First, write put the information into an equation
2*(x+9) = 3*(x-7)
distribute the 2 and 3
2x + 18 = 3x - 21
Combine like terms (Xs and munbers) (in this case, it's easier to add 15 to both sides and subtract 2x from both sides so both sides end up being positive)
33 = x
x=33
Answer:
Step-by-step explanation:
The given function is .
The domain refers to all values of x for which this function is defined.
Recall that: the domain of 
is 
And we know is the reciprocal of .
Therefore the complement of the domain of which is ![(-\infty,-1]\cup [1,+\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C-1%5D%5Ccup%20%5B1%2C%2B%5Cinfty%29)
is the domain of
Answer:
=38+x
Step-by-step explanation:
18-2+2x-24÷(x+2) =
16+2x-24-24÷(x+2)
24+16+2x
40+x+2
38+x
Answer:
Region D.
Step-by-step explanation:
Here we have two inequalities:
y ≤ 1/2x − 3
y < −2/3x + 1
First, we can see that the first inequality has a positive slope and the symbol (≤) so the values of the line itself are solutions, this line is the solid line in the graph.
And we have that:
y ≤ 1/2x − 3
y must be smaller or equal than the solid line, so here we look at the regions below the solid line, which are region D and region C.
Now let's look at the other one:
y < −2/3x + 1
y = (-2/3)*x + 1
is the dashed line in the graph.
And we have:
y < −2/3x + 1
So y is smaller than the values of the line, so we need to look at the region that is below de dashed line.
The regions below the dashed line are region A and region D.
The solution for the system:
y ≤ 1/2x − 3
y < −2/3x + 1
Is the region that is a solution for both inequalities, we can see that the only region that is a solution for both of them is region D.
Then the correct option is region D.
Answer:
y = 2cos5x-9/5sin5x
Step-by-step explanation:
Given the solution to the differential equation y'' + 25y = 0 to be
y = c1 cos(5x) + c2 sin(5x). In order to find the solution to the differential equation given the boundary conditions y(0) = 1, y'(π) = 9, we need to first get the constant c1 and c2 and substitute the values back into the original solution.
According to the boundary condition y(0) = 2, it means when x = 0, y = 2
On substituting;
2 = c1cos(5(0)) + c2sin(5(0))
2 = c1cos0+c2sin0
2 = c1 + 0
c1 = 2
Substituting the other boundary condition y'(π) = 9, to do that we need to first get the first differential of y(x) i.e y'(x). Given
y(x) = c1cos5x + c2sin5x
y'(x) = -5c1sin5x + 5c2cos5x
If y'(π) = 9, this means when x = π, y'(x) = 9
On substituting;
9 = -5c1sin5π + 5c2cos5π
9 = -5c1(0) + 5c2(-1)
9 = 0-5c2
-5c2 = 9
c2 = -9/5
Substituting c1 = 2 and c2 = -9/5 into the solution to the general differential equation
y = c1 cos(5x) + c2 sin(5x) will give
y = 2cos5x-9/5sin5x
The final expression gives the required solution to the differential equation.
