m□ebd=4 x-8 and m□ebc=5 x+20
This is solvable only if e b is the initial side and b d and b c lies on opposite side of each other and lies on a line i.e c,b,d are Collinear.
∠ebd and ∠ebc will form a linear pair.The meaning of linear pair is that angles forming on one side of a straight line through a common vertex which are adjacent is 180°.
i.e
∠ ebd + ∠ebc = 180°
4 x- 8 + 5x + 20= 180°
adding like terms
⇒ 9 x +12 =180°
⇒ 9 x = 180° - 12
⇒ 9 x = 168°
⇒ x =( 168/9)°=(56/3)°
now m□ebc =5 x +20
= 5 × 56/3 + 20
= 280/3 + 20
=340/3
m□ebc=( 340/3)°
So, solution set is x =(56/3)° and m□ebc =(340/3)°
Answer:
0 and -1
Step-by-step explanation:
Answer:
x=6
y=-5
Step-by-step explanation:
3x+2y=8 equation 1
x-2y=16 equation 2
equation 2+equation1 ; 4x = 24
x = 6
substitution x in equation1 ;
18+2y=8
2y = -10
y = -5
40 millilitres in total. 32 millilitres of alcohol and 8 of water
Answer:
Differential equation will be ![\frac{dV}{dt}=-KV^{\frac{2}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-KV%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D)
Step-by-step explanation:
Let V is the volume of the raindrop and surface area of the drop is S.
Since volume of the raindrop reduces at the rate directly proportional to the surface area.
![\frac{dV}{dt}\alpha S](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%5Calpha%20S)
Or
where k is the proportionality constant.
We know volume of the spherical drop V = ![\frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
![r=(\frac{3V}{4\pi})^{\frac{1}{3}}](https://tex.z-dn.net/?f=r%3D%28%5Cfrac%7B3V%7D%7B4%5Cpi%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D)
Since S = ![4\pi r^{2}](https://tex.z-dn.net/?f=4%5Cpi%20r%5E%7B2%7D)
Therefore, ![\frac{dV}{dt}=-k\times 4\pi r^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-k%5Ctimes%204%5Cpi%20r%5E%7B2%7D)
![\frac{dV}{dt}=-4k\pi(\frac{3V}{4\pi})^{\frac{2}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-4k%5Cpi%28%5Cfrac%7B3V%7D%7B4%5Cpi%7D%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D)
![\frac{dV}{dt}=-k(4\pi)^{1-\frac{2}{3}}(3V)^{\frac{2}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-k%284%5Cpi%29%5E%7B1-%5Cfrac%7B2%7D%7B3%7D%7D%283V%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D)
![\frac{dV}{dt}=-{k}{(4\pi)^\frac{1}{3}}\times (3)^{\frac{2}{3}}V^{\frac{2}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-%7Bk%7D%7B%284%5Cpi%29%5E%5Cfrac%7B1%7D%7B3%7D%7D%5Ctimes%20%283%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7DV%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D)
![\frac{dV}{dt}=-{k}{(4\pi)^\frac{1}{3}}\times (9)^{\frac{1}{3}}V^{\frac{2}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-%7Bk%7D%7B%284%5Cpi%29%5E%5Cfrac%7B1%7D%7B3%7D%7D%5Ctimes%20%289%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7DV%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D)
![\frac{dV}{dt}=-{k}{(36\pi)^\frac{1}{3}}\times V^{\frac{2}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-%7Bk%7D%7B%2836%5Cpi%29%5E%5Cfrac%7B1%7D%7B3%7D%7D%5Ctimes%20V%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D)
Since coefficient of
is a constant.
Then ![(36\pi )\frac{1}{3}=K](https://tex.z-dn.net/?f=%2836%5Cpi%20%29%5Cfrac%7B1%7D%7B3%7D%3DK)
![\frac{dV}{dt}=-KV^\frac{2}{3}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-KV%5E%5Cfrac%7B2%7D%7B3%7D)
Therefore, differential equation for the volume of the raindrop as function of time will be ![\frac{dV}{dt}=-KV^{\frac{2}{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D-KV%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D)