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Question Number 226149 Answers: 0 Comments: 0

Question Number 226144 Answers: 2 Comments: 1

6^x + 6^y = 42 x + y =3 Find x and y ?

$$\mathrm{6}^{{x}} \:+\:\mathrm{6}^{{y}} \:=\:\mathrm{42} \\ $$$${x}\:+\:{y}\:=\mathrm{3} \\ $$$${Find}\:{x}\:{and}\:{y}\:? \\ $$

Question Number 226118 Answers: 2 Comments: 0

Question Number 226117 Answers: 3 Comments: 0

Question Number 226115 Answers: 0 Comments: 3

can we find the black area (the shadow of earth)on the moon during moon eclipse after time t of its starting? Given: Moon radius=R_m Earth rafius=R_e Sun radius=R_s ω_e and ω_m given(both anticlockwise) ω_m ⇒Earth is center ω_e ⇒Sun is center Sun Earth distance(center)=L_(S−E) Earth Moon distance(center)=L_(E−M)

$${can}\:{we}\:{find}\:{the}\:{black}\:{area} \\ $$$$\left({the}\:{shadow}\:{of}\:{earth}\right){on}\:{the}\:{moon}\:{during} \\ $$$${moon}\:{eclipse}\:{after}\:{time}\:{t}\:{of} \\ $$$${its}\:{starting}? \\ $$$${Given}: \\ $$$${Moon}\:{radius}={R}_{{m}} \\ $$$${Earth}\:{rafius}={R}_{{e}} \\ $$$${Sun}\:{radius}={R}_{{s}} \\ $$$$\omega_{{e}} \:{and}\:\omega_{{m}} \:{given}\left({both}\:{anticlockwise}\right) \\ $$$$\omega_{{m}} \Rightarrow{Earth}\:{is}\:{center} \\ $$$$\omega_{{e}} \Rightarrow{Sun}\:{is}\:{center} \\ $$$${Sun}\:{Earth}\:{distance}\left({center}\right)={L}_{{S}−{E}} \\ $$$${Earth}\:{Moon}\:{distance}\left({center}\right)={L}_{{E}−{M}} \\ $$

Question Number 226112 Answers: 2 Comments: 0

Question Number 226111 Answers: 1 Comments: 0

Show that the equation (1/(sinθ+cosθ)) + (1/(sinθ−cosθ)) = 1 may be express in the form a(sinθ)^2 +bsinθ+c=0 where a b and c are constants to be found.

$${Show}\:{that}\:{the}\:{equation} \\ $$$$\frac{\mathrm{1}}{{sin}\theta+{cos}\theta}\:+\:\frac{\mathrm{1}}{{sin}\theta−{cos}\theta}\:=\:\mathrm{1} \\ $$$${may}\:{be}\:{express}\:{in}\:{the}\:{form} \\ $$$${a}\left({sin}\theta\right)^{\mathrm{2}} +{bsin}\theta+{c}=\mathrm{0}\:{where}\:{a}\:{b}\: \\ $$$${and}\:{c}\:{are}\:{constants}\:{to}\:{be}\:{found}. \\ $$

Question Number 226053 Answers: 1 Comments: 0

The coefficient of x^2 in the expansion of (1+ (2/p)x)^5 + (1+px)^6 is 70. Find the possible values of the constant p.

$${The}\:{coefficient}\:{of}\:{x}^{\mathrm{2}} \:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\mathrm{1}+\:\left(\mathrm{2}/{p}\right){x}\right)^{\mathrm{5}} \:+\:\left(\mathrm{1}+{px}\right)^{\mathrm{6}} \:{is}\:\mathrm{70}. \\ $$$${Find}\:{the}\:{possible}\:{values}\:{of}\:{the} \\ $$$${constant}\:{p}. \\ $$

Question Number 226045 Answers: 0 Comments: 0

Question Number 226015 Answers: 1 Comments: 0

Question Number 225993 Answers: 0 Comments: 0

Question Number 225934 Answers: 1 Comments: 0

{ ((⌈ ((8−2x)/3) ⌉ ; x≥ 0)),((⌊ ((3x−1)/4) ⌋ ; x<0)) :}. − −1)+

$$\:\: \begin{cases}{\lceil\:\frac{\mathrm{8}−\mathrm{2}{x}}{\mathrm{3}}\:\rceil\:;\:{x}\geqslant\:\mathrm{0}}\\{\lfloor\:\frac{\mathrm{3}{x}−\mathrm{1}}{\mathrm{4}}\:\rfloor\:;\:{x}<\mathrm{0}}\end{cases}. \\ $$$$\left.\:\: − −\mathrm{1}\right)+\: \\ $$$$ \\ $$

