Answer:
1/32v²sin2θ
Step-by-step explanation:
Given the expression r(theta) = 1/16v²sinθcosθ
According to double angle of trigonometry identity;
Sin2θ = sin(θ+θ)
Sin2θ = sinθcosθ + cosθsinθ
Sin2θ = 2sinθcosθ
sinθcosθ = sin2θ/2 ... **
Substituting equation ** into the question
1/16v²sinθcosθ = 1/16v²(sin2θ/2)
1/16v²sinθcosθ = 1/2×1/16v²(sin2θ)
1/16v²sinθcosθ = 1/32v²sin2θ
Hence using the double angle identity, the equivalent expression is 1/32v²sin2θ
Answer:
Show a picture so I can help
For f(x), which has a vertex at (2,0), the y-intercept at (0,4) is above this vertex, so the parabola opens upward. This means that the vertex is the only point that touches the x-axis, so there is only 1 x-intercept.
For h(x), the graph does not have any x-intercepts.
For g(x) = x^2 + x - 2 = (x+2)(x-1), this intersects the x-axis at x = -2 and x = 1, so there are 2 x-intercepts.
From least to greatest: h(x), f(x), g(x).
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