Is xy/y^2 a polynomial
1 answer:
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Answer: answer witten in the picture attached to this......
Answer:
The ordered pair make both inequalities true
Step-by-step explanation:
we have
------> inequality A
------> inequality B
we know that
If a ordered pair satisfy both inequalities
then
the ordered pair is a solution of both inequalities
we're going to verify every case
<u>case A)</u> point
Substitute the values of x and y in both inequalities
<u>Inequality A</u>
------> is not true
therefore
The ordered pair does not make both inequalities true
It's not necessary to verify the B inequality
<u>case B)</u> point
Substitute the values of x and y in both inequalities
<u>Inequality A</u>
------> is true
<u>Inequality B</u>
------> is true
therefore
The ordered pair make both inequalities true
<u>case C)</u> point
Substitute the values of x and y in both inequalities
<u>Inequality A</u>
------> is not true
therefore
The ordered pair does not make both inequalities true
It's not necessary to verify the B inequality
<u>case D)</u> point
Substitute the values of x and y in both inequalities
<u>Inequality A</u>
------> is not true
therefore
The ordered pair does not make both inequalities true
It's not necessary to verify the B inequality