Which angle is an x-intercept for the function y = cos(1/2x)

2 answers:

3 0

For the function
y = cos(1/2 x)
The x-intercept can be calculated by equating the function to zero and solving for x. So,
y = cos(1/2 x) = 0
1/2 x = arc cos 0
1/2 x = 90° +180°n
x =2 (90° +180°n)
x = 180° + 360°n
or converting to radians
x = (180° + 360° n)(π/180°)
x = π + 2π n
where n is any whole number

if n = 0
x = π

Therefore, the x-intercept is π or π+2π n

Alisiya [41]2 years ago

3 0

PLATO ANSWER: Pi (π)

Hope I helped, Have a great day!!!

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Pavel [41]

X 0 y was 30

at x 12 y is about 55

12-0 = 12

55-30 = 25

25/12 = 2.08 inches per year average

so answer should be D

5 0

3 years ago

PLEASE HELP

Sholpan [36]

1) We have that the equation is $x^2=20y$ , hence y=x^2/20. The standard equation of such an equation is y= $\frac{1}{4p} x^2$ . Hence, p=5 in this case. The focus is at (0,5) and the directrix is at y=-5 (a tip is that the directrix is always "opposite" the focus point of a parabola; if the directrix is at x=-7 for example, the focus is at (7,0)).
2) Similarly, we have that the equation is $x=3y^2 \\ \frac{1}{4p} =3$ . Thus, p=1/12. In this case, the parabola opens along the x-axis and the focus is at (1/12, 0). Also, the directrix is at x=-1/12. Hence the correct answer is B.
3) We are given that the parabola has a p of 9. Also, the focus lies along the y-axis, hence the parabola is opening along the y-axis. Finally, the focus is on the positive half, so the parabola is opening upwards. The equation for this case is y= $y=\frac{1}{4p} x^2= \frac{1}{36 } x^2$ .
4) Similarly as above. The directrix is superfluous, we only need the p-value. THe same comments about the parabola apply and if we substitute p=8 in the formula: $y= \frac{1}{4p} x^2$ we get y= $\frac{1}{32} x^2$ .
5) This is somewhat different, even though we do not need the directrix again. The focus lies on the x-axis, thus the parabola opens in this direction. The focus lies on the positive part of the axis, thus the parabola opens to the right. We also are given p=7. Hence, the equation we need is of the form $x= \frac{1}{4p} y^2$ . Substituting p=7, we get $x= \frac{1}{28} y^2$ .
6) The equation of a prabola with a vertex at (0,0) is of the form y=-ax^2. The minus sign is needed since the parabola is downwards. Since we are given anothe point, we can calculate a. We have to take y=-74 and x=14 feet (since left to right is 28, we need to take half). $-a= \frac{y}{x^2} = \frac{-74}{14^2} =-0.378$ . Thus a=0.378. Hence the correct expressions is y=-0.378* $x^2$