132 feet/s
132/3.3 m/s
40 m/s
in 10 seconds
40 * 10
400 m
1. (a) The orbital period can be calculated using the equation
T = 2πrr r
gR2
e
where r = Re + h
where re = 6378 km is the earth’s radius, r is the satellites distance
from the earth’s center and h = 205 km is the satellite’s orbital altitude, and g = 9.81m/s2
is the gravitational acceleration. With these
given values the orbital period is
Torbit = 5312.5s = 1.4757h
Answer:
![AB=\left[\begin{array}{cc}32&1&18&-22\\\end{array}\right]](https://tex.z-dn.net/?f=AB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D32%261%2618%26-22%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
To multiply matrices, we need to take the dot product of each row and column.
First, the dot product of (1,5) and (2,6) is what goes in the top-leftmost section of the resulting matrix. So the dot product would be (1*2)+(5*6)=2+30=32.
Second, the dot product of (1,5) and (6,-1) is what goes in the top-rightmost section of the resulting matrix. So the dot product would be (1*6)+(5*-1)=6-5=1.
Third, the dot product of (-3,4) and (2,6) is what goes in the bottom-leftmost section of the resulting matrix. So the dot product would be (-3*2)+(4*6)=-6+24=18.
Fourth, the dot product of (-3,4) and (6,-1) is what goes in the bottom-rightmost section of the resulting matrix. So the dot product would be (-3*6)+(4*-1)=-18-4=-22.
Therefore, the resulting matrix is ![\left[\begin{array}{cc}32&1&18&-22\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D32%261%2618%26-22%5C%5C%5Cend%7Barray%7D%5Cright%5D)
The first thing to do is isolate the x so get rid of the -2.7 but since is a negative you are going to add it to 10.3 so>>> 2.7+10.3 =13 x≥13 is the answer.