scrutinizer

1 Questions
125 Answers

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1036

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+259

How to convert Mbps in MBps

Hi, I wanted to use the sci calc, but don't know how the command line should look like for converting data speed of 9.98 Mbps in MBps. Could you give me an advise?

Thanks

scrutinizer 30 Nov 2014

+259

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Using the formulas of the sum trnasformation into the product we get (cos(80) + cos(40)) - cos(20) = 2*[cos(120/2)]*[cos(40/2)] - cos(20) = 2*[cos (60)]*[cos(20)] - cos(20) = cos (20)[2cos 60 - 1] =
= cos(20)[2*0.5 - 1] = cos(20)*0 = 0

cos(80)+cos(40)-cos(20)

scrutinizer2 Sep 2013

+259

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You ought to rely on the fact that the sum of the angles of quadrilaterial is equal to 360 degrees. Let x be the value of one angle, y be the one of the other, z - the 3rd and w - the 4th. We have the firrst equation x + y + z + w = 360. Taken separately x = y - 11, z = 2y, w = 1.5z. Got the system {x + y + z + w = 360, x = y - 11, z = 2y, w = 1.5z}. Let's solve it for any chosen variable, y for example. In terms of y w = 1.5*2y = 3y, z = 2y, x = y - 11. Inserting them into the 1st equation, y - 11 + y + 2y + 3y = 360 => 7y = 371 = > y = 53 (degrees). Inserting into the subsequent equations we find x = 53 - 11 = 42, z = 2*53 = 106, w = 3*53 = 159 degrees. As you could possibly notice the found values satisfy the requirement of the four angles to form an ascending sequance.

[input]solve(x + y + z + w = 360, x = y - 11, z = 2y, w = 1.5z, x,y,z,w)[/input]

scrutinizer2 Sep 2013

+259

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If you have two variables then to find them you should have TWO equations system.

scrutinizer2 Sep 2013

+259

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[input]solve(180/3 (3*x/2)^2 x= 960,x)[/input]

scrutinizer31 Agu 2013

+259

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1- 8/9 = 9/9 - 8/9 = 1/9

scrutinizer31 Agu 2013

+259

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Let x be the saving of one brother and y be the saving of other brother. Their combined saving is, then, x + y = 55. Since one brother has 15 bucks more that the other we have x = y + 15, or y = x +15. We got here two equations system {x + y = 55, x = y + 15}. Inserting the expression for x into the first equation we find the value of y, inserting the latter into the equation for x we find it.

solve(x + y = 55, x = y + 15, x, y)

scrutinizer30 Agu 2013

+259

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3x + (16 ÷ 2x ) – 14 = -2x => 3x + (8 ÷ x ) – 14 = -2x, multiplying both sides by x to break free of denominator, we arrive at the quadratic equation 3*x^2 +2*x^2 - 14x + 8 = 0 =>

5*x^2 - 14x + 8 = 0. D = (-14/2)^2 - 5*8 = 9 => x = [7+-3]/5 x1 = 2, x2 = 4/5

solve(3x + 16 / (2x) - 14 = -2x, x)

scrutinizer30 Agu 2013

+259

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acos(cos). Click Help section for explanations

For ex., [input]acos(0.5)[/input]

scrutinizer30 Agu 2013

+259

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You should use the cosine theorem. Let's construct a triangle in a fashion like in the fig.below.

Triangle.png

Here, the angle B = 55 grades, AB = 58 cm, BC = 70 cm, and AC the unknown side. Using the cosine theorem where the angle B is located in-between AB and AC we get

BC^2 = 58^2 + 70^2 - 2*58*70*cos (55 grades) = ...

scrutinizer30 Agu 2013

+259

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6/7 x 5/21 = (6 x 5) / 3 * 7^2 = (2 x 5)/49 = 10/49

scrutinizer30 Agu 2013