24. Which set of ordered pairs is not a function?

Mathematics

1 answer:

Inessa [10]3 years ago

4 0

Answer:

The set of ordered pairs that is not a function is

a. {(1,-5), (2,-4),(1,-4)}

The reason is that it is not a function is because the function is supposed to be 1 to 1. Each x value for y value

Step-by-step explanation:

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I need help on this, please hurry and thank you!

inna [77]

Answer:

$A.\ a_n=n^2+1$

Step-by-step explanation:

$Check:$

$a_n=n^2+1\\\\a_1=1^2+1=1+1=2\qquad CORRECT\\\\a_2=2^2+1=4+1=5\qquad CORRECT\\\\a_3=3^2+1=9+1=10\qquad CORRECT\\\\a_4=4^2+1=16+1=17\qquad CORRECT\\\\a_5=5^2+1=25+1=26\qquad CORRECT$

Used PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

First Power, next Addition

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3 years ago

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How can you tell whether the parabola had a reflection?

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When it doesn’t look like a U anymore and is upside down
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3 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$