Answer: Nominal
Step-by-step explanation:
A categorical variable whose values are purely qualitative and unordered is called a Nominal variable. Nominal variables are qualitative variables that does not have a particular rank, order or value. An example of nominal variables are colour (red,blue etc), gender (male, female), skin and hair colour etc.
Answer:
The answer is 3 feet.
Step-by-step explanation:
First, the height is 7 right. Then take away 1 foot since is said 1 foot more than twice the width. Then divide by 2 meaning twice. Meaning 7-1=6 6/2=3. There you go! Hope these formulas helped!
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → A = π - (B + C)
→ B = π - (A + C)
→ C = π - (A + B)
Use Sum to Product Identity: sin A - sin B = 2 cos [(A + B)/2] · sin [(A - B)/2]
Use the following Cofunction Identity: cos (π/2 - A) = sin A
<u>Proof LHS → RHS:</u>
LHS: sin A - sin B + sin C
= (sin A - sin B) + sin C




![\text{Factor:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Given:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)\bigg]\\\\\\.\qquad \qquad =2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\cos \bigg(\dfrac{\pi}{2} -\dfrac{(A+B)}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BGiven%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%20-%28A%2BB%29%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%3D2%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D%20-%5Cdfrac%7B%28A%2BB%29%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Cofunction:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ \sin \bigg(\dfrac{A-B}{2}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BCofunction%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%20%5Csin%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B2%7D%5Cbigg%29%2B%5Csin%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B2%7D%5Cbigg%29%5Cbigg%5D)
![\text{Sum to Product:}\qquad 2\sin \bigg(\dfrac{C}{2}\bigg)\bigg[ 2\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad \qquad =4\sin \bigg(\dfrac{A}{2}\bigg)\cdot \cos \bigg(\dfrac{B}{2}\bigg)\cdot \sin \bigg(\dfrac{C}{2}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cqquad%202%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29%5Cbigg%5B%202%5Csin%20%5Cbigg%28%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BB%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D4%5Csin%20%5Cbigg%28%5Cdfrac%7BA%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BB%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Csin%20%5Cbigg%28%5Cdfrac%7BC%7D%7B2%7D%5Cbigg%29)

Answer:
A. Different volumes of water are moved different distances.
Step-by-step explanation:
Integration is used to find work required to pump water out of a tank because different volumes of water are moved different distances and to sum it all we the tool required is integration. Moreover, work done is a path function or an inexact differential. It does depend upon the path followed by the process.
hence the correct answer is A.