqrt{\mathrm{2}{n}\pi}\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{{e}} \\ $$

Question Number 118740 Answers: 1 Comments: 1

f(x+2)+f(x−1)=2x^2 +14 f(x)=?

$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{14} \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 118712 Answers: 4 Comments: 0

Prove the following inequalities: 1)(((n+1)/2))^n >n! for ∀n∈N^∗ ,n>1 2)∣sinnx∣≤n∣sinx∣ for ∀n∈N^∗

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequalities}: \\ $$$$\left.\mathrm{1}\right)\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} >\mathrm{n}!\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} ,\mathrm{n}>\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\mid\mathrm{sinnx}\mid\leqslant\mathrm{n}\mid\mathrm{sinx}\mid\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} \\ $$

Question Number 118651 Answers: 1 Comments: 2

Question Number 118636 Answers: 1 Comments: 2

Question Number 118634 Answers: 0 Comments: 3

Prove that the equation of the circle passing through the points of intersection of these two curves: y=1+(c/x) ; y=x^2 (c < (2/(3(√3))) ) is (x−(c/2))^2 +(y−1)^2 =1+(c^2 /4) .

$${Prove}\:{that}\:{the}\:{equation}\:{of}\:{the}\:{circle} \\ $$$${passing}\:{through}\:{the}\:{points}\:{of} \\ $$$${intersection}\:{of}\:{these}\:{two}\:{curves}: \\ $$$$\:\:{y}=\mathrm{1}+\frac{{c}}{{x}}\:;\:\:{y}={x}^{\mathrm{2}} \:\:\:\:\:\left({c}\:<\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\right)\: \\ $$$${is}\:\:\:\left({x}−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}\:\:. \\ $$

Question Number 118482 Answers: 1 Comments: 0

If the tangents at the end of a focal chord of parabola meet the tangent at the vertex in C,D.prove that CD substends a right angle at the focus

$$\mathrm{If}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{focal}\:\:\mathrm{chord}\:\mathrm{of} \\ $$$$\mathrm{parabola}\:\mathrm{meet}\:\mathrm{the} \\ $$$$\mathrm{tangent}\:\mathrm{at}\:\mathrm{the}\:\:\mathrm{vertex} \\ $$$$\mathrm{in}\:\mathrm{C},\mathrm{D}.\mathrm{prove}\:\mathrm{that}\:\mathrm{CD} \\ $$$$\mathrm{substends}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angle} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{focus} \\ $$

Question Number 118452 Answers: 2 Comments: 3

Question: 2^x +2^(2x+1) +1=y^2 solve this equation if x,y𝛆Z

$$\boldsymbol{{Question}}: \\ $$$$\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{2}^{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}} +\mathrm{1}=\boldsymbol{{y}}^{\mathrm{2}} \:\:\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{equation}}\:\boldsymbol{{if}} \\ $$$$\boldsymbol{{x}},\boldsymbol{{y}\epsilon}\mathbb{Z} \\ $$

Question Number 118435 Answers: 2 Comments: 0

If 4 (x^9 )^(1/(4 )) −9 (x^9 )^(1/(8 )) + 4 = 0 , then (x^9 )^(1/(4 )) + (x^(−9) )^(1/(4 )) =?

$${If}\:\mathrm{4}\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:−\mathrm{9}\:\sqrt[{\mathrm{8}\:}]{{x}^{\mathrm{9}} }\:+\:\mathrm{4}\:=\:\mathrm{0}\:,\:{then}\: \\ $$$$\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{−\mathrm{9}} }\:=? \\ $$

Question Number 118391 Answers: 2 Comments: 0

show that if ^ a^2 +b^2 can be divised by 7, a+b can also be divised by 7.

$${show}\:{that}\:{if}\:\:^{} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{can}\:{be}\:{divised} \\ $$$${by}\:\mathrm{7},\:{a}+{b}\:{can}\:{also}\:{be}\:{divised}\:{by}\:\mathrm{7}. \\ $$

Question Number 118371 Answers: 0 Comments: 1

old problem question 118120 tan tan x =tan 3x −tan 2x let t=tan x (1) tan t =((t^5 +2t^3 +t)/(3t^4 −4t^2 +1)) for t≥0 we get (approximating) t_0 =0 t_1 ≈1.28941477 t_2 ≈4.17629616 t_3 ≈7.49316173 t_4 ≈10.7303610 t_5 ≈13.9285293 ... x=nπ+arctan t let n=0 to stay in the first period 0≤t<+∞ ⇒ 0≤arctan t <(π/2) ⇒ (1) has infinite solutions for 0≤x<(π/2) graphically this is easy to see, plot these: f_1 (t)=tan t f_2 (t)=((t(t^4 +2t^2 +1))/(3t^4 −4t^2 +1))=(t/3)+((2t(5t^2 +1))/(3(3t^4 −4t^2 +1))) ⇒ g(t)=(1/3)t is asymptote of f_1 (t) and obviously tan t =at with a∈R has infinite solutions

