Step-by-step explanation:
yes I am how about you, I always get stuck on my assignments
X*80+y*60= 50*74
x +y =50 | * ( -60)
80x + 60y= 50*74
-60x - 60y=-50*60
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20x= 50*(74-60)=50*14
x=50*14/20=35 pounds of 80 cents tea
y=50-35=15 pounds of 60 cents tea
<span>1) Write the point-slope form of the equation of the horizontal line that passes through the point (2, 1). y = 1/2x
2)Write the point-slope form of the equation of the line that passes through the points (6, -9) and (7, 1).
m = (-9 - 1) / (6 - 7) = -10/-1 = 10
y + 9 = 10 (x - 6)
y = 10x - 69
3) A line passes through the point (-6, 6) and (-6, 2). In two or more complete sentences, explain why it is not possible to write the equation of the given line in the traditional version of the point-slope form of a line.
4)Write the point-slope form of the equation of the line that passes through the points (-3, 5) and (-1, 4).
m = (5 - 4) / (-3 - -1) = 1/-2
y - 5 = (-1/2) (x +3)
y = (-1/2)x + 7/2
5) Write the point-slope form of the equation of the line that passes through the points (6, 6) and (-6, 1).
m = (6-1)/(6 - -6) = 5 / 12
y - 6 = (5/12) (x-6)
y = (5/12)x + 17 / 2
6) Write the point-slope form of the equation of the line that passes through the points (-8, 2) and (1, -4).
m = (2 - -4) / (-8 -1) = 6 / -7
y - 2 = (-6/7) (x + 8)
y = (-6/7)x - 50 / 7
7) Write the point-slope form of the equation of the line that passes through the points (5, -9) and (-6, 1).
m = (-9 - 1) / (5 - -6) = -10 / 11
y + 9 = (-10 / 11) (x - 5)
y = (-10 / 11)x -49/11
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Answer:
For example, the years 1600, 2000, and 2400 are century leap years since those numbers are divisible by 400, while 1700, 1800, 1900, 2100, 2200, and 2300 are common years despite being divisible by 4.
Step-by-step explanation:
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2
abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.
d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
