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lorasvet [3.4K]
3 years ago
11

Vhich of the following functions have graphs that oscillate? Select all that apply

Mathematics
1 answer:
gladu [14]3 years ago
6 0

Answer:

Only options A and B are correct, i.e. y = sin∅ and y = cos∅.

Step-by-step explanation:

The graphs that oscillate, should be periodic in nature and continuous functions.

We know all trigonometric functions are periodic in nature. But not all trigonometric functions are continuous.

The function is continuous if its values are defined at all angles and it should not be undefined for any particular angle.

We know the following facts:-

tan(90°) = ∞

sec(90°) = ∞

csc(0°) = ∞

cot(0°) = ∞

<u>Only sin and cos don't have values ∞ at any angle.</u> It means only sin and cos are continuous functions.

Hence, only options A and B are correct, i.e. y = sin∅ and y = cos∅.

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Solve the following equation simultaneously 1/x-5/y=7, 2/x+1/y=3​
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Answer:

  (x, y) = (1/2, -1)

Step-by-step explanation:

Subtracting twice the first equation from the second gives ...

  (2/x +1/y) -2(1/x -5/y) = (3) -2(7)

  11/y = -11 . . . . simplify

  y = -1 . . . . . . . multiply by y/-11

Using the second equation, we can find x:

  2/x +1/-1 = 3

  2/x = 4 . . . . . . . add 1

  x = 1/2 . . . . . . . multiply by x/4

The solution is (x, y) = (1/2, -1).

_____

<em>Additional comment</em>

If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.

A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).

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