$$\mathrm{old}\:\mathrm{problem}\:{question}\:\mathrm{118120} \\ $$$$\mathrm{tan}\:\mathrm{tan}\:{x}\:=\mathrm{tan}\:\mathrm{3}{x}\:−\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\mathrm{let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\mathrm{tan}\:{t}\:=\frac{{t}^{\mathrm{5}} +\mathrm{2}{t}^{\mathrm{3}} +{t}}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{for}\:{t}\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{approximating}\right) \\ $$$${t}_{\mathrm{0}} =\mathrm{0} \\ $$$${t}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{28941477} \\ $$$${t}_{\mathrm{2}} \approx\mathrm{4}.\mathrm{17629616} \\ $$$${t}_{\mathrm{3}} \approx\mathrm{7}.\mathrm{49316173} \\ $$$${t}_{\mathrm{4}} \approx\mathrm{10}.\mathrm{7303610} \\ $$$${t}_{\mathrm{5}} \approx\mathrm{13}.\mathrm{9285293} \\ $$$$... \\ $$$${x}={n}\pi+\mathrm{arctan}\:{t} \\ $$$$\mathrm{let}\:{n}=\mathrm{0}\:\mathrm{to}\:\mathrm{stay}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{period} \\ $$$$\mathrm{0}\leqslant{t}<+\infty\:\Rightarrow\:\mathrm{0}\leqslant\mathrm{arctan}\:{t}\:<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{1}\right)\:\mathrm{has}\:\mathrm{infinite}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{graphically}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see},\:\mathrm{plot}\:\mathrm{these}: \\ $$$${f}_{\mathrm{1}} \left({t}\right)=\mathrm{tan}\:{t} \\ $$$${f}_{\mathrm{2}} \left({t}\right)=\frac{{t}\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}}{\mathrm{3}}+\frac{\mathrm{2}{t}\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow\:{g}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{3}}{t}\:\mathrm{is}\:\mathrm{asymptote}\:\mathrm{of}\:{f}_{\mathrm{1}} \left({t}\right) \\ $$$$\mathrm{and}\:\mathrm{obviously}\:\mathrm{tan}\:{t}\:={at}\:\mathrm{with}\:{a}\in\mathbb{R}\:\mathrm{has} \\ $$$$\mathrm{infinite}\:\mathrm{solutions} \\ $$

Question Number 118346 Answers: 0 Comments: 0

Let 30 furniture sets arrive at two city stations A and B ,15 sets for each station All funiture sets need to be delivered to two furniture stores C and D,and 10 sets must be delivered to store C,and to store D−20.It is known that delivery one furniture set from the station A to the stores C and D costs 1 and 3 monetary units,and from station B−2 and 5 units respectively,.It is necessary to draw out such a transportation plan that the cost of transportation is the lowest

$$\mathrm{Let}\:\mathrm{30}\:\mathrm{furniture}\:\mathrm{sets}\:\mathrm{arrive}\:\mathrm{at}\:\mathrm{two}\:\mathrm{city} \\ $$$$\mathrm{stations}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:,\mathrm{15}\:\mathrm{sets}\:\mathrm{for}\:\mathrm{each}\:\mathrm{station} \\ $$$$\mathrm{All}\:\mathrm{funiture}\:\mathrm{sets}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{delivered} \\ $$$$\mathrm{to}\:\mathrm{two}\:\mathrm{furniture}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D},\mathrm{and}\:\mathrm{10} \\ $$$$\mathrm{sets}\:\mathrm{must}\:\mathrm{be}\:\mathrm{delivered}\:\mathrm{to}\:\mathrm{store}\:\mathrm{C},\mathrm{and} \\ $$$$\mathrm{to}\:\mathrm{store}\:\mathrm{D}−\mathrm{20}.\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{delivery} \\ $$$$\mathrm{one}\:\mathrm{furniture}\:\mathrm{set}\:\mathrm{from}\:\mathrm{the}\:\mathrm{station}\:\mathrm{A}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D}\:\mathrm{costs}\:\mathrm{1}\:\mathrm{and}\:\mathrm{3}\:\mathrm{monetary} \\ $$$$\mathrm{units},\mathrm{and}\:\mathrm{from}\:\mathrm{station}\:\mathrm{B}−\mathrm{2}\:\mathrm{and}\:\mathrm{5}\:\mathrm{units} \\ $$$$\:\mathrm{respectively},.\mathrm{It}\:\mathrm{is}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{draw} \\ $$$$\mathrm{out}\:\mathrm{such}\:\mathrm{a}\:\mathrm{transportation}\:\mathrm{plan}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{cost}\:\mathrm{of}\:\mathrm{transportation}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lowest} \\ $$

Question Number 118322 Answers: 2 Comments: 2

solve (1/x)+(1/y)+(1/z)=(3/4) with x,y,z∈N

$${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$

Question Number 118298 Answers: 0 Comments: 1

b, x, y, c are consecutive terms of a G.P. ∴ x = br, y = br^2 , c = br^3 A.M. between b and c is a ∴ ((b+c)/2) = a ⇒ 2a = b+c 2abc = (b+c)bc = b^2 c + bc^2 = b^2 (br^3 ) + b(br^3 )^2 = b^3 r^3 + b^3 r^6 = (br)^3 + (br^2 )^3 = x^3 + y^3 ∴ x^3 + y^3 = 2abc ←

