Answer:
14π
Step-by-step explanation:
C = 2πr
D=2r
D=14
14=2r
r=7
C=2π(7)
C=14π or 43.98
Answer:
Cylinder
Step-by-step explanation:
The given geometric shape is of a cylinder.
By the polynomial remainder theorem,
![x+1](https://tex.z-dn.net/?f=x%2B1)
will be a factor of
![f(x)=x^4+ax^3+bx^2+cx+d](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E4%2Bax%5E3%2Bbx%5E2%2Bcx%2Bd)
if the remainder upon division is 0, and this remainder is given by
![f(-1)](https://tex.z-dn.net/?f=f%28-1%29)
:
![f(-1)=(-1)^4+a(-1)^3+b(-1)^2+c(-1)+d](https://tex.z-dn.net/?f=f%28-1%29%3D%28-1%29%5E4%2Ba%28-1%29%5E3%2Bb%28-1%29%5E2%2Bc%28-1%29%2Bd)
![0=1-a+b-c+d](https://tex.z-dn.net/?f=0%3D1-a%2Bb-c%2Bd)
![a+c=1+b+d](https://tex.z-dn.net/?f=a%2Bc%3D1%2Bb%2Bd)
Since
![a,c\in\{1,\ldots,5\}](https://tex.z-dn.net/?f=a%2Cc%5Cin%5C%7B1%2C%5Cldots%2C5%5C%7D)
, it follows that
![a+c\in\{2,\ldots,10\}](https://tex.z-dn.net/?f=a%2Bc%5Cin%5C%7B2%2C%5Cldots%2C10%5C%7D)
. But notice that if
![a+c=2](https://tex.z-dn.net/?f=a%2Bc%3D2)
, then we have
![2=1+b+d\implies 1=b+d](https://tex.z-dn.net/?f=2%3D1%2Bb%2Bd%5Cimplies%201%3Db%2Bd)
and since
![b,d\in\{1,\ldots,5\}](https://tex.z-dn.net/?f=b%2Cd%5Cin%5C%7B1%2C%5Cldots%2C5%5C%7D)
, the equation above requires that either
![b=0](https://tex.z-dn.net/?f=b%3D0)
or
![d=0](https://tex.z-dn.net/?f=d%3D0)
, which is impossible. So
![a+c\in\{3,\ldots,10\}](https://tex.z-dn.net/?f=a%2Bc%5Cin%5C%7B3%2C%5Cldots%2C10%5C%7D)
.
So we have 8 cases to check:
(1) Notice that if
![a+c=10](https://tex.z-dn.net/?f=a%2Bc%3D10)
, we have
![b+d=9](https://tex.z-dn.net/?f=b%2Bd%3D9)
. This is only possible for
![(b,d)\in\{(4,5),(5,4)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%284%2C5%29%2C%285%2C4%29%5C%7D)
.
(2) If
![a+c=9](https://tex.z-dn.net/?f=a%2Bc%3D9)
, then
![b+d=8](https://tex.z-dn.net/?f=b%2Bd%3D8)
, and so we can have
![(b,d)\in\{(3,5),(4,4),(5,3)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%283%2C5%29%2C%284%2C4%29%2C%285%2C3%29%5C%7D)
.
(3) If
![a+c=8](https://tex.z-dn.net/?f=a%2Bc%3D8)
, then
![b+d=7](https://tex.z-dn.net/?f=b%2Bd%3D7)
, and so
![(b,d)\in\{(2,5),(3,4),(4,3),(5,2)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%282%2C5%29%2C%283%2C4%29%2C%284%2C3%29%2C%285%2C2%29%5C%7D)
.
(4) If
![a+c=7](https://tex.z-dn.net/?f=a%2Bc%3D7)
, then
![(b,d)\in\{(1,5),(2,4),(3,3),(4,2),(5,1)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%281%2C5%29%2C%282%2C4%29%2C%283%2C3%29%2C%284%2C2%29%2C%285%2C1%29%5C%7D)
.
(5) If
![a+c=6](https://tex.z-dn.net/?f=a%2Bc%3D6)
, then
![(b,d)\in\{(1,4),(2,3),(3,2),(4,1)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%281%2C4%29%2C%282%2C3%29%2C%283%2C2%29%2C%284%2C1%29%5C%7D)
.
