Recent content by username12345

Ah, got it... a = \frac{F_x}{m} = \frac{229}{310} = 0.74 Thanks.

Homework Statement A worker drags a crate across a factoryt floor by pulling on a rope tied to the crate. The worker exerts a force of 450 N on the rope, which is inclined 38 degrees to the horizontal, and the floor exerts a horizontal frictional force on 125 N that opposes the motion. You...

Yes, but why divide by -2? My question was more, how would they get that answer instead? If I am calculating a different (but equivalent) answer then how?

Homework Statement Find a Cartesian equation of the plane P containing A (2, 0, −3) , B(1, −1, 6) and C(5, 5, 0) , and determine if point D(3, 2, 3) lies on P. Homework Equations vector cross product ax + by + cz = 0 The Attempt at a Solution Take the cross product of AB and...

Ok, that explanation helped, thankyou.

Ok, got it, thanks.

Do you mean like this... 3x^2 + y = x^2 - 2y + 6 3y = -2x^2 + 6 y = \frac{-2x^2 + 6}{3} y = \frac{-2x^2}{3} + 2 ?

The explanation above is good. So, if say the gradient vector was i + 2j, then a vector such that Dv f = 0 would be +- (2i - j) ?

Homework Statement Sketch the level curve of the surface z = \frac{x^2 - 2y + 6}{3x^2 + y} belonging to height z = 1 indicating the points at which the curves cut the y−axis. Homework Equations The Attempt at a Solution I put 1 = \frac{x^2 - 2y + 6}{3x^2 + y} but then don't...

Homework Statement Let f (x, y) = e^x^2 + 3e^y . At the point (0, 1) find: (a) a vector u such that the directional derivative D_u f is maximum and write down this maximum value, (b) a vector v such that D_v f = 0 Homework Equations grad f / directional derivative formulaThe Attempt at a...

Homework Statement Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources. At what distance from source A is there constructive interference between points A and B...

Best answer I can get is 1/2 or -1/2. Anyway test on this starts in 45 minutes so just hope this question doesn't come up.

You mean into this? \mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim cosh(2x)}{{\lim 4 + \lim x^3 + \lim x^2}} so, \mathop {\lim }\limits_{x \to 0 } \frac{\lim 1 - \lim 2 + \lim sinh^2(x+1)}{{\lim 4 + \lim x^3 + \lim x^2}} = \frac{1 - 2 + \lim sinh^2(x+1)}{4} I can't get that to...

I tried subbing x = 0 into the above in the original equation and get: \frac{1 - 2e^2 + 4 - \frac{2}{e^2}}{16x^3 + 4x^2} Now should I divide each term by x^3? Can someone please show me the method? I am getting confused now.

sinh(0) = 0 so use LHopitals rule : \lim_{x\to 0} \frac{ \sinh x}{x} = \lim_{x\to 0} \frac{ \cosh x}{1} = \frac{cosh 0}{1} = \frac{1}{1} = 1 I don't know what you mean, sorry.