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Brut [27]
3 years ago
12

We learned that division expressions that have the same quotient and remainder are not necessarily equal to each other explain h

ow this is possible
Mathematics
2 answers:
Misha Larkins [42]3 years ago
7 0

Division expressions that have the same quotient and remainder are not necessarily equal to each other. This is possible because each expressions have different numbers so the calculation can be different

<h3>Further explanation </h3>

Division is breaking the number up into the equal number of parts.

We learned that division expressions that have the same quotient and remainder are not necessarily equal to each other. For example when we get 20 divided by 4 then If we take 20 things and put them into four equal sized groups, there will be 5 things in each group. Therefore the answer is 5 and hence it will not be any remainder (remainder is 0).

Another example is when we get 21 divided by 4 then If we take 21 things and put them into four equal sized groups, there will be 5 things in each group and the remainder of 1.

Third example is when we get 26 divided by 5 then If we take 26 things and put them into four equal sized groups, there will be 5 things in each group and the remainder of 1.

Division expressions that have the same quotient and remainder are not necessarily equal to each other. This is possible because each expressions have different numbers so the calculation can be different. The difference is that the remainders are not the same in every expressions.  The first remainder is 1 part out of 4, when the third remainder is 1 part out of 5.

<h3>Learn more</h3>
  1. Learn more about division brainly.com/question/471768
  2. Learn more about quotient brainly.com/question/8921605
  3. Learn more   about remainder brainly.com/question/5423

<h3>Answer details</h3>

Grade:    5

Subject:  Math

Chapter:  expressions

Keywords:  division, quotient, remainder, expressions, equal

olasank [31]3 years ago
6 0
<span>Let's say you divide 25 by 4.  You will get a quotient 6, but have that remainder of 1.  You could also divide 37 by 6, and likewise get a quotient of 6, with a remainder of 1.  The difference is that the remainders are not truly the same.  The first remainder is 1 part out of 4, but the second remainder is 1 part out of 6</span>
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Which of the following is associated with the famous impossible Delian problem?
Natalka [10]

THE ANSWER ABOVE IS INCORRECT!! If you are doing Apex, the answer is doubling a cube. It can't be done with a straightedge or compass so it is impossible and was believed to be Apollo wanted something of his doubled with was a cube.

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2 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.

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What is the correct answer?
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Given:

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f(x)=(x+3)^2(x-5)^6

To find:

The zeros of the given function.

Solution:

The general form of polynomial is

P(x)=a(x-c_1)^{m_1}(x-c_2)^{m_2}...(x-c_n)^{m_n}       ...(i)

where, a is a constant, c_1,c_2,...,c_n are zeros of respective multiplicities m_1,m_2,...,m_n.

We have,

f(x)=(x+3)^2(x-5)^6

On comparing this with (i), we get

c_1=-3,m_1=2

c_2=5,m_2=6

It means, -3 is a zero with multiplicity 2 and 5 is a zero with multiplicity 6.

Therefore, the correct option is B.

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Answer:

Step-by-step explanation:

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2 years ago
Find the area of circle with a radius of 2 in exact terms​
ss7ja [257]

Answer:

12.56 (13 if rounded)

Step-by-step explanation:

A=pi x r squared

A=2 squared multiplied by pi

A=4 times pi

plug into calculator should be about 12.56

7 0
3 years ago
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