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3 years ago

What decimal is equivalent to - 60\22

Mathematics

1 answer:

VashaNatasha [74]3 years ago

8 0

Answer:

-2.72 repeating (put a line over the numbers after the decimal point)

Step-by-step explanation:

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10. (a) Consider the following matrices: A = ( 2 ) B = (3) and C = (-3) w = Find the det(A). [1] (ii) Is the matrix A singular?

Viefleur [7K]

i) We have to find the determinant of A.

We can do this as:

$\det(A)=|\begin{bmatrix}{3} & {6} \\ {1} & {-2}\end{bmatrix}|=3*(-2)-6*1=-6-6=-12$

ii) We have to find if the matrix A is singular.

Singular matrix have determinant equal to 0.

This is not the case for A, as its determinant is -12. Then, A is not a singular matrix.

iii) We have to find the values of x and y so that AB = C.

We have to write the matrix multiplication and we will obtain a system of linear equations:

We can now solve the system of equations by adding 3 times the second equation to the first equation:

$\begin{gathered} 3(x-2y)+(3x+6y)=3(-3)+(-3) \\ 3x-6y+3x+6y=-9-3 \\ 6x+0y=-12 \\ x=\frac{-12}{6} \\ x=-2 \end{gathered}$

We can now use the second equation to find the value of y:

$\begin{gathered} x-2y=-3 \\ x+3=2y \\ y=\frac{x+3}{2} \\ y=\frac{-2+3}{2} \\ y=\frac{1}{2} \end{gathered}$

The values are x = -2 and y = 1/2.

iv) When we want to multiply two matrices, the required condition is that the number of columns of the matrix on the left is equal the number of rows of the matrix on the right.

In the case of A(2x2) and B(2x1), when we do A*B this condition is satisfied.

But when we try to multiply BA, the number of columns of B is not equal to the number of rows of A, so the matrix mulitplication is not possible.

Answer:

i) det(A) = -12

ii) A is not singular because singular matrices have determinant equal to zero.

iii) x = -2 and y = 1/2.

iv) Is not possible because the number of columns of the first matrix has to be equal to the number of rows of the second matrix.

6 0

1 year ago

A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e

My name is Ann [436]

Answer:

$\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)} \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

Step-by-step explanation:

If the system contains $n$ atoms, we can arrange $r$ quanta of energy in

$\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$

ways.

$\mathbf{a)}$

In this case,

$n = 2, r=1$ .

Therefore,

$\binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2$

which means that we can arrange 1 quanta of energy in 2 ways.

$\mathbf{b)}$

In this case,

$n = 20, r=10$ .

Therefore,

$\binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756$

which means that we can arrange 10 quanta of energy in 184 756 ways.

$\mathbf{c)}$

In this case,

$n = 2 \times 10^{23}, r = 10^{23}$ .

Therefore, we obtain that the number of ways is

$\binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}$

3 0

3 years ago

Find the y intercept and x intercept for the following liner equation -4x-10y=-40

Vinvika [58]

The y intercept is 4, and the x intercept is 0.

5 0

3 years ago

Linda received 4/5 of the votes in the votes in the election for student council president.what percent of votes did Linda recei

Mariana [72]

Answer:

80% 0.8

Step-by-step explanation:

To find the percentage out of a fraction you would divide 4÷5 to get 0.8, the decimal answer, then to turn that to a percent you would move the point two places to the right to get 80%.

6 0

3 years ago

Is 54 divided by 6 equal 0.09

Thepotemich [5.8K]

No the answer is 9 because 54 divided by 9 is 9

4 0

3 years ago

Read 2 more answers