Yes, it’s D because you would have want to compare both fractions by making them have the same denominator. Since 60 is the least common denominator for 10 and 12 (the denominators), you would convert the fractions by multiplying the numerator and denominator of 5/10 by 6 to get 60. 5/10 would turn into 30/60. Then you would multiply 6/12 both by 5, and then it would turn into 30/60. Now that you have equal fractions, you can guess that they are the same. Therefore, 30/60=30/60! An easier way of doing it would be just simplifying them into 1/2 and comparing them + seeing that they are both 1/2. Hope this helps!
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Caramelatte
Answer:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Let's find the answer.
Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.
Equations:
Eq. 1 : x1 + 2x2 + 5x3 = 5
Eq. 2 : x1 + x2 + x3 = 6
E1. 3 : 4x1 + 6x2 + 5x3 = 7
Coefficients for x1 ; x2 ; x3
From eq. 1 : 1 ; 2 ; 5
From eq. 2 : 1 ; 1 ; 1
From eq. 3 : 4 ; 6 ; 5
So matrix A is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D)
And the vector of vriables (X) is:
![\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D)
Now we can find the resulting vector (B) using the 'resulting values' from each equation:
![\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
In conclusion, AX=B is:
![\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%265%5C%5C1%261%261%5C%5C4%266%265%5Cend%7Barray%7D%5Cright%5D%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx1%5C%5Cx2%5C%5Cx3%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C6%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
Answer:
-3,-1 so a
Step-by-step explanation:
X^2 - 8x + 13 = 0
subtract 13 from both sides
x^2 - 8x = -13
-8/2 = -4, (-4)^2 = 16, add 16 to both sides
x^2 - 8x + 16 = -13 + 16
factor the left side
(x - 4)^2 = 3
Answer:
you add an image
Step-by-step explanation: