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Thepotemich [5.8K]
3 years ago
13

Sketch one cycle of the cosine function y= cos 2θ

Mathematics
1 answer:
Liula [17]3 years ago
7 0
Based on function transformations, the graph of y= cos(2θ), is the same as the graph of <span>y= cos(θ), with a component B of 2, in which case, it means the period is the only thing that differs

well the period for this function will then be  </span>\bf \cfrac{2\pi }{B}\iff \cfrac{2\pi }{2}\implies \pi

so, the graph of is, the same as y= cos(θ), just shrunk by half horizontally
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D: 50

Step-by-step explanation:

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Which point is on the graph of the equation y = -3x + 4?
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The points that fits the equation would be option D.

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How do I find the zeros of x^4-10x^3-66x^2-94x-39
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Answer: x = -1, -3, 7 + √62, 7 - √62


Step-by-step explanation:

q                                    p

x⁴ - 10x³ - 66x² - 94x - 39

\frac{p}{q} = +/- \frac{1*3*13*39}{1}

possible rational factors: 1, -1, 3, -3, 13, -13, 39, -39

Use synthetic division <em>or long division</em> to see which factor will leave a remainder of 0.

try x + 1 = 0  ⇒   x = -1

-1   |  1     -10     -66     -94     -39

    <u>|  ↓     -1        11        55      39 </u>

       1     -11      -55     -39       0

(x + 1)(x³ - 11x² - 55x - 39)        

next, try x + 3 = 0  ⇒   x = -3   <em>for the new polynomial</em>

-3   |  1     -11     -55     -39

     <u>|  ↓     -3       42     39</u>

        1     -14      -13       0  

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x + 1 = 0  ⇒  x = -1

x + 3 = 0   ⇒ x = -3

x² - 14x - 13 = 0

x = \frac{-(b) +/- \sqrt{(b)^{2} -4(a)(c)}}{2(a)}

  = \frac{-(-14) +/- \sqrt{(-14)^{2} -4(1)(-13)}}{2(1)}

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  = \frac{14 +/- \sqrt{248}}{2}

  = \frac{14 +/- 2\sqrt{62}}{2}

  = 7 +/- \sqrt{62}


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Expression equivalent to 96+60
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60+96

156

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