14.
Angles 4 and 6 are supplementary, because they are on the same line. Supplementary angles add up to 180 degrees, and a line must be 180 degrees.
15.
Angles 1 and 8 are congruent, because they are alternate exterior angles
16.
m = y2 - y1 / x2 - x1
m = 7 - 2 / 4 - 5
m = 5 / -1
m = -5
17.
m = 3 - 3 / 7 - (-5)
m = 0 / 12
m = 0
18.
m = 1 - (-2) / 5 - (-4)
m = 3 / 9
m = 1/3
19.
A = (0, 3) - B = (3,0)
m = 0 - 3 / 3 - 0
m = -3 / 3
<em>m = -1</em>
C = (0, -2) - D = (4, 2)
m = 2 - (-2) / 4 - 0
m = 4 / 4
<em>m = 1</em>
Perpendicular, because the slopes are opposite reciprocals.
20.
E = (1, 2) - F = (0, 0)
m = 0 - 2 / 0 - 1
m = -2 / -1
<em>m = 2</em>
G = (1, -3) - H = (3, 0)
m = 0 - (-3) / 3 - 1
<em>m = 3 / 2</em>
Neither, because the slopes are different.
21.
I = (0, 1) - J = (2, -4)
m = -4 - 1 / 2 - 0
<em>m = -5/2</em>
K = (-1, -2) - L = (4, 0)
m = 0 - (-2) / 4 - (-1)
<em>m = 2/5</em>
Perpendicular, because the slopes are opposite reciprocals.
22.
M = (-2, 2) - N = (2, 2)
Horizontal line
<em>m = 0</em>
O = (3, 0) - P = (-3, 0)
Horizontal line
<em>m = 0
</em>Parallel, because the slopes are the same.
<em>
</em>23.
Angle 2 is congruent to angle 1 because of the alternate exterior angle theorem.
Angle 1 is congruent to angle 3 because of the vertical angle theorem.
Angle 2 is congruent to angle 3 because of substitution.
Line l is parallel to line m because the corresponding angles are congruent.
A research paper recommends using Poisson process to model the number of failures in commercial water pipes The paper also gives estimates the failure rate in units of failures per 100 miles of pipe per day; for four different types of pipe and for many different years_ For example, for cast iron pipe in 2005, the authors' estimate is the failure rate is 0.0864 failures per 100 miles per day: Suppose a town had 2500 miles of cast iron pipe underground in 2005 What is the probability of at least two failures
Answer:
The number of deserters is 34.
Step-by-step explanation:
We have to calculate the number of desertors in a group of 1500 soldiers.
The sergeant divides in groups of different numbers and count the lefts over.
If he divide in groups of 5, he has on left over. The amount of soldiers grouped has to end in 5 or 0, so the total amount of soldiers has to end in 1 or 6.
If he divide in groups of 7, there are three left over. If we take 3, the number of soldiers gruoped in 7 has to end in 8 or 3. The only numbers bigger than 1400 that end in 8 or 3 and have 7 as common divider are 1428 and 1463.
If we add the 3 soldiers left over, we have 1431 and 1466 as the only possible amount of soldiers applying to the two conditions stated until now.
If he divide in groups of 11, there are three left over. We can test with the 2 numbers we stay:
As only 1466 gives a possible result (no decimals), this is the amount of soldiers left.
The deserters are 34:
I hope this helps you
-4x-2y+4x+8y= -12-24
6y= -36
y= -6
-4x-2. (-6)= -12
-4x+12= -12
-4x= -24
x=6
(6, -6)
