Answer:
Step-by-step explanation:
You have to divide -5 from each side so -5/-5x which would be x and divide -5 by 40 which is -8 so x is greater than or equal to -8
Answer: (3,4)
Step-by-step explanation:
Answer:
A= linear, B= linear
Step-by-step explanation:
because im learning this rn too
Answer:
b. The system is not in echelon form because the system is not organized in descending "stair step" pattern so that the index of the leading variables increases from the top to bottom.
Step-by-step explanation:
The given linear system has a equation which is not in echelon form. The echelon is a system which divides the data into rigid hierarchal groups. The given linear equation in not in echelon form as the leading variables are increased from top to bottom indicating descending stair step pattern.
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).