scrutinizer

1 Questions
125 Answers

0

1038

1

+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 Nov 30, 2014

+259

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Guys, please learn how to properly present your questions so that persons you are seeking to help you would be able to adequately understand what actually you're asking, OK?

Now for this task: here we have an equations' system with 2 equations and 2 variables. Let's x be the length of the one part of the cable, y - the length of the rest part. Hence the general length is

x + y = 84. Since one part is 18 m longer that the other so x = y + 18 or y = x + 18. Inserting the expression for any of these variables into the very first equation and solving it for y or x respectively we then can find the other variable ( x or y respectively)

solve(x + y = 84, x = y + 18)

scrutinizerAug 26, 2013

+259

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Could you complete your question? An equation should have both sides separated by the "equal" sign unless the task is merely to simplfy this expression. But even in this case your posted expression lacks correct arrangement: (csc((cot(180-x)-tan(x))/(sec(90+x)* sec( cos, sin,tan, cot,...???)(90+x))

Simplification:

[csc((cot(180-x)-tan(x))/(sec(90+x)*sec(90+x)] = 1/[sin((cot(180-x)-tan(x)) : 1/(cos(90+x))^2)]= 1/[sin((-cot(x) - tan(x)): 1/(-sin x)^2)] = 1/[sin(-(cot(x) + tan(x))*(sin x)^2)] =

= 1/[sin(-(cos x / sin x + sin x / cos x)*(sin x)^2)] = 1/[sin(-((cos x)^2 + (sin x)^2) / sin x cos x )*(sin x)^2)] = 1/[sin(-1 / sin x cos x )*(sin x)^2)] = we got smth sharmless and mindless which bears witness that here cannot be sec(90 + a)*sec(90 + a). So specifize it, please.

scrutinizerAug 26, 2013

+259

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(-125)^(2/3) = sqrt[3]{-125^2} = sqrt[3]{125^2} = sqrt[3]{(5^3)^2} = sqrt[3]{(5^2)^3} = 5^2 = 25

scrutinizerAug 26, 2013

+259

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650+530+52+-28+-75 = 650+530+52-28-75

scrutinizerAug 26, 2013

+259

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If I correctly understood your question you need to find some angle "theta" or "phi" from the expression cos (phi). In order to achieve that you have to solve a trigonometric equation (simple one, by the way) where \phi (or "theta") = +- arccos \phi (or theta) + 2\pi n

scrutinizerAug 26, 2013

+259

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You ought to take into concideration such basic notions as sine and cosine lines. Since the right triangle is given with C obviously being the right angle and B and C - acute angles (their sum is 90), let the triangle be formed of hypotenuse=the radius of the trigonometric circle in a rectangular coordinate system = 1 and the sides AC and BC being cathetuses. If to assume that the cathetus AC is lying on Ox axis, then AC is the cosine line of A, BC is the sine line of A and the right triangle ABC stands vertically. If to rotate it so that BC became the horizontal cathetus and AC vertical one, then BC would be the cosine line of B, AC - the sine line of B. So, it seems that sin A = cos B = 5/13 that is the answer to your question.

scrutinizerAug 23, 2013

+259

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As easy as ABC : 5y - x^2 ; x = -3 y = -5 => 5*(-5) - (-3)^2 = -25 - 9 = -34

scrutinizerAug 23, 2013

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The task - as it was displayed here by the topic starter - was to express a in terms of other variables not solve for all variables.

scrutinizerAug 22, 2013

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(3x^3)^4 = (3^4 *x^12) = 81x^12

scrutinizerAug 22, 2013

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v^2 = u^2 + 2as => 2as = v^2 - u^2 => a = [v^2 - u^2]/2s

scrutinizerAug 21, 2013