Assume (a,b) has a minimum element m.
m is in the interval so a < m < b.
a < m
Adding a to both sides,
2a < a + m
Adding m to both sides of the first inequality,
a + m < 2m
So
2a < a+m < 2m
a < (a+m)/2 < m < b
Since the average (a+m)/2 is in the range (a,b) and less than m, that contradicts our assumption that m is the minimum. So we conclude there is no minimum since given any purported minimum we can always compute something smaller in the range.
This statement is false.
If he were to double the amount that Emily has to get franks, franks would have more.
Look :)
Emily's = 300
Frank's = 300 x 2 = 600
now...
Emily's = 300
Frank's = 600
So frank has more! NOT fewer.
Answer:
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Step-by-step explanation:
Given
Required
Determine 
We know that:
This gives:
Collect like terms
Take square roots
By tan identity
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For x > 2, you would circle the 2, but you would not shade it in because it is greater than/ less than, not greater than or equal to/less than or equal to. From there on, you would move it to the right, because here x is greater than 2, and the numbers to the right of 2 are greater than two.
For x < 2, you would again circle the 2, but don't shade it in. And then you would make a line heading towards the left, because the numbers to the left of 2 are smaller than 2, and here, x is smaller than 2, since 2 is greater.
A value that is not the solution to either inequality is 2. 2 is not greater than 2, and 2 is not less than 2.
