16 divided by 1/8 - 1
first, you would make 16 into a fraction, making the new equation 16/1 divided by 1/8 -1. you'd then flip 1/8 to 8/1 making the new equation 16 divided by 8 - 1. 16 divided by 8 is 2, and 2 - 1 = 1.
the answer is 1.
Answer: THE ANSER IS A. No, the domain value 5 corresponds to two range values, -8 and 5.
Find the domain and range.
Domain:
{
6
, 5 ,
1
}
Range:
{
−
7
, −
8 ,
4 ,
5
}
Step-by-step explanation: IT'S NOT C OR D
Determine if the relation is a function.
The relation is not a function.
Answer:
1/2
Step-by-step explanation:
3/8 + 2/8 is 5/8, 1/2 is equal to 4/8, so 5/8 is only 1/8 away from 1/2 while 5/8 is also 3/8 away from 1 or 8/8 therefore, it is closer to 1/2
So I'm going to assume that this question is asking for <u>non extraneous solutions</u>, or solutions that are found in the equation <em>and</em> are valid solutions when plugged back into the equation. So firstly, subtract 2 on both sides of the equation:

Next, square both sides:

Next, subtract x and add 2 to both sides of the equation:

Now we are going to be factoring by grouping to find the solution(s). Firstly, what two terms have a product of 6x^2 and a sum of -5x? That would be -3x and -2x. Replace -5x with -2x - 3x:

Next, factor x^2 - 2x and -3x + 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

Now you can rewrite the equation as 
Now, apply the Zero Product Property and solve for x as such:

Now, it may appear that the answer is C, however we need to plug the numbers back into the original equation to see if they are true as such:

Since both solutions hold true when x = 2 and x = 3, <u>your answer is C. x = 2 or x = 3.</u>