Question

HELPPPP!! I seriously need help

Attachment below

Answer 1

@electrokid can you help me please?

Answer 2

\[f(x)=A\left(1+{r\over100}\right)^x\]

Answer 3

for D, A = 23 and rate is 6

Answer 4

so, plugging it in,
\[f(x)=23\left(1+{6\over100}\right)^x\\
\boxed{f(x)=23(1.06)^x}\]

Answer 5

What do I do for B? @electrokid

Answer 6

from the information in table1,
notice that we can write the function as:
\[f(x)=22(a)^x\] where 'a' is unknown.
now, also given that when x = 1, f(x)=21.56
so,plugging this in,
\[21.56=22(a)^1\\{21.56\over22}=a=?\]solve that fraction and you get the value for a.
plug that value in the above.

ISSUE: the question says that there is a growth. However, the value for f(x) has decreased instead of increasing!

Answer 7

I got 0.98 after solving that.. yea it decreased. @electrokid

Answer 8

I thought "A" was the number outside of the parenthesis, and was the Y-intercept?

Answer 9

good, so,\[f(x)=22(0.98)^x\]
so, the rate of growth is given by:
\[1+{r\over100}=0.98\]
solve for r
now, 'r' will be negative indicating a negative growth!

Answer 10

wouldn't 98/100 give you 0.98? @electrokid

Answer 11

@Phi I'm working on the same problem, and she and I are both stuck.. can you help?

Answer 12

@jdoe0001?