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Question Number 188296 Answers: 1 Comments: 0

Question Number 188295 Answers: 0 Comments: 0

let S be the sets be the sequences of lenght 2018 whose terms are in the sets {1,2,3,4,5,6,10} and sum to 3860. prove that the cardinality of S is at most 2^(3860) ∙( ((2018)/(2048)))^(2018)

$$ \\ $$$$\:\:\:\:\boldsymbol{{let}}\:\:\boldsymbol{{S}}\:\boldsymbol{{be}}\:\boldsymbol{{the}}\:\boldsymbol{{sets}}\:\boldsymbol{{be}}\:\boldsymbol{{the}}\:\boldsymbol{{sequences}}\:\boldsymbol{{of}}\:\boldsymbol{{lenght}}\:\mathrm{2018}\:\:\: \\ $$$$\:\:\:\boldsymbol{{whose}}\:\boldsymbol{{terms}}\:\boldsymbol{{are}}\:\boldsymbol{{in}}\:\boldsymbol{{the}}\:\boldsymbol{{sets}}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{10}\right\}\:\boldsymbol{{and}}\:\boldsymbol{{sum}}\:\boldsymbol{{to}}\:\mathrm{3860}.\:\:\: \\ $$$$\:\:\:\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{cardinality}}\:\boldsymbol{{of}}\:\boldsymbol{{S}}\:\boldsymbol{{is}}\:\boldsymbol{{at}}\:\boldsymbol{{most}}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{3860}} \centerdot\left(\:\frac{\mathrm{2018}}{\mathrm{2048}}\right)^{\mathrm{2018}} \\ $$$$ \\ $$$$\:\:\:\: \\ $$

Question Number 188294 Answers: 0 Comments: 0

Question Number 188463 Answers: 2 Comments: 0

Question Number 188286 Answers: 1 Comments: 0

when sin(x)+cos(x)=a find sec(x)+csc(x)=?

$${when}\:\:\:\:\:\:{sin}\left({x}\right)+{cos}\left({x}\right)={a} \\ $$$${find}\:\:\:\:\:\:\:\:\:{sec}\left({x}\right)+{csc}\left({x}\right)=? \\ $$

Question Number 188284 Answers: 1 Comments: 1

a friend shared this challenging problem to me. 2^y ×y^2 +(2y)^(2y) =272 no inspection approach! thank you all.

$${a}\:{friend}\:{shared}\:{this}\:{challenging}\: \\ $$$${problem}\:{to}\:{me}. \\ $$$$\mathrm{2}^{\boldsymbol{{y}}} ×\boldsymbol{{y}}^{\mathrm{2}} +\left(\mathrm{2}\boldsymbol{{y}}\right)^{\mathrm{2}\boldsymbol{{y}}} \:=\mathrm{272} \\ $$$$\boldsymbol{{no}}\:\boldsymbol{{inspection}}\:\boldsymbol{{approach}}! \\ $$$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{all}}. \\ $$

Question Number 188281 Answers: 1 Comments: 0

1) ((18log10000+(√(43x)))/(12)) 2) ((√(43x))/(18log1000)) 3) 18log10000+(√(43x)) 4) ((18log10000)/( (√(43x)))) 5) log(sinx)+sin(log100) 6) log(sinx)+12 7) (√(x^2 −x+90)) 8) log(sin(π/4))+(1/( (√5))) Which two of the following questions can be polynomials?

$$\left.\mathrm{1}\right)\:\:\frac{\mathrm{18}{log}\mathrm{10000}+\sqrt{\mathrm{43}{x}}}{\mathrm{12}} \\ $$$$\left.\mathrm{2}\right)\:\frac{\sqrt{\mathrm{43}{x}}}{\mathrm{18}{log}\mathrm{1000}} \\ $$$$\left.\mathrm{3}\right)\:\:\mathrm{18}{log}\mathrm{10000}+\sqrt{\mathrm{43}{x}}\:\:\: \\ $$$$\left.\mathrm{4}\right)\:\:\:\frac{\mathrm{18}{log}\mathrm{10000}}{\:\sqrt{\mathrm{43}{x}}} \\ $$$$\left.\mathrm{5}\right)\:\:\:{log}\left({sinx}\right)+{sin}\left({log}\mathrm{100}\right) \\ $$$$\left.\mathrm{6}\right)\:\:\:{log}\left({sinx}\right)+\mathrm{12} \\ $$$$\left.\mathrm{7}\right)\:\:\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{90}} \\ $$$$\left.\mathrm{8}\right)\:\:{log}\left({sin}\frac{\pi}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$Which two of the following questions can be polynomials?

Question Number 188280 Answers: 0 Comments: 2

Prove that 1+2+3+4+..... = −(1/(12))

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+.....\:=\:−\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Question Number 188278 Answers: 1 Comments: 0

Question Number 188277 Answers: 0 Comments: 0

Question Number 188270 Answers: 1 Comments: 1

Question Number 188267 Answers: 2 Comments: 0

Question Number 188268 Answers: 1 Comments: 2

Question Number 188263 Answers: 0 Comments: 0

Question Number 188262 Answers: 1 Comments: 0

solve the equation; {: ((x + y +z = 30(√2))),((x − y − z = 7,5)),((x + y − z = (√(22)))) } x ; y ; z = ?? they form funny positions

$$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{equation}};\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\left.\begin{matrix}{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:+\boldsymbol{{z}}\:=\:\:\mathrm{30}\sqrt{\mathrm{2}}}\\{\boldsymbol{{x}}\:−\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\:\mathrm{7},\mathrm{5}}\\{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:−\:\boldsymbol{{z}}\:=\:\sqrt{\mathrm{22}}}\end{matrix}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}\:;\:\boldsymbol{{y}}\:;\:\boldsymbol{{z}}\:=\:?? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{they}\:{form}\:{funny}\:{positions}\: \\ $$$$ \\ $$

