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erastova [34]
2 years ago
12

What is 1 plus 1? any guesses? any guesses? any guesses?

Mathematics
2 answers:
lord [1]2 years ago
8 0

Answer:

ummmmmmmmmmmmmmmmmmmm OH 22 HAH

Step-by-step explanation:

Alex17521 [72]2 years ago
7 0

Answer:

Obviously, it's 2

Step-by-step explanation:

First, 1 is 1x1, so branch out under both 1's and put 1 x 1. Then, you add 1 + 1 + 1 + 1, which is 4. Then you do 4 - 2 since it is the two 1's in the original problem, and you get 2.

This is SUPER extra...

As well as mathematically incorrect lol

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In parallelogram ABCD, the length of BC=3x-1, CD=2x, and AD=x+5. Find the perimeter of ABCD
yarga [219]

Answer:

Step-by-step explanation:

BC=AD

3x-1=x+5

3x-x=5+1

2x=6

x=3

BC=3×3-1=8

CD=2x=2×3=6

perimeter=2(8+6)=2(14)=28

6 0
3 years ago
Classify the decimal forms of 3/12 and and 2/9 as repeating or as repeating or terminating.​
andre [41]

Answer:

3/12: terminating; 2/9: repeating

Step-by-step explanation:

3/12=0.25 terminating as it has a finite number of digits after the decimal point.

2/9=0.222.... repeating decimal as its number is repeated indefinitely.

6 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
3 years ago
Twenty seven less than eight times a number is equal to five. What is the number?
Korvikt [17]
<h3><u>The unknown number is equal to 4.</u></h3>

8x - 27 = 5

Add 27 to both sides.

8x = 32

Divide both sides by 8

x = 4

6 0
3 years ago
Read 2 more answers
DOES ANYONE KNOW THE ANSWER ???? Please I can’t fail this test !
Zigmanuir [339]

Answer:

The last option

Step-by-step explanation:

You take the middle 2 numbers, add them up, and divide by 2.

6 0
2 years ago
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