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Question Number 16990    Answers: 1   Comments: 4

If for positive integers a and b, a + b = (a/b) + (b/a), find a^2 + b^2 .

$$\mathrm{If}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:{a}\:\mathrm{and}\:{b}, \\ $$$${a}\:+\:{b}\:=\:\frac{{a}}{{b}}\:+\:\frac{{b}}{{a}},\:\mathrm{find}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} . \\ $$

Question Number 16969    Answers: 0   Comments: 0

Question Number 16935    Answers: 1   Comments: 0

A two-digit number has the property that the square of its tens digit plus ten times its units digit is equal to the square of its units digit plus ten times its tens digit. Find all two digit numbers which have this property, and are prime numbers.

$$\mathrm{A}\:\mathrm{two}-\mathrm{digit}\:\mathrm{number}\:\mathrm{has}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{its}\:\mathrm{tens}\:\mathrm{digit}\:\mathrm{plus} \\ $$$$\mathrm{ten}\:\mathrm{times}\:\mathrm{its}\:\mathrm{units}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{square}\:\mathrm{of}\:\mathrm{its}\:\mathrm{units}\:\mathrm{digit}\:\mathrm{plus}\:\mathrm{ten}\:\mathrm{times} \\ $$$$\mathrm{its}\:\mathrm{tens}\:\mathrm{digit}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{two}\:\mathrm{digit} \\ $$$$\mathrm{numbers}\:\mathrm{which}\:\mathrm{have}\:\mathrm{this}\:\mathrm{property},\:\mathrm{and} \\ $$$$\mathrm{are}\:\mathrm{prime}\:\mathrm{numbers}. \\ $$

Question Number 16909    Answers: 1   Comments: 0

(√x^x^6 ) = 729 , Find x

$$\sqrt{\mathrm{x}^{\mathrm{x}^{\mathrm{6}} } }\:=\:\mathrm{729}\:,\:\:\:\:\mathrm{Find}\:\mathrm{x} \\ $$

Question Number 16906    Answers: 1   Comments: 0

x^x = 100, find x.

$$\mathrm{x}^{\mathrm{x}} \:=\:\mathrm{100},\:\mathrm{find}\:\mathrm{x}. \\ $$

Question Number 16881    Answers: 2   Comments: 0

The quadratic polynomials p(x) = a(x − 3)^2 + bx + 1 and q(x) = 2x^2 + c(x − 2) + 13 are equal for all values of x. Find the values of a, b, and c.

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{polynomials} \\ $$$${p}\left({x}\right)\:=\:{a}\left({x}\:−\:\mathrm{3}\right)^{\mathrm{2}} \:+\:{bx}\:+\:\mathrm{1}\:\mathrm{and} \\ $$$${q}\left({x}\right)\:=\:\mathrm{2}{x}^{\mathrm{2}} \:+\:{c}\left({x}\:−\:\mathrm{2}\right)\:+\:\mathrm{13}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a}, \\ $$$${b},\:\mathrm{and}\:{c}. \\ $$

Question Number 16829    Answers: 0   Comments: 1

Solve for x 6(4^x + 9^x ) = (13.6)^x

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$\mathrm{6}\left(\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{9}^{\mathrm{x}} \right)\:=\:\left(\mathrm{13}.\mathrm{6}\right)^{\mathrm{x}} \\ $$

Question Number 16840    Answers: 0   Comments: 3

6/2(2+1)

$$\mathrm{6}/\mathrm{2}\left(\mathrm{2}+\mathrm{1}\right) \\ $$

Question Number 16823    Answers: 0   Comments: 2

solve for g. 6(4^x + g^x ) = 13.6^x

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{g}. \\ $$$$\mathrm{6}\left(\mathrm{4}^{\mathrm{x}} \:+\:\mathrm{g}^{\mathrm{x}} \right)\:=\:\mathrm{13}.\mathrm{6}^{\mathrm{x}} \\ $$

Question Number 16835    Answers: 1   Comments: 0

Question Number 16794    Answers: 1   Comments: 0

Question Number 16723    Answers: 1   Comments: 0

Solve: 2^x + 48 = 16x

$$\mathrm{Solve}:\:\mathrm{2}^{\mathrm{x}} \:+\:\mathrm{48}\:=\:\mathrm{16x} \\ $$

Question Number 16720    Answers: 0   Comments: 0

let x=tanθ ,so θ=tan^(−1) x given that tan^(−1) (√(1+x2−1/x))

$${let}\:{x}={tan}\theta\:,{so}\:\theta=\mathrm{tan}^{−\mathrm{1}} {x}\:{given}\:{that}\: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{1}+{x}\mathrm{2}−\mathrm{1}/{x}} \\ $$

