Answer:
False
Step-by-step explanation:
the <u>dot product</u> between two vectors A and B is:
A·B=AB
where
is the angle between the vectors, if they are parallel, this angle is zero. so
and so the dot product is:
A·B =AB
and since 
the dot product is equal to
A·B=AB
The dot product of parallel vectors is NOT zero
Answer:

Step-by-step explanation:
We have been given an equivalence equation
. We are asked to find all the square root of the given equivalence equation.
Upon converting our given equivalence equation into an equation, we will get:
Add 53 on both sides:


Take square root of both sides:

Therefore, the square root for our given equation would be
.
Let's put this on the usual Cartesian grid just so we can talk about it without drawing a picture. We'll use map conventions, right is east, up is north.
The ball starts at (0,0). 10.3 feet northwest means we have an isosceles right triangle whose diagonal is 10.3 feet. It's isosceles because northwest means equal parts north and west.
The sides of these triangles are in ratio

so the coordinates after the first putt are

The negative sign indicates west, which doesn't really matter for this problem. The distance from the origin to this point is 10.3 as required.
Now a second putt of 3.8 feet north puts us at

The squared distance to the origin is exactly

A little calculator work tells us

Third choice.
The answer is $5.15! Hope this helps