Answer:
First map, 1cm:10km
Distance between cleveland and cincinnati :
![40 \times 10km = 400km](https://tex.z-dn.net/?f=40%20%5Ctimes%2010km%20%3D%20400km)
Second map, 1cm: 50km
![\frac{400km}{50km} = 8](https://tex.z-dn.net/?f=%20%5Cfrac%7B400km%7D%7B50km%7D%20%20%3D%208)
Therefore, the distance between the two places in second map will be 8 cm.
Data:
P = 5.300,00
A = 7.000,00
r = 9% = 0,09
n = 4
t = ?
We have the following compound interest formula:
![A = P*(1+ \frac{r}{n})^{nt}](https://tex.z-dn.net/?f=A%20%3D%20P%2A%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D%20)
![7000 = 5300*(1+ \frac{0.09}{4})^{4t}](https://tex.z-dn.net/?f=7000%20%3D%205300%2A%281%2B%20%5Cfrac%7B0.09%7D%7B4%7D%29%5E%7B4t%7D%20)
![7000 = 5300*(1+ 0.0225)^{4t}](https://tex.z-dn.net/?f=7000%20%3D%205300%2A%281%2B%200.0225%29%5E%7B4t%7D)
![7000 = 5300*(1.0225)^{4t}](https://tex.z-dn.net/?f=7000%20%3D%205300%2A%281.0225%29%5E%7B4t%7D)
![(1.0225)^{4t} = \frac{7000}{5300}](https://tex.z-dn.net/?f=%281.0225%29%5E%7B4t%7D%20%3D%20%20%5Cfrac%7B7000%7D%7B5300%7D%20)
![(1.0225)^{4t} = 1.32075](https://tex.z-dn.net/?f=%281.0225%29%5E%7B4t%7D%20%3D%201.32075)
Now, take the natural logarithm of both sides:
![ln(1.0225^{4t}) = ln(1.32075)](https://tex.z-dn.net/?f=ln%281.0225%5E%7B4t%7D%29%20%3D%20ln%281.32075%29)
![4*t*ln(1.0225) = = ln(1.32075)](https://tex.z-dn.net/?f=4%2At%2Aln%281.0225%29%20%3D%20%20%3D%20ln%281.32075%29)
![4*t = \frac{ln(1.32075)}{ln(1.0225)}](https://tex.z-dn.net/?f=4%2At%20%3D%20%20%5Cfrac%7Bln%281.32075%29%7D%7Bln%281.0225%29%7D%20)
![4t = 12.50301769...](https://tex.z-dn.net/?f=4t%20%3D%2012.50301769...)
![4t \approx 12.50302](https://tex.z-dn.net/?f=4t%20%5Capprox%2012.50302)
![t \approx \frac{12.50302}{4}](https://tex.z-dn.net/?f=t%20%5Capprox%20%20%5Cfrac%7B12.50302%7D%7B4%7D%20)
![\boxed{t \approx 3.1258years}](https://tex.z-dn.net/?f=%5Cboxed%7Bt%20%5Capprox%203.1258years%7D)
<span>Therefore, we have: (answer)
</span>
3 years 1 month and 16 days
Thank you for posting your question here, to solve the compound inequality of 5x + 11 ≥ –9 and 10x – 3 ≤ 27 the answer is A which is " x ≥ –4 and x ≤ 3". Below is the solution:
5x + 11 ≥ -9 and 10x - 3
≤ 27<span>
5x ≥ -9 - 11 and 10x ≤ 27 + 3
5x ≥ -20 and 10x ≤ 30
x ≥ -20/5 and x ≤ 30/10
<span>x ≥ -4 and x ≤ 3</span></span>
Answer:
15units
Step-by-step explanation:
BC is: 5x=5×3=15 units. so,Length of BC=15 units