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Ma physics teacher gave me this a while back, took me a while to solve. M not dhat smart but i want to find another way of solving it. x=yz + (1/2)y(z)2. Make z d subject of 4mula....thankz

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Since 1/2 and 2 cancel.

x = yz + yz

x = 2yz

z = x/2y

So... why is the physics teacher giving math problems?

-edit-

forgot spoilers.

Edited by curr3nt

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Itz actually a physics 4mula, i just changed the letters so that one might not notice cuz itz very common. Your answer would've been right if the '2' meant 'x2', it actually means 'raised to the pwr of 2'.....sorry, my mistake

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Is this the formula?

x = (y * z) + (y * z^2)/2

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z= -1+sqrt[1+(2x/y)], and z= z= -1-sqrt[1+(2x/y)]

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The closest physics formula I can find is s = vt + at2/2 but v <> a so how can they both be y?

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x = yz + yz2/2

2x/y = z2 + 2z

2x/y + 1 = z2 + 2z + 1

2x/y + 1 = (z + 1)2

±√(2x/y + 1) = z + 1

z = ±√(2x/y + 1) - 1

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Why do you insist on doing other people's homework?

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@Curr3nt, thankz alot, whew, so there's another way...bt stil kinda d same (based on quadratics), bt stil diff @thechad, lol

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Why do you insist on doing other people's homework?

I insist? Normally I do not insist. As for this one... it seemed not to be actual homework. I choose this time to believe he was checking his work.

And if he actually lied then I figure class will get quite a bit interesting if he doesn't actually learn how to work with equations.

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I think this is a Physics formula

S = ut + 1\2 at2

Here, both u and a are same, that is initial spped and acceleration are same.

This is possible only when the the object is still(not moving)

So, S = 0*t + 1/2 * 0 * t2 = 0

or

at time, t=0

So, S = u * 0 + 1/2 *a * 02 = 0

So, the subject can go like, AT TIME 0(INITIALLY)

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