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schepotkina [342]
3 years ago
15

Could x = -5 be a solution to the inequality -6 < 3x + 9 < 21?

Mathematics
2 answers:
____ [38]3 years ago
8 0

<em>So you think that x = -5 is the solution for the inequality? Let's check it out.</em>

<em>(Check)</em>

<em>- Substitute x = -5 in the inequality</em>

<em />-6<em />

<em>If it was the symbol ≤ you'd be correct, but it's < so It's a bit wrong.</em>

<em />

<em>Now let's solve the inequality. Separate the inequality by parts.</em>

<em />\left \{ {{-6<em />

<em>Therefore, the inequality is true only when x > -5 or x < 4.</em>

<em />

<em>(Check)</em>

<em>- Substitute x = -4 in the inequality.</em>

<em />-6<em />

<em>The inequality is true for first case.</em>

<em>- Substitute x = 3 in the inequality</em>

<em />-6<em />

<em>The inequality is true for second case.</em>

<em />

<em>So the answer for inequality is -5 < x < 4</em>

<em />

<em>That means only {-4, -3, -2, ..., 3} can make the inequality true.</em>

musickatia [10]3 years ago
6 0

Answer:

<h2><u><em>x=−5<x<4</em></u></h2>

Step-by-step explanation:

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Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p
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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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(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).

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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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brainly.com/question/1421592

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