Jill is playing a game in which a number cube is rolled two times. The sum of the rolls is recorded. What is the sample space fo
r the game? (PLEASE HELP)!!!!! a. 1,2,3,4,5,6,
b. 2,3,4,5,6,7,8,9,10,11,12
c. 1,2,3,4,5,6,7,8,9,10,11,12
d. 1,2,3,4,5,6,7,8,9,10,12,15,16,18,20,24,25,30,36
When you roll a cube once, the smallest number it can show is 1, and the biggest one is 6.
So when you roll it twice, the smallest possible sum of both results is 2, and the biggest one is 12. . . . . . Just like list 'b' .
k is the spring stiffness. The unstretched length of the spring is L.
When the mass is added, the spring stretches to an equilibrium position of L+s, where mg = ks. When the mass is displaced a distance x (where x is positive if the displacement is down and negative if it's up), the spring is stretched a total distance s + x.
There are two forces on the mass: weight and force from the spring. Sum of the forces in the downward direction:
∑F = ma
mg − k(s + x) = ma
mg − ks − kx = ma
Since mg = ks:
-kx = ma
Acceleration is second derivative of position, so:
-kx = m d²x/dt²
Let's find k:
F = kx
400 = 2k
k = 200
We know that m = 50. Substituting:
-200x = 50 d²x/dt²
-4x = d²x/dt²
d²x/dt² + 4x = 0
This is a linear second order differential equation of the form:
x" + ω² x = 0
The solution to this is:
x = A cos (ωt) + B sin (ωt)
Here, ω² = 4, so ω = 2.
x = A cos (2t) + B sin (2t)
We're given initial conditions that x(0) = 0 and x'(0) = -10 (remember that down is positive and up is negative).
4b. The midline of a triangle is half the length of the baseline it is parallel to. Hence the segment FH is twice the length of CD, or 2·16 = 32.
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6. The angle bisector MP divides the triangles segments so those on one side of it are proportional to corresponding ones on the other side of it. Here, that means ...
x/(x+3) = 6/10
10x = 6x + 18 . . . . multiply by 10(x+3)
4x = 18 . . . . . . . . . subtract 6x
x = 4.5 . . . . . . . . . divide by 4
Then segment LM is x+3 = 4.5+3 = 7.5, and LN = x+6 = 4.5+6 = 10.5.