Answer:
<em>First even integer: 6</em>
Step-by-step explanation:
<u>Inequalities</u>
Assume x is the first even integer. The next integer is x+2, and the last integer ix x+4.
The condition states that the sum of the first and the second number is 15 less than three times the third. This takes us to the inequality:

Operating:

Subtracting 2 and 2x:

Simplifying:

Solving:
x>5
There are infinitely many solutions. For example, for x=6 (first even number into the solution interval):
First integer: 6
Second integer: 8
Third integer: 10
There are other solutions, like 20,22,24 but the first set is 6,8,10.
<span>(2^1/2x2^3/4)^2
</span><span> ((2^1/2)(2^3/4))^2
</span> ((2^1/2)^2)((2^3/4)^2)
(2)(2^3/2)
(4*2^3)^(1/2)
(2*2*2^3)^(1/2)
(*2^5/2)
The answer for this case is
b. <span>sqrt 2^5</span>
Answer: -1 and 4
Explanation: the others aren’t because they have variables in one and numbers in the other
The answer is D because if you plug in the ordered pair in their correct places the equation becomes 4+6=10.
Answer:
Is there a picture to go with this because otherwise it's not enough info...
Step-by-step explanation: