40/24+9/24= 40+9/24 = 49/24
The equation<span> of a </span>line<span> is typically written as </span>y<span>=mx+b where m is the </span>slope<span> and b is the </span>y<span>-intercept. If you a </span>point<span> that a </span>line passes through<span>, and its </span>slope<span>, this page will show you how to find the </span>equation<span> of </span>the line<span>. Fill the </span>point<span> that </span>the line passes through... ( , ) Example: (3,2<span>) ...and the </span>slope<span> of </span>the line. m= Example: m=<span>3, or ... hope this helpps!!!!</span>
We have to find the slope of the line that passes through this points:
p1(-2,15), p2(-8,-5)
the slope of a line through two points is calculated like this:
m = (y2 - y1)/(x2 - x1)
where x1,y1,x2,y2 are the coordinates of the two points, so we have:
m = (-5 - 15)/(-8 - (-2))
m = (-20)/(-8 + 2)
m = -20/-6
m = 10/3
that is the line slope
The length of the new sidewalk between the Dormitory to the recreation center is 143 yards
We need to understand the concept of Bearing and Distance in order to be able to determine how long is the new sidewalk.
<h3>What are Bearing and Distance?</h3>
The Bearing of an angle measures the movement of an angle in a clockwise direction starting from the North Pole. However, the distance shows how the length of the sides of the angle can also be measured.
From the diagram attached below:
- The adjacent of the sidewalk way = 80 yards.
- The hypotenuse shows the new sidewalk way = x( unknown)
- The angle θ = 56°
Using the relation from trigonometry:
x ≅ 143 yards
Learn more about Bearing and distance here:
brainly.com/question/17007896
Answer:
130.826 kilojoules
Step-by-step explanation:
The Work required to pump water = pgV
Where, p = density of water = 1000 kg/m³
g = acceleration due to gravity = 9.81ms-¹
V = volume of water
Since, radius r = 5 ft and height h = 9ft
Volume of water in the cylindrical tank = (2/3)πr²h
V = (2/3)π *5²*9 = 471.24 ft³ = 471.24 * 0.0283m³ = 13.336m³
Work required = 1000*9.81*13.336
W = 130.826 KJ
Therefore, the work required to pump water 2/3 of the volume of the cylindrical tank = 130.826 kilojoules.
