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STatiana [176]
3 years ago
10

Help me with this please

Mathematics
1 answer:
Gre4nikov [31]3 years ago
6 0

9514 1404 393

Answer:

   B.  √6

Step-by-step explanation:

The circles are not tangent to one another. If they were, the distance between their centers would be the sum of their radii: 1 +1 = 2.

__

The center of the first circle is (√3, √3), and the center of the second is the origin. The distance between these two centers is given by the distance formula:

  d = √((x2 -x1)^2 +(y2 -y1)^2)

  d = √((√3 -0)^2 +(√3 -0)^2) = √(3+3) = √6 . . . . matches choice B

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I will use a quadratic formula:

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The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)

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<h3>Which trigonometric functions are positive in which quadrant?</h3>
  • In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
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(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

y= 3\cos(2(x + \pi/2)) - 2

The options are:

  1. y= 3\sin(2(x + \pi/4)) - 2
  2. y= -3\sin(2(x + \pi/4)) - 2
  3. y= 3\cos(2(x + \pi/4)) - 2
  4. y= -3\cos(2(x + \pi/2)) - 2

Checking all the options one by one:

  • Option 1: y= 3\sin(2(x + \pi/4)) - 2

y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

  • Option 2: y= -3\sin(2(x + \pi/4)) - 2

This option if would be true, then from option 1 and this option, we'd get:
-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0

which isn't true for all values of x.

Thus, this option is not same as the given function.

  • Option 3: y= 3\cos(2(x + \pi/4)) - 2

The given function is y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2

This option's function simplifies as:

y= 3\cos(2(x + \pi/4)) - 2 = 3\cos(2x + \pi/2) -2 = -3\sin(2x) - 2

Thus, this option isn't true since \sin(2x) \neq \cos(2x) always (they are equal for some values of x but not for all).

  • Option 4: y= -3\cos(2(x + \pi/2)) - 2

The given function simplifies to:y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2

The given option simplifies to:

y= -3\cos(2(x + \pi/2)) - 2 = -3\cos(2x + \pi ) -2\\y = 3\cos(2x) -2

Thus, this function is not same as the given function.

Thus, the function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

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