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Darya [45]
3 years ago
6

Fiona wrote the linear equation y = x – 5. when henry wrote his equation, they discovered that his equation had all the same sol

utions as fiona’s. which
equation could be henry’s?
Mathematics
2 answers:
djverab [1.8K]3 years ago
6 0
I would go with A but I am not 100% sure so you may want to check however you can.
AleksAgata [21]3 years ago
4 0

Answer with explanation:

The options are

      (A) x- \frac{5 y}{4} =\frac{25}{4}(B) x- \frac{5 y}{2} =\frac{25}{4}(C) x-\frac{5 y}{4} =\frac{25}{2}(D) x-\frac{5 y}{2} =\frac{25}{2}

Equation wrote by ,Fiona is:

    y = x -5

The equation which has same solution that Fiona has written,will be coincident with the equation of line that Fiona has written.

And,Equation of that line will be:

   Multiply equation 1, by any real number , k,where, k≠0.

⇒k y= k x- 5 k

This should be ,equation that had  the same solutions as Fiona’s,where k can take any real number but not 0.

None of the Option

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P = 2(L + W)
P = 14
W = L - 5

14 = 2(L + L - 5)
14 = 2(2L - 5)
14 = 4L - 10
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24/4 = L
6 = L.......the length is 6 inches

W = L - 5
W = 6 - 5
W = 1 <=== the width is 1 inch
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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)

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>
  • In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
  • In second quadrant(π/2 < θ < π), only sin and cosec are positive.
  • In the third quadrant (π < θ < 3π/2), only tangent and cotangent 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

Learn more about sine to cosine conversion here:

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3x = 26-8

3x = 18

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x = 6

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