A cross-country team had a total of 10 practices last week, with each practice being in the morning or the afternoon. During eac
1 answer:
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From the problem, the vertex = (0, 0) and the focus = (0, 3)
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm
Answer:
In February, 423 daytime minutes is used
Step-by-step explanation:
Let the base plan charges be x
And cost per daytime minute be y
In December,
x + 510y = 92.25------------------(1)
In January,
x + 397y = 77.56---------------------(2)
Subtracting eq(2) from eq(1)
x + 510y = 92.25
x + 397y = 77.56
-------------------------------
0 + 113y = 14.69
-------------------------------
y = \frac{14.69}{113}
y = 0.13----------------------------------(3)
Substituting (3) in (1)
x + 510(0.13) = 92.25
x + 66.3 = 92.25
x = 92.25 - 66.3
x = 25.95
So In February
base plan + (daytime minute)(cost per daytime minute) = 80.9
25.95 + (daytime minute)(0.13) = 80.9
(daytime minute)(0.13) = 80.9 - 25.95
(daytime minute)(0.13) = 54.95
(daytime minute) =
daytime minutes = 422.69
daytime minute
Answer:
∅1=15°,∅2=75°,∅3=105°,∅4=165°,∅5=195°,∅6=255°,∅7=285°,
∅8=345°
Step-by-step explanation:
Data
r = 8 sin(2θ), r = 4 and r=4
iqualiting; 8.sin(2∅)=4; sin(2∅)=1/2, 2∅=asin(1/2), 2∅=30°, ∅=15°
according the graph 2, the cut points are:
I quadrant:
0+15° = 15°
90°-15°=75°
II quadrant:
90°+15°=105°
180°-15°=165°
III quadrant:
180°+15°=195°
270°-15°=255°
IV quadrant:
270°+15°=285°
360°-15°=345°
No intersection whit the pole (0)
