Answer:
About 5 years ago my girlfriend and I were taking a summer camping trip in southern Alberta. I was in the passenger seat trying to find the campground on a map when she drove over the crest of a hill, was blinded by the sun drove into the ditch. It was a seriously steep ditch but we weren't going very fast so all was fine. I looked over at her and laughed before returning my attention to the map, assuming she could safely bring the vehicle back onto the road.
The next thing I knew, the SUV launched onto the pavement and she lost control and we began swerving. I remember feeling the wheels on the driver's side lift off the ground, then the impact as I was slammed into the door and glass exploded into my face. We barrel rolled and we rolled over-front for a seriously long way.
At some point during the chaos I looked over at her to make sure she was "ok", and just as I did so I watched as she was thrown into the ground through her door window and the corner of the roof just above her seat was crushed inward. The way it looked to me was that she had just been crushed between the ground and the roof of the vehicle. I passed out at that point.
When I came to I had somehow already unbuckled myself from my seat and the vehicle was on its roof. I crawled over to her seat and was in absolute shock to see that she was still in one piece. I removed her from her seat and got us out through the windshield before carrying her for about a half a kilometer down the road, still in shock and fueled entirely by adrenaline.
We were found by a driver who had gone past the wreckage and we were eventually taken to a hospital. I broke 3 ribs on my right side and dislocated my right shoulder, she was severely concussed and scraped up but otherwise mostly okay. However she was 7 weeks pregnant at the time and we found out while in hospital that her body had rejected it under the immense and sudden stress.
We are still together and have 3 amazing and beautiful children, but she still holds onto a lot of guilt surrounding the accident and the loss.
Explanation:
Which of the following is a country that cruise ships commonly sail under the flag of?
O United States
O Canada
O Panama
O South Africa
< Previous
Answer:
Panama
hope this helps!
Manuel is working on a project in Visual Studio. He wants to keep this program showing on the entire desktop, but he also needs to have several other applications open so that he can research the project.
Answer:
d. Task View
Explanation:
Based on the scenario being described within the question it can be said that the best feature for this would be the Windows 10 task view. This is a task switcher and virtual desktop system included in the Windows 10 operating system, and allow the individual user to quickly locate, manage, open or hide different windows/tasks. Such as having several projects open in different monitors running at the same time.
Who must yeild at T-intersections?
Answer: the driver on the street
Explanation:
Consider sending a 1,600-byte datagram into a link that has an mtu of 500 bytes. suppose the original datagram is stamped with the identification number 291. how many fragments are generated? what are the values in the various fields in the ip datagram(s) generated related to fragmentation?
Explanation:
Step one
The maximum size of data field in each fragment = 480
(because there are 20 bytes IP header) Thus the number of required
fragments [tex]=\frac{1600-20}{480} \\\\= \frac{1580}{480} \\\\=3.29\\\\[/tex]
thus the number of required fragment is 4
Step two
Each fragment will have identification number 291. each fragment except the last one will be of size 500 bytes (including IP header). the offset of the fragments will be 0, 60, 120, 180. each of the first 3 fragments will have
flag = 1; the last fragment will have flag =0
Using the appropriate formula, the number of fragments which would be present in the datagram to be sent would be 4
The minimum length of IP header = 20 bytes
Maximum transmission unit (mtu) = 500 bytes
Hence, the payload would be calculated thus :
mtu - header ;Payload = 500 - 20 = 480Hence, the maximum size of data field per Fragment = 480 bytes
The number of fragments required :
[tex]\frac{datagram \: size - Header \: size}{payload} [/tex][tex] Number \: of \: fragments = \frac{1600 - 20}{480} = 3.29[/tex]
Hence, the number of fragments is 4
Size per Fragment would be 500 bytes each ; the last Fragment would be about 100 bytes Each Fragment would bear the identification number 291.Learn more : https://brainly.com/question/16289731
A customer seeks to buy a new computer for a private use at home.The customer primarily needs the computer to use the Microsoft PowerPoint application for the purpose of practice presentation skills.As a sales person what size hard disc would you recommend and why?
Answer:
The most common size for desktop hard drives is 3.5 inches, they tend to be faster and more reliable, and have more capacity. But they also make more noise.
Explanation:
If you are continually deleting and installing programs or creating content, the disc must have good reliability.
Keep in mind that larger hard drives are also a little slower, so it is preferable to opt for two smaller ones. Large hard drives are partitioned so there is no problem gettin
chbdg good performance, but if you put everything on one big disk and it breaks, you will lose everything.
If you buy 2 small disks, check that the motherboard does not limit the speed of a second hard disk.
the blue course at the ski club is 5.8 km long. The red course is 10.3 km long. Mark skied the blue course 15 times and Katy skied the red course 15 times. How much further did Katy ski than Mark?
Answer:
67.5km
Explanation:
mark = 5.8 x 15 = 87km
Katy = 10.3 x 15 = 154.5km
154.5 - 87 = 67.5km
the hose is 2 inches in diameter. What circumference does the plug need to be? remember to type just a number
Answer:
6.28
Explanation:
This problem bothers on the mensuration of flat shapes, a circle.Given data
Diameter d= [tex]2in[/tex]
Radius r = [tex]\frac{d}{2} = \frac{2}{2} = 1in[/tex]
We know that the expression for the circumference of a circle is given as
[tex]C= 2\pi r[/tex]
Substituting our given data and solving for C we have
[tex]C= 2*3.142*1\\C= 6.28[/tex]