Question Number 16699    Answers: 0   Comments: 8

Related to Q16675: Find the number of intersection points of graph sin x=(x/(10)). Let′s see sin x = (x/n) with n>1. For n≤1 there is one intersection point. Let x=2kπ+t with k∈N ∧ t∈[0,2π] sin x=sin t cos x=cos t we find the point on f(x)=sin x where its tangent is g(x)=(x/n). f′(x)=cos x=cos t g′(x)=(1/n) cos t=(1/n) t=cos^(−1) (1/n) sin t=(n/(√(n^2 +1))) so that f(x) intersects with g(x), ((sin x)/x)≥(1/n) ⇒n sin x≥x ⇒n sin t≥2kπ+t ⇒k≤((n sin t −t)/(2π))=(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π)) k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ number of intersecting points is m=2×2(k_(max) +1)−1=4k_(max) +3 for n=10 k_(max) =⌊(((n^2 /(√(n^2 +1)))−cos^(−1) (1/n))/(2π))⌋ =⌊((((10^2 )/(√(10^2 +1)))−cos^(−1) (1/(10)))/(2π))⌋=⌊1.35⌋=1 ⇒m=4×1+3=7 for n=20 k_(max) =⌊((((20^2 )/(√(20^2 +1)))−cos^(−1) (1/(20)))/(2π))⌋=⌊2.94⌋=2 ⇒m=4×2+3=11

$$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16675}: \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{points} \\ $$$$\mathrm{of}\:\mathrm{graph}\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{x}}{\mathrm{10}}. \\ $$$$ \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{see}\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{x}}{\mathrm{n}}\:\mathrm{with}\:\mathrm{n}>\mathrm{1}. \\ $$$$\mathrm{For}\:\mathrm{n}\leqslant\mathrm{1}\:\mathrm{there}\:\mathrm{is}\:\mathrm{one}\:\mathrm{intersection}\:\mathrm{point}. \\ $$$$ \\ $$$$\mathrm{Let}\:\mathrm{x}=\mathrm{2k}\pi+\mathrm{t}\:\mathrm{with}\:\mathrm{k}\in\mathbb{N}\:\wedge\:\mathrm{t}\in\left[\mathrm{0},\mathrm{2}\pi\right] \\ $$$$\mathrm{sin}\:\mathrm{x}=\mathrm{sin}\:\mathrm{t} \\ $$$$\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\mathrm{x}\:\mathrm{where}\:\mathrm{its} \\ $$$$\mathrm{tangent}\:\mathrm{is}\:\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{n}}. \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{cos}\:\mathrm{x}=\mathrm{cos}\:\mathrm{t} \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{cos}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{t}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{sin}\:\mathrm{t}=\frac{\mathrm{n}}{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{intersects}\:\mathrm{with}\:\mathrm{g}\left(\mathrm{x}\right), \\ $$$$\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\geqslant\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{x}\geqslant\mathrm{x} \\ $$$$\Rightarrow\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\geqslant\mathrm{2k}\pi+\mathrm{t} \\ $$$$\Rightarrow\mathrm{k}\leqslant\frac{\mathrm{n}\:\mathrm{sin}\:\mathrm{t}\:−\mathrm{t}}{\mathrm{2}\pi}=\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$ \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{intersecting}\:\mathrm{points}\:\mathrm{is} \\ $$$$\mathrm{m}=\mathrm{2}×\mathrm{2}\left(\mathrm{k}_{\mathrm{max}} +\mathrm{1}\right)−\mathrm{1}=\mathrm{4k}_{\mathrm{max}} +\mathrm{3} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{10} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{n}^{\mathrm{2}} }{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{n}}}{\mathrm{2}\pi}\rfloor \\ $$$$=\lfloor\frac{\frac{\mathrm{10}^{\mathrm{2}} }{\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{10}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{1}.\mathrm{35}\rfloor=\mathrm{1} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{1}+\mathrm{3}=\mathrm{7} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{20} \\ $$$$\mathrm{k}_{\mathrm{max}} =\lfloor\frac{\frac{\mathrm{20}^{\mathrm{2}} }{\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{1}}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{20}}}{\mathrm{2}\pi}\rfloor=\lfloor\mathrm{2}.\mathrm{94}\rfloor=\mathrm{2} \\ $$$$\Rightarrow\mathrm{m}=\mathrm{4}×\mathrm{2}+\mathrm{3}=\mathrm{11} \\ $$

Question Number 16632    Answers: 0   Comments: 0

e^(xy) + y^2 = sin(x + y) Express the variable y interms of x only.

$$\mathrm{e}^{\mathrm{xy}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{sin}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$$$\mathrm{Express}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{y}\:\mathrm{interms}\:\mathrm{of}\:\mathrm{x}\:\mathrm{only}. \\ $$

Question Number 16600    Answers: 1   Comments: 8

Solve simultaneously x^2 + y^2 = 61 .............. equation (i) x^3 − y^3 = 91 .............. equation (ii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{61}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{91}\:\:\:\:\:\:\:\:\:..............\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$

