Creo q la repuesta es f(10) = 59049
The initial investment = $250
<span>annual simple interest rate of 3% = 0.03
</span>
Let the number of years = n
the annual increase = 0.03 * 250
At the beginning of year 1 ⇒ n = 1 ⇒⇒⇒ A(1) = 250 + 0 * 250 * 0.03 = 250
At the beginning of year 2 ⇒ n = 2 ⇒⇒⇒ A(2) = 250 + 1 * 250 * 0.03
At the beginning of year 3 ⇒ n = 3 ⇒⇒⇒ A(2) = 250 + 2 * 250 * 0.03
and so on .......
∴ <span>The formula that can be used to find the account’s balance at the beginning of year n is:
</span>
A(n) = 250 + (n-1)(0.03 • 250)
<span>At the beginning of year 14 ⇒ n = 14 ⇒ substitute with n at A(n)</span>
∴ A(14) = 250 + (14-1)(0.03*250) = 347.5
So, the correct option is <span>D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50
</span>
Answer:
273 frogs.
Step-by-step explanation:
The equation of growth in the number of frogs N = 137(1.09)^t where t is the time in years.
The year 2022 is 8 years after 2014 so the prediction for N in 2022 is:
N = 137(1.09)^8 = 273.
Answer:
59sq meters
Step-by-step explanation:
A=L*W
A/W=L
165/22=7.5
7.5=Length
P=L+L+W+W
7.5+7.5+22+22=59sq meters
Answer:
z=13
Step-by-step explanation:
SU 1
------ = ------
WV 2
z 1
----- = ----------
z+13 2
Using cross products
2z = z+13
Subtract z from each side
2z-z = z+13-z
z = 13