(6) If
![a+c=5](https://tex.z-dn.net/?f=a%2Bc%3D5)
, then
![(b,d)\in\{(1,3),(2,2),(3,1)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%281%2C3%29%2C%282%2C2%29%2C%283%2C1%29%5C%7D)
.
(7) If
![a+c=4](https://tex.z-dn.net/?f=a%2Bc%3D4)
, then
![(b,d)\in\{(1,2),(2,1)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%281%2C2%29%2C%282%2C1%29%5C%7D)
.
(8) If
![a+c=3](https://tex.z-dn.net/?f=a%2Bc%3D3)
, then
![(b,d)\in\{(1,1)\}](https://tex.z-dn.net/?f=%28b%2Cd%29%5Cin%5C%7B%281%2C1%29%5C%7D)
.
At the same time, we have 8 cases to consider to find how many options there are for
![(a,c)](https://tex.z-dn.net/?f=%28a%2Cc%29)
.
(1)
![a+c=10](https://tex.z-dn.net/?f=a%2Bc%3D10)
. We have only one choice of
![(a,c)=(5,5)](https://tex.z-dn.net/?f=%28a%2Cc%29%3D%285%2C5%29)
.
(2)
![a+c=9](https://tex.z-dn.net/?f=a%2Bc%3D9)
. This is the same as when
![b+d=9](https://tex.z-dn.net/?f=b%2Bd%3D9)
, which we found to be 2 choices.
(3) Same as
![b+d=8](https://tex.z-dn.net/?f=b%2Bd%3D8)
; 3 choices.
(4) Same as
![b+d=7](https://tex.z-dn.net/?f=b%2Bd%3D7)
; 4 choices.
(5) 5.
(6) 4.
(7) 3.
(8) 2.
In total, there are
![2\times1+3\times2+4\times3+5\times4+4\times5+3\times4+2\times3+1\times2](https://tex.z-dn.net/?f=2%5Ctimes1%2B3%5Ctimes2%2B4%5Ctimes3%2B5%5Ctimes4%2B4%5Ctimes5%2B3%5Ctimes4%2B2%5Ctimes3%2B1%5Ctimes2)
![=2(2\times1+3\times2+4\times3+5\times4)](https://tex.z-dn.net/?f=%3D2%282%5Ctimes1%2B3%5Ctimes2%2B4%5Ctimes3%2B5%5Ctimes4%29)
![=2\displaystyle\sum_{n=1}^4n(n+1)](https://tex.z-dn.net/?f=%3D2%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E4n%28n%2B1%29)
![=80](https://tex.z-dn.net/?f=%3D80)
ways to choose
![a,b,c,d](https://tex.z-dn.net/?f=a%2Cb%2Cc%2Cd)
such that
![x+1](https://tex.z-dn.net/?f=x%2B1)
is a factor of
![x^4+ax^3+bx^2+cx+d](https://tex.z-dn.net/?f=x%5E4%2Bax%5E3%2Bbx%5E2%2Bcx%2Bd)
, so the answer is B.
Note the symmetry of the sum above. You can easily give a slightly briefer combinatorial argument for this answer, but I figured a more brute-force approach would be easier to follow.
I think its False
(got it from google)
Answer:
We start with the equation:
(x^2 - 1)/(x + 1) = 2
We want to rewrite this in standard form.
We can multiply both sides by (x + 1) to get:
((x^2 - 1)/(x + 1))*(x + 1) = 2*(x + 1)
x^2 - 1 = 2*(x + 1) = 2*x + 2
Now we can move all the terms to one side to get:
x^2 - 1 - 2 - 2*x = 0
x^2 - 2*x - 3 = 0
And this is a standard quadratic equation, that we can solve by using the Bhaskara's formula, this will give us the solutions:
![x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4*1*(-3)} }{2*1} = \frac{2 \pm 4}{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-%28-2%29%20%5Cpm%20%5Csqrt%7B%28-2%29%5E2%20-%204%2A1%2A%28-3%29%7D%20%7D%7B2%2A1%7D%20%20%3D%20%5Cfrac%7B2%20%5Cpm%204%7D%7B2%7D)
Then the two solutions are:
x = (2 + 4)/2 = 3
x = (2 - 4)/2 = -1