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AlgebraQuestion and Answers: Page 363

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

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 =?