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Question Number 183293 Answers: 1 Comments: 0

Help! A beam being lifted by two forces where F1makes an angle of 23° degrees with the y axis acts in the second quadrant F2 acts in the first quadrant making an angle of 32° degrees with the y axis and the resultant force is 67 N, determine F1 and F2.

$$\: \\ $$$$\:\mathrm{Help}! \\ $$$$\: \\ $$$$\:\mathrm{A}\:\mathrm{beam}\:\mathrm{being}\:\mathrm{lifted}\:\mathrm{by}\:\mathrm{two}\:\mathrm{forces}\:\mathrm{where}\: \\ $$$$\:\mathrm{F1makes}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{23}°\:\mathrm{degrees}\:\mathrm{with}\:\mathrm{the}\:\mathrm{y} \\ $$$$\:\mathrm{axis}\:\mathrm{acts}\:\mathrm{in}\:\mathrm{the}\:\mathrm{second}\:\mathrm{quadrant}\:\mathrm{F2}\:\mathrm{acts}\:\mathrm{in} \\ $$$$\:\mathrm{the}\:\mathrm{first}\:\mathrm{quadrant}\:\mathrm{making}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{32}° \\ $$$$\:\mathrm{degrees}\:\mathrm{with}\:\mathrm{the}\:\mathrm{y}\:\mathrm{axis}\:\mathrm{and}\:\mathrm{the}\:\mathrm{resultant} \\ $$$$\:\mathrm{force}\:\mathrm{is}\:\mathrm{67}\:\mathrm{N},\:\mathrm{determine}\:\mathrm{F1}\:\mathrm{and}\:\mathrm{F2}. \\ $$$$\: \\ $$

Question Number 183285 Answers: 0 Comments: 0

Question Number 183281 Answers: 2 Comments: 1

Question Number 183279 Answers: 1 Comments: 0

∫ ((sin x+cos x)/((tan x−cot x)^3 )) dx =?

$$\:\:\int\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\left(\mathrm{tan}\:{x}−\mathrm{cot}\:{x}\right)^{\mathrm{3}} }\:{dx}\:=? \\ $$

Question Number 183272 Answers: 0 Comments: 0

Question Number 183270 Answers: 1 Comments: 5

Question Number 183269 Answers: 0 Comments: 0

Question Number 183247 Answers: 0 Comments: 0

∫_0 ^(π/2) cos(t)ln(tant)dt

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left({t}\right){ln}\left({tant}\right){dt} \\ $$

Question Number 183246 Answers: 3 Comments: 0

Question Number 183243 Answers: 1 Comments: 0

∫ (((√(sin x)) −(√(cos x)))/( (√(sin x)) + (√(cos x)))) dx =?

$$\:\:\int\:\frac{\sqrt{\mathrm{sin}\:{x}}\:−\sqrt{\mathrm{cos}\:{x}}}{\:\sqrt{\mathrm{sin}\:{x}}\:+\:\sqrt{\mathrm{cos}\:{x}}}\:{dx}\:=? \\ $$

Question Number 183241 Answers: 1 Comments: 0

Find the equation of the line which is tangent to the parabola y^2 =12x and forms an angle of 45° with the line y=3x−4.

$${Find}\:{the}\:{equation}\:{of}\:{the}\:{line}\:{which} \\ $$$${is}\:{tangent}\:{to}\:{the}\:{parabola}\:{y}^{\mathrm{2}} =\mathrm{12}{x} \\ $$$${and}\:{forms}\:{an}\:{angle}\:{of}\:\mathrm{45}°\:{with}\: \\ $$$${the}\:{line}\:{y}=\mathrm{3}{x}−\mathrm{4}. \\ $$

Question Number 183240 Answers: 1 Comments: 0

find the value of cofficent μ in the following system from the determinat: 2x_1 +μx_2 +x_3 =0 (μ−1)x_1 −x_2 +2x_3 =0 4x_1 +x^2 +4x^3 =0

$${find}\:{the}\:{value}\:{of}\:{cofficent}\:\mu\:{in}\:{the}\:{following} \\ $$$${system}\:{from}\:{the}\:{determinat}: \\ $$$$\mathrm{2}{x}_{\mathrm{1}} +\mu{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\mu−\mathrm{1}\right){x}_{\mathrm{1}} −{x}_{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{4}{x}_{\mathrm{1}} +{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} =\mathrm{0} \\ $$

