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Brrunno [24]
4 years ago
8

Which of the following lists of ordered pairs is a function?

Mathematics
2 answers:
likoan [24]4 years ago
4 0
A. (1, 2), (0, 3), (1, 4), (6, 5)
It's not a function, becouse:
(1, 2): 1 ⇒ 2 and (1, 4) - 1 ⇒ 4

B. (9, 4), (5, 0), (1, -4), (-2, -6)
It's a function

C. (4, 2), (4, -2), (9, 3), (16, 4)
It's not a function, becouse:
(4, 2): 4 ⇒ 2 and (4, -2): 4 ⇒ -2

D. (9, 4), (5, 0), (5, 3), (-2, -6)
It's not a function, becouse:
(5, 0): 5 ⇒ 0 and (5, 3): 5 ⇒ 3
Zigmanuir [339]4 years ago
4 0
So a function is basically the x values don't repeat with any differnt y values
(x,y)

the first number should never repeat with a differnt y value

A. 1 repeats with2 and 4, not a function
B.. no repeat, it's a function
C. 4 repeats with 2 and -2,not a function
D. 5 repeats with 0 and 3, not a function


answer is B
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Step-by-step explanation:

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Find the next two terms. 162, 54, 18, ?, ?
morpeh [17]

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6 and 2

Step-by-step explanation:

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3 years ago
I need help Asap please show me work
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392

Step-by-step explanation:

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Use the Quotient Property to Simplify Square Roots Number 129
Inessa05 [86]

Using the quotient property we would have

\begin{gathered} \sqrt{\frac{75r^9}{8^8}}=\frac{\sqrt{75r^9}}{\sqrt{8^8}} \\  \end{gathered}

This follows the rule

\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}

From the simplified expression as shown above

\frac{\sqrt{75r^9}}{\sqrt{8^8}}=\frac{\sqrt{3\times25\times r^9}}{\sqrt{(2^3)^8}}

Thus;

\frac{\sqrt{3\times25r^9}}{\sqrt{(2^3)^8}}=\frac{\sqrt{25}\times\sqrt{3}\times\sqrt{r^9}}{\sqrt{2^{^3\times8}}}=\frac{5\sqrt{3r^9}}{\sqrt{2^{24}}}

Therefore

\begin{gathered} Using\text{ fraction index law we could simplify the denominator} \\ \frac{5\sqrt{3r^9}}{2^{\frac{24}{2}}}=\frac{5\sqrt{3r^9}}{2^{12}} \end{gathered}

We can not simplify the 3 and the r raised to power of 9 as their power is not even, hence the final answer is given below

\frac{5\sqrt{3r^9}}{2^{12}}=\frac{5\sqrt{3r^9}}{4096}

The final answer is :

\frac{5\sqrt{3r^9}}{4096}

8 0
1 year ago
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