Answer:
Number of year to get $10,000 as interest = 7.5 year (Approx)
Step-by-step explanation:
Given:
Initial investment = $20,000
Interest on investment = $10,000
Rate of interest = 5.6% = 0.056
Find:
Number of year to get $10,000 as interest:
Computation:
![Interest\ on\ investment = Initial\ investment[(1+r)^n-1]\\\\10,000 = 20,000[(1+0.056)-1]\\\\0.5=(1.056)^n-1\\\\1.5 =(1.056)^n\\\\n=7.44](https://tex.z-dn.net/?f=Interest%5C%20on%5C%20investment%20%3D%20Initial%5C%20investment%5B%281%2Br%29%5En-1%5D%5C%5C%5C%5C10%2C000%20%3D%2020%2C000%5B%281%2B0.056%29-1%5D%5C%5C%5C%5C0.5%3D%281.056%29%5En-1%5C%5C%5C%5C1.5%20%3D%281.056%29%5En%5C%5C%5C%5Cn%3D7.44)
Number of year to get $10,000 as interest = 7.44
Number of year to get $10,000 as interest = 7.5 year (Approx)
52 + (-77) =
52 - 77 = -25
----------------------------
52 - (-77) =
52 + 77 = 129
---------------------------
your answer is: 52 - (-77)
44/35 = 1 9/35
So, your answer is 1 9/35
set h=0 and solve for <span>t
</span>So: 0 = 32t - 16^2
<span>32t−16<span>t2</span>=0</span><span>16t(2−t)=0</span><span><span>t=2
</span></span>