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**0001T**- Replies: 1

i am trying to solve this [t is time and T is temperature] to find the time t,

6450*Pi*(0.00185^2)*(0.0625)*836

dt = --------------------------------------------- dT

(1.2^2)*7.5 - 108*Pi*0.00185*0.25*T

between the limits 0 and 180degC

can someone help me:D

this is a bit embarassing...because i had forgotten the basics:(

how do i solve this

-5db = 20log M(f)

I see.

Thank you so much.

Stanley_Marsh wrote:

I dont think you did wrong , cuz 4(3,0,4)=(12,0,16)

so if i present the surface normal as (12,0,16) it is not considered as wrong?

thanks

**0001T**- Replies: 5

Hi, I just got to know about this forum & I got this little problem...

I am trying to solve this cross product; 3 vertices p1=(0,0,3), p2=(4,0,0) & p3=(0,4,3) given, I need to find the normal of the surface bounded by these 3 vertices.

I was told answer is (3,0,4) but my attempts kept leading me to (12,0,16)

here's what I did.

normal = (p1-p2) x (p3-p2) = [(0,0,3)-(4,0,0)]x[(4,0,0)-(0,4,3)] = (-4,0,3)X(4,-4,-3)

normal = (12,0,16)

where did i go wrong?

Thank you

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