Answer:
−p3q2r−p3qr2−p2q3r−p2q2r2−p2qr3+pq3r2+pq2r3
Step-by-step explanation:
(p2qr+pq2r+pqr2)((−p)(q)+qr+−pr)
(p2qr)((−p)(q))+(p2qr)(qr)+(p2qr)(−pr)+(pq2r)((−p)(q))+(pq2r)(qr)+(pq2r)(−pr)+(pqr2)((−p)(q))+(pqr2)(qr)+(pqr2)(−pr)
−p3q2r+p2q2r2−p3qr2−p2q3r+pq3r2−p2q2r2−p2q2r2+pq2r3−p2qr3
−p3q2r−p3qr2−p2q3r−p2q2r2−p2qr3+pq3r2+pq2r3
Digit numbers.
Let the numbers 00 to 89 represent that the train is on time.
Let the numbers between 90 and 99 represent that the train is late.
Randomly select 6 numbers, with repetition allowed.
Count the number of times the train is late.
Repeat this simulation multiple times.
You will most likely obtain a result of between 0 and 2 times that the train is late.
if you're talking about the whole number, 100.96. if you're referring to decimal place values; 96.70.
underline the digit that you round to
circle the digit to the right
five or more goes up
four or less stays the same
everything behind becomes a zero.
Answer:
ST = 7.07 units
Step-by-step explanation:
* Lets explain how to find the length of a segment
- The length of a segment whose endpoints are 
and 
can be founded by the rule of the distance
* Lets solve the problem
∵ The line segment is ST
∵ S is (-3 , 10)
∵ T is (-2 , 3)
- Assume that S is and T is
∴ 
and 
∴ 
and 
- By using the rule above
∴ 
∴ 
∴ 
∴ 
∴ ST = 7.07 units
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content…
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side.
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day.
