Equation: 450 = 128 + 150 + n
work:
1st step:
128+150 = 278
2nd step:
450-278 = 172
answer: 172
3rd step (checking) :
450 = 128 + 150 + 172
450 = 278 + 172
450 = 450
Answer:
A - 4/13ths is a larger number
B- 3/10ths is a ratoinal mumner
Answer:
50
Step-by-step explanation:
75 dived into 3 parts can also be written as 75/3. 75/3 is 25 making each part worth 25. 2 segments are included in the marked portion so its 50 or (25 + 25)
If nCk represents the number of ways k parts can be chosen from a pool of n, the probability of interest is the complement of the probability of selecting all good parts.
1 - (167C3)/(170C3) = 42,085/804,440 ≈ 0.0523
_____
nCk = n!/(k!(n-k)!)
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:
‹1, -1, 1› × ‹0, 1, 1›
You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.
So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...
In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0
That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:
a - b + c = 0
b + c = 0
This is two equations, three unknowns, so you can solve it with one free parameter:
b = -c
a = c - b = -2c
The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›
The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:
|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6
Then we divide that vector by its magnitude to yield one solution:
‹ -2/√6 , -1/√6 , 1/√6 ›
And take the negative for the other:
‹ 2/√6 , 1/√6 , -1/√6 ›
