If there is one table (t=1) then 6 chairs (c=6) can be placed around the table, 2 along the length on each side and 1 at each end.
When t=2, and the tables are end to end (joined at their width) c=10, that is, 4 chairs on each side of the double table and 1 at each end. Each time a table is added c increases by 4 so we can write c=4t+2 the constant 2 being the single chair at each end. If the tables are separated then c=6t.
4 and a half ducks.
That's pretty much the answer.
Answer:
0.54
Step-by-step explanation:
1.2/100*45
Answer:
(3, 5)
Step-by-step explanation:
y = 2x - 1
3x + 2y = 19
In the second equation, substitute 2x - 1 in for y, since y equals 2x - 1 in the first equation. Solve for x.
3x + 2(2x - 1) = 19
3x + 4x - 2 = 19
7x - 2 = 19
7x - 2 + 2 = 19 + 2
7x = 21
7x/7 = 21/7
x = 3
Pick the first equation to solve for y since it says y =. Substitute 3 in for x.
y = 2x - 1
y = 2(3) - 1
y = 6 - 1
y = 5
