Answer:
The equation of a line that is parallel to line a is y=½x + 2, or what is the same -½x+y-2=0
Step-by-step explanation:
A line is a polynomial function of the first degree that has the following form:
y = m * x + b
where m is the slope of the line and b is the intercept with the Y axis. The slope is the inclination of the line with respect to the abscissa axis.
In this case you know that the slope of line a is ½.
Parallel lines are two or more lines in a plane that never intersect. Two lines are parallel if they have the same slope and different y-intercepts.
You have:
- y+½x=2 → y= -½x +2
- -2x+y+2=0 → y=2x -2
- 2x-2=y
The line that has the same slope as line a is the last one, so <u><em>the equation of a line that is parallel to line a is y=½x + 2, or what is the same -½x+y-2=0</em></u>
Answer:
upper control limit = 13.34
Step-by-step explanation:
The c-chart is used to check the change in the number of recorded calls per day
First find out the center line
c = sum of recorded calls/number of days
c = (3 + 0 + 8 + 9 + 6 + 7 + 4 + 9 + 8)/9
c = 54/9
c = 6
The upper control limit for 3σ is
upper control limit = c + 3√c
upper control limit = 6 + 3√6
upper control limit = 13.34
Therefore, the upper control limit for the 3-sigma c-chart is 13.34.
Answer:
See solutions for detail.
Step-by-step explanation:
a. 
is the instantaneous rate of change of volume given with respect to time, t.
The volume's rate of change is written as a function of time.
-
is the rate of change in the height of water in the tank with respect to time, t.
b. 
is the only constant. Water flows into the constant at a constant rate, say 
per minute.
c. is positive. Volume water in the take is increasing from time to time.
-The volume at time t=1 is greater than the volume at t=0, hence, it's a positive rate of change.
d. is a positive rate. The initial height of water in the tank is zero.
-The final height at time t is 0.25h. The height is increasing with time.
Hence, it is positive.
The answers to the questions
First find the decimal equivalent of square root 3: SQRT(3) = 1.732 ( roughly)
If the base and height were each 3, then the hypotenuse would need to be:
3^2 + 3^2 = C^2
9 + 9 = C^2
18 = C^2
C = SQRT(18) = 4.24
This is larger than sqrt(3), so this cannot be a right triangle.
If one leg was 3 and the other leg was sqrt(3) then the hypotenuse would be:
3^2 + 1.73^2 = C^2
9 + 3 = C^2
12 = C^2
C = SQRT(12) = 3.46
This is larger than 3, this cannot be a right triangle.
The answer is b) no.
