Problem 1
<h3>Answer: D) 2pi</h3>
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Explanation:
The given table we have is
If you were to graph all those points and draw a cosine curve through them, then the function you should get is
The B coefficient is 1, so that leads to a period of
T = 2pi/B
T = 2pi/1
T = 2pi
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Problem 2
<h3>Answer: C)
343.134 yards</h3>
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Explanation:
Draw out a triangle with these properties
- side a = 270
- side b = 170
- angle C = 100
We'll use the law of cosines to find side c
c^2 = a^2 + b^2 - 2*a*b*cos(C)
c^2 = 270^2 + 170^2 - 2*270*170*cos(100)
c^2 = 117740.902710
c = sqrt(117740.902710)
c = 343.133943
c = 343.134
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Problem 3
<h3>Answer: A) 2</h3>
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Explanation:
If you started with y = sin(x) and applied the transformations stated, then you should end up with 
From here, plug in x = 0 to find the y intercept.
Vertical asymptotes are when y tends to infinity, which in other words means you need a fraction where you're dividing by zero. So to find the vertical asymptotes, solve the denominator of the fraction and the solutions are the asymptotes since that is where it will be dividing by zero.
Horizontal asymptotes are the opposite and is where x tends to infinity -- This can be done by dividing through with the xs still there, since you are effectively just looking at multiples of infinity and infinity squared and so on. When you have an x to the power you can forget about all the other x's on the same half of the fraction, since infinity squared is infinitely bigger than infinity - yes it's abstract and a bit stupid I know.
If the degree of the polynomial on the denominator is larger than the numerator, there is an asymptote at y=0 (since the bottom becomes infinitely larger than the top). If it is lower than the numerator, you have to divide through by long division and you end up with an oblique asymptote i.e. an asymptote where the dotted line would follow a linear equation e.g. y=x
For example
The vertical asymptote can be found by solving 2x^2-8=0 i.e. x=+-2 so our vertical asymptotes are at 2 and -2.
The horizontal asymptote, think of all the x's as infinity. SInce there are x^2s, we dont need to worry about the numbers or the other x's
Therefore we are left with 6x^2/2x^2, which gives us 3. Therefore our horizontal asymptote would be at y=3.
Answer:
The coordinates of b are: (x, y) = (3, -4)
Step-by-step explanation:
Given
Given that the midpoint of the ab is:
The coordinates of a are:
To determine
The coordinates of b = ?
Let b (x, y) be the coordinates of b.
Given that the midpoint of the ab is m(-2, -1)
Thus,
(x - 7) / 2 = -2, (y + 2) /2 = -1
now solving
(x - 7) / 2 = -2
(x - 7) = -2 × 2
x - 7 = -4
adding 7 in both sides
x - 7 + 7 = -4 + 7
x = 3
and solving
(y + 2) /2 = -1
(y + 2) = -1 × 2
y + 2 = -2
subtracting 2 from both sides
y + 2 - 2 = -2 - 2
y = -4
Thus,
The coordinates of b are: (x, y) = (3, -4)
