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MrRissso [65]
3 years ago
15

What expression is equivalent to (3^2)^-2

Mathematics
2 answers:
alex41 [277]3 years ago
8 0
Answer: 1/81

- First you simplify it to 3^-4

- Then express it with a positive exponent which goes to 1/3^4

- Then evaluate it to 1/81

hope this helps! :)
MariettaO [177]3 years ago
6 0

Answer:

1/81

Step-by-step explanation:

You just tap the calculator

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The width of a rectangle is 8 inches less than the length. The perimeter is 76 inches. Find the length and the width.
REY [17]
L=30in
w Width

in
P Perimeter

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Width is 8 and perimeter is 76
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Statistics..
ch4aika [34]
1 of 2 type of lettuces
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3 years ago
1.
Harlamova29_29 [7]

Answer:

a) x = 1

Step-by-step explanation:

Given: 2x + 3 = 5

1. Subtract 3 from both sides:

2x + 3 - 3 = 5 - 3

⟹ 2x = 2

2. Divide both sides by 2:

2x ÷ 2 = 2 ÷ 2

⟹ x = 1

3. Check your work:

2(1) + 3 = 5

2 + 3 = 5

⟹ 5 = 5 ✔

Learn more about solving algebraic equations here:

brainly.com/question/27597890

brainly.com/question/25229258

8 0
2 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
Multiply and simplify: b3 • b • b4 • b2
Luden [163]
Are those supposed to be exponents? When multiplying exponents you add them together

b^3 * b^1 * b^4 * b^2  
3 + 1 + 4 + 2 = 10

Answer: b^10
5 0
2 years ago
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