I would guess positive linear considering it’s a positive graph that looks basically linear.
Answer:
In standard form it is x^4 - 12x^3y + 54x^2y^2 - 108x y^3 + 81y^4.
Step-by-step explanation:
(3y)^4 + 4C1(3y)^3(-x) + 4C2(3y)^2(-x)^2 + 4C3(3y)(-x)^3 + (-x)^4
= 81y^4 - 108y^3x + 54y^2x^2 - 12yx^3 + x^4
Answer:
6 cubic cm
Step-by-step explanation:
3 · 3 · 2 = 18
18 · 1/3 = 6
Answer:
a) weight of the car = 2816,1 lbs
b) 2773 lbs
Step-by-step explanation:
The equilibrium force is 490 lbs. That force keep the car at rest, then
∑ Fy = 0 and ∑Fx = 0
Forces acting on the car:
The external force 490 lbs
weight of the car uknown
Normal force
sin∠10° = 0,174
cos∠10° = 0.985
∑Fx = 0 mg*sin10°- 490 = 0 ∑Fy = 0 mg*cos10° - N = 0
mg*0,174= 490
mg = 490 / 0,174
mg = 2816,1 lbs
weight of the car = 2816,1 lbs
The Normal force
mg*cos10° - N = 0 2816,1 * 0,985 = N
N = 2773 lbs
Then equal force in magnitude and in opposite direction will car exets on the driveway
Answer:
2,637,070
Step-by-step explanation:
2891+66= 2963
2963x 890= 2,637,070
Correction
890+ 66= 956
2891 x 2 = 5782
956 x 2= 1912
5782 + 1912= 7694
