All categories

4 years ago

In the figure, m and n are rays.

Mathematics

1 answer:

suter [353]4 years ago

4 0

Is there a picture to go along with this??

You might be interested in

10 x ( -9 + (-6) ) <br> --------------------------

babymother [125]

Answer:

This graph will help you.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Can y’all help me on 23?!

tia_tia [17]

So if the ratio of red to blue is 8:3 and there are 15 blue cars then there are 40 red cars.
If the ratio of red to white is 2:10 then then there are 200 white cars.

Explanation:
15 divided by 3 is 5. So you multiply 8 by 5 to get 40. 40:15 reduces to 8:3.

With 40 red cars the ratio of 2:10 would be made into 40:200.

The fractions/ratios have to remain proportional.

5 0

3 years ago

The difference between half a number and 11 is -25

kenny6666 [7]

Answer:

1/2n - 11 = -25

Hope this helps you! Have a good night!

Also, n = -28

1/2n - 11 = -25

1/2n - 11 + 11 = -25 + 11

1/2n = -14

1/2n / 1/2 = -14 / 1/2

n = -28

Check:

1/2(-28) - 11 = -25

-14 - 11 = -25

-25 = -25

6 0

3 years ago

Determine the recursive function that defines the sequence.

Strike441 [17]

<h2>Answer:</h2>

Option: C is the correct answer.

C. $f(1)=4\\\\f(n)=5\cdot f(n-1)\ ;\ n\geq 2$

<h2>Step-by-step explanation:</h2>

Recursive Formula--

It is the formula which is used to represent the nth term of a sequence in terms of (n-1)th term of the sequence.

Here we are given a table of values by:

n f(n)

1 4

2 20

3 100

i.e. when n=1 we have:

$f(1)=4$

Also,

$f(2)=20\\\\i.e.\\\\f(2)=5\cdot 4\\\\i.e.\\\\f(2)=5\cdot f(1)$

Also,

$f(3)=100\\\\i.e.\\\\f(3)=5\cdot 20\\\\i.e.\\\\f(3)=5\cdot f(2)$

Hence, the recursive formula is:

$f(n)=5\cdot f(n-1)\ for\ n\geq 2$

5 0

4 years ago

Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.

Scrat [10]

Answer:

S = [0.2069,0.7931]

Step-by-step explanation:

Transition Matrix:

$P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

$P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]$

Transition matrix P raised to the power 3 (at k = 3)

$P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]$

Transition matrix P raised to the power 4 (at k = 4)

$P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]$

Transition matrix P raised to the power 5 (at k = 5)

$P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]$

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

6 0

4 years ago