First subtract the weight the dog weighs now to the weight he was:
48.9
- 29.7
------------
19.2
Then take the number you got which is the weight the dog gain in total, and divide it by the months the dog gained weight to get the average weight gained each month:
19.2/8 = 2.4
The average weight gain each month was 2.4 pounds.
Hope this helped you!
Step-by-step explanation:
118 is the answer 238-120 = 118
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 
]
You are a rhombus sometimes although not correctly referred to as a diamond
Answer:
3.33 and 1/3
Step-by-step explanation:
"Dense" here means that there are infinite irrational numbers between two rational numbers. Also, there are infinite rational numbers between two rational numbers. That's the meaning of dense. Actually, that can be apply to all real numbers, there always is gonna be a number between other two.
But, to demonstrate that irrationals are dense, we have to based on an interval with rational limits, because the theorem about dense sets is about rationals, and the dense irrational set is a deduction from it. That's why the best option is 2, because that's an interval with rational limits.
