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3 years ago

How to find y in the equation y+2x=3 then graph the answer

1 answer:

7 0

Go on the graph look at the y axis and x axis go to the x axis look for 2 and put the point and then go to postive 3 and put a point and that’s your answer.

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Urgent urgent urgent PLEASE help❤️❤️

Georgia [21]

Answer:

1. c(m)=7

2. c(M)=420

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The extreme water slide forms a 20 angle with the tower it leans against 80 feet above the ground find the length of the slide x

GarryVolchara [31]

Answer:

85.1 ft

Step-by-step explanation:

The extreme water slide forms a 20 angle with the tower it leans against 80 feet above the ground find the length of the slide x

We solve using the Trigonometric function of Cosine

cos y = adjacent/hypotenuse

y = angle = 20°

Adjacent = 80 ft

Hypotenuse = ?

cos 20 = 80/Hypotenuse

Hypotenuse = 80/cos 20

x = 85.134221798 ft

Approximately = 85.1 ft

Length of the slide = 85.1 ft

3 0

3 years ago

find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration

Natasha_Volkova [10]

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

$A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]$

now for given values:

$\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\$

$\to (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} - \frac{1}{4}] =0 \\\\\to (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\$

$\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\$

$= max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}$

In point b:

Its

spectral radius is less than 1 hence matrix is convergent.

In point c:

$\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) = \left(\begin{array}{c}3&1&2\end{array}\right) , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\ \to x^{(k+1)} = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right] \\\\$

after solving the value the answer is

$\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]$

4 0

3 years ago

Two cars simultaneously left Points A and B and headed towards each other, and met after 2 hours and 45 minutes. The distance be

Oksana_A [137]

<h2>Hello!</h2>

The answers are:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

To calculate the speed of the cars, we need to write two equations, one for each car, in order to create a relation between the two speeds and be able to calculate one in function of the other.

So,

Tet be the first car speed "x" and the second car speed "y", writing the equations we have:

For the first car:

$x_{FirstCar}=x_o+v*t$

For the second car:

We know that the speed of the second car is the speed of the first car plus 14 mph, so:

$x_{SecondCar}=x_o+(v+14mph)*t$

Now, from the statement that both cars met after 2 hours and 45 minutes, and the distance between to cover (between A and B) is 264 miles, so, we can calculate the relative speed between them:

If the cars are moving towards each other the relative speed will be:

$RelativeSpeed=FirstCarSpeed-(-SecondCarspeed)\\\\RelativeSpeed=x-(-x-14mph)=2x+14mph$