Please help me on these questions above

tiny-mole [99]

B. (k+10f)(k-5f) you can tell in 2 ways. 1. if the second sign is neg then the signs in the factors are going to be one neg and one positive so that knocks out C and D. Then you look at the second term and since the 5 is positive then the 10 has to be positive so you end up with a positive 5kf.

The second way is to FOIL them out. When you check B you get $k^{2} -5kf+10kf- 50f^{2}$ combine your like terms and you have your original expression of $k^{2} +5kf-50 f^{2}$

Hope that helps

3 0

3 years ago

Sandra wants to buy a dress that is on sale for 20% off. Sandra also gets a 5% discount for being a member of the store's buyer'

Paraphin [41]

0.25c - that’s 25% off the cost (c) of the dress

7 0

3 years ago

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an equivalence relation 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.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get