In order to have infinitely many solutions with linear equations/functions, the two equations have to be the same;
In accordance, we can say:
(2p + 7q)x = 4x [1]
(p + 8q)y = 5y [2]
2q - p + 1 = 2 [3]
All we have to do is choose two equations and solve them simultaneously (The simplest ones for what I'm doing and hence the ones I'm going to use are [3] and [2]):
Rearrange in terms of p:
p + 8q = 5 [2]
p = 5 - 8q [2]
p + 2 = 2q + 1 [3]
p = 2q - 1 [3]
Now equate rearranged [2] and [3] and solve for q:
5 - 8q = 2q - 1
10q = 6
q = 6/10 = 3/5 = 0.6
Now, substitute q-value into rearranges equations [2] or [3] to get p:
p = 2(3/5) - 1
p = 6/5 - 1
p = 1/5 = 0.2
Answer:
Bsjwkwkwkkenenrnrnrbtbbtbfbfnrntnt rnrnrkkrjrjr
Step-by-step explanation:
bdbrbrbr
Answer:
Step-by-step explanation:
The sample space should show all the possible outcomes.
The first option includes all 4 colors and both head and tails.
The second option is missing tails as an outcome.
The third option is missing blue.
The fourth option is missing tails.
So it must be the first one.
Maybe it would be B
hope it helps
