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storchak [24]
3 years ago
5

What temperature is ten degrees higher than minus seven degrees

Mathematics
2 answers:
Eduardwww [97]3 years ago
8 0
The correct answer is 3 degrees.

-7 + 10   Ten degrees higher (+10) than minus seven degrees (-7)
3   When adding a positive to a negative, the answer will be the difference between the two numbers. Whether the answer is positive or negative depends on which number is larger; for instance, in this case, 3 is positive because 10 is larger than 7.

Therefore, the answer is 3 degrees.

Hope this helps!
ch4aika [34]3 years ago
6 0
+3 is the answer, it's negative 7 plus 10
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Select all that apply. Which of the following expressions have a quotient of -1/7?
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Answer:

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Step-by-step explanation:

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2 years ago
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If 0.142857142857......is expressed in the form m/n, then what is the value<br>of (2m + n)?​
frosja888 [35]

Answer:

The value will be: 1,285,713

Step-by-step explanation:

We have the number:

N = 0.142857...

Such that we have 6 decimals after the decimal point that repeat infinitely.

Then let's start by multiplying our number by 1,000,000 (same number of zeros as decimals after the decimal point)

Then we get:

1,000,000*N = 142,857.142857...

Now let's subtract N from both sides:

1,000,000*N - N =  142,857.142857... - 0.142857...

999,999*N = 142,857

Now let's isolate N

N = 142,857/999,999

Here is our number expressed in the form m/n, where:

m = 142,857

n = 999,999

Then the value of (2*m + n) is:

(2*142,857 + 999,999) = 1,285,713

5 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.

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Step-by-step explanation:

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gizmo_the_mogwai [7]

Answer:

A. 4

Step-by-step explanation:

1. Use the SLOPE FORMULA: M= rise/run= y2-y1/x2-x1

2. Plug in the two coordinates that were plotted on the equation: (1,1) & (3,9)

3. Label the coordinates

( 1 , 1 ) = ( x1, y1 )& ( 3, 9 ) = ( x2, y2 )

4. SOLVE & PLUG IN

m = 9 - 1 / 3 - 1 = 8 / 2

5. Simplify 8/2 (Divide by 2)

6. answer: 4/1 = 4

8 0
2 years ago
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