The greatest common factor is the largest factor that both terms share.
First, start with the coefficients, 14 and 28. The largest factor they both share is 14. 14 x 1 = 14, 14 x 2 = 28.
Next we move to the variables, abc and a²b²c³. The largest factor the both share is abc, all to the power of 1.
So, your GCF and final answer is 14abc
ΔTXV is an <span>isosceles triangle because ∡T = ∡X therefore TV = VX.
3x - 24 = 2x + 1 |subtract 2x from both sides
x - 24 = 1 |add 24 to both sides
x = 25
</span>
The first one and the third one are correct. because for the when you plug 1 on the first piece wise function you get 3.
f(3) =1. and f(1)=3 so f(1)>f(3)
m x H = ![\left[\begin{array}{ccc}-25&37.5&-12.5\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-25%2637.5%26-12.5%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Step 1; Multiply 5 with this matrix ![\left[\begin{array}{ccc}-1&2\\4&8\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%262%5C%5C4%268%5C%5C%5Cend%7Barray%7D%5Cright%5D)
and we get a matrix ![\left[\begin{array}{ccc}-5&10\\20&40\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%2610%5C%5C20%2640%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Multiply the fraction 
with the matrix and we get ![\left[\begin{array}{ccc}-\frac{2m}{5} &\frac{4m}{5} \\\frac{8m}{5} &\frac{16m}{5} \\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B2m%7D%7B5%7D%20%26%5Cfrac%7B4m%7D%7B5%7D%20%5C%5C%5Cfrac%7B8m%7D%7B5%7D%20%26%5Cfrac%7B16m%7D%7B5%7D%20%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step2; Now equate corresponding values of the matrices with each other.
-5 = 
and so on. By equating we get the value of m as 
Step 3; Add the matrices to get the value of matrix m.
Adding the three matrices on the RHS we get ![\left[\begin{array}{ccc}2&9&-9\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%269%26-9%5C%5C%5Cend%7Barray%7D%5Cright%5D)
.
Step 4; Adding the matrices on the LHS we get the resulting matrix as H +
![\left[\begin{array}{ccc}4&6&-8\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%266%26-8%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
. Equating the matrices from step 3 and 4 we get the value of H as ![\left[\begin{array}{ccc}-2&3&-1\\\9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%263%26-1%5C%5C%5C9%5Cend%7Barray%7D%5Cright%5D)
Step 5; Now to find the value of m x H we need to multiply the value of with the matrix
Step 6; Multiplying we get the matrix m x H = [ -25 
]
