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const2013 [10]
3 years ago
12

Loommon

Mathematics
1 answer:
Sloan [31]3 years ago
6 0

Hello!

We have the point (-5,3). We will plug this into each equation and see which ones are true.

3=-10+13

True

3=5-2

True

3=-15-5

False

3=2.5+6

False

3=10-2

False  

Therefore, the correct answers are A and B.

I hope this helps!

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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 <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 A=J^{-1}AJ. 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 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.

<em>Transitive</em>: 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

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

Recall that if P and Q are invertible, then QP is invertible and (QP)^{-1}=P^{-1}Q^{-1}. So, if we denote R=QP we obtained that

A=R^{-1}CR. Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0
4 years ago
The school that Julia goes to is selling tickets to the annual dance competition. On the first day
ollegr [7]

We have a system of equations in two variables, namely, a and c. Use the substitution to find a and c.

3 0
4 years ago
Read 2 more answers
EasyElectrics supermarket orders light bulbs from suppliers, AA Electronics and AAA Electronics. EasyElectrics purchases 30% of
saw5 [17]

Answer: (a) 0.006

               (b) 0.027

Step-by-step explanation:

Given : P(AA) = 0.3 and P(AAA) = 0.70

Let event that a bulb is defective be denoted by D and not defective be D';

Conditional probabilities given are :

P(D/AA) = 0.02 and P(D/AAA) = 0.03

Thus P(D'/AA) = 1 - 0.02 = 0.98

and P(D'/AAA) = 1 - 0.03 = 0.97

(a) P(bulb from AA and defective) = P ( AA and D)

                                                       = P(AA) x P(D/AA)

                                                       = 0.3 x 0.02 = 0.006

(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)

                         = P(AA) x P(D/AA) + P(AAA) x P(D/AAA)

                         = 0.3(0.02) + 0.70(0.03)

                         = 0.027

3 0
4 years ago
I really need help...Will mark brainiest
alexandr1967 [171]

Answer:

9a2b2=4c2

the 2s mean squared btw

Step-by-step explanation:

6 0
3 years ago
What are the domain and range of fx) = (1/5)^x
a_sh-v [17]

Answer: the domain is all real numbers. The Range is all real numbers greater than zero.

Step-by-step explanation:

this si the same as the graph  y=5^-x. Which decreases from left to right and approaches but never touches the x-axis.

6 0
3 years ago
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