Answer:
![\displaystyle x=-\frac{1}{6}, 3](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%3D-%5Cfrac%7B1%7D%7B6%7D%2C%203)
Step-by-step explanation:
Hi there!
![y = 6x^2 - 17x - 3](https://tex.z-dn.net/?f=y%20%3D%206x%5E2%20-%2017x%20-%203)
Factor by grouping:
![y = 6x^2 - 18x+1x - 3\\y = 6x(x- 3)+1x - 3\\y = 6x(x- 3)+(x - 3)\\y = (6x+1)(x- 3)](https://tex.z-dn.net/?f=y%20%3D%206x%5E2%20-%2018x%2B1x%20-%203%5C%5Cy%20%3D%206x%28x-%203%29%2B1x%20-%203%5C%5Cy%20%3D%206x%28x-%203%29%2B%28x%20-%203%29%5C%5Cy%20%3D%20%286x%2B1%29%28x-%203%29)
Let y=0. Apply the zero product property:
![0 = (6x+1)(x- 3)](https://tex.z-dn.net/?f=0%20%3D%20%286x%2B1%29%28x-%203%29)
![6x+1=0\\6x=-1\\\\\displaystyle x=-\frac{1}{6}](https://tex.z-dn.net/?f=6x%2B1%3D0%5C%5C6x%3D-1%5C%5C%5C%5C%5Cdisplaystyle%20x%3D-%5Cfrac%7B1%7D%7B6%7D)
AND
![x-3=0\\x=3](https://tex.z-dn.net/?f=x-3%3D0%5C%5Cx%3D3)
I hope this helps!
Hi,
You just have to change z^25 to z^24 or z^27.
![125x^{18}y^3z^{24}=(5x^6*y*z^8)^3](https://tex.z-dn.net/?f=%20125x%5E%7B18%7Dy%5E3z%5E%7B24%7D%3D%285x%5E6%2Ay%2Az%5E8%29%5E3%20)
Answer:
5 years and 5 months
Step-by-step explanation:
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<u>Compound Interest Formula</u>
![\large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}](https://tex.z-dn.net/?f=%5Clarge%20%5Ctext%7B%24%20%5Csf%20A%3DP%281%2B%5Cfrac%7Br%7D%7Bn%7D%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 = $17,474.00
- P = $7,790.00
- r = 15% = 0.15
- n = 12
- t = number of years
Substitute the given values into the formula and solve for t:
![\implies \sf 17474=7790\left(1+\dfrac{0.15}{12}\right)^{12t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%2017474%3D7790%5Cleft%281%2B%5Cdfrac%7B0.15%7D%7B12%7D%5Cright%29%5E%7B12t%7D)
![\implies \sf \dfrac{17474}{7790}=\left(1.0125}\right)^{12t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Cdfrac%7B17474%7D%7B7790%7D%3D%5Cleft%281.0125%7D%5Cright%29%5E%7B12t%7D)
![\implies \sf \ln\left(\dfrac{17474}{7790}\right)=\ln \left(1.0125}\right)^{12t}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Cln%5Cleft%28%5Cdfrac%7B17474%7D%7B7790%7D%5Cright%29%3D%5Cln%20%5Cleft%281.0125%7D%5Cright%29%5E%7B12t%7D)
![\implies \sf \ln\left(\dfrac{17474}{7790}\right)=12t \ln \left(1.0125}\right)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Cln%5Cleft%28%5Cdfrac%7B17474%7D%7B7790%7D%5Cright%29%3D12t%20%5Cln%20%5Cleft%281.0125%7D%5Cright%29)
![\implies \sf t=\dfrac{\ln\left(\frac{17474}{7790}\right)}{12 \ln \left(1.0125}\right)}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20t%3D%5Cdfrac%7B%5Cln%5Cleft%28%5Cfrac%7B17474%7D%7B7790%7D%5Cright%29%7D%7B12%20%5Cln%20%5Cleft%281.0125%7D%5Cright%29%7D)
![\implies \sf t=5.419413037...\:years](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20t%3D5.419413037...%5C%3Ayears)
Therefore, the money was in the account for 5 years and 5 months (to the nearest month).
The answer is D. The number of AP tests increases as GPA increases.
Answer:
wild
Step-by-step explanation: