Answer:
A quick hack is often to partially express some function in terms of a Taylor approximation about x0, since higher order terms of x go to zero if we are considering limits for (x−x0)→0. To really answer your question we need to know what the original question was, that is, about which point do you want the expansion? Let us assume around 0. Then we have the Maclaurin series:
cos(x)=1−12x2+O(x4)
You can add more terms if you need to. Now we write:
ln(1+(−12x2))=…
Do you know the standard Maclaurin series for this function?
Hint: it is of the form ln(1+u)
Step-by-step explanation:
Answer:
120 customers on the 2nd day
Step-by-step explanation:
100/1ST DAY
120/2nd day
140/3rd day
160/ 4TH DAY
Answer:
2/10
Step-by-step explanation:
you have to match the bottom number on both fractions then do what ever you did to the top number
so with this to get the 5 to a 10 you have to times it by 2
so you do the same to the 1
This is a simple Pythagorean Theorem problem. 7 is the hypotenuse so you would do c^2-a^2=b^2. In this case 7^2-2^2=b^2. 7^2 is 49 and 2^2 is 4 so you have 49-4=b^2. 49-4 is 45. The square root of 45 is 6.70. So x=6.70
