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Answer:
There are two pairs of solutions: (2,7) and (-1,4)
Step-by-step explanation:
We will use substitution.
y = x^2 + 3
y = x +5
Since the second equation is equal to y, replace y in the first equation with the second equation.
y = x^2 + 3
x + 5 = x^2 + 3
Rearrange so that one side is equal to 0.
5 - 3 = x^2 - x
2 = x^2 - x
0 = x^2 - x - 2
You may use quadratic formula or any form of factoring to find the zeros (x values that make the equation equal to 0).
a = 1, b = -1, c = -2
Zeros = and
Zeros = 2 and -1
Now that you have your x values, plug them into the equations to find their corresponding y values.
y = x^2 + 3
y = (2)^2 + 3
y = 7
Pair #1: (2,7)
y = x^2 + 3
y = (-1)^2 + 3
y = 4
Pair #2: (-1,4)
Therefore, there are two pairs of solutions: (2,7) and (-1,4).
<h3>Answer: Choice B</h3>
Explanation:
Cosine is positive in quadrants I and IV, but quadrant IV isn't shaded in so we can rule out choice A.
Sine is positive in quadrants I and II. So far it looks like choice B could work. In fact, it's the answer because sin(pi/6) = 1/2 and sin(5pi/6) = 1/2. So if 0 ≤ sin(x) < 1/2, then we'd shade the region between theta = 0 and theta = pi/6; as well as the region from theta = 5pi/6 to theta = pi.
Choice C is ruled out because tangent is positive in quadrants I and III, but quadrant III isn't shaded.
Choice D is ruled out for similar reasoning as choice A. Recall that