If <em>a</em> is fixed and <em>b</em>,<em>c</em> are unknowns then the equation <em>b</em>+<em>c</em>=10-<em>a</em> has 11-<em>a</em> solutions. They are pairs (b,c): (0,10-a), (1,9-a), (2,8-a), ... (10-a,0). As <em>a</em> runs from 0 to 10 we have total number of solutions (11-0)+(11-1)+...(11-1)=11+10+...+1=(1+11)*11/2=66.
I've had a problem similar to this before. He should start out with 0 flowers. He would have 2 flowers at the first one, so he would put one on the first grave then have one for the second. When he gets to the second, he puts down one flower and the last flower on the 3rd grave.
I mean, this isn't a question but its true?
Answer:
Step-by-step explanation:
What did you include in your response? Check all that apply.
There would be an open circle at (2, 1). <u>Yes</u>
There would be a closed circle at (2, 3). <u>Yes</u>
There would be an open circle at (4, 3). <u>Yes</u>
There would be a closed circle at (4, −4). <u>Yes</u>
Endpoints that are not included in the domain of a particular piece of a function are represented by an open circle. <u>Yes</u>
