Answer:
Step-by-step explanation:
Area of shaded region = Area of quadrant - area of triangle
= (1/4)πr² - (1/2)bh
![=\frac{1}{4}*3.14*5*5-\frac{1}{2}*5*5\\\\=19.625 - 12.5\\\\= 7.125\\\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%7D%2A3.14%2A5%2A5-%5Cfrac%7B1%7D%7B2%7D%2A5%2A5%5C%5C%5C%5C%3D19.625%20-%2012.5%5C%5C%5C%5C%3D%207.125%5C%5C%5C%5C)
= 7 cm²
Compare
to
. Then in applying the LCT, we have
![\displaystyle\lim_{n\to\infty}\left|\frac{\frac1{\sqrt{n^2+1}}}{\frac1n}\right|=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cleft%7C%5Cfrac%7B%5Cfrac1%7B%5Csqrt%7Bn%5E2%2B1%7D%7D%7D%7B%5Cfrac1n%7D%5Cright%7C%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%20n%7B%5Csqrt%7Bn%5E2%2B1%7D%7D%3D1)
Because this limit is finite, both
![\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7B%5Csqrt%7Bn%5E2%2B1%7D%7D)
and
![\displaystyle\sum_{n=1}^\infty\frac1n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1n)
behave the same way. The second series diverges, so
![\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac1%7B%5Csqrt%7Bn%5E2%2B1%7D%7D%3D1%2B%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1n)
is divergent.
Yes, the two rectangles are similar, because rectangle 2 is a dilation of rectangle 1.
<h3>
Are the two rectangles similar?</h3>
We know that rectangle 1 has dimensions L and W.
And rectangle 2 is made by multiplying the dimensions of rectangle 1 by a factor k > 0.
Then, rectangle 2 is just a dilation of rectangle 1, this means that in fact, the two rectangles are similar by definition.
Then:
Dimensions of rectangle 1:
- Length = L
- Width = W.
- Perimeter = 2*(W + L)
- Area = W*L
For rectangle 2:
- Length = k*L
- Width = k*W
- Perimeter = 2*(k*L + k*W) = k*(2*(L + W))
- Area = (k*L)*(k*W) = k²(L*W)
Above we can see that the perimeter of rectangle 2 is k times the perimeter of rectangle 1, and the area of rectangle 2 is k squared times the area of the rectangle 1.
If you want to learn more about similar figures:
brainly.com/question/14285697
#SPJ1
Because the balls or spheres are tightly packed in a cube, their radius is approximately equal to 2 ft. The volume of a sphere is calculated through the equation,
V = 4/3πr³
Substituting the known radius,
V = 4/3π(2 ft)³
V = 33.51 ft³
The 9 spheres will have a total volume of 301/59 ft³ which is also equal to 521152.52 in³.