The distance between the points is 16
The area of D is given by:
The average value of f over D is given by:
![\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Cfrac%7B343%7D%7B3%7D%20%7D%20%20%5Cint%5Climits%5E7_0%20%20%5Cint%5Climits%5E%7Bx%5E2%7D_0%20%7B4x%5Csin%28y%29%7D%20%5C%2C%20dydx%20%20%3D%20-%5Cfrac%7B3%7D%7B343%7D%20%20%5Cint%5Climits%5E7_0%20%7B4x%5Ccos%28x%5E2%29%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D-%5Cfrac%7B3%7D%7B343%7D%20%5Cint%5Climits%5E%7B49%7D_0%20%7B2%5Ccos%28t%29%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%20%5Cleft%5B%5Csin%28t%29%5Cright%5D_0%5E%7B49%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%5Csin49)
The correct option is B. Reread the directions for each section.
If you have time left after completing the test don't reread the directions for each section.
<h3>What is meant
previewing a test?</h3>
In order to find the various problem kinds and their point values during the test preview, you must read the complete test. Mark the questions that you can answer quickly and easily.
The benefits of previewing the test are-
- You will probably be given credit for the answers even if you accidentally copied them from the scratch paper.
- Your effort on the scratch paper can earn you some points if a thoughtless mistake causes you to get the answer wrong.
- If you do make a mistake, it will be simpler to find it when the instructor goes through the test.
- By doing this, you can avoid making the same errors on your next exam.
Therefore, resolve each issue by re-entering the solution into the equation or performing the opposing operation needed to provide the appropriate response. Do not leave the testing room until the bell has rung or after you have gone over each problem twice.
To know more about Previewing, here
brainly.com/question/1144128
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The complete question is-
All of the following are suggestions of things to do if you have time left after completing the test except
A. Look at your answer sheet to make sure its filled properly
B. Reread the directions for each section
C. Return to the questioms you were unsure of
D. Make sure your answers correspond to the correct questions
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.
Answer:
64x^3-241x^2+296x-125
Step-by-step explanation:
