Answer:
-x¹⁴ / 5040
-½ < x < ½
Step-by-step explanation:
f(x) = e^(-x²)
The Taylor series for eˣ centered at 0 is:
eˣ = ∑ (1/n!) xⁿ
Substitute -x²:
e^(-x²) = ∑ (1/n!) (-x²)ⁿ
e^(-x²) = ∑ (1/n!) (-1)ⁿ x²ⁿ
The 14th degree term occurs at n=7.
(1/7!) (-1)⁷ x¹⁴
-x¹⁴ / 5040
ln(1 + x) = ∑ₙ₌₁°° (-1)ⁿ⁺¹ xⁿ / n
If we substitute 4x²:
ln(1 + 4x²) = ∑ₙ₌₁°° (-1)ⁿ⁺¹ (4x²)ⁿ / n
Using ratio test:
lim(n→∞)│aₙ₊₁ / aₙ│< 1
lim(n→∞)│[(-1)ⁿ⁺² (4x²)ⁿ⁺¹ / (n+1)] / [(-1)ⁿ⁺¹ (4x²)ⁿ / n]│< 1
lim(n→∞)│-1 (4x²) n / (n+1)│< 1
4x² < 1
x² < ¼
-½ < x < ½
Answer:
72 sq. mi
Step-by-step explanation:
Breaking this down, we have 2 right triangles with sides of 3, 4, and 5 miles, and 3 rectangles with dimensions 3 x 5, 4 x 5, and 5 x 5 miles. Remember that the area of a triangle is 1/2 x b x h , where b and h are the triangle's base and height. The base and height of the triangles at the bases of the figure are 3 and 4, so each triangle has an area of 1/2 x 3 x 4 = 1/2 x 12 = 6 sq. mi, or 6 + 6 = 12 sq. mi together.
Onto the rectangles, we can find their area by multiplying their length by their width. Since the width of these rectangles is the same for all three - 5 mi - we can make our lives a little easier and just "glue" the lengths together, giving us a longer rectangle with a length of 3 + 4 + 5 = 12 mi. Multiplying the two, we find the area of the rectangles to be 5 x 12 = 60 sq. mi.
Adding this area to the triangle's area gives us a total area of 12 + 60 = 72 sq. mi.
Answer:
4mm
Step-by-step explanation:
you have to add both sides after substituting each of the 4 sides by 1mm
which gives you the total of 4
