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)
Answer:
Step-by-step explanation:
Answer:
There is nothing but your question and you should try on your own
Step-by-step explanation:
Answer:
Step-by-step explanation:
Take the beginning number and add or subtract each transaction to get a new balance. For example,
349.45
- 23.42 = 326.03
- 14.95 = 311.08
+ 276.50 = 587.58
- 219.93 = 367.65
- 76.84 = 290.81
6(x + 2) = 2(x + 10)
6x + 12 = 2x + 20
6x - 2x = -12 + 20
4x = 8
x = 8/4
x = 2
3(y - 4) + 2 = 5y - 28
3y - 12 + 2 = 5y - 28
3y -10 = 5y - 28
3y - 5y = 10 - 28
-2y = -18
y = -18/-2
y = 9
5(2z - 3) - 3z = 3(z + 3)
10z - 15 - 3z = 3z + 9
7z - 15 = 3z + 9
7z - 3z = 15 + 9
4z = 24
z = 24/4
z = 6
