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Lina20 [59]
3 years ago
10

WILL GIVE BRAINLIEST...Which of the following is equivalent to -2i(6 - 7i)

Mathematics
2 answers:
Sav [38]3 years ago
8 0

Answer:

A: ( 0 - 2i ) ( 6 - 7i )

Step-by-step explanation:

-2i ( 6 - 7i )

= -14 - 12i

( 0 - 2i ) ( 6 - 7i )

= -14 - 12i

NeTakaya3 years ago
6 0

Answer:

(0-2i)(6-7i)

Step-by-step explanation:

The answer to this is pretty straightforward, as if you add a 0 to -2i, it will not change the expression.

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A runner jogged 5 1/4 miles in 45 minutes. At this rate how far will she run in 2 hours?
Anon25 [30]
<span>(21/2) / (3/4) 

21/2 * 4/3 
=14 miles </span>
5 0
3 years ago
Mrs. Booker was being chased by a horde of zombies. She had to run 8 yards to make it to a safe zone. How far did she run in fee
Vikentia [17]

Answer:

24 feet

Step-by-step explanation:

Given that,

Mrs. Booker had to run 8 yards to make it to a safe zone.

We need to convert this distance in feet.

The conversion from yards to feet is as follows :

1 yard = 3 feet

So,

8 yards = 24 feet

Hence, Mrs. Booker had to run 24 feet to make it to a safe zone.

5 0
2 years ago
Heathers rabbit has a mass of 5.190 kilograms. Jeff's rabbit has a mass of 5.195 kilograms.Write and inequality statement compar
aleksklad [387]

Answer:

5.190 < 5.195

Step-by-step explanation:

5.195 is bigger than 5.190, so 5.195>5.190

8 0
2 years ago
Read 2 more answers
Linear Algebra question! Please help!
kozerog [31]

Answers:

  1. false
  2. false
  3. true
  4. false
  5. True

==================================================

Explanation:

Problem 1

This is false because the A and B should swap places. It should be (AB)^{-1} = B^{-1}A^{-1}.

The short proof is to multiply AB with its inverse (AB)^{-1}  and we get: (AB)*(AB)^{-1} = (AB)*(B^{-1}A^{-1}) = A(B*B^{-1})*A^{-1} = A*A^{-1} = I

The fact we get the identity matrix proves that we have the proper order at this point. The swap happens so that B matches up its corresponding inverse B^{-1} and the two cancel each other out.

Keep in mind matrix multiplication is <u>not</u> commutative. So AB is not the same as BA.

-------------------------

Problem 2

This statement is true if and only if AB = BA

(A+B)^2 = (A+B)(A+B)

(A+B)^2 = A(A+B) + B(A+B)

(A+B)^2 = A^2 + AB + BA + B^2

(A+B)^2 = A^2 + 2AB + B^2 ... only works if AB = BA

However, in most general settings, matrix multiplication is <u>not</u> commutative. The order is important when multiplying most two matrices. Only for special circumstances is when AB = BA going to happen. In general,  AB = BA is false which is why statement two breaks down and is false in general.

-------------------------

Problem 3

This statement is true.

If A and B are invertible, then so is AB.

This is because both A^{-1} and B^{-1} are known to exist (otherwise A and B wouldn't be invertible) and we can use the rule mentioned in problem 1. Make sure to swap the terms of course.

Or you can use a determinant argument to prove the claim

det(A*B) = det(A)*det(B)

Since A and B are invertible, their determinants det(A) and det(B) are nonzero which makes the right hand side nonzero. Therefore det(A*B) is nonzero and AB has an inverse.

So if we have two invertible matrices, then their product is also invertible. This idea can be scaled up to include things like A^4*B^3 being also invertible.

If you wanted, you can carefully go through it like this:

  1. If A and B are invertible, then so is AB
  2. If A and AB are invertible, then so is A*AB = A^2B
  3. If A and A^2B are invertible, then so is A*A^2B = A^3B

and so on until you build up to A^4*B^3. Therefore, we can conclude that A^m*B^n is also invertible. Be careful about the order of multiplying the matrices. Something like A*AB is different from AB*A, the first of which is useful while the second is not.

So this is why statement 3 is true.

-------------------------

Problem 4

This is false. Possibly a quick counter-example is to consider these two matrices

A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}

both of which are invertible since their determinant is nonzero (recall the determinant of a diagonal matrix is simply the product along the diagonal entries). So it's not too hard to show that the determinant of each is 1, and each matrix shown is invertible.

However, adding those two mentioned matrices gets us the 2x2 zero matrix, which is a matrix of nothing but zeros. Clearly the zero matrix has determinant zero and is therefore not invertible.

There are some cases when A+B may be invertible, but it's not true in general.

-------------------------

Problem 5

This is true because each A pairs up with an A^{-1} to cancel out (similar what happened with problem 1). For more info, check out the concept of diagonalization.

5 0
2 years ago
Jack bought 8 tokens for $4.40 at this rate how many tokens can she buy with $6.05
Ipatiy [6.2K]

Answer:

11

Step-by-step explanation:

First, find how much a token costs.

4.40 ÷ 8 = .55

Then divide 6.05 by .55 to get 11.

3 0
3 years ago
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