Yo sup??
We can solve this questions by trying option verification.
We observe that 0 is common among all the options so 0 has to be a zero of the function.
when x=1
g(1)=1-1+4-4=0
Therefore 1 is also a zero of the function.
Since 1 is not there is option 2 and 3, hence our answer is either option 1 or 4
when x=2
g(2)=16-8+16-8=0
Therefore 2 is also a zero of the function.
Hence the correct answer is option D ie 0,1,2
Hope this helps.
The question is:
Check whether the function:
y = [cos(2x)]/x
is a solution of
xy' + y = -2sin(2x)
with the initial condition y(π/4) = 0
Answer:
To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.
Let us do that.
y = [cos(2x)]/x
y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]
Now,
xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x
= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)
= -2sin(2x)
Which is the right hand side of the differential equation.
Hence, y is a solution to the differential equation.
First you divide 17.5/ 3 and then the answer multiply by 15 and get 87.5 i think
Answer: No, he can't pass the course.
Explanation:
Since we have given that
Marks on tests 1, 2, and 4 are given by
82, 76, 90 respectively.
Unfortunately, he cut the third test and receive a 0.
According to question, we have given that passing grade for the course is 70, and his one test left to take ,
Let the marks obtained in fifth test be x
So,
And it is not possible to get 102 over 100.
so, he can't still pass the course.
Hence, No, he can't pass the course.
What picture ? Did you attached it
