The equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12) is y = -5x + 1
<h3>Equation of a line</h3>
A line is the distance between two points
Given the equation of a line expressed as 10x + 2y = -2. Determine the slope
2y = -10x -2
y = -5x - 1
Slope of the line is -5
The equation of a line in point-slope form is y - y1 = m(x-x1)
Substitute the point and the slope of the parallel line
y - 12 = -5(x - 0)
y - 12 = -5x
y = -5x + 12
Hence the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12) is y = -5x + 12
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If I read the question correctly, the information about the 80x80 fencing and the grass region are all irrelevant information.
In order to find the area of the side walk we are going to use the formula A=pi(r^2) where r = the radius. The diameter of the pool is 60 feet and the the sidewalk is 5 feet bigger all around meaning we had 5 feet to both sides of this 60 feet giving you an overall diameter of 70 feet (or 60 + 5*2). To find the radius take half of the diameter. 70/2 = 35.
Now we plug in 35 to the equation. A = Pi(35^2) = 1225pi. Leave your answer like this as we are not finished and will need to round at the end for an accurate answer. This is NOT the end, that number is the area of the sidewalk AND the pool.
The sidewalk is not a full circle, just a border so we now have to find the area of the actual pool and subtract it from this 1225pi for the sidewalk's area. Same formula, Area = pi(r^2). The diameter of the pool is 60 meaning the radius is 30 (or 60/2). Plug into the equation, pi(30^2) = 900pi.
Now do 1225pi - 900pi to get 325pi. Break out your calculator and hit the approximation button. 325pi is approximately 1021.018 if rounded to 3 decimal places.
The area of the 5-foot wide sidewalk encircling the pool is approximately 1021.018 square feet.
Use the std eqn of a circle with center at (h,k) and radius r:
(x-0)^2 + (y+2)^2 = 9^2, or just x^2 + (y+2)^2 = 9^2.
The potential energy, E, of the penny is given by E=mgh. The energy, Q, required to raise the temperature of an object by an amount ΔT is given by Q=mcΔT. We can equate these two to get the result but we must use proper units and include the 60%:
(0.6)mgh=mcΔT
We see we can divide out the mass from each side
0.6gh=cΔT, then 0.6gh/c=ΔT
(0.6)9.81(m/s²)50m/385(J/kg°C) = 0.7644°C
since this is the change in temperature and it started at 25°C we get
T=25.7644°C
As you can see the result does not depend on mass. The more massive the copper object the more potential energy it will have to contribute to the heat energy, but the more stuff there will be to heat up, and the effect is that the mass cancels.
Answer:
1800 m/s if I'm not mistaken! :))
Step-by-step explanation:
