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topjm [15]
3 years ago
5

Hard math question!

Mathematics
1 answer:
luda_lava [24]3 years ago
8 0
First, let me show you some notation.

To show a matrix is an inverse of another matrix, we write A^{-1}

-1 is not an exponent. It just shows that a matrix is an inverse of another matrix.

For a 2x2 matrix, we can get the inverse by first making b and c negatives and swap the positions of a and d.

Then multiply each entry in the matrix by 1 divided by the determinant.

\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]^{-1} = 
  \frac{1}{ad - bc}\left[\begin{array}{ccc}d&{-b}\\{-c}&a\end{array}\right] =  \\  \\ \\ \left[\begin{array}{ccc}d(\frac{1}{ad-bc})&{-b}(\frac{1}{ad-bc}) \\ {-c}(\frac{1}{ad-bc}) &a(\frac{1}{ad-bc}) \end{array}\right]

I hope this helped!
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How to simplify fractions
serious [3.7K]

Step-by-step explanation:

\frac{2}{3} + \frac{2}{9} \\\\\frac{6+2}{9}\\\\\frac{8}{9}

5 0
3 years ago
Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw
PolarNik [594]

Answer:

Using <u>backward reasoning</u> we want to show that <em>"We can never get nine 0's"</em>.

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

<u>There were 9 </u><u>0's</u><u>:</u>

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

<u>There were 9 </u><u>1's</u><u>:</u>

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

<u />

5 0
3 years ago
the sum of three numbers is 6. the third number is the sum of the first and second numbers. the first number is one more than th
Monica [59]
Your numbers would be 1, 2, and 3.

1+2=3 

1+2+3=6
6 0
3 years ago
Read 2 more answers
Jane has a bag of candy, she gives 8 to her friend. Then jane's grandmother buys 5 more pieces than jane had originally. Now jan
goblinko [34]

Answer:

12 Pieces

Step-by-step explanation:

Number of candies in Jane's bag = x

she gives 8 to her friend;

x - 8

Then jane's grandmother buys 5 more pieces than jane had originally; add (x + 5)

x - 8 + x + 5

Now jane has 27 pieces of candy

x - 8 + x + 5 = 27

Solve for x;

x - 8 + x + 5 = 27

2x - 3 = 27

2x = 27 -3

2x = 24

x = 12

4 0
3 years ago
Can someone please give me the answers to this? ... please ...
Darina [25.2K]

Step-by-step explanation:

anyway, without any further information about ground consistency, friction and stuff, we have to assume ideal circumstances.

so, the same energy that made the ball flying through the air makes it also bounce or roll across the ground carrying it exactly the same distance at the same time.

the distances with an ideal, reflective ground will be the same.

it is the same energy released with the same inertia.

6 0
3 years ago
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