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harina [27]
3 years ago
12

A linear equation exists for the following sets of ordered pairs: {(-1, 1), (0, 3), (-1/3, 7/3)}. The rule for the linear equati

on states that y is three more than twice x. Set up a linear equation that satisfies the rule, and use the set of ordered pairs to prove that your equation satisfies the set of coordinates.
Mathematics
2 answers:
olya-2409 [2.1K]3 years ago
8 0
The rule of the linear equation which is 'y is three more than twice x' can be translated into y = 2x + 3. Thus, we simply use the points and subtitute it to the y and x variables. 

1st point: (-1, 1)
1 = 2(-1) + 3
1 = -2 + 3
1=1 

2nd point: (0, 3)
3 = 2(0) + 3 
3 =3

3rd point: (-1/3, 7/3)
7/3 = 2(-1/3) + 3
7/3 = -2/3 + 3
 7/3 = 7/3

All points prove the equation satifsfies the set of coordinates.
marshall27 [118]3 years ago
4 0
From the statement:
y is three more than twice x

We can generate the equation:
y = 3 + 2x

or
y = 2x + 3

Testing the equation with the given ordered pairs:
For (-1,1)
1 = 2(-1) + 3
1 = 1 (correct!)

For (0,3)
3 = 2(0) + 3
3 = 3 (correct!)

For (-1/3, 7/3)
7/3 = 2(-1/3) + 3
7/3 = 7/3 (correct)
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Simplify (8^4/10^4)^-1/4using rational exponents
Vlad1618 [11]

Answer:

   (\frac{8^4}{10^4} )^{-\frac{1}{4} }  = \frac{5}{4}

Step-by-step explanation:

<u><em>Explanation</em></u>

Given  

       (\frac{8^4}{10^4} )^{-\frac{1}{4} }

By using algebra formulas

  (\frac{a}{b} )^{m} = \frac{a^m}{b^n}

(\frac{8^4}{10^4} )^{-\frac{1}{4} } = \frac{(8^{4})^{\frac{-1}{4} }  }{(10^{4} )^{\frac{-1}{4} } }

(a^{m} )^{n} = a^{mn}

\frac{(8^{4})^{\frac{-1}{4} }  }{(10^{4} )^{\frac{-1}{4} } } = \frac{8^{4X\frac{-1}{4} } }{10^{4 X\frac{-1}{4} } }

          = \frac{8^{-1} }{10^{-1} }

           = \frac{10}{8} \\= \frac{5}{4}

<u><em>Final answer:-</em></u>

 (\frac{8^4}{10^4} )^{-\frac{1}{4} }  = \frac{5}{4}

7 0
3 years ago
Can you guys help me with this question?
IRINA_888 [86]
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So the second blank would be 42.
Since 7*6=42.

The third blank would equal to 56.
Since 7*8=56
5 0
4 years ago
How do you work out 5 2/3 - 2 3/4 ???
loris [4]
Hi,
i worked out the problem and I attached the picture  

6 0
3 years ago
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