564. multiply 80 by 7, then add 4
Answer:
In linear functions, rate of change is constant: as x goes up, y will go up a consistent amount. In exponential functions, the rate of change increases by a consistent multiplier
Step-by-step explanation:
Answer:
The value of home after t years is $249,000
Step-by-step explanation:
Given as :
The original value of home = p = $249,000
The value of home depreciate every year at rate = r = 7.1%
Let The value of home after t years = A
<u>Now, According to question</u>
The value of home after t years = original value of home × ![(1-\dfrac{\textrm rate}{100})^{\textrm time}](https://tex.z-dn.net/?f=%281-%5Cdfrac%7B%5Ctextrm%20rate%7D%7B100%7D%29%5E%7B%5Ctextrm%20time%7D)
Or, A = p × ![(1-\dfrac{\textrm r}{100})^{\textrm t}](https://tex.z-dn.net/?f=%281-%5Cdfrac%7B%5Ctextrm%20r%7D%7B100%7D%29%5E%7B%5Ctextrm%20t%7D)
Or, A = $249,000 × ![(1-\dfrac{\textrm 7.1}{100})^{\textrm t}](https://tex.z-dn.net/?f=%281-%5Cdfrac%7B%5Ctextrm%207.1%7D%7B100%7D%29%5E%7B%5Ctextrm%20t%7D)
Or, A = $249,000 × ![(0.929)^{\textrm t}](https://tex.z-dn.net/?f=%280.929%29%5E%7B%5Ctextrm%20t%7D)
So,The value of home after t years = A = $249,000 ![(0.929)^{\textrm t}](https://tex.z-dn.net/?f=%280.929%29%5E%7B%5Ctextrm%20t%7D)
Hence,The value of home after t years is $249,000
. Answer
Answer:
2/6561
Step-by-step explanation:
Geometric sequence formula : ![a_n=a_1(r)^n^-^1](https://tex.z-dn.net/?f=a_n%3Da_1%28r%29%5En%5E-%5E1)
where an = nth term, a1 = first term , r = common ratio and n = term position
given ratio : 1/3 , first term : 2 , given this we want to find the 9th term
to do so we simply plug in what we are given into the formula
recall formula : ![a_n=a_1(r)^n^-^1](https://tex.z-dn.net/?f=a_n%3Da_1%28r%29%5En%5E-%5E1)
define variables : a1 = 2 , r = 1/3 , n = 9
plug in values
a9 = 2(1/3)^(9-1)
subtract exponents
a9 = 2(1/3)^8
evaluate exponent
a9 = 2 (1/6561)
multiply 2 and 1/6561
a9 = 2/6561