Question Number 183239 Answers: 0 Comments: 0

determine eigenvalues and digonalize by row operation [(4,(−9),6,(12)),(9,(−1),4,6),(2,(−11),8,(16)),((−1),( 3),0,(−1)) ]

$${determine}\:{eigenvalues}\:{and}\:{digonalize} \\ $$$${by}\:{row}\:{operation} \\ $$$$\begin{bmatrix}{\mathrm{4}}&{−\mathrm{9}}&{\mathrm{6}}&{\mathrm{12}}\\{\mathrm{9}}&{−\mathrm{1}}&{\mathrm{4}}&{\mathrm{6}}\\{\mathrm{2}}&{−\mathrm{11}}&{\mathrm{8}}&{\mathrm{16}}\\{−\mathrm{1}}&{\:\:\:\:\mathrm{3}}&{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix} \\ $$$$ \\ $$

Question Number 183236 Answers: 1 Comments: 2

dererminer la valeur x?

$${dererminer}\:\:{la}\:{valeur}\:{x}? \\ $$

Question Number 183227 Answers: 4 Comments: 1

Question Number 183216 Answers: 1 Comments: 2

Question Number 183211 Answers: 1 Comments: 0

y^((iv)) +16y^((iii)) +9y^((ii)) +256y^((i)) +256y=0 M.m

$$\mathrm{y}^{\left(\mathrm{iv}\right)} +\mathrm{16y}^{\left(\mathrm{iii}\right)} +\mathrm{9y}^{\left(\mathrm{ii}\right)} +\mathrm{256y}^{\left(\mathrm{i}\right)} +\mathrm{256y}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 183210 Answers: 1 Comments: 0

(d^3 y/dx^3 )+4(d^2 y/dx^2 )+(dy/dx)−6y=0 M.m

$$\frac{\mathrm{d}^{\mathrm{3}} \mathrm{y}}{\mathrm{dx}^{\mathrm{3}} }+\mathrm{4}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{6y}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 183209 Answers: 1 Comments: 0

Solve the Differential equation below (d^3 y/dx^3 )+8(d^2 y/dx^2 )+12(dy/dx)=0 M.m

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{Differential}\:\mathrm{equation}\:\mathrm{below} \\ $$$$\frac{\mathrm{d}^{\mathrm{3}} \mathrm{y}}{\mathrm{dx}^{\mathrm{3}} }+\mathrm{8}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{12}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 183205 Answers: 2 Comments: 0

Find the coefficient of x^(11) in (2x^2 +x−3)^6 .

$$\:\:{Find}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{11}} \:{in}\: \\ $$$$\:\left(\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{3}\right)^{\mathrm{6}} \:. \\ $$

Question Number 183202 Answers: 2 Comments: 0

ax^3 +bx^2 +c=0 x_1 = x_2 = x_3 =

$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{c}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} = \\ $$$${x}_{\mathrm{2}} = \\ $$$${x}_{\mathrm{3}} = \\ $$

Question Number 183295 Answers: 2 Comments: 1

Question Number 183185 Answers: 1 Comments: 0

Question Number 183181 Answers: 2 Comments: 1

prove that (0/0)=1

$${prove}\:{that}\:\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{1} \\ $$

Question Number 183174 Answers: 0 Comments: 2

In the given figure E is the mid point of AB. IF the area of ΔEBF is 8cm^2 .find the area of the parallelogram ABCD.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{given}\:\mathrm{figure}\:\mathrm{E}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}\:\mathrm{point}\:\mathrm{of}\:\mathrm{AB}. \\ $$$$\:\mathrm{IF}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{EBF}\:\mathrm{is}\:\mathrm{8cm}^{\mathrm{2}} .\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallelogram}\:\mathrm{ABCD}. \\ $$

Question Number 183165 Answers: 1 Comments: 0

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