If the width is 15, and the scale factor from width to length is 3:2, then you will have to divide 15 by 3 to get your original width.
15/3 = 5.
5 is your original width.
Answer:
A. 26 1/2
Step-by-step explanation:
change denominator so they can be the same
3/4*3 = 9/12
1/3*4 = 4/12
Add the whole numbers : 10 +15=25
9/12+4/12= 13/12 so 1 1/12
26 1/12
We want to find the domain for the graphed function. We will see that the correct option is A: -12 ≤ x ≤ 14
<h3>What is the domain of a function?</h3>
For a function f(x) we define the domain as the set of possible inputs that we can use in the function.
In this case, we need to see the values in the horizontal axis that the graph covers.
We can see that it goes from -12 to 14. We also can see that there are two jumps, at x = -6 and at x = 8, but these values belong to the graph (denoted by the black dot) meaning that these are in the domain.
Then the domain is:
-12 ≤ x ≤ 14
If you want to learn more about domains, you can read:
brainly.com/question/1770447
Answer:
-9/14
Step-by-step explanation:
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.
