1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
qaws [65]
3 years ago
5

2000000 in standard form​

Mathematics
1 answer:
algol133 years ago
3 0

Answer:

2 \times  {10}^{6}

You might be interested in
What is the answer to 4-4+4-5+5/1
MrMuchimi

Answer:

4

Step-by-step explanation:

5/1=5

4-4+4-5+5=4

4 0
3 years ago
Read 2 more answers
925 round to the nearest hundred
yarga [219]
925 rounded to the nearest hundred is 900 becuase first we identify the hundreds digit which in this case is 9. Second,we identify the next smallest place value (the digit to the right of the hundreds place) which in this case is 2. Is that digit greater than or equal to five? No - we round Down. The hundreds digit is stays the same but every digit after it becomes a zero.
4 0
3 years ago
Read 2 more answers
6th grade math help me pleasee
JulsSmile [24]

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

1. Kris

2. Katie

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

6 0
2 years ago
Pleaseeeeee help this is due today and I don’t know how to do thissss! Tug of War Activity
Hoochie [10]

Answer:

i

Step-by-step explanation:

7 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
Other questions:
  • Write the trigonometry ratio for the triangle below.
    13·2 answers
  • If 29 more than 2 times a number is -15, what is the number?<br> -7<br> -22<br> 7<br> 22
    7·1 answer
  • Write the partial fraction decomposition of the given rational expression. startfraction 5 x plus 2 over x superscript 4 baselin
    10·2 answers
  • I'm not the greatest at math so I got stuck<br> x=<br><br> 80<br> 90<br> 100
    13·1 answer
  • Juanita has a pail with a capacity of 3.4 liters. How many milliliters will the Pail hold?
    14·1 answer
  • Determine whether or not the following question can be considered as a statistical question.
    13·1 answer
  • 7676fhfghfhfghfhfhfh6
    6·1 answer
  • HARD POINT TUESDAY <br><br><br> ANSWER THIS EQUATION <br><br> -4x - 2y= -12<br> 4x + 8y = -24
    8·1 answer
  • 5) Determine the Distance between (-3, 1) and (5,-6). Round your answer 10 points
    13·1 answer
  • Pls help with steps also
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!