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Question Number 68835 Answers: 1 Comments: 4

reposting this because it′s not yet solved correctly. I criticize that most of you people do not test if your solutions fit the given equations in many cases x^2 +1−(√(x^3 +x))=6x please determine (1) how many real solutions (2) how many complex solutions we can expect (3) solve it (4) test your solutions

$$\mathrm{reposting}\:\mathrm{this}\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{yet}\:\mathrm{solved} \\ $$$$\mathrm{correctly}.\:\mathrm{I}\:\mathrm{criticize}\:\mathrm{that}\:\mathrm{most}\:\mathrm{of}\:\mathrm{you}\:\mathrm{people} \\ $$$$\mathrm{do}\:\mathrm{not}\:\mathrm{test}\:\mathrm{if}\:\mathrm{your}\:\mathrm{solutions}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equations}\:\mathrm{in}\:\mathrm{many}\:\mathrm{cases} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\mathrm{1}−\sqrt{{x}^{\mathrm{3}} +{x}}=\mathrm{6}{x} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{determine} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{how}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{how}\:\mathrm{many}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{expect} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{solve}\:\mathrm{it} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{test}\:\mathrm{your}\:\mathrm{solutions} \\ $$

Question Number 68831 Answers: 1 Comments: 3

The square ABCD has side equal to 1 and the distance AP is (1/8). Calculate the side of the equilateral triangle PMN inscribed in the square.

$$\mathrm{The}\:\mathrm{square}\:{ABCD}\:\mathrm{has}\:\mathrm{side}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{distance}\:{AP}\:\:\mathrm{is}\:\:\frac{\mathrm{1}}{\mathrm{8}}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:{PMN}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{square}. \\ $$

Question Number 68830 Answers: 0 Comments: 0

please can someone check question 68728

$${please}\:{can}\:{someone}\:{check}\:{question}\:\:\:\mathrm{68728} \\ $$

Question Number 68825 Answers: 1 Comments: 0

Question Number 68808 Answers: 1 Comments: 0

Question Number 69018 Answers: 0 Comments: 1

Question Number 68783 Answers: 1 Comments: 5

Question Number 68782 Answers: 0 Comments: 0

Let d_n be the determinant of the n×n matrix whose entries, from left to right and then from top to bottom, are cos 1, cos 2, ..., cos n^2 . (For example, d_3 = determinant (((cos 1 cos 2 cos 3)),((cos 4 cos 5 cos 6)),((cos 7 cos 8 cos 9))). The argument of cos is always in radians not degrees.) Evalue lim_(n→∞) d_(n.)

$$\mathrm{Let}\:{d}_{{n}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{of}\:\mathrm{the}\:{n}×{n} \\ $$$$\mathrm{matrix}\:\mathrm{whose}\:\mathrm{entries},\:\mathrm{from}\:\mathrm{left}\:\mathrm{to}\:\mathrm{right} \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{from}\:\mathrm{top}\:\mathrm{to}\:\mathrm{bottom},\:\mathrm{are} \\ $$$${cos}\:\mathrm{1},\:{cos}\:\mathrm{2},\:...,\:{cos}\:{n}^{\mathrm{2}} .\:\left(\mathrm{For}\:\mathrm{example},\right. \\ $$$${d}_{\mathrm{3}} =\begin{vmatrix}{{cos}\:\mathrm{1}\:\:{cos}\:\mathrm{2}\:\:{cos}\:\mathrm{3}}\\{{cos}\:\mathrm{4}\:\:{cos}\:\mathrm{5}\:\:{cos}\:\mathrm{6}}\\{{cos}\:\mathrm{7}\:\:{cos}\:\mathrm{8}\:\:{cos}\:\mathrm{9}}\end{vmatrix}. \\ $$$$\mathrm{The}\:\mathrm{argument}\:\mathrm{of}\:{cos}\:\mathrm{is}\:\mathrm{always}\:\mathrm{in}\:\mathrm{radians} \\ $$$$\left.\mathrm{not}\:\mathrm{degrees}.\right)\: \\ $$$$\mathrm{Evalue}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{d}_{{n}.} \\ $$

Question Number 68775 Answers: 1 Comments: 0

Question Number 68774 Answers: 0 Comments: 0

Question Number 68773 Answers: 0 Comments: 0

Question Number 68768 Answers: 1 Comments: 1

((2x−1))^(1/3) +((x−1))^(1/3) = 1

$$\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:=\:\mathrm{1} \\ $$

Question Number 68767 Answers: 0 Comments: 1

Question Number 68761 Answers: 1 Comments: 1

Two arcs having their centers on a circle are cutting each other at a single point inside the circle and thus dividing the circle in four regions. If the arcs cut each other in a:b & c:d ratios what is the ratio between four regions of the circle when the circle has radius R,the arc divided in a:b has radius r_1 and the arc divided in c:d has radius r_2 .

Question Number 68740 Answers: 1 Comments: 1