Answer:
see the prove below
Step-by-step explanation:
Let bisector of angle MCN is KC. Points N,M,K are on the one side from the line AB
Let MCB=x. So the NCA is x as well ( the angle of ray reflection NVA is equal to the angle of ray putting on the line MVB).
The value of angle ACB=180 degrees, because AB is a straight line.
KC is MCN bisector, so KCN=KCM =y.
So ACB= KCN+KCM+MCB+NCA=x+x+y+y=2x+2y=180
=> x+y=90 degrees => KC is perpendiculat to line AB.
The statement is proved
First determine how many square feet there are in the front lawn.
(15.0 ft) x (20.0 ft) = 300. ft^2
Now find how many snow flakes land in the entire lawn.
(300. ft^2) x (1450 snow flakes/min/ft^2) = 435,000 snow flakes/min
Now find the total mass of snow landing on the lawn per hour.
(435,000 snow flakes/min) x (1.70 mg/snow flake) x (1 kg/1,000,000 mg) x (60 min/1hour) = 44.4 kg of snow per hour.
We split [2, 4] into
subintervals of length
,
![[2,4]=\left[2,2+\dfrac2n\right]\cup\left[2+\dfrac2n,2+\dfrac4n\right]\cup\left[2+\dfrac4n,2+\dfrac6n\right]\cup\cdots\cup\left[2+\dfrac{2(n-1)}n,4\right]](https://tex.z-dn.net/?f=%5B2%2C4%5D%3D%5Cleft%5B2%2C2%2B%5Cdfrac2n%5Cright%5D%5Ccup%5Cleft%5B2%2B%5Cdfrac2n%2C2%2B%5Cdfrac4n%5Cright%5D%5Ccup%5Cleft%5B2%2B%5Cdfrac4n%2C2%2B%5Cdfrac6n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B2%2B%5Cdfrac%7B2%28n-1%29%7Dn%2C4%5Cright%5D)
so that the right endpoints are given by the sequence
![x_i=2+\dfrac{2i}n=\dfrac{2(n+i)}n](https://tex.z-dn.net/?f=x_i%3D2%2B%5Cdfrac%7B2i%7Dn%3D%5Cdfrac%7B2%28n%2Bi%29%7Dn)
for
. Then the Riemann sum approximating
![\displaystyle\int_2^42x\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_2%5E42x%5C%2C%5Cmathrm%20dx)
is
![\displaystyle\sum_{i=1}^nf(x_i)\dfrac{4-2}n=\frac8{n^2}\sum_{i=1}^n(n+i)=\frac8{n^2}\left(n^2+\frac{n(n+1)}2\right)=\frac{12n+4}n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Enf%28x_i%29%5Cdfrac%7B4-2%7Dn%3D%5Cfrac8%7Bn%5E2%7D%5Csum_%7Bi%3D1%7D%5En%28n%2Bi%29%3D%5Cfrac8%7Bn%5E2%7D%5Cleft%28n%5E2%2B%5Cfrac%7Bn%28n%2B1%29%7D2%5Cright%29%3D%5Cfrac%7B12n%2B4%7Dn)
The integral is given exactly as
, for which we get
![\displaystyle\int_2^42x\,\mathrm dx=\lim_{n\to\infty}\frac{12n+4}n=12](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_2%5E42x%5C%2C%5Cmathrm%20dx%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B12n%2B4%7Dn%3D12)
To check: we have
![\displaystyle\int_2^42x\,\mathrm dx=x^2\bigg|_2^4=4^2-2^2=16-4=12](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_2%5E42x%5C%2C%5Cmathrm%20dx%3Dx%5E2%5Cbigg%7C_2%5E4%3D4%5E2-2%5E2%3D16-4%3D12)
I'm assuming no air friction here, first use conservation of energy to find velocity, so initially all the energy is from the spring, (1/2)kx^2, then it is converted to kinetic energy, (1/2)mv^2. so .5kx^2=.5mv^2. the .5s cancel, kx^2=mv^2. solve for v, v=sqrt(kx^2/m) plug in your values for the variables and find v. then use vf^2=vi^2+2ax in the u direction and solve for vf, vf=sqrt(vi^2+2ax). use that velocity in vf=vi+at and solve for t. vi=0 and a=g=9.81, so vf/a=t. plug in the values from the previous equation and the acceleration due to gravity to find the time it takes to hit the ground vertically. Now use v=x/t and solve for x. x=vt, plug in the velocity of the ball horizontally, which we found with the conservation of energy part, and y which we found with the previous equation, and you get your x, the distance it will take for the ball to hit the ground.