Question Number 16589    Answers: 1   Comments: 0

x^2 −∣3x+2∣+x ≥ 0 find range of values of x agreeing with above inequality.

$$\:\:\:\mathrm{x}^{\mathrm{2}} −\mid\mathrm{3x}+\mathrm{2}\mid+\mathrm{x}\:\geqslant\:\mathrm{0} \\ $$$$\:\mathrm{find}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{agreeing} \\ $$$$\mathrm{with}\:\mathrm{above}\:\mathrm{inequality}. \\ $$

Question Number 16542    Answers: 0   Comments: 1

if (a/(∣Z_2 −Z_3 ∣))=(b/(∣Z_3 −Z_1 ∣))=(c/(∣Z_1 −Z_2 ∣)) Then.. find.. (a^2 /(Z_2 −Z_3 )) + (b^2 /(Z_3 −Z_1 )) + (c^2 /(Z_1 −Z_2 )) a) 0 b) 1 c) 2 d)N.O.T

$${if}\:\frac{{a}}{\mid{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} \mid}=\frac{{b}}{\mid{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} \mid}=\frac{{c}}{\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid}\:\:{Then}.. \\ $$$${find}..\:\frac{{a}^{\mathrm{2}} }{{Z}_{\mathrm{2}} −{Z}_{\mathrm{3}} }\:+\:\frac{{b}^{\mathrm{2}} }{{Z}_{\mathrm{3}} −{Z}_{\mathrm{1}} }\:+\:\frac{{c}^{\mathrm{2}} }{{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} }\: \\ $$$$ \\ $$$$\left.{a}\right)\:\mathrm{0} \\ $$$$\left.{b}\right)\:\mathrm{1} \\ $$$$\left.{c}\right)\:\mathrm{2} \\ $$$$\left.{d}\right){N}.{O}.{T} \\ $$

Question Number 16540    Answers: 2   Comments: 0

if ∣Z∣=1 Then ((1+Z)/(1+Z^ )) is equal to... a) Z b) Z^ c) Z+Z^ d) N.O.T

$${if}\:\mid{Z}\mid=\mathrm{1}\:{Then}\:\frac{\mathrm{1}+{Z}}{\mathrm{1}+\bar {{Z}}}\:\:{is}\:{equal}\:{to}... \\ $$$$\left.{a}\right)\:{Z}\:\: \\ $$$$\left.{b}\right)\:\:\bar {{Z}} \\ $$$$\left.{c}\right)\:{Z}+\bar {{Z}} \\ $$$$\left.{d}\right)\:{N}.{O}.{T} \\ $$

Question Number 16519    Answers: 2   Comments: 0

Solve: ∣2 − x∣ − 2 ∣x + 1∣ < 1

$$\mathrm{Solve}: \\ $$$$\mid\mathrm{2}\:−\:\mathrm{x}\mid\:−\:\mathrm{2}\:\mid\mathrm{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{1} \\ $$

Question Number 16373    Answers: 1   Comments: 1

if Σ_(k=0) ^(200) i^k +Π_(p=1) ^(50) i^p =x+iy then..(x,y)is... a. (0,1) b. (1,−1) c. (2,3) d. (4,8)

$${if}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}{i}^{{k}} +\underset{{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}{i}^{{p}} ={x}+{iy}\:{then}..\left({x},{y}\right){is}... \\ $$$${a}.\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${b}.\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${c}.\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${d}.\:\left(\mathrm{4},\mathrm{8}\right) \\ $$

Question Number 16238    Answers: 0   Comments: 0

we have a^5 +b^5 =1 and u^5 +v^5 =1 find value a^3 u^5 +b^3 v^5 =?

$${we}\:{have}\:{a}^{\mathrm{5}} +{b}^{\mathrm{5}} =\mathrm{1}\:{and}\:{u}^{\mathrm{5}} +{v}^{\mathrm{5}} =\mathrm{1} \\ $$$${find}\:{value}\:\:{a}^{\mathrm{3}} {u}^{\mathrm{5}} +{b}^{\mathrm{3}} {v}^{\mathrm{5}} =? \\ $$

Question Number 16116    Answers: 0   Comments: 0

Question Number 16115    Answers: 0   Comments: 0

Question Number 16374    Answers: 1   Comments: 0

The Value of the sum.. Σ_(n=1) ^(13) (i^n +i^(n+1) ), Where i=(√(−1 )) is.. (a.) i (b.) i−1 (c.) −i (d.) 0

$${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\ $$

Question Number 16047    Answers: 0   Comments: 1

number of real solution of equation a^(2008) + b^(2008) = 2008ab−2006

$${number}\:{of}\:{real}\:{solution}\:{of}\:{equation}\:\:\:\:{a}^{\mathrm{2008}} \:+\:\:{b}^{\mathrm{2008}} \:\:=\:\mathrm{2008}{ab}−\mathrm{2006}\:\:\:\:\:\:\:\:\: \\ $$

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