Take 36 and add it to 60 it gives you 96 then divide it by 2 and the answer is 48
There are 52 weeks in a year. In order to figure out how much you have to save each week to save a total of $2000, you have to divide $2000 by 52.
2000÷52= 38.462
So to save $2000 you would have to set aside about $38.50 each week.
Answer:
(x, y) = (-3, -6)
Step-by-step explanation:
The second equation can be used to write an expression for y:
... -5x -21 = y . . . . . . . add y-21 to both sides
This expression can be substituted for y in the first equation:
... 6 = -4x +(-5x -21)
... 27 = -9x . . . . . add 21, collect terms
... -3 = x . . . . . . . divide by -9
Using this value of x in the expression for y, we find ...
... -5(-3) -21 = y = -6
The solution is x = -3, y = -6.
Answer:
7.2 + 0.8w
Step-by-step explanation:
combine like terms. Like terms have same variable with same power.
2.2w and (-1.4w) are like terms & 4.8 and 2.4 are like terms.
4.8 + 2.2w - 1.4w + 2.4 = 4.8 + 2.4 + 2.2w - 1.4w
= 7.2 + 0.8w
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.
