Answer:
Mid-Atlantic Ridge Theory
Step-by-step explanation:
hope this helps. :)
Use the double angle identity:
sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)
Now rewrite
sin(2<em>x</em>) sin(<em>x</em>) + cos(<em>x</em>) = 0
as
2 sin²(<em>x</em>) cos(<em>x</em>) + cos(<em>x</em>) = 0
Factor out cos(<em>x</em>) :
cos(<em>x</em>) (2 sin²(<em>x</em>) + 1) = 0
Consider the two cases,
cos(<em>x</em>) = 0 OR 2 sin²(<em>x</em>) + 1 = 0
Solve for cos(<em>x</em>) and sin²(<em>x</em>) :
cos(<em>x</em>) = 0 OR sin²(<em>x</em>) = -1/2
Squaring a real number always gives a non-negative number, so the second case doesn't offer any real solutions. We're left with
cos(<em>x</em>) = 0
Cosine is zero for odd multiples of <em>π</em>/2, so we have
<em>x</em> = (2<em>n</em> + 1) <em>π</em>/2
where <em>n</em> is any integer.
Answer:
![Probability = \frac{1}{20}](https://tex.z-dn.net/?f=Probability%20%3D%20%5Cfrac%7B1%7D%7B20%7D)
Step-by-step explanation:
Given
![People = 6](https://tex.z-dn.net/?f=People%20%3D%206)
![Selection = 3](https://tex.z-dn.net/?f=Selection%20%3D%203)
Required
Determine the probability of selecting the oldest 3
First, we need to determine the total possible selection.
Since it's a selection, we make use of combination as follows:
![^nC_r = \frac{n!}{(n-r)!r!}](https://tex.z-dn.net/?f=%5EnC_r%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-r%29%21r%21%7D)
In this case:
![n = 6](https://tex.z-dn.net/?f=n%20%3D%206)
![r = 3](https://tex.z-dn.net/?f=r%20%3D%203)
So:
![^nC_r = \frac{n!}{(n-r)!r!}](https://tex.z-dn.net/?f=%5EnC_r%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-r%29%21r%21%7D)
![^6C_3 = \frac{6!}{(6-3)!3!}](https://tex.z-dn.net/?f=%5E6C_3%20%3D%20%5Cfrac%7B6%21%7D%7B%286-3%29%213%21%7D)
![^6C_3 = \frac{6!}{3!3!}](https://tex.z-dn.net/?f=%5E6C_3%20%3D%20%5Cfrac%7B6%21%7D%7B3%213%21%7D)
![^6C_3 = \frac{6*5*4*3!}{3!3*2*1}](https://tex.z-dn.net/?f=%5E6C_3%20%3D%20%5Cfrac%7B6%2A5%2A4%2A3%21%7D%7B3%213%2A2%2A1%7D)
![^6C_3 = \frac{6*5*4}{3*2*1}](https://tex.z-dn.net/?f=%5E6C_3%20%3D%20%5Cfrac%7B6%2A5%2A4%7D%7B3%2A2%2A1%7D)
![^6C_3 = \frac{5*4}{1}](https://tex.z-dn.net/?f=%5E6C_3%20%3D%20%5Cfrac%7B5%2A4%7D%7B1%7D)
![^6C_3 = 20](https://tex.z-dn.net/?f=%5E6C_3%20%3D%2020)
In the combination, there can be only 1 set of oldest people.
So:
![Probability = \frac{1}{20}](https://tex.z-dn.net/?f=Probability%20%3D%20%5Cfrac%7B1%7D%7B20%7D)
could you maybe ask this again and provide an image of some sort or some more context? im not sure
Answer:
no solution
Step-by-step explanation:
7m-(5m-12)= 8+2m+4
2m+12= 2m+4
no solution