Answer:
Step-by-step explanation:
at the beginning of year 4, only 3 years have elapsed, the 4th year hasn't started yet, since it's at the beginning, so at the beginning of year 4 we can say only 4-1 years have elapsed.
![A=700\left( 1 + \frac{0.05}{1} \right)^{1\cdot 3}\implies A = 700(1+0.05)^3\implies A(4)=700(1+0.05)^{4-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A(n)=700(1+0.05)^{n-1}~\hfill](https://tex.z-dn.net/?f=A%3D700%5Cleft%28%201%20%2B%20%5Cfrac%7B0.05%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%203%7D%5Cimplies%20A%20%3D%20700%281%2B0.05%29%5E3%5Cimplies%20A%284%29%3D700%281%2B0.05%29%5E%7B4-1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20A%28n%29%3D700%281%2B0.05%29%5E%7Bn-1%7D~%5Chfill)
Answer:
70.3km
Step-by-step explanation:
When the bearings are visualized and the points connected as a triangle, The angle x would be the difference of the both bearings from x
i.e 117-41 = 76
Using cosine rule,
YZ²= 43² + 67² - 2 x 43 x 67
cos76
YZ= √4944
= 70.3
For num 4 the answer to six is 22 the reason is this formula so write this down for n aswell
n(n+1)/2+1 the example could be as we know for 1 cut we always will have 2 slices so 1(1+1)/2+1 is 2 and for 6 cuts 6(6+1)/2+1 is 22 therefore the answer to 6 cuts is 22 and to n cuts is the formula n(n+1)/2+1
