Actually the position function with respect to time under constant acceleration is:
a=g
v=⌠g dt
v=gt+vi
s=⌠v
s=gt^2/2+vit+si
So if vi and si are zero then you just have:
s=gt^2/2
Notice that it is not gt^2 but (g/2) t^2
So the first term in any quadratic is half of the acceleration times time squared because of how the integration works out...
Anyway....
sf=(a/2)t^2+vit+si
(sf-si)-vit=a(t^2)/2
2(sf-si)-2vit=at^2
a=(2(sf-si)-2vit)/t^2 and if si and vi equal zero
a=(2s)/t^2
Answer:
a) Ellipse
b) Sphere
c) Great circle
d) Triangle
e) Circle
Step-by-step explanation:
A) Diagonal cross section of cylinder
Ellipse
B) Shape created when a semi circle is rotated about y axis
Sphere
C) Diagonal cross section through the widest part of sphere
Great Circle
D) Perpendicular cross section of cone
Triangle
E) Cross section parallel to the base of cone
Circle.
Answer:
$1.35 will be your answer
Step-by-step explanation:
18 x 0.075 = 1.35
Hello,
In order to exist ,k>=0
1) sin (180°+x)=-sin (x) ==>sin(203°)=-√k
2) cos² x +sin² x=1==>cos x=√(1+sin² x)=√(1+k)
3)tan (-x)=-tan (x) =-sin (x)/cos (x) =- √k/√(1+k)=√(k/(1+k))
I THINK SSS, SAS AND ASA WOULD PROVE TRIANGLES ARE CONGRUENT.
