I guess the would help you
consider first number is = x
formula,
(n / 2)(first number + last number) = sum,
7/2(x+x+6)= 945
2x+6= 945×2/7
2x+6 = 270
2x= 270-6
x=264/2= 132
Answer the smallest number is 132.
Answer:
the solution is (-4, 3)
Step-by-step explanation:
if you were to graph y = -7/4x - 4 you could plot your first point at (0, -4) and get your second point by going up 7 units (from (0, -4) and to the left by 4
if you were to graph y = -1/4x + 2 you could plot your first point at (0, 2) and get your second point by going up 1 unit (from (0, 2) and to the left by 4
the second point plotted for each equation will be the point of intersection, (-4, 3)
Split up the integration interval into 4 subintervals:
![\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac%5Cpi8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi8%2C%5Cdfrac%5Cpi4%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi4%2C%5Cdfrac%7B3%5Cpi%7D8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%7B3%5Cpi%7D8%2C%5Cdfrac%5Cpi2%5Cright%5D)
The left and right endpoints of the
-th subinterval, respectively, are


for
, and the respective midpoints are

We approximate the (signed) area under the curve over each subinterval by

so that

We approximate the area for each subinterval by

so that

We first interpolate the integrand over each subinterval by a quadratic polynomial
, where

so that

It so happens that the integral of
reduces nicely to the form you're probably more familiar with,

Then the integral is approximately

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.
Answer:
The answer is 400 when you round it off to the nearest hundred.