All categories

3 years ago

Which of the sets of ordered pairs represents a function?

Mathematics

2 answers:

pishuonlain [190]3 years ago

6 0

The answer is:

D - Neither A nor B

-Neither of the ordered pairs shows a straight linear line, and thus, they're not a function.

Edit 3/4/19: D is <em>not</em> the correct answer! I took the test, and it was the only question I had gotten wrong. Neither this answer or the answer above is correct.

alexandr1967 [171]3 years ago

3 0

Answer:

The correct option is 4. Neither A nor B represents a function.

Step-by-step explanation:

The given sets of ordered pairs are

$A=\{(1,-5), (8,-5), (8,7), (2,9)\}$

$B=\{(7,-4), (7,-2), (6,-3), (-9,5)\}$

A set of ordered pairs represents a function if there exist unique outputs for all inputs. It means for each values of x there exist, a unique value of y.

In set A the value of y-coordinates are -5 and 7 at $x=8$ .

At x=8, there exist more than one value of y, so the set A is not a function.

In set B the value of y-coordinates are -4 and -2 at $x=7$ .

At x=7, there exist more than one value of y, so the set B is not a function.

Therefore neither A nor B represents a function and option 4 is correct.

You might be interested in

In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.

DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

$(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2$

$+ 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5$

$5C_0(\frac{1}{ax})^5(2ax^2)^0 =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1 = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\$

$5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}$

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

6 0

3 years ago

You are given a positive number p, but you are not told what the specific value of p is. A vertical light beam comes from high

Zarrin [17]

Answer:

(x, y) = (0, p)

Step-by-step explanation:

In the given equation, p is the distance from the vertex (0, 0) to the focus of the parabola. Any vertical "light beam" will bounce off the curve and pass through the focus at (0, p).

4 0

3 years ago

What is 117.456 rounded to the nearest tenth

mylen [45]

I believe, the answer is 117.5

5 0

3 years ago

Read 2 more answers

Please answer a and b, I will give brainliest!!!!!

aniked [119]

Answer:

Step-by-step explanation: