We'll be assuming this is with simple interest.
now, a year has 12 months, thus in 18 months there are 18/12 years.
![\bf 1050=1020\left[ 1+r\left( \frac{3}{2} \right) \right]\implies \cfrac{1050}{1020}=1+r\left( \frac{3}{2} \right) \implies \cfrac{35}{34}=1+r\left( \frac{3}{2} \right) \\\\\\ \cfrac{35}{34}-1=\cfrac{3}{2}r\implies \cfrac{1}{34}=\cfrac{3}{2}r\implies \cfrac{2\cdot 1}{3\cdot 34}=r\implies \cfrac{1}{51}=r \\\\\\ r\%=100\cdot \cfrac{1}{51}\implies r\approx\stackrel{\%}{1.96078431372549}](https://tex.z-dn.net/?f=%5Cbf%201050%3D1020%5Cleft%5B%201%2Br%5Cleft%28%20%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20%5Ccfrac%7B1050%7D%7B1020%7D%3D1%2Br%5Cleft%28%20%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Ccfrac%7B35%7D%7B34%7D%3D1%2Br%5Cleft%28%20%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B35%7D%7B34%7D-1%3D%5Ccfrac%7B3%7D%7B2%7Dr%5Cimplies%20%5Ccfrac%7B1%7D%7B34%7D%3D%5Ccfrac%7B3%7D%7B2%7Dr%5Cimplies%20%5Ccfrac%7B2%5Ccdot%201%7D%7B3%5Ccdot%2034%7D%3Dr%5Cimplies%20%5Ccfrac%7B1%7D%7B51%7D%3Dr%0A%5C%5C%5C%5C%5C%5C%0Ar%5C%25%3D100%5Ccdot%20%5Ccfrac%7B1%7D%7B51%7D%5Cimplies%20r%5Capprox%5Cstackrel%7B%5C%25%7D%7B1.96078431372549%7D)
W=15
Subtract 80 from both sides then you divide by -2w.
So if 2 is the first term then if n=1 then n+1=sequence so far but
second term is 9 so if n=2 then the 2nd term=9
you must use trial and error to find
so it is
7n-5 so if n=1 then
7(1)-5=7-5=2
if n=2 then
7(2)-5=14-5=9
sucess
the equation is 7n-5
For question 9 - It is stating that AE is equal to 38. So the length of the entire line is 38. We then need to look at which figures are actually useful to us, which in this case is AC which is 10. Therefore the answer is as simple as 38-10, which gives us an answer of 28 for the other half of the line, notated as CE.
The answer would be 1.11t or t=1.11 same thing
