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Volgvan
3 years ago
5

WILL GIVE BRAINLIST! Which set of ordered pairs shows a functional relationship?

Mathematics
2 answers:
MaRussiya [10]3 years ago
3 0

Answer:

C

Step-by-step explanation:

Cloud [144]3 years ago
3 0
It is D not x value repeats
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How do you Solve 9/11 + 3/4
Mazyrski [523]

Answer:

69

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44

Step-by-step explanation:

6 0
4 years ago
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What is the radical part of both ^3 square root 54 and ^3 square root 128 when the expressions are simplified?
Elodia [21]
The radical parts of the two expressions above will be as follows;
a]
^3 sqrt 54=(54)^(1/3)
=(2×27)^(1/3)
=(2)^(1/3)×(27)^(1/3)
=(2)^(1/3)×(3³)^(1/3)
=3(2)^(1/3)
or
3(^3√2)
Hence the answer is 3(2)^(1/3)=3(^3√2)

a] ^3sqrt 128=(128)^(1/3)
=(2×64)^(1/3)
=2^(1/3)×64^(1/3)
=2^(1/3)×(4³)^(1/3)
=2^(1/3)×4^(3×1/3)
=2^(1/3)×4
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6 0
4 years ago
Find any domain restrictions on the given rational equation:
Maurinko [17]

Answer:

\dfrac{x}{2 \cdot x + 14} + \dfrac{x - 4}{6} = \dfrac{3}{x^2} + 2\cdot x - 35; Domain \ restriction \ x \neq 0 \   or \  -7

Step-by-step explanation:

The given rational equation is presented here as follows;

\dfrac{x}{2 \cdot x + 14} + \dfrac{x - 4}{6} = \dfrac{3}{x^2} + 2\cdot x - 35

A domain restriction are the limits to the ranges of input values (x-values) of a function

The three main types of domain restrictions are the reciprocal function, the log function, and the root function

The form of restriction in the given rational are reciprocal form, which are;

\dfrac{x}{2 \cdot x + 14}, and \dfrac{3}{x^2}, from which the function is undefined when;

2·x + 14 = 0, therefore when x = -7, or x² = 0, when x = 0

Therefore, the domain restrictions are that the function is defines for all <em>x</em>, except x = -7 and x = 0

The domain restrictions are x ≠ -7 and x ≠ 0.

5 0
3 years ago
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Sergio [31]
Hi,

f(x)=x+7\\&#10;&#10;g(x)= \dfrac{1}{x-13} \\&#10;&#10;(fog)(x)=f(g(x))=f(\dfrac{1}{x-13})=\dfrac{1}{x-13}+7\\&#10;&#10;&#10;
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8 0
3 years ago
Simplify for this question !
PolarNik [594]

Answer:

=6x^9-3 [12/2=6]

=6x^6

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3 years ago
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