For the answer, see the picture attached.
Answer:
1 and 4 are correct. 2 and 3 are not.
Step-by-step explanation:
1.
When x = 0 where does the horse start?
y = 1.5*sin(0 + 0.5)*2*pi + 1.5
y = 1.5*sin(0.5*2pi) + 1.5
y = 1.5*sin(pi) + 1.5 But sin pi = 0
y = 0 + 1.5 So the horse is starting at the midpoint of it's travel.
2.
This one is a trick question. You can reason it without exact answers. At some point the sin(x + 0.5)*2pi will equal 1. When it does 1.5 * 1 + 1.5 = 3.0 At some other point sin(x + 0.5)*2pi = -1. When that happens the whole thing goes to 0. So the total of the distance traveled is 6 not three.
3.
You can figure this one out by letting x = 0.01 When it does then the value of the function is
y = 1.5*sin [(0.01 + 0.5)*2pi] + 1.5
y = 1.5*sin(0.51*2*pi) + 1.5
y = 1.5*sin(3.204424) + 1.5
y = 1.5*(- .0623) + 1.5
y = -0.9418 + 1.5
y = 1.4058 so it is going downward The value is getting smaller.
4.
The horse starts out in the middle of the pole. What does x need to equal so that x + 0.5 = 1 ? And why 1. The answer to why 1 is that then the sine function will equal sin(2*pi)
That happens when x = 0.5 which is 1/2 a minute. If it takes 1/2 a minute to execute 1 complete cycle, then in 5 minutes the cycle will be executed 10 times. This one is correct.
The slope of the line would be positive 5/1 instead of positive 3/1. the y-intercept stays the same.
The cost function is an illustration of a linear function
The equation of the cost function is C(x) = 35 + 0.5x
The given parameters are:
-- the cost of renting a cooler
--- the rate per bottle
Assume the number of bottles in a cooler is x.
The cost function for the bottles in a cooler would be:
So, we have:
Evaluate the products
Hence, the equation is
Read more about linear functions at:
brainly.com/question/15602982
Answer:
If k = −1 then the system has no solutions.
If k = 2 then the system has infinitely many solutions.
The system cannot have unique solution.
Step-by-step explanation:
We have the following system of equations
The augmented matrix is
![\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C1%261%261%26k%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)
The reduction of this matrix to row-echelon form is outlined below.
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%263%26-2%26k%5E2-4%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%260%260%26k%5E2-k-2%5Cend%7Barray%7D%5Cright%5D)
The last row determines, if there are solutions or not. To be consistent, we must have k such that
Case k = −1:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-1-2%5C%5C0%260%260%26%28-1%29%5E2-%28-1%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-3%5C%5C0%260%260%26-2%5Cend%7Barray%7D%5Cright%5D)
If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.
Case k = 2:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%262-2%5C%5C0%260%260%26%282%29%5E2-%282%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%5D)
This gives the infinite many solution.
