Answer:
They either both have to be positive or negative.
Step-by-step explanation:
1 / 1 = 1
-1 / -1 = 1
This gets you positive, when both dividend and divisors are positive or negative.
-1 / 1 = -1
1 / -1 = -1
This gets you negative, when both dividend and divisors are different signs.
Answer:
(2,-17) should be the minimum.
Step-by-step explanation:
The minimum of a quadratic function occurs at 
. If a is positive, the minimum value of the function is 
occurs at
Find the value of
x = 2
evaluate f(2).
replace the variable x with 2 in the expression.
simplify the result.
The final answer is -17
Use the x and y values to find where the minimum occurs.
HOPE THIS HELPS!
Answer: you would times everything I don’t remember how I did this but I did it by multiplying everything hope I helped and you would add the cm with the answer you got
Here are a few things you'll need to know for this question:
- Domain: <u>The list of x-values that are possible on a line.</u>
- Range: <u>The list of y-values that are possible on a line.</u>
- Interval Notation: <u>Shows the domain/range using the endpoints</u>. Brackets mean that the endpoint is included, parentheses mean the endpoint is excluded. Ex: (2,10]. 2 is excluded, 10 is included.
- Closed Circles: <u>The endpoint is included.</u>
- Open Circles: <u>The endpoint is excluded.</u>
So firstly, let's look at the domain. We see that there is a closed circle at x = -2 and an open circle at x = 5. Using what we know, <u>the interval notation of the domain is [-2,5).</u>
Next, let's look at the range. We see that there is a closed circle at y = -5 and an open circle at y = 2. Using what we know, <u>the interval notation of the range is [-5,2).</u>