Question Number 118286 Answers: 2 Comments: 0

What condition should be satisfied by the vectors a and b for the following relations to hold true :(a)∣a+b∣=∣a−b∣ ;(b)∣ a+b∣>∣ a−b∣;(c)∣a+ b∣< ∣a−b∣

Question Number 118280 Answers: 0 Comments: 1

(26) OA^(→) = a^→ = (((4.8)),((3.6)) ) , OB^(→) = b^→ = ((( 8)),((15)) ) a^→ . b^→ = (4.8)(8) + (3.6)(15) = 92.4^(→ ) a = (√(4.8^2 +3.6^2 )) = (√(23.04+12.96)) = (√(36)) = 6 b = (√(8^2 +15^2 )) = (√(64+225)) = (√(289)) = 17 a.b = 6 . 17 = 102 cos AOB = ((a^→ .b^→ )/(a.b)) = ((92.4)/(102)) = 0.9059 = cos 25.06° ∴ ∠AOB = 25.06°

$$\left(\mathrm{26}\right)\:\:\:\overset{\rightarrow} {\mathrm{OA}}\:=\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\mathrm{4}.\mathrm{8}}\\{\mathrm{3}.\mathrm{6}}\end{pmatrix}\:\:,\:\:\overset{\rightarrow} {\mathrm{OB}}\:=\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\mathrm{8}}\\{\mathrm{15}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\:\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{4}.\mathrm{8}\right)\left(\mathrm{8}\right)\:+\:\left(\mathrm{3}.\mathrm{6}\right)\left(\mathrm{15}\right)\:=\:\mathrm{92}.\overset{\rightarrow\:} {\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\sqrt{\mathrm{4}.\mathrm{8}^{\mathrm{2}} +\mathrm{3}.\mathrm{6}^{\mathrm{2}} }\:=\:\:\sqrt{\mathrm{23}.\mathrm{04}+\mathrm{12}.\mathrm{96}}\:=\:\sqrt{\mathrm{36}}\:=\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{64}+\mathrm{225}}\:=\:\sqrt{\mathrm{289}}\:=\:\mathrm{17} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{6}\:.\:\mathrm{17}\:=\:\mathrm{102} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\mathrm{AOB}\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{\mathrm{92}.\mathrm{4}}{\mathrm{102}}\:=\:\mathrm{0}.\mathrm{9059}\:=\:\mathrm{cos}\:\mathrm{25}.\mathrm{06}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\angle\mathrm{AOB}\:=\:\mathrm{25}.\mathrm{06}° \\ $$

Question Number 118277 Answers: 0 Comments: 0

(23) Let a^→ = ((( 7)),((24)) ) and b^→ = ((( 3)),((−4)) ) a^→ .b^→ = (7)(3) + (24)(−4) = 21−96 = −75 a = (√(7^2 +24^2 )) = (√(49+576)) = (√(625)) = 25 b = (√(3^2 +(−4)^2 )) = (√(9+16)) = (√(25)) = 5 a.b = 25 . 5 = 125 Let θ be the angle between a^→ and b^→ . cosθ = ((a^→ .b^→ )/(a.b)) = ((−75)/(125)) = −0.6 = cos126.87° ∴ θ = 126.87°

$$\left(\mathrm{23}\right)\:\mathrm{Let}\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\:\:\mathrm{7}}\\{\mathrm{24}}\end{pmatrix}\:\:\:\mathrm{and}\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{7}\right)\left(\mathrm{3}\right)\:+\:\left(\mathrm{24}\right)\left(−\mathrm{4}\right)\:=\:\mathrm{21}−\mathrm{96}\:=\:−\mathrm{75} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{49}+\mathrm{576}}\:=\:\sqrt{\mathrm{625}}\:=\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{9}+\mathrm{16}}\:=\:\sqrt{\mathrm{25}}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{25}\:.\:\mathrm{5}\:=\:\:\mathrm{125}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\overset{\rightarrow} {\mathrm{a}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{b}}. \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{cos}\theta\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{−\mathrm{75}}{\mathrm{125}}\:=\:−\mathrm{0}.\mathrm{6}\:=\:\mathrm{cos126}.\mathrm{87}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\theta\:=\:\mathrm{126}.\mathrm{87}°\:\:\:\: \\ $$

Question Number 118275 Answers: 1 Comments: 1

Question Number 118263 Answers: 0 Comments: 0

Question Number 118227 Answers: 1 Comments: 0

If x = (√(42−(√(42−(√(42−...)))))) y = (√(x+(√(x+(√(x+...)))))) z=(√(y.(√(y.(√(y.(√(y...)))))))) . Find x+y+z .

$${If}\:{x}\:=\:\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−\sqrt{\mathrm{42}−...}}} \\ $$$${y}\:=\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}} \\ $$$${z}=\sqrt{{y}.\sqrt{{y}.\sqrt{{y}.\sqrt{{y}...}}}}\:.\:{Find}\:{x}+{y}+{z}\:. \\ $$

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