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.
Answer:
wow. this is a hard one. i would choose that c would be 2 and d would be 6.
Step-by-step explanation:
well, what i did was first went ahead and reflected the x-axis. then, i made the first translation, c. then, i went ahead and found out the translation of d. YOU COUNT THE SQUARES.
Answer:
C times 23
Step-by-step explanation:
Answer:
domain:{-2, infinity) range:(negative infinity , positive infinity)
Step-by-step explanation:
Domain is all x points and range is all y points.
Answer: C
Step-by-step explanation:
Simplify to 4x^2 + 12x + 5 = 0 so that all the terms are on one side.
Do a part of the quadratic formula to see.
You only need to do the
part. If it is negative, that means there are irrational solutions. If it is positive, it has two solutions. If it is 0, it has 1 solution.

It is positive so it has two solutions.