25 times :) hope this helped!
After a little manipulation, the given diff'l equation will look like this:
e^y * dy = (2x + 1) * dx.
x^2
Integrating both sides, we get e^y = 2------- + x + c, or e^y = x^2 + x + c
2
Now let x=0 and y = 1, o find c:
e^1 = 0^2 + 0 + c. So, c = e, and the solution is e^y = x^2 + x + e.
X^2-7x+38=5x+3
x^2-7x+38-5x-3=5x+3-5x-3
x^2-12x+35=0
Factoring:
(x-5)(x-7)=0
Two solutions:
x-5=0→x-5+5=0+5→x=5
x-7=0→x-7+7=0+7→x=7
Answer: x=5 and x=7
Answer: Options B. 7 and D. 5
X>_ 0
If you don’t understand that then x is greater than or equal to 0
