We keep the 2x^2 because we can only subtract something from that if it also is squared. From there, we subtract the 7x from 2x^2-11. Since there is no previous x's in the 2x^2-11, we just make it -7x. So, without even continuing the problem, we see that A) is the correct answer because it is the only one with -7x.
Answer:
A(t) = amount remaining in t years
= A0ekt, where A0 is the initial amount and k is a constant to be determined.
Since A(1690) = (1/2)A0 and A0 = 80,
we have 40 = 80e1690k
1/2 = e1690k
ln(1/2) = 1690k
k = -0.0004
So, A(t) = 80e-0.0004t
Therefore, A(430) = 80e-0.0004(430)
= 80e-0.172
≈ 67.4 g
Step-by-step explanation:
Are you sure about your solves??
I think correct answer is 6,
I don't know but the standard way is in the photo:
35,030,000 is the correct answer, hope this helps.
Answer:
f{0) is greater than g(0) and f(2) is greater than g(2).
Step-by-step explanation:
f{0) is greater than g(0) = f(0)=8 and g(0)=2
f(2) is greater than g(2) = f(2)=8 and g(2)= -4
