Answer:
a) ⅓ units²
b) 4/15 pi units³
c) 2/3 pi units³
Step-by-step explanation:
4y = x²
2y = x
4y = (2y)²
4y = 4y²
4y² - 4y = 0
y(y-1) = 0
y = 0, 1
x = 0, 2
Area
Integrate: x²/4 - x/2
From 0 to 2
(x³/12 - x²/4)
(8/12 - 4/4) - 0
= -⅓
Area = ⅓
Volume:
Squares and then integrate
Integrate: [x²/4]² - [x/2]²
Integrate: x⁴/16 - x²/4
x⁵/80 - x³/12
Limits 0 to 2
(2⁵/80 - 2³/12) - 0
-4/15
Volume = 4/15 pi
About the x-axis
x² = 4y
x² = 4y²
Integrate the difference
Integrate: 4y² - 4y
4y³/3 - 2y²
Limits 0 to 1
(4/3 - 2) - 0
-2/3
Volume = ⅔ pi
C..The median of Nadia’s data is equal to the median of Ben’s data. D. <span>Nadia had the highest score on a test</span>
we are given that
both sides are equal
so, their length must be equal
![HK=\sqrt{(p-1)^2+(5-2)^2}](https://tex.z-dn.net/?f=%20HK%3D%5Csqrt%7B%28p-1%29%5E2%2B%285-2%29%5E2%7D%20%20)
![HK=\sqrt{(p-1)^2+9}](https://tex.z-dn.net/?f=%20HK%3D%5Csqrt%7B%28p-1%29%5E2%2B9%7D%20%20)
now, we can find other side
![JK=\sqrt{(p-7)^2+(5-2)^2}](https://tex.z-dn.net/?f=%20JK%3D%5Csqrt%7B%28p-7%29%5E2%2B%285-2%29%5E2%7D%20%20)
![JK=\sqrt{(p-7)^2+9}](https://tex.z-dn.net/?f=%20JK%3D%5Csqrt%7B%28p-7%29%5E2%2B9%7D%20%20)
now, we can set them equal
![HK=JK](https://tex.z-dn.net/?f=%20HK%3DJK%20%20)
![\sqrt{(p-1)^2+9}=\sqrt{(p-7)^2+9}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28p-1%29%5E2%2B9%7D%3D%5Csqrt%7B%28p-7%29%5E2%2B9%7D%20%20)
now, we can take square both sides
![(p-1)^2+9=(p-7)^2+9](https://tex.z-dn.net/?f=%20%28p-1%29%5E2%2B9%3D%28p-7%29%5E2%2B9%20%20)
now, we can solve for p
![(p-1)^2=(p-7)^2](https://tex.z-dn.net/?f=%20%28p-1%29%5E2%3D%28p-7%29%5E2%20%20)
![p^2-2p+1=p^2-14p+49](https://tex.z-dn.net/?f=%20p%5E2-2p%2B1%3Dp%5E2-14p%2B49%20%20)
![-2p+1=-14p+49](https://tex.z-dn.net/?f=%20-2p%2B1%3D-14p%2B49%20%20)
![12p=48](https://tex.z-dn.net/?f=%2012p%3D48%20%20)
..............Answer