When we divide the interval from 1 to 5 into 4 subintervals, each has a width of 1. Simpson's rule has us evaluate the integral as
... integral = (1/3)(f(1) +4f(2) +2f(3) +4f(4) +f(5)) = (1/3)(10 +4·25 +2·46 +4·73 +106)
... integral = (1/3)(600) = 200 . . . . . by Simpson's rule
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The integral evaluates to
![\displaystyle \int_{1}^{5}{\left(3x^2+6x+1\right)}\,dx=\left[x^3+3x^2+x\right]\limits_{1}^{5}=(5^3-1^3)+3(5^2-1^2)+(5-1)\\=124+3*24+4=200](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B1%7D%5E%7B5%7D%7B%5Cleft%283x%5E2%2B6x%2B1%5Cright%29%7D%5C%2Cdx%3D%5Cleft%5Bx%5E3%2B3x%5E2%2Bx%5Cright%5D%5Climits_%7B1%7D%5E%7B5%7D%3D%285%5E3-1%5E3%29%2B3%285%5E2-1%5E2%29%2B%285-1%29%5C%5C%3D124%2B3%2A24%2B4%3D200)
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Simpson's rule uses a quadratic interpolation, so evaluates quadratics exactly.
Step-by-step explanation:
<h3><u>To</u><u> </u><u>Solve</u><u>:</u><u>-</u></h3>

<h3><u>SOL</u><u>UTION</u><u>:</u><u>-</u></h3>





Solve the following system using elimination:
{7 x + 2 y = -19 | (equation 1)
{2 y - x = 21 | (equation 2)
Add 1/7 × (equation 1) to equation 2:
{7 x + 2 y = -19 | (equation 1)
{0 x+(16 y)/7 = 128/7 | (equation 2)
Multiply equation 2 by 7/16:
{7 x + 2 y = -19 | (equation 1)
{0 x+y = 8 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{7 x+0 y = -35 | (equation 1)
{0 x+y = 8 | (equation 2)
Divide equation 1 by 7:
{x+0 y = -5 | (equation 1)
{0 x+y = 8 | (equation 2)
Collect results:
Answer: {x = -5, y = 8
this is approximately 6 to 200
divide both by 2
3 to 100
3 to 10^2
3 out of 100
Answer:

Step by step
Remember, a reference angle is a positive acute angle that represents an angle θ of any measurement.
a)
, then its reference angle is 
b)
is in the second cuadrant, then its reference angle is 
c)
, then it is in the first cuadrant and the reference angle is
.
d)
and it is in the third cuadrant, then its reference angle is 