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Mathematics

1 answer:

fiasKO [112]3 years ago

4 0

Answer:

Option A

Option B

Option E

Option G

Option A and E go together

and

Option B and G go together.

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You want to walk from home to a clothing store that is 3/4 miles away. You stop for a rest after 1/2 miles. How much farther do

dmitriy555 [2]

Answer:

1/4 miles.

Step-by-step explanation:

If I want to walk from my house to a clothing store 3/4 of a mile away, this implies that the location is 0.75 miles from my home.

Now, if after walking 1/2 mile I decide to stop, I will have traveled 0.5 miles in total. In this way, when I get back on my way, I will have to travel 0.25 miles to reach my destination (0.75 - 0.50).

Therefore, since 1 divided by 4 equals 0.25, I have to walk 1/4 mile to reach my destination.

5 0

3 years ago

Explain why √3x+2 = -16 has no solution.

pogonyaev

The equation you wrote has no solutions since the left hand side is equal to 6x-10 and the right hand side is equal to 6x+8. Setting them equal we get -10=8 which is obviously wrong.

8 0

3 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