Just use the coefficients for each variable and put it in a matrix. Is the second question set equal to something? Because then you would put it in place as the question mark. (Did the pic come through?)
So the 3x3 matrix is made up of the coefficients and then to augment the matrix, you add another column at the end with the solutions. Good luck!
Answer:
Step-by-step explanation:
Remember that our original exponential formula was y = a b x. You will notice that in these new growth and decay functions, the b value (growth factor) has been replaced either by (1 + r) or by (1 - r). The growth "rate" (r) is determined as b = 1 + r.
An exponential function of a^x (a>0) is always ln(a)*a^x, as a^x can be rewritten in e^(ln(a)*x). By deriving, the term (ln(a)) gets multiplied with a^x. The derivative shows, that the rate of change is similiar to the function itself. For 0<a<1, ln(a) becomes negative and so is the rate of change.
Linear models are used when a phenomenon is changing at a constant rate, and exponential models are used when a phenomenon is changing in a way that is quick at first, then more slowly, or slow at first and then more quickly.
She will have negative eight dollars. Thats all I can say
A. each mug will be 1/9 because 4/9 divide by 4 is 1/9
b 3/9 or 1/3 because 1/9 times 3 is 3/9
~ JZ
<span>Martin deposits $200
in a savings account that earns 5% annual interest.
year interest balance
1 200 * 5% 200(1.05)
2 200(1.05) * 5% 200(1.05)^2
3 200(1.05)^2*5% 200(1.05)^3
y 200(1.05)^y
=> m = 200 (1.05)^y
four years later,
cary deposits $200 in an account earning the same interest.
</span>
<span><span>year interest balance
5 200 * 5% 200(1.05)
6 200(1.05) * 5% 200(1.05)^2
7 200(1.05)^2*5% 200(1.05)^3
y 200(1.05)^(y-4)
=> c = 200(1.05)^ (y-4)
</span>
Answer:
Martin: 200(1.05)^y
Cary: 200(1.05)^(y–4)</span>
