Answer:
![NO = 16\ units](https://tex.z-dn.net/?f=NO%20%3D%2016%5C%20units)
Step-by-step explanation:
Given
![NO = 2x](https://tex.z-dn.net/?f=NO%20%3D%202x)
![OP = 8](https://tex.z-dn.net/?f=OP%20%3D%208)
Required
Determine NO
Since O is a segment on NP, we have that
![NP = NO + OP](https://tex.z-dn.net/?f=NP%20%3D%20NO%20%2B%20OP)
Substitute values for these parameters
![3x = 2x + 8](https://tex.z-dn.net/?f=3x%20%3D%202x%20%2B%208)
Collect Like Terms
![3x - 2x = 8](https://tex.z-dn.net/?f=3x%20-%202x%20%3D%208)
![x = 8](https://tex.z-dn.net/?f=x%20%3D%208)
Substitute 8 for x in ![NO = 2x](https://tex.z-dn.net/?f=NO%20%3D%202x)
![NO = 2 * 8](https://tex.z-dn.net/?f=NO%20%3D%202%20%2A%208)
![NO = 16\ units](https://tex.z-dn.net/?f=NO%20%3D%2016%5C%20units)
A. The coordinates of the midpoint of CD in terms of p and q is [(4 + p) / 2 , (5 + q) / 2]
B. The coordinates of D, Given that the midpoint of CD is (7, 1) is (10 , -3)
<h3>A. How to determine the mid point</h3>
- Coordinate of C = (4, 5)
- Coordinate of D = (p, q)
- Mid point =?
Mid point = (X , Y)
X = (x₁ + x₂) / 2
X = (4 + p) / 2
Y = (y₁ + y₂) / 2
Y = (5 + q) / 2
Thus,
Mid point = (X , Y)
Mid point = [(4 + p) / 2 , (5 + q) / 2]
<h3>B. How to determine the coordinates of D</h3>
- Mid point = (7, 1)
- Coordinates of D =?
Mid point = (7, 1) = (X , Y)
X = (4 + p) / 2
7 = (4 + p) / 2
Cross multiply
7 × 2 = 4 + p
14 = 4 + p
Collect like terms
p = 14 - 4
p = 10
Y = (5 + q) / 2
1 = (5 + q) / 2
Cross multiply
1 × 2 = 5 + q
2 = 5 + q
Collect like terms
q = 2 - 5
q = -3
Coordinates of D = (p, q)
Coordinates of D = (10 , -3)
Learn more about coordinate geometry:
brainly.com/question/4976351
#SPJ1
Answer:
D
Step-by-step explanation:
The Pythagorean Theorem only applies to right triangles where the sum of the squares of each side length is equal to the length of the hypotenuse.
The other options can be solved with the Law of Sines or the Law of Cosines, depending on what angles and sides are given.
The result of
is ![x^3 + 13x^2+ 55x+ 75](https://tex.z-dn.net/?f=x%5E3%20%2B%2013x%5E2%2B%2055x%2B%2075)
<h3>Composite functions</h3>
Composite functions are functions that are gotten by combining multiple functions
<h3>The functions</h3>
The functions are given as:
![f(x) = x^2 + 8x + 15](https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E2%20%2B%208x%20%2B%2015)
![g(x) = x+5](https://tex.z-dn.net/?f=g%28x%29%20%3D%20x%2B5)
To calculate f(x) * g(x), we have:
![f(x) \times g(x) = [x^2 + 8x + 15] \times [x + 5]](https://tex.z-dn.net/?f=f%28x%29%20%5Ctimes%20g%28x%29%20%3D%20%5Bx%5E2%20%2B%208x%20%2B%2015%5D%20%5Ctimes%20%5Bx%20%2B%205%5D)
Expand the above expression
