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.
Answer:
B.) David should pull his goalie
Step-by-step explanation:
There is no picture given for us to answer the question!
all you have to do is find pairs of factor for both 15 and 56
e.g. 1,15 3,5
1.56 2, 28 4, 14
so three possibilities could be
1/1 x 15/56
3/4 x 5/14
1/2 x 15/28
Answer:
Step-by-step explanation:
$15,000×0.08 = $1,200 interest for 360 days
$1,200 × 30/360 = $100 interest
