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hichkok12 [17]
4 years ago
15

Hi>iiiii>iiiiiiiiiiiiii

Mathematics
1 answer:
Bezzdna [24]4 years ago
7 0

Answer:

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The correct answer is D but i don’t understand
iren [92.7K]
The answer depends on the way you solved it.

I am assuming you take the base on which you perceived to be: . 5 and height which is 2.


So you must've ended with 1

But in actuality, you need to use the equation above and plug in 1.5 and 1
Subtract them and you should get 1.25

Do the same to the other side, you should get 6.

.125 x 6  = .875

8 0
3 years ago
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Y+3= 2/3(x - 2)<br> Solve the system by graphing
Serggg [28]

Answer:

y=2/3x-13/3

Step-by-step explanation:

i hope this helps

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3 years ago
Who is truly worthy of Brainliest? Answer this \/\/\/\/
Leno4ka [110]

Answer:

Intercepts are the points where a graph touches the axes. The X-intercept is that point on the X-axis, and the Y-intercept that point on the Y-axis. In the graph on the left, the X-intercept is (4,0), and the Y-intercept is (0,2)

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3 years ago
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Dayquan decided to run some laps around the track. If the diameter of the track is 75 feet, how far around the track did he run?
SpyIntel [72]

Answer: 235.5 feet

Step-by-step explanation:

In this case, we have to calculate the circumference of the track. The circumference will be calculated as:

= 2πr

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5 0
3 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
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