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lukranit [14]
3 years ago
11

7,125,000,000 what is this to the nearest billion

Mathematics
1 answer:
Luda [366]3 years ago
8 0
7,000,000,000 I believe??? Comment if I'm wrong
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0.01x - 0.3y = 1
dusya [7]

Answer:

x-30y=100 and 10y=x-20

Step-by-step explanation:


8 0
4 years ago
Read 2 more answers
Find the midpoint of the line segment with the endpoints A and B.
zavuch27 [327]

Answer:

<h2>( 6 , 7 )</h2>

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

M = (  \frac{x1 + x2}{2} , \:  \frac{y1 + y2}{2} )\\

From the question we have

m = ( \frac{2 + 10}{2} \:  , \:  \frac{6 + 8}{2} ) \\  =  (\frac{12}{2}  \: , \:  \frac{14}{2} ) \:  \:  \:  \:  \:  \:  \:

We have the final answer as

<h3>( 6 , 7 )</h3>

Hope this helps you

3 0
3 years ago
Can someone help please
elena-s [515]

Answer:

I'm sorry I can help you maybe a little bit if that's okay. Btw what grade are you in?

Step-by-step explanation:

3 0
4 years ago
What is the value of x?<br> 68<br> 4x + 12<br> 17<br> 15<br> 32<br> 8
Marianna [84]

Answer:

k12

Step-by-step explanation:

4 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 <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
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