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denis23 [38]
2 years ago
8

Does this set of ordered pairs represent a function? {(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)} A. The relation is a function. E

ach input value is paired with more than one output value. B. The relation is a function. Each input value is paired with one output value. C. The relation is not a function. Each input value is paired with only one output value. D. The relation is not a function. Each input value is paired with more than one output value.
Mathematics
1 answer:
Aleks04 [339]2 years ago
7 0

The correct option regarding whether the relation is a function is:

B. The relation is a function. Each input value is paired with one output value.

<h3>When does a relation represent a function?</h3>

A relation represent a function if <u>each value of the input is paired with one value of the output</u>.

In this problem, when the input - output mappings are given by:

{(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)}.

Which means that yes, each input value is paired with one output value, hence the relation is a function and option B is correct.

A similar problem, in which we verify if a relation is a function, is given at brainly.com/question/12463448

#SPJ1

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Answer:

See Explanation

Step-by-step explanation:

\frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ LHS = \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}   \\  \\  =  \frac{ \sin5A \:\cos A -  \cos5A \:  \sin A}{\sin A \:\cos A }  \\  \\  =  \frac{ \sin(5A -A )}{\sin A \:\cos A}  \\  \\ =  \frac{ \sin 4A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =  \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A}  \\  \\ =4\cos 2A \\  \\  = RHS \\  \\ thus \\  \\  \frac{ \sin5A}{\sin A}  -  \frac{ \cos5A}{\cos A}  = 4\cos2A \\  \\ hence \: proved

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3 years ago
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Evaluate the expression when x=3 , y=−4 , and z=−6.<br> z−2x/y
hichkok12 [17]

Answer:

9/2

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3 years ago
A segment in the complex plane has a midpoint at –1 + i. If the segment has an endpoint at –5 – 7i, what is the other endpoint?
morpeh [17]

Answer:

The other end point is: s+ti = 3+9i

Step-by-step explanation:

Mid-Point(M) in the complex plane states that the midpoint of the line segment joining two complex numbers a+bi and s+ti is the  average of the numbers at the endpoints.

It is given by:    M = \frac{a+s}{2} +(\frac{b+t}{2})i

Given: The midpoint = -1 + i and the segment has an endpoint at -5 - 7i

Find the other endpoints.

Let a + bi = -5 -7i  and let other endpoint s + ti (i represents imaginary )

Here, a = -5 and b = -7 to find s and t.

then;

-1+i = \frac{-5+s}{2} + ( \frac{-7+t}{2})i     [Apply Mid-point formula]

On comparing both sides

we get;

-1 = \frac{-5+s}{2}  and  1 = \frac{-7+t}{2}

To solve for s:

-1 = \frac{-5+s}{2}

or

-2 = -5+s

Add 5 to both side we have;

-2+5 = -5+s+5

Simplify:

3 = s or

s =3

Now, to solve for t;

1 = \frac{-7+t}{2}

2 =-7+t

Add 7 to both sides we get;

2+7 = -7+t+7

Simplify:

9 = t

or

t =9

Therefore, the other end point (s+ti) is, 3+9i




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