Answer:
<u>20.5 Inches</u>
Step-by-step explanation:
I just took the review so this is a verified answer.
Answer:
It's actually D (2x - 1)(x+1)
Step-by-step explanation:
Answer:
Step-by-step explanation:
this one is tricky so b/c the lines are at 90 degees we know that the arcs are similar so set up the parts of the arcs equal to 360 degrees
360 = 130 + 130 + x + x
see each part of the circle ?
then use your mad algebra skilz
100 = 2x
50 = x
:) nice.. when I do the math , it seems easy , huh :DDD
Splitting up the interval [0, 6] into 6 subintervals means we have
![[0,1]\cup[1,2]\cup[2,3]\cup\cdots\cup[5,6]](https://tex.z-dn.net/?f=%5B0%2C1%5D%5Ccup%5B1%2C2%5D%5Ccup%5B2%2C3%5D%5Ccup%5Ccdots%5Ccup%5B5%2C6%5D)
and the respective midpoints are
![\dfrac12,\dfrac32,\dfrac52,\ldots,\dfrac{11}2](https://tex.z-dn.net/?f=%5Cdfrac12%2C%5Cdfrac32%2C%5Cdfrac52%2C%5Cldots%2C%5Cdfrac%7B11%7D2)
. We can write these sequentially as
![{x_i}^*=\dfrac{2i+1}2](https://tex.z-dn.net/?f=%7Bx_i%7D%5E%2A%3D%5Cdfrac%7B2i%2B1%7D2)
where
![0\le i\le5](https://tex.z-dn.net/?f=0%5Cle%20i%5Cle5)
.
So the integral is approximately
![\displaystyle\int_0^6x^2\,\mathrm dx\approx\sum_{i=0}^5({x_i}^*)^2\Delta x_i=\frac{6-0}6\sum_{i=0}^5({x_i}^*)^2=\sum_{i=0}^5\left(\frac{2i+1}2\right)^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E6x%5E2%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D0%7D%5E5%28%7Bx_i%7D%5E%2A%29%5E2%5CDelta%20x_i%3D%5Cfrac%7B6-0%7D6%5Csum_%7Bi%3D0%7D%5E5%28%7Bx_i%7D%5E%2A%29%5E2%3D%5Csum_%7Bi%3D0%7D%5E5%5Cleft%28%5Cfrac%7B2i%2B1%7D2%5Cright%29%5E2)
Recall that
![\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Eni%5E2%3D%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D6)
![\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Eni%3D%5Cfrac%7Bn%28n%2B1%29%7D2)
![\displaystyle\sum_{i=1}^n1=n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En1%3Dn)
so our sum becomes
![\displaystyle\sum_{i=0}^5\left(\frac{2i+1}2\right)^2=\sum_{i=0}^5\left(i^2+i+\frac14\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D0%7D%5E5%5Cleft%28%5Cfrac%7B2i%2B1%7D2%5Cright%29%5E2%3D%5Csum_%7Bi%3D0%7D%5E5%5Cleft%28i%5E2%2Bi%2B%5Cfrac14%5Cright%29)
Number of students who have sent a text message only is given by the difference between those that have sent a text message and those that have done both.
i.e. Number of students who has sent a text message only = 58 - 12 = 46
Number of students who have uploaded a selfie only is given by the
difference between those that have uploaded a selfie and those that
have done both.
i.e. Number of students who have uploaded a selfie only = 21 - 12 = 9
The total number of students surveyed is 70.
Let the number of students who have neither sent a text message nor taken a selfie today be x, then
46 + 9 + 12 + x = 70
67 + x = 70
x = 70 - 67 = 3
Therefore, 3 students neither sent a text message nor taken a selfie today.
The number of students that have sent a text message or have taken a selfie today is given by the sum of the number of students that have sent a text message and the number of students that have taken a selfie less the number of people that have done both.
i.e. number of students that have sent a text message or have taken a selfie today = 58 + 21 - 12 = 67.