Recent content by elmosworld403

Yes sorry X/ (X-2) > 2 and form (P(x))/(Q(x)) > 0

Solve x/ x-2 >2 by first rewriting it in the form P(x) / Q(x) >0 How do I go about first rewriting it into that form?

Kk so i don't have to do the zeros. I redid the question. -(4-2x)>8 -4+2X>8 2x>12 X>6 4-2x>8 -2x>4 x>-2 Interval notation (-2,∞)

l 4-2x l >8 4-2x>0 4-2x>8 -2x>-4 -2x>4 x>2 x>-2 4-2x<0 4-2x<8 -2x<-4 -2x<4 X<2 X<-2 -2< X < 2 ANSWER- (-2,2) interval notation Do I have to keep doing the 4-2x>0 4-2x<0 Even tho it doesn't say it in the question? It is also...

1. A student on a skateboard, with a combined mass of 78.2 kg, is moving east at 1.60 m/s. As he goes by, the student skilfully scoops his 6.4-kg backpack from the bench where he had left it. What will be the velocity of the student immediately after the pickup? Me solving...

I asked my teacher today. It wasn't anything complicated about technique, etc. She just wanted me to explain how increasing time can decrease the net force during an impulse. Fnet▲T= impulse. The curler can catch the stone and move with it as the momentum decreases just like how catchers in...

Question: Unit is on Momentum and Impulse Experienced Curlers know how to safely stop a moving stone. What do they do and why? At first I thought sweeping helps the moving stone stop. But looking around the internet sweeping actually decreases the friction between the stone and the ice by...

Holy Thanks Dude, my mind is blown right now. Didn't think that the highest point would not take into account the vertical since its velocity is zero. So the at the highest point only the momentum of the horizontal is left.:approve: Until gravity takes over? then the vertical has a velocity...

Idk Horizontal Distance=? Velocity=? Time=? Vertical Acceleration= 9.81 m/s squared. Vinitial=? VFinal=? Distance=? Time=? The question only gives me the momentum and weight and the angle. Momentum= 800 kg m/s weight= 2 kg angle= 30 degrees

I checked the answer it was 800 kg- m/s But the picture is confusing me isn't that the momentum of the cannonball at a 30 degree angle taking both the vertical and horizontal into account?

Horizontal velocity doesn't change at all. Vertical velocity does from gravity.

A 2.00 kg cannonball is fired out of a cannon at an angle of 30.0 to the horizontal. When the cannonball reaches the top of its path, its momentum has a magnitude of 800 kg-m/s. What is the horizontal component of the cannonball's momentum when it left the cannon? Change in Momentum...