Question:
A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.
Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling a number greater than 2.
If there is more than one element in the set, separate them with commas.
Answer:
![S = \{1,2,3,4,5,6\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7B1%2C2%2C3%2C4%2C5%2C6%5C%7D)
![Greater2 = \{3,4,5,6\}](https://tex.z-dn.net/?f=Greater2%20%3D%20%5C%7B3%2C4%2C5%2C6%5C%7D)
Step-by-step explanation:
Given
A roll of a 6 sided number cube
Solving (a): The sample space
This implies that we list out all number on the number cube.
So:
![S = \{1,2,3,4,5,6\}](https://tex.z-dn.net/?f=S%20%3D%20%5C%7B1%2C2%2C3%2C4%2C5%2C6%5C%7D)
Solving (b): Outcomes greater than 2
This implies that we list out all number on the number cube greater than 2 i.e. 3 to 6.
So:
![Greater2 = \{3,4,5,6\}](https://tex.z-dn.net/?f=Greater2%20%3D%20%5C%7B3%2C4%2C5%2C6%5C%7D)
You didnt give enough information but 40
The answer would be B. All you would do is subtract 6.4 - 2.06 and you would annex a zero after the four. Then you would get 4.34
Answer:
Step-by-step explanation:
d(x)=√((x-2)2+(y-0)²)
=√((x-2)²+y²)
=√((x-2)²+(x-1)²)
=√(x²-4x+4+x²-2x+1)
=√(2x²-6x+5)
D=d²(x)=2x²-6x+5
![b.\\\frac{dD}{dx}=4x-6\\\frac{dD}{dx}=0,gives \\4x-6=0\\x=\frac{3}{2}\\\frac{d^2D}{dx^2}=4 >0 at x=\frac{3}{2}\\so~D~or~d^2~or~d~is~minimum~at~x=\frac{3}{2}\\so~y=1(\frac{3}{2} )-1=1/2=0.5\\so~nearest~point~is~(1.5,0.5)](https://tex.z-dn.net/?f=b.%5C%5C%5Cfrac%7BdD%7D%7Bdx%7D%3D4x-6%5C%5C%5Cfrac%7BdD%7D%7Bdx%7D%3D0%2Cgives%20%5C%5C4x-6%3D0%5C%5Cx%3D%5Cfrac%7B3%7D%7B2%7D%5C%5C%5Cfrac%7Bd%5E2D%7D%7Bdx%5E2%7D%3D4%20%3E0%20at%20x%3D%5Cfrac%7B3%7D%7B2%7D%5C%5Cso~D~or~d%5E2~or~d~is~minimum~at~x%3D%5Cfrac%7B3%7D%7B2%7D%5C%5Cso~y%3D1%28%5Cfrac%7B3%7D%7B2%7D%20%29-1%3D1%2F2%3D0.5%5C%5Cso~nearest~point~is~%281.5%2C0.5%29)
Answer:
The work done is 202.50Nm
Step-by-step explanation:
Given
![F =450N](https://tex.z-dn.net/?f=F%20%3D450N)
![x_1 = 30cm](https://tex.z-dn.net/?f=x_1%20%3D%2030cm)
![x_2 = 60cm](https://tex.z-dn.net/?f=x_2%20%3D%2060cm)
Required
The work done
First, we calculate the spring constant (k)
![F = kx_1](https://tex.z-dn.net/?f=F%20%3D%20kx_1)
![450N = k *30cm](https://tex.z-dn.net/?f=450N%20%3D%20k%20%2A30cm)
![k = \frac{450N}{30cm}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B450N%7D%7B30cm%7D)
![k =15N/cm](https://tex.z-dn.net/?f=k%20%3D15N%2Fcm)
So:
![F = kx_1](https://tex.z-dn.net/?f=F%20%3D%20kx_1)
![F(x) = 15x](https://tex.z-dn.net/?f=F%28x%29%20%3D%2015x)
The work done using Hooke's law is:
![W =\int\limits^a_b {F(x)} \, dx](https://tex.z-dn.net/?f=W%20%3D%5Cint%5Climits%5Ea_b%20%7BF%28x%29%7D%20%5C%2C%20dx)
This gives:
![W =\int\limits^{60}_{30} {15x} \, dx](https://tex.z-dn.net/?f=W%20%3D%5Cint%5Climits%5E%7B60%7D_%7B30%7D%20%7B15x%7D%20%5C%2C%20dx)
Rewrite as:
![W =15\int\limits^{60}_{30} {x} \, dx](https://tex.z-dn.net/?f=W%20%3D15%5Cint%5Climits%5E%7B60%7D_%7B30%7D%20%7Bx%7D%20%5C%2C%20dx)
Integrate
![W =15 \frac{x^2}{2}|\limits^{60}_{30}](https://tex.z-dn.net/?f=W%20%3D15%20%5Cfrac%7Bx%5E2%7D%7B2%7D%7C%5Climits%5E%7B60%7D_%7B30%7D)
This gives:
![W =15 *\frac{60^2 - 30^2}{2}](https://tex.z-dn.net/?f=W%20%3D15%20%2A%5Cfrac%7B60%5E2%20-%2030%5E2%7D%7B2%7D)
![W =15 *\frac{2700}{2}](https://tex.z-dn.net/?f=W%20%3D15%20%2A%5Cfrac%7B2700%7D%7B2%7D)
![W =15 *1350](https://tex.z-dn.net/?f=W%20%3D15%20%2A1350)
![W =20250N-cm](https://tex.z-dn.net/?f=W%20%3D20250N-cm)
Convert to Nm
![W =\frac{20250Nm}{100}](https://tex.z-dn.net/?f=W%20%3D%5Cfrac%7B20250Nm%7D%7B100%7D)
![W =202.50Nm](https://tex.z-dn.net/?f=W%20%3D202.50Nm)