1. Using your straightedge, draw a reference line, if one is not provided.
2. Copy the side of the square onto the reference line, starting at a point labeled A'.
3. Construct a perpendicular at point B' to the line through ab2.
4. Place your compass point at B', and copy the side of the square onto the perpendicular b'g. Label the end of the segment copy as point C.
5. With your compass still set at a span representing AB, place the compass point at C and swing an arc to the left.
6. Holding this same span, place the compass point at A' and swing an arc intersecting with the previous arc. Label the point of intersection as D.
7. Connect points A' to D, D to C, and C to B' to form a square.
70 x 9.50 = 665 so 1000 - 665 =335 and 335 / 9.50 = 35 so, it needs 35 more tickets to get 1000.
Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:
![\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac2n%5Cright%5D%2C%5Cleft%5B%5Cdfrac2n%2C%5Cdfrac4n%5Cright%5D%2C%5Cleft%5B%5Cdfrac4n%2C%5Cdfrac6n%5Cright%5D%2C%5Cldots%2C%5Cleft%5B%5Cdfrac%7B2%28n-1%29%7Dn%2C2%5Cright%5D)
Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,
where 
. Each interval has length 
.
At these sampling points, the function takes on values of
We approximate the integral with the Riemann sum:
Recall that
so that the sum reduces to
Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:
Just to check:
Answer:
8
Step-by-step explanation:
i did 24 divided by 3
hope this helps
-mercury
Answer:
It will be worth $1732.30.
Step-by-step explanation:
After 1 year : 6000 x .78 = 4680
2 yr - 4680 x .78 = 3650.4
3 yr - 3650.4 x .78 = 2847.312
4 yr - 2847.312 x .78 = 2220.90336
5 yr - 2220.90336 x .78 = 1732.304621
1732.30
I think this is what it was asking
