Partitioning the interval
![[6,11]](https://tex.z-dn.net/?f=%5B6%2C11%5D)
into

equally-spaced subintervals gives

rectangles of width

and of heights determined by the right endpoints of each subinterval.
If

, then

,

, and so on, up to

. Because we're using the right endpoints, the approximation will consider

The definite integral is then approximated by

You have




To check that this is correct, let's make sure the sum converges to the exact value of the definite integral. As

, you have the sum converging to

.
Meanwhile,
![\int_6^{11}(1-5x)\,\mathrm dx=\left[x-\dfrac52x^2\right]_{x=6}^{x=11}=-\dfrac{415}2](https://tex.z-dn.net/?f=%5Cint_6%5E%7B11%7D%281-5x%29%5C%2C%5Cmathrm%20dx%3D%5Cleft%5Bx-%5Cdfrac52x%5E2%5Cright%5D_%7Bx%3D6%7D%5E%7Bx%3D11%7D%3D-%5Cdfrac%7B415%7D2)
so we're done.
Is 2in the radius? if so, the diameter is 4 in
Answer:
m1 = 62.5
m2 = 62.5
m3 = 55°
m4 = 75°
m5 = 50°
Step-by-step explanation:
m1: Sides 1 and 2 are congruent so angle will be the exact same as m2:
180° - 55° = 125 | 125 / 2 = <u>62.5°</u>
m2: same as m1; <u>62.5°</u>
m3: equivalent to <u>55°</u> due to its location
m4: angle of a line is 180° so 180° - 105° = <u>75°</u>
m5: angles of triangle add to 180° so 180° - (75° + 55°) = <u>50°</u>