![f(x) \times g(x) = x^3 + 8x^2 + 15x + 5x^2 + 40x + 75](https://tex.z-dn.net/?f=f%28x%29%20%5Ctimes%20g%28x%29%20%3D%20x%5E3%20%2B%208x%5E2%20%2B%2015x%20%2B%205x%5E2%20%2B%2040x%20%2B%2075)
Collect like terms
![f(x) \times g(x) = x^3 + 8x^2 + 5x^2+ 15x + 40x + 75](https://tex.z-dn.net/?f=f%28x%29%20%5Ctimes%20g%28x%29%20%3D%20x%5E3%20%2B%208x%5E2%20%2B%205x%5E2%2B%2015x%20%20%2B%2040x%20%2B%2075)
Evaluate the like terms
![f(x) \times g(x) = x^3 + 13x^2+ 55x+ 75](https://tex.z-dn.net/?f=f%28x%29%20%5Ctimes%20g%28x%29%20%3D%20x%5E3%20%2B%2013x%5E2%2B%2055x%2B%2075)
Hence, the result of
is ![x^3 + 13x^2+ 55x+ 75](https://tex.z-dn.net/?f=x%5E3%20%2B%2013x%5E2%2B%2055x%2B%2075)
Read more about composite functions at:
brainly.com/question/10687170
![\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bmatrix%7D1%5C%5C1%261%5C%5C1%262%261%5C%5C1%263%263%261%5C%5C1%264%266%264%261%5Cend%7Bbmatrix%7D)
The rows add up to
![1,2,4,8,16](https://tex.z-dn.net/?f=1%2C2%2C4%2C8%2C16)
, respectively. (Notice they're all powers of 2)
The sum of the numbers in row
![n](https://tex.z-dn.net/?f=n)
is
![2^{n-1}](https://tex.z-dn.net/?f=2%5E%7Bn-1%7D)
.
The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When
![n=1](https://tex.z-dn.net/?f=n%3D1)
,
![(1+x)^1=1+x=\dbinom10+\dbinom11x](https://tex.z-dn.net/?f=%281%2Bx%29%5E1%3D1%2Bx%3D%5Cdbinom10%2B%5Cdbinom11x)
so the base case holds. Assume the claim holds for
![n=k](https://tex.z-dn.net/?f=n%3Dk)
, so that
![(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k](https://tex.z-dn.net/?f=%281%2Bx%29%5Ek%3D%5Cdbinom%20k0%2B%5Cdbinom%20k1x%2B%5Ccdots%2B%5Cdbinom%20k%7Bk-1%7Dx%5E%7Bk-1%7D%2B%5Cdbinom%20kkx%5Ek)
Use this to show that it holds for
![n=k+1](https://tex.z-dn.net/?f=n%3Dk%2B1)
.
![(1+x)^{k+1}=(1+x)(1+x)^k](https://tex.z-dn.net/?f=%281%2Bx%29%5E%7Bk%2B1%7D%3D%281%2Bx%29%281%2Bx%29%5Ek)
![(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)](https://tex.z-dn.net/?f=%281%2Bx%29%5E%7Bk%2B1%7D%3D%281%2Bx%29%5Cleft%28%5Cdbinom%20k0%2B%5Cdbinom%20k1x%2B%5Ccdots%2B%5Cdbinom%20k%7Bk-1%7Dx%5E%7Bk-1%7D%2B%5Cdbinom%20kkx%5Ek%5Cright%29)
![(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}](https://tex.z-dn.net/?f=%281%2Bx%29%5E%7Bk%2B1%7D%3D1%2B%5Cleft%28%5Cdbinom%20k0%2B%5Cdbinom%20k1%5Cright%29x%2B%5Cleft%28%5Cdbinom%20k1%2B%5Cdbinom%20k2%5Cright%29x%5E2%2B%5Ccdots%2B%5Cleft%28%5Cdbinom%20k%7Bk-2%7D%2B%5Cdbinom%20k%7Bk-1%7D%5Cright%29x%5E%7Bk-1%7D%2B%5Cleft%28%5Cdbinom%20k%7Bk-1%7D%2B%5Cdbinom%20kk%5Cright%29x%5Ek%2Bx%5E%7Bk%2B1%7D)
Notice that
