Answer : value of x and y in the matrix equation is:
x = -3
y = +4, -4
Step-by-step explanation :
The matrix expression is:
![\left[\begin{array}{ccc}x+4\\y^2+1\end{array}\right]+\left[\begin{array}{ccc}-9x\\-17\end{array}\right]=\left[\begin{array}{ccc}28\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B4%5C%5Cy%5E2%2B1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-9x%5C%5C-17%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D28%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
First we have to add left hand side matrix.
![\left[\begin{array}{ccc}(x+4)+(-9x)\\(y^2+1)+(-17)\end{array}\right]=\left[\begin{array}{ccc}28\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28x%2B4%29%2B%28-9x%29%5C%5C%28y%5E2%2B1%29%2B%28-17%29%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D28%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Now we have to add left hand side terms.
![\left[\begin{array}{ccc}x+4-9x\\y^2+1-17\end{array}\right]=\left[\begin{array}{ccc}28\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B4-9x%5C%5Cy%5E2%2B1-17%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D28%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}4-8x\\y^2-16\end{array}\right]=\left[\begin{array}{ccc}28\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4-8x%5C%5Cy%5E2-16%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D28%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Now we have to equating left hand side matrix to right hand matrix, we get:
Therefore, the value of x and y in the matrix equation is -3 and +4, -4 respectively.
5/54 or approximately 0.092592593
There are 6^3 = 216 possible outcomes of rolling these 3 dice. Let's count the number of possible rolls that meet the criteria b < y < r, manually.
r = 1 or 2 is obviously impossible. So let's look at r = 3 through 6.
r = 3, y = 2, b = 1 is the only possibility for r=3. So n = 1
r = 4, y = 3, b = {1,2}, so n = 1 + 2 = 3
r = 4, y = 2, b = 1, so n = 3 + 1 = 4
r = 5, y = 4, b = {1,2,3}, so n = 4 + 3 = 7
r = 5, y = 3, b = {1,2}, so n = 7 + 2 = 9
r = 5, y = 2, b = 1, so n = 9 + 1 = 10
And I see a pattern, for the most restrictive r, there is 1 possibility. For the next most restrictive, there's 2+1 = 3 possibilities. Then the next one is 3+2+1
= 6 possibilities. So for r = 6, there should be 4+3+2+1 = 10 possibilities.
Let's see
r = 6, y = 5, b = {4,3,2,1}, so n = 10 + 4 = 14
r = 6, y = 4, b = {3,2,1}, so n = 14 + 3 = 17
r = 6, y = 3, b = {2,1}, so n = 17 + 2 = 19
r = 6, y = 2, b = 1, so n = 19 + 1 = 20
And the pattern holds. So there are 20 possible rolls that meet the desired criteria out of 216 possible rolls. So 20/216 = 5/54.
No, because with the parentheses, you distribute the 2 to both the x and the 5, making that equation 2x-10
Answer:
E=20H
Step-by-step explanation:
E =20 when H=1
there fore the amount he receives per working time will be 20*H
