Mexine can stack 20 boxes in 15 minutes. how long will it take her to stack 100 boxes
1 answer:
You might be interested in
Answer:
See Below.
Step-by-step explanation:
We are given the isosceles triangle ΔABC. By the definition of isosceles triangles, this means that ∠ABC = ∠ACB.
Segments BO and CO bisects ∠ABC and ∠ACB.
And we want to prove that ΔBOC is an isosceles triangle.
Since BO and CO are the angle bisectors of ∠ABC and ∠ACB, respectively, it means that ∠ABO = ∠CBO and ∠ACO = ∠BCO.
And since ∠ABC = ∠ACB, this implies that:
∠ABO = ∠CBO =∠ACO = ∠BCO.
This is shown in the figure as each angle having only one tick mark, meaning that they are congruent.
So, we know that:
∠ABC is the sum of the angles ∠ABO and ∠CBO. Likewise, ∠ACB is the sum of the angles ∠ACO and ∠BCO. Hence:
Since ∠ABO =∠ACO, by substitution:
Subtracting ∠ABO from both sides produces:
So, we've proven that the two angles are congruent, thereby proving that ΔBOC is indeed an isosceles triangle.
The unit disk can be parameterized by the function
where
and 
. The squared distance between any point in this region 
and the point (1, 1) is
The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.
![=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7D%5Br%5E2-2r%28%5Ccos%5Ctheta-%5Csin%5Ctheta%29%2B2%5Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D)
where
The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.