Answer:
mee I'm in 8th
Step-by-step explanation:
plzz help will be very thankful
Answer:
Not a function.
Does not pass vertical line test.
B.
Step-by-step explanation:
Something called a vertical line test is used to determine if a relation that has been graphed is a function or not.
We say if it passes then it is a function.
It will pass if you are able to draw infinitely many vertical lines covering the whole graph and each vertical line either touches your relation once or none.
If a single vertical line that you draw touches more than once, then it isn't a function.
When I say draw, I don't mean you should physically do it, but more so imagine it.
Now this particular relation is not a function because I can find a vertical line that touches more than once. Take the vertical line x=5 for example.
It will touch at (5,-6) and (5,6). You cannot have an x assigned to more than one y.
Answer:
For an equal number of servings, the apples will have six times as much sugar.
Step-by-step explanation:
process of elimination. All the other answer choices are incorrect and this one checks out.
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 ›