![\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}](https://tex.z-dn.net/?f=%5Cdbinom%20k%5Cell%2B%5Cdbinom%20k%7B%5Cell%2B1%7D%3D%5Cdfrac%7Bk%21%7D%7B%5Cell%21%28k-%5Cell%29%21%7D%2B%5Cdfrac%7Bk%21%7D%7B%28%5Cell%2B1%29%21%28k-%5Cell-1%29%21%7D)
![\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}](https://tex.z-dn.net/?f=%5Cdbinom%20k%5Cell%2B%5Cdbinom%20k%7B%5Cell%2B1%7D%3D%5Cdfrac%7Bk%21%28%5Cell%2B1%29%7D%7B%28%5Cell%2B1%29%21%28k-%5Cell%29%21%7D%2B%5Cdfrac%7Bk%21%28k-%5Cell%29%7D%7B%28%5Cell%2B1%29%21%28k-%5Cell%29%21%7D)
![\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}](https://tex.z-dn.net/?f=%5Cdbinom%20k%5Cell%2B%5Cdbinom%20k%7B%5Cell%2B1%7D%3D%5Cdfrac%7Bk%21%28%5Cell%2B1%29%2Bk%21%28k-%5Cell%29%7D%7B%28%5Cell%2B1%29%21%28k-%5Cell%29%21%7D)
![\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}](https://tex.z-dn.net/?f=%5Cdbinom%20k%5Cell%2B%5Cdbinom%20k%7B%5Cell%2B1%7D%3D%5Cdfrac%7Bk%21%28k%2B1%29%7D%7B%28%5Cell%2B1%29%21%28k-%5Cell%29%21%7D)
![\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}](https://tex.z-dn.net/?f=%5Cdbinom%20k%5Cell%2B%5Cdbinom%20k%7B%5Cell%2B1%7D%3D%5Cdfrac%7B%28k%2B1%29%21%7D%7B%28%5Cell%2B1%29%21%28%28k%2B1%29-%28%5Cell%2B1%29%29%21%7D)
![\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}](https://tex.z-dn.net/?f=%5Cdbinom%20k%5Cell%2B%5Cdbinom%20k%7B%5Cell%2B1%7D%3D%5Cdbinom%7Bk%2B1%7D%7B%5Cell%2B1%7D)
So you can write the expansion for
![n=k+1](https://tex.z-dn.net/?f=n%3Dk%2B1)
as
![(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}](https://tex.z-dn.net/?f=%281%2Bx%29%5E%7Bk%2B1%7D%3D1%2B%5Cdbinom%7Bk%2B1%7D1x%2B%5Cdbinom%7Bk%2B1%7D2x%5E2%2B%5Ccdots%2B%5Cdbinom%7Bk%2B1%7D%7Bk-1%7Dx%5E%7Bk-1%7D%2B%5Cdbinom%7Bk%2B1%7Dkx%5Ek%2Bx%5E%7Bk%2B1%7D)
and since
![\dbinom{k+1}0=\dbinom{k+1}{k+1}=1](https://tex.z-dn.net/?f=%5Cdbinom%7Bk%2B1%7D0%3D%5Cdbinom%7Bk%2B1%7D%7Bk%2B1%7D%3D1)
, you have
![(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}](https://tex.z-dn.net/?f=%281%2Bx%29%5E%7Bk%2B1%7D%3D%5Cdbinom%7Bk%2B1%7D0%2B%5Cdbinom%7Bk%2B1%7D1x%2B%5Ccdots%2B%5Cdbinom%7Bk%2B1%7Dkx%5Ek%2B%5Cdbinom%7Bk%2B1%7D%7Bk%2B1%7Dx%5E%7Bk%2B1%7D)
and so the claim holds for
![n=k+1](https://tex.z-dn.net/?f=n%3Dk%2B1)
, thus proving the claim overall that
![(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n](https://tex.z-dn.net/?f=%281%2Bx%29%5En%3D%5Cdbinom%20n0%2B%5Cdbinom%20n1x%2B%5Ccdots%2B%5Cdbinom%20n%7Bn-1%7Dx%5E%7Bn-1%7D%2B%5Cdbinom%20nnx%5En)
Setting
![x=1](https://tex.z-dn.net/?f=x%3D1)
gives
![(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n](https://tex.z-dn.net/?f=%281%2B1%29%5En%3D%5Cdbinom%20n0%2B%5Cdbinom%20n1%2B%5Ccdots%2B%5Cdbinom%20n%7Bn-1%7D%2B%5Cdbinom%20nn%3D2%5En)
which agrees with the result obtained for part (c).