Answer:
The integral for the arc of length is:
By using Simpon’s rule we get: 1.5355453
And using technology we get: 2.3020
The approximation is about 33% smaller than the exact result.
Explanation:
The formula for the length of arc of the function f(x) in the interval [a,b] is:
![\displaystyle\int_a^b \sqrt{1+[f'(x)]^2}dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_a%5Eb%20%5Csqrt%7B1%2B%5Bf%27%28x%29%5D%5E2%7Ddx)
We need the derivative of the function:
And we need it squared:
![[f'(x)]^2=\frac{1}{x^2}](https://tex.z-dn.net/?f=%5Bf%27%28x%29%5D%5E2%3D%5Cfrac%7B1%7D%7Bx%5E2%7D)
Then the integral is:
Now, the Simposn’s rule with n=4 is:
In this problem:
So, the Simposn’s rule formula becomes:
Then simplifying a bit:
Then we just do those computations and we finally get the approximation via Simposn's rule:
While when we do the integral by using technology we get: 2.3020.
The approximation with Simpon’s rule is close but about 33% smaller:
Answer:
The length of the hypotenuse is 17.
Step-by-step explanation:
I found this out by doing:
15 x 15 = 225
8 x 8 = 64
225 + 64 = 289
Hope this helps you! :)
Answer:
5: c=65 d=125 e=60
6: c= 58 d= 121 e=63
7. a=55 b=64 c=61
8. a=61 b=55 c=64
Step-by-step explanation: :)
Hello!
Looking at the image above, we can see that the two “legs” of the triangle both have a length of 7 units. Consequently, this triangle can be classified as an isosceles triangle. An isosceles triangle has two sides of equal length (in this case, KJ and KL) and two corresponding angles of equal measure (in this case, ∠J and ∠L).
∠J has a measure of 25 degrees. Consequently, with our knowledge of isosceles triangles, we can conclude that ∠L has a measure of 25 degrees as well.
I hope this helps!
