Answer:
Hope the picture will help you..............
Answer:
x = ![\frac{1}{a-b}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Ba-b%7D)
Step-by-step explanation:
Given
ax = bx + 1
Collect the terms in x on the left side by subtracting bx from both sides
ax - bx = 1 ← factor out x from each term on the left side
x(a - b) = 1 ← divide both sides by (a - b)
x = ![\frac{1}{a-b}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Ba-b%7D)
Answer:
15
Step-by-step explanation:
If each necklace takes 30 beads, she can make 15 necklaces until she runs out of beads. 450÷30=15. Hope this helps!
Answer:
![-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}](https://tex.z-dn.net/?f=-%5Ctheta%20cos%5Cthsta%2Bsin%5Ctheta%20%3D%20%5Cfrac%7By%5E%7B2%7D%20%7D%7B2%7D%20%2B%20ln%20y%20%2B%20%5Cpi%20%20-%20%5Cfrac%7B1%7D%7B2%7D)
Step-by-step explanation:
Given the initial value problem
subject to y(π) = 1. To solve this we will use the variable separable method.
Step 1: Separate the variables;
![\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Ctheta%7D%28%5Cfrac%7Bdy%7D%7Bd%5Ctheta%7D%20%29%20%3D%5Cfrac%7B%20ysin%5Ctheta%7D%7By%5E%7B2%7D%2B1%20%7D%20%5C%5C%5Cfrac%7B1%7D%7B%5Ctheta%7D%28%5Cfrac%7Bdy%7D%7Bsin%5Ctheta%20d%5Ctheta%7D%20%29%20%3D%5Cfrac%7B%20y%7D%7By%5E%7B2%7D%2B1%20%7D%20%5C%5C%5Cfrac%7B1%7D%7B%5Ctheta%7D%28%5Cfrac%7B1%7D%7Bsin%5Ctheta%20d%5Ctheta%7D%20%29%20%3D%20%5Cfrac%7B%20y%7D%7Bdy%28y%5E%7B2%7D%2B1%20%29%7D%20%5C%5C%5C%5C%5Ctheta%20sin%5Ctheta%20d%5Ctheta%20%3D%20%5Cfrac%7B%20%28y%5E%7B2%7D%2B1%29dy%7D%7By%7D%20%5C%5Cintegrating%5C%20both%20%5C%20sides%5C%5C%5Cint%5Climits%20%5Ctheta%20sin%5Ctheta%20d%5Ctheta%20%3D%5Cint%5Climits%20%20%5Cfrac%7B%20%28y%5E%7B2%7D%2B1%29dy%7D%7By%7D%20%5C%5C-%5Ctheta%20cos%5Ctheta%20-%20%5Cint%5Climits%20%28-cos%5Ctheta%29d%5Ctheta%20%3D%20%5Cint%5Climits%20ydy%20%2B%20%5Cint%5Climits%20%5Cfrac%7Bdy%7D%7By%7D)
![-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}](https://tex.z-dn.net/?f=-%5Ctheta%20cos%5Cthsta%2Bsin%5Ctheta%20%3D%20%5Cfrac%7By%5E%7B2%7D%20%7D%7B2%7D%20%2B%20ln%20y%20%2BC%5C%5CGiven%20%5C%20the%5C%20condition%5C%20y%28%5Cpi%20%29%20%3D%201%5C%5C-%5Cpi%20cos%5Cpi%20%2Bsin%5Cpi%20%20%3D%20%5Cfrac%7B1%5E%7B2%7D%20%7D%7B2%7D%20%2B%20ln%201%20%2BC%5C%5C%5C%5C%5Cpi%20%2B%200%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2B%20C%20%5C%5CC%20%3D%20%5Cpi%20%20-%20%5Cfrac%7B1%7D%7B2%7D)
The solution to the initial value problem will be;
![-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}](https://tex.z-dn.net/?f=-%5Ctheta%20cos%5Cthsta%2Bsin%5Ctheta%20%3D%20%5Cfrac%7By%5E%7B2%7D%20%7D%7B2%7D%20%2B%20ln%20y%20%2B%20%5Cpi%20%20-%20%5Cfrac%7B1%7D%7B2%7D)