The second matrix ![\left[\begin{array}{ccc}-6&3&12\\0&15&-24\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-6%263%2612%5C%5C0%2615%26-24%5Cend%7Barray%7D%5Cright%5D)
represents the triangle dilated by a scale factor of 3.
Step-by-step explanation:
Step 1:
To calculate the scale factor for any dilation, we divide the coordinates after dilation by the same coordinated before dilation.
The coordinates of a vertice are represented in the column of the matrix. Since there are three vertices, there are 2 rows with 3 columns. The order of the matrices is 2 × 3.
Step 2:
If we form a matrix with the vertices (-2,0), (1,5), and (4,-8), we get
![\left[\begin{array}{ccc}-2&1&4\\0&5&-8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%261%264%5C%5C0%265%26-8%5Cend%7Barray%7D%5Cright%5D)
The scale factor is 3, so if we multiply the above matrix with 3 throughout, we will get the matrix that represents the vertices of the triangle after dilation.
Step 3:
The matrix that represents the triangle after dilation is given by
![3\left[\begin{array}{ccc}-2&1&4\\0&5&-8\end{array}\right] = \left[\begin{array}{ccc}3(-2)&3(1)&3(4)\\3(0)&3(5)&3(-8)\end{array}\right] = \left[\begin{array}{ccc}-6&3&12\\0&15&-24\end{array}\right]](https://tex.z-dn.net/?f=3%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%261%264%5C%5C0%265%26-8%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%28-2%29%263%281%29%263%284%29%5C%5C3%280%29%263%285%29%263%28-8%29%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-6%263%2612%5C%5C0%2615%26-24%5Cend%7Barray%7D%5Cright%5D)
This is the second option.
Answer:
Me
Step-by-step explanation:
Step-by-step explanation:
4(5x-9)=-2(x+7)
4*5x-4*9=-2*x-2*7
20x-36=-2x-14
20x+2x=-14+36
22x=22
x=22/22
x=1
Answer:
Step-by-step explanation:
<u>Given equation:</u>
<u>Solve for cubic units³</u>
There are 30 cubic inches³.
<u>Word form:</u>
- Three times Two Times Five
- equals Six times Five
- equals Thirty.
There are Thirty cubic inches³.
Answer:
sample size n would be 149305 large
Value of n (149305) is too high, this will be the practical problem with attempting to find this confidence interval
Step-by-step explanation:
Given that;
standard deviation α = 150 min
confidence interval = 99%
since; p( -2.576 < z < 2.576) = 0.99
so z-value for 99% CI is 2.576
E = 1 minutes
Therefore
n = [(z × α) / E ]²
so we substitute
n = [(2.576 × 150) / 1 ]²
n = [ 386.4 ]²
n = 149304.96 ≈ 149305
Therefore sample size would be 149305 large
Value of n is too high, that would be the practical problem with attempting to find this confidence interval
