Answer: it gets smaller, 13 times smaller
Step-by-step explanation:
Size of the figure/13
Answer:
See below for answers and explanations
Step-by-step explanation:
<u>Part A</u>
The average rate of change of a function over the interval ![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
is equal to 
, hence:
Therefore, the average rate of change of 
over the interval ![[5,9]](https://tex.z-dn.net/?f=%5B5%2C9%5D)
is 
.
<u>Part B</u>
Do the same thing as in Part A:
Therefore, the average rate of change of 
over the interval ![[0.25,1]](https://tex.z-dn.net/?f=%5B0.25%2C1%5D)
is 
.
<u>Part C</u>
To interpret our answer from Part B in terms of the real world it represents, we say that between 0.25 seconds and 1 second, the ball falls at a rate of 4 feet per second (since our average rate of change is negative).
The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
Answer:
E
Step-by-step explanation:
The problem says that triangle BDC lies in the plane k, which means that whatever angle is formed by another point beyond this plane with any of the three segments that form BDC (BD, DC, and BC) is the same as the angle formed by the line connecting the point and the plane.
Here, we're given that AD⊥DC, which means AD forms a 90° angle with DC. Then, since DC is already on the plane, we already know for sure that AD is definitely perpendicular to plane k.
Thus, the answer is E (none of these).
