a year has 12 months, driving 1000 miles a month and changing the oil every 4000 miles
let 'x' represent the number of oil changes in one year:
x=(12*1000)/4000
x=3
So 3 oil changes would be obtained throughout the year.
Step-by-step explanation:
f(x) = (x + 2)(x - 3)(x + 12) = 0
(x + 2)(x - 3)(x + 12) = 0
so, the zeros of the function :
=> x1 = -2
=> x2= 3
=> x3 = -12
Answer:
3.25 years (nearest hundredth)
Step-by-step explanation:
<u>Compound Interest Formula</u>
![\large \text{$ \sf A=P\left(1+\frac{r}{n}\right)^{nt} $}](https://tex.z-dn.net/?f=%5Clarge%20%5Ctext%7B%24%20%5Csf%20A%3DP%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D%20%24%7D)
where:
- A = final amount
- P = principal amount
- r = interest rate (in decimal form)
- n = number of times interest applied per time period
- t = number of time periods elapsed
Given:
- A = $4916
- P = $4000
- r = 6.4% = 0.064
- n = 4 (quarterly)
Substitute the given values into the formula and solve for t:
![\implies \sf 4916=4000\left(1+\frac{0.064}{4}\right)^{4t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%204916%3D4000%5Cleft%281%2B%5Cfrac%7B0.064%7D%7B4%7D%5Cright%29%5E%7B4t%7D)
![\implies \sf 4916=4000\left(1.016\right)^{4t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%204916%3D4000%5Cleft%281.016%5Cright%29%5E%7B4t%7D)
![\implies \sf \dfrac{4916}{4000}=\left(1.016\right)^{4t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Cdfrac%7B4916%7D%7B4000%7D%3D%5Cleft%281.016%5Cright%29%5E%7B4t%7D)
![\implies \sf \dfrac{1229}{1000}=\left(1.016\right)^{4t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Cdfrac%7B1229%7D%7B1000%7D%3D%5Cleft%281.016%5Cright%29%5E%7B4t%7D)
![\implies \sf \ln \left(\dfrac{1229}{1000}\right)=\ln \left(1.016\right)^{4t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Cln%20%5Cleft%28%5Cdfrac%7B1229%7D%7B1000%7D%5Cright%29%3D%5Cln%20%5Cleft%281.016%5Cright%29%5E%7B4t%7D)
![\implies \sf \ln \left(\dfrac{1229}{1000}\right)=4t \ln \left(1.016\right)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Cln%20%5Cleft%28%5Cdfrac%7B1229%7D%7B1000%7D%5Cright%29%3D4t%20%5Cln%20%5Cleft%281.016%5Cright%29)
![\implies \sf t=\dfrac{\ln \left(\frac{1229}{1000}\right)}{4\ln \left(1.016\right)}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20t%3D%5Cdfrac%7B%5Cln%20%5Cleft%28%5Cfrac%7B1229%7D%7B1000%7D%5Cright%29%7D%7B4%5Cln%20%5Cleft%281.016%5Cright%29%7D)
![\implies \sf t=3.247594892...](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20t%3D3.247594892...)
Therefore, it will take 3.25 years (nearest hundredth) for the account to grow to $4916.
Learn more about compound interest here:
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Answer:
The answer is D
Step-by-step explanation:
The only logical explanation
Answer:
C or A
Step-by-step explanation:
I think this is write