How do I work it out ?

Mathematics

1 answer:

algol [13]2 years ago

4 0

Answer:

x - y = 5

Explanation:

As you can see in the first problem the ration is x to y = 8 to 3. This means for every 8 x values, there are 3 y values. Therefore, x - y should be the same as 8 - 3, which is 5.

You might be interested in

Which of these statements is true

Eva8 [605]

Answer:

B: The y-intercept is (0,2).

Step-by-step explanation:

See attached graph.

7 0

1 year ago

Read 2 more answers

Answer the very simple question below!

zalisa [80]

Answer: 14

Step-by-step explanation: just go to the opposite side of the interger and plz Brainlest would be much appreciated.

5 0

2 years ago

Read 2 more answers

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