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:
Three consecutive numbers are x, x+1 and x+2.
Four times the first integer is 4x
The sum of the second and third is (x+1)+(x+2)=2x+3.
So, we have
Subtract 2x from both sides:
Divide both sides by 2:
So, you can't have three consecutive integers such that four times the first is 18 more than the sum of the other two: the three numbers would be 10.5, 11.5, 12.5.
In fact, you have
and
it can be used to calculate the length of a circular bracelet
Answer:
<u>∗ = 0.4x³</u>
Step-by-step explanation:
(15y + ∗)² = 225y²+12x³y+0.16x⁶
<u>Note:</u>
225y² = 15y * 15y = (15y)²
12x³y = 2 * 15y * 0.4x³
0.16x⁶ = 0.4x³ * 0.4x³ = (0.4x³)²
So, by factoring the right hand side:
225y²+12x³y+0.16x⁶ = (15y + 0.4x³)²
By comparing the left hand side with (15y + 0.4x³)²
<u>So, ∗ should be replaced with the monomial 0.4x³</u>
