Answer:
The answer to the question is
The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is (-∞, 4)
Step-by-step explanation:
To apply look for the interval, we divide the ordinary differential equation by (t-4) to
y'' + 
y' + 
y = 
Using theorem 3.2.1 we have p(t) = , q(t) = , g(t) =
Which are undefined at 4. Therefore the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution, that is where p, q and g are continuous and defined is (-∞, 4) whereby theorem 3.2.1 guarantees unique solution satisfying the initial value problem in this interval.
Answer:
11:22 AM
Step-by-step explanation:
By means of a statement we can know that:
After an hour and 8 minutes, he stays 8 minutes behind
Which means that in 2 hours, it will take 16 minutes
If it is 10:30, it means that it is 6:30 hours later, that is to say 6.5.
therefore 6.5 x 8 = 52
Which means that at 10:30 the clock is 52 minutes late
So the correct time is 11:22 AM
Answer:
The smaller number is 8
The bigger number is 11
Step-by-step explanation:
let the first number be A
let the second number be B
A + 2B = 30..... equation 1
A + B = 19...... equation 2
Subtract equation 2 from 1.
A + 2B = 30
- A + B = 19
-------------------
0 + B = 11
B = 11
Substitute B = 11 into equation 2
A + B = 19
A + 11 = 19
collect like terms
A = 19-11
A = 8.
A= 8 and B = 11
A = 7
B = 14
For A:
(7(sqrt)2)sin(45) = 7
7/sin(30) = 14
Answer:
y= -2x........eqn i
y=5x+3......eqn ii
substituting the value of Y from eqn i in eqn ii, we get,
-2x=5x+3
-7x=3
x= -3/7
so, y= -2x= -2× (-3/7)
so, y= 6/7
