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vredina [299]
2 years ago
12

Write an equation for the line below.

Mathematics
1 answer:
Dmitry_Shevchenko [17]2 years ago
4 0

Answer:

  x = 3

Step-by-step explanation:

A vertical line has the form x = constant. The graph shows the constant you want is 3.

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What is h=F/g and solve for “F”
Tasya [4]

Answer:

gh=F

Step-by-step explanation:

h=F/g

multiply g to the other side

gh=F

4 0
3 years ago
Nick types 275 words in 2.5 minutes. At this rate, how many hours would it take for him to type 9900 words?
ioda
3/5 of an hour (0.6 hours)



First, find how many words Nick types in an hour by multiplying 275 * 60.

Nick types 16,500 words per hour.

Now just divide 9,900 by 16,500.

It takes Nick 3/5 of an hour to type 9,900 words. (0.6 hours)



Please consider marking this answer as Brainliest to help me advance.

5 0
3 years ago
If we multiply a positive and a negative number, is the result positive or<br> negative?
Marina86 [1]

Answer:

Negative

Step-by-step explanation:

It's negative because my math teacher said so a few years ago. Sorry for the not so great explanation... I didn't pay attention that day.

3 0
2 years ago
PLEASE HELP!!!! What is the measure of angle BCD
faust18 [17]

Answer:

82° because

98° + 70° = 168°

360° - 168° = 192°

192° is for two angles, so it can't be 192°, 220°

it's not VERY obtuse angle. So it remains only 82°

3 0
3 years ago
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
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