The first term, a, is 2. The common ratio, r, is 4. Thus,
a_(n+1) = 2(4)^(n).
Check: What's the first term? Let n=1. Then we get 2(4)^1, or 8. Is that correct? No.
Try this instead:
a_(n) = a_0*4^(n-1). Is this correct? Seeking the first term (n=1), does this formula produce 2? 2*4^0 = 2*1 = 2. YES.
The desired explicit formula is a_(n) = a_0*4^(n-1), where n begins at 1.
D because I’ve done this test before
Answer:
Infinite Solutions
Step-by-step explanation:
x + 2y = 10
6y = 3x - 30
To solve for x and y we use substitution method
Let's solve the first equation for x
x + 2y = 10
Subtract 2y on both sides
x = 10 - 2y
Now plug in x in second equation
6y = -3x + -30
6y = -3 (10-2y) - 30
6y = -30 + 6y - 30
6y = 6y
Both sides are the same, so both x and y have infinite solutions.