Question Number 225932 Answers: 1 Comments: 0

If, ((by+cz)/(b^2 +c^2 ))=((cz+ax)/(c^2 +a^2 ))=((ax+by)/(a^2 +b^2 )) then prove that, (x/a)=(y/b)=(z/c)

$$\:{If},\:\frac{{by}+{cz}}{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{cz}+{ax}}{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}+{by}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\:{then}\:{prove}\:{that},\:\frac{{x}}{{a}}=\frac{{y}}{{b}}=\frac{{z}}{{c}} \\ $$

Question Number 225939 Answers: 2 Comments: 0

Question Number 225938 Answers: 6 Comments: 0

Question Number 225840 Answers: 1 Comments: 1

Question Number 225837 Answers: 3 Comments: 0

Show that, log(√(7(√(7(√(7(√(7....α)))))))) =1

$${Show}\:{that},\:{log}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}....\alpha}}}}\:=\mathrm{1} \\ $$

Question Number 225810 Answers: 0 Comments: 0

Prove that in any triangle: ((4R)/r) ≥ ((w_a w_b w_c )/(h_a h_b h_c )) ∙ ((1/a) + (1/b))∙((√a) + (√b))^2

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{any}\:\mathrm{triangle}: \\ $$$$\frac{\mathrm{4R}}{\mathrm{r}}\:\geqslant\:\frac{\mathrm{w}_{\boldsymbol{\mathrm{a}}} \:\mathrm{w}_{\boldsymbol{\mathrm{b}}} \:\mathrm{w}_{\boldsymbol{\mathrm{c}}} }{\mathrm{h}_{\boldsymbol{\mathrm{a}}} \:\mathrm{h}_{\boldsymbol{\mathrm{b}}} \:\mathrm{h}_{\boldsymbol{\mathrm{c}}} }\:\centerdot\:\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\right)\centerdot\left(\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\right)^{\mathrm{2}} \\ $$

Question Number 225613 Answers: 7 Comments: 0

Question Number 225503 Answers: 2 Comments: 0

find x∈C for r∈R\{0} (−r)^x =r

$${find}\:{x}\in\mathbb{C}\:{for}\:{r}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\left(−{r}\right)^{{x}} ={r} \\ $$

Question Number 225392 Answers: 2 Comments: 1

Question Number 225330 Answers: 1 Comments: 0

if (fogoh)(x)=cos^2 (x+9) then f(x)=? , g(x)=? , h(x)=?

$${if}\:\:\left({fogoh}\right)\left({x}\right)={cos}^{\mathrm{2}} \left({x}+\mathrm{9}\right) \\ $$$${then}\:\:\:{f}\left({x}\right)=?\:,\:\:{g}\left({x}\right)=?\:\:,\:{h}\left({x}\right)=? \\ $$

Question Number 225323 Answers: 1 Comments: 0

Find: (((3 + 2 (5)^(1/4) )/(3 - 2 (5)^(1/4) )))^(1/4) . (((5)^(1/4) - 1)/( (5)^(1/4) + 1)) = ?

$$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{3}\:+\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}\:-\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} .\:\:\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}\:-\:\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\mathrm{1}}\:\:=\:? \\ $$

Question Number 225282 Answers: 2 Comments: 4

Q225146 here if the wedge and the ground had no friction it would start going → what is acc. at time t? μ=kx

$${Q}\mathrm{225146} \\ $$$${here}\:{if}\:{the}\:{wedge}\:{and}\:{the} \\ $$$${ground}\:\:{had}\:{no}\:{friction} \\ $$$${it}\:{would}\:{start}\:{going}\:\rightarrow \\ $$$${what}\:{is}\:{acc}.\:{at}\:{time}\:{t}? \\ $$$$\mu={kx} \\ $$

Question Number 225230 Answers: 1 Comments: 3

tg^4 10°+tg^4 50°+tg^4 70°=?

$$\:\:\:{tg}^{\mathrm{4}} \mathrm{10}°+{tg}^{\mathrm{4}} \mathrm{50}°+{tg}^{\mathrm{4}} \mathrm{70}°=? \\ $$

Question Number 225240 Answers: 1 Comments: 0

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