Answer:
![A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-4%262%5C%5C2%266%26-6%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given
Required
Find the standard matrix
The standard matrix (A) is given by
Where
![T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=T%28x%29%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
becomes
![Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=Ax%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
The x on both sides cancel out; and, we're left with:
![A = [T(e_1)\ T(e_2)\ T(e_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D)
Recall that:
In matrix:
is represented as: ![\left[\begin{array}{c}a\\b\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D)
So:
![T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]](https://tex.z-dn.net/?f=T%28e_1%29%20%3D%20%281%2C2%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
![T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_2%29%20%3D%20%28-4%2C6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-4%5C%5C6%5Cend%7Barray%7D%5Cright%5D)
![T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_3%29%20%3D%20%282%2C-6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-6%5Cend%7Barray%7D%5Cright%5D)
Substitute the above expressions in
Hence, the standard of the matrix A is:
Okay this is simple once you get use to it. What you first need to do is figure out the formula. I don’t have a calculator on me so I will just tell you the formula so you can get the answer
: 1/3 times 8 times 6 times 10
<span>625(5xy)^-3/ (5x)^2 y^7
625
= -------------------- / 25x^2y^7
125 x^3y^3
= 5/x^3y^3 / </span>25x^2y^7
= 5/x^3y^3 * (1/ 25x^2y^7)
= 1 / 5x^5y^10
answer
1
-----------------
5x^5y^10
This chart represents a proportional relationship. The rate of change will be 4.
<h3>What is the constant of proportionality?</h3>
Firstly, there is a proportional relationship.
Two values x and y are said to be in a proportional relationship if x=ky, where x and y are variables and k is a constant.
The constant k is called the constant of proportionality.
Given;
x y
1 4
2 8
3 12
4 16
Here, x = ky
put the first value
k = 1/4
This chart represents a proportional relationship.
So the rate of change will be
Learn more about proportional ;
brainly.com/question/20048815
#SPJ1
Answer: 1 hour 15 minutes
Step-by-step explanation:
From the question, we are informed that Laura and Alicia both exercise 5 days a week and that Laura exercises for 30 minutes each day. For the 5 days, she'll exercise for:
= 5 × 30 minutes
= 150 minutes
= 2 hours 30 minutes
Alicia exercises for 45 minutes each day. Fir the 5 days, she'll exercise for:
= 5 × 45 minutes
= 225 minutes
= 3 hours 45 minutes
We then calculate the difference in their exercise per week which will be:
= 3 hours 45 minutes - 2 hours 30 minutes
= 1 hour 15 minutes
