I think the answer to the problem is 80 because when you put a number in the tenths place all you do is add a extra zero
The picture is very blurry for me can you take another
Answer:
y=7
x=-7
Step-by-step explanation:
2(-3x - 7y )= 2(-28)
3(2x+ 3y)=3(7)
-6x - 14y = -56
6x + 9y = 21
Add the two equations
-1(0x - 5y)= -1(- 35)
5y = 35
Divide both sides by 5
y=7
Substitution
6x + 9(7) = 21
6x + 63 = 21
Subtract 63 from both sides
6x = -42
Divide both sides by 6
x= - 7
Answer:
The difference is a binomial with a degree of 6
Step-by-step explanation:
Given:
a^3b + 9a^2b^2 − 4ab^5 and a^3b − 3a^2b^2 + ab^5
Let A = a^3b + 9a^2b^2 - 4ab^5
B= a^3b - 3a^2b^2+ab^5
The difference between A and B is
A - B = a^3b + 9a^2b^2 - 4ab^5 - (a^3b - 3a^2b^2 + ab^5)
Open parenthesis
A - B= a^3b + 9a^2b^2 - 4ab^5 - a^3b + 3a^2b^2 - ab^5
= a^3b - a^3b + 9a^2b^2 + 3a^2b^2 - 4ab^5 - ab^5
= 12a^2b^2 - 5ab^5
The fist term 12a^2b^2 has 2+2=4 as a degree
The second term 5ab^5 has 1 +5 =6 as a degree
Therefore,
the answer is: The difference is a binomial with a degree of 6
Answer:
<em>The solution of the system is:
</em>
Step-by-step explanation:
The given system of equations is.......

So, the augmented matrix will be: ![\left[\begin{array}{cccc}-1&-3&|&-17\\2&-6&|&-26\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-1%26-3%26%7C%26-17%5C%5C2%26-6%26%7C%26-26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now, we will transform the augmented matrix to the reduced row echelon form using row operations.
<u>Row operation 1 :</u> Multiply the 1st row by -1. So..........
![\left[\begin{array}{cccc}1&3&|&17\\2&-6&|&-26\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C2%26-6%26%7C%26-26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 2:</u> Add -2 times the 1st row to the 2nd row. So.......
![\left[\begin{array}{cccc}1&3&|&17\\0&-12&|&-60\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C0%26-12%26%7C%26-60%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 3:</u> Multiply the 2nd row by
. So.......
![\left[\begin{array}{cccc}1&3&|&17\\0&1&|&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%26%7C%2617%5C%5C0%261%26%7C%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
<u>Row operation 4:</u> Add -3 times the 2nd row to the 1st row. So........
![\left[\begin{array}{cccc}1&0&|&2\\0&1&|&5\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%7C%262%5C%5C0%261%26%7C%265%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now, from this reduced row echelon form of the augmented matrix, we can get that
and 
So, the solution of the system is: 