If you have two vectors A and B,
Dot product is a scalar quantity dealing with how much of one vector is in the same direction as the other vector, or the projection of one onto the other. You can see that from the cosine part of this form-
![A~*~B = [A][B]cos(\theta)](https://tex.z-dn.net/?f=A~%2A~B%20%3D%20%5BA%5D%5BB%5Dcos%28%5Ctheta%29)
The cross product is a vector perpendicular to both A and B. It deals with how much of one vector is perpendicular to the other vector. You can see that in the sine part of this form -
Answer:
A=8
Step-by-step explanation:
If you use the formula 30-60-90 which is a-radical 3a-2a
since the hypotenuse is 2a it equals 16
therefore if its 2a all you need to do is divide it where then you get 8.
35-4x>2
-4x>2-35
-4x>-33
-x>-33/4
x<33/4 (x<8.25)
Solution: x<8.25. or x∈(-∞ , 8.25)
<span>The number of x-intercepts that appear on the graph of the function
</span>f(x)=(x-6)^2(x+2)^2 is two (2): x=6 (multiplicity 2) and x=-2 (multiplicity 2)
Solution
x-intercepts:
f(x)=0→(x-6)^2 (x+2)^2 =0
Using that: If a . b =0→a=0 or b=0; with a=(x-6)^2 and b=(x+2)^2
(x-6)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x-6)^2] = sqrt(0)→x-6=0
Adding 6 both sides of the equation:
x-6+6=0+6→x=6 Multiplicity 2
(x+2)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x+2)^2] = sqrt(0)→x+2=0
Subtracting 2 both sides of the equation:
x+2-2=0-2→x=-2 Multiplicity 2