Question Number 188327 Answers: 3 Comments: 0

if the roots of 2x^2 −xn = 2x + m is 5, then find : 4n + m − 5

$${if}\:{the}\:{roots}\:{of}\:\:\mathrm{2}{x}^{\mathrm{2}} \:−{xn}\:=\:\mathrm{2}{x}\:+\:{m}\:\:{is}\:\mathrm{5}, \\ $$$$\:{then}\:{find}\::\:\mathrm{4}{n}\:+\:{m}\:−\:\mathrm{5}\: \\ $$$$\: \\ $$

Question Number 188260 Answers: 0 Comments: 0

ABCD is a rectangle such that ∣AB∣>∣BC∣ and O is the mid−point of DC, if ∣OB∣=0.1m and ∠BOC =𝛉, find an expression for the perimeter of the rectangle in terms of 𝛉. Find also, the values of R and 𝛃 for which the perimeter is Rcos(𝛉−𝛃). Deduce, the greatest possible value oc the perimeter.

$$\boldsymbol{{ABCD}}\:{is}\:{a}\:{rectangle}\:{such}\:{that} \\ $$$$\:\mid\boldsymbol{{AB}}\mid>\mid\boldsymbol{{BC}}\mid\:{and}\:{O}\:{is}\:{the}\:{mid}−{point} \\ $$$$\:{of}\:\boldsymbol{{DC}},\:{if}\:\mid\boldsymbol{{OB}}\mid=\mathrm{0}.\mathrm{1}{m}\:{and}\:\angle\boldsymbol{{BOC}}\:=\boldsymbol{\theta}, \\ $$$$\:{find}\:{an}\:{expression}\:{for}\:{the}\:{perimeter}\:{of} \\ $$$$\:{the}\:{rectangle}\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}.\:{Find}\:{also},\: \\ $$$${the}\:{values}\:{of}\:\boldsymbol{{R}}\:{and}\:\boldsymbol{\beta}\:{for}\:{which}\:{the}\: \\ $$$${perimeter}\:{is}\:\boldsymbol{{Rcos}}\left(\boldsymbol{\theta}−\boldsymbol{\beta}\right).\:{Deduce},\:{the}\: \\ $$$${greatest}\:{possible}\:{value}\:{oc}\:{the}\:{perimeter}. \\ $$

Question Number 188259 Answers: 1 Comments: 0

Question Number 188258 Answers: 0 Comments: 0

Solve y=x(y′)^2 −(1/(y′))

$${Solve}\: \\ $$$${y}={x}\left({y}'\right)^{\mathrm{2}} −\frac{\mathrm{1}}{{y}'} \\ $$$$ \\ $$

Question Number 188251 Answers: 0 Comments: 0

Question Number 188250 Answers: 1 Comments: 0

Question Number 188248 Answers: 0 Comments: 0

(x^3 −y−3x)[(x^3 −3x)^2 −y^2 ]=200 (x^3 +y−3x)[(x^3 −3x)^2 +y^2 ]=600 solved in R

$$\left({x}^{\mathrm{3}} −{y}−\mathrm{3}{x}\right)\left[\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} \right]=\mathrm{200} \\ $$$$\left({x}^{\mathrm{3}} +{y}−\mathrm{3}{x}\right)\left[\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \right]=\mathrm{600} \\ $$$${solved}\:{in}\:{R} \\ $$

Question Number 188247 Answers: 1 Comments: 0

Prove that (1) 5555^(2222) +2222^(5555) divisible by 7 (2) 3^(105) +4^(105) divisible by 7

$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{5555}^{\mathrm{2222}} +\mathrm{2222}^{\mathrm{5555}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{3}^{\mathrm{105}} +\mathrm{4}^{\mathrm{105}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\: \\ $$

Question Number 188239 Answers: 3 Comments: 0

The perimeter of a triangle is 16 units. How many triangles with integer sides can be made?

$$ \\ $$The perimeter of a triangle is 16 units. How many triangles with integer sides can be made?

Question Number 188226 Answers: 1 Comments: 0

If Ω = Σ_(cyc) ((sin(A − (π/6)))/(cos(B − (π/6))cos(C − (π/6)))) in △ABC Solve for real numbers: x^4 − 4Ωx^3 + 6Ωx^2 − 4Ωx + 1 = 0

$$\mathrm{If}\:\:\:\Omega\:=\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{sin}\left(\mathrm{A}\:−\:\frac{\pi}{\mathrm{6}}\right)}{\mathrm{cos}\left(\mathrm{B}\:−\:\frac{\pi}{\mathrm{6}}\right)\mathrm{cos}\left(\mathrm{C}\:−\:\frac{\pi}{\mathrm{6}}\right)}\:\:\:\mathrm{in}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{x}^{\mathrm{4}} \:−\:\mathrm{4}\Omega\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{6}\Omega\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4}\Omega\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 188224 Answers: 1 Comments: 0

If Ω = Σ_(n=1) ^∞ (Π_(k=2) ^∞ ((k^3 − 1)/(k^3 + 1)))^n Solve for complex numbees: z^4 + 3z^3 + Ωz^2 + 3z + 1 = 0

$$\mathrm{If}\:\:\:\Omega\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\underset{\boldsymbol{\mathrm{k}}=\mathrm{2}} {\overset{\infty} {\prod}}\:\frac{\mathrm{k}^{\mathrm{3}} \:−\:\mathrm{1}}{\mathrm{k}^{\mathrm{3}} \:+\:\mathrm{1}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbees}: \\ $$$$\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{3z}^{\mathrm{3}} \:+\:\Omega\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{3z}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

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