The answer to the question is $4.7
Solving for the midpoint:
x - coordinate: (7.6 + 4.6)/2 = 6.1
y - coordinate: (10.1 + 3.1)/2 = 6.6
Midpoint: (6.1 , 6.6)
Solving for the distance:
Distance formula: sqrt( (x2 - x1)^2 + (y2 - y1)^2 )
D = sqrt( (7.6 - 4.6)^2 + (10.1 - 3.1)^2 <span>)
D = sqrt( 3^2 + 7^2)
D = sqrt(58)
D = 7.62 units
Distance = 7.62 units</span>
Answer: D) 13y^25 and 2y^25
Like terms involve the same variables, and each of those variables must have the same exponents.
Another example of a pair of like terms would be 5x^3y^2 and 7x^3y^2. Both involve the variable portion "x^3y^2" which we can replace with another variable, say the variable z. That means 5x^3y^2 becomes 5z and 7x^3y^2 becomes 7z. After getting to 5z and 7z, it becomes more clear we have like terms.
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that 
. Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that 
. In this equality we can perform a right multiplication by 
and obtain 
. Then, in the obtained equality we perform a left multiplication by P and get 
. If we write 
and 
we have 
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have 
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and 
. So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.
She should choose bank r because we don't know how far she is from work.
