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3 years ago

Is it a function or not a function? Please help me??

Mathematics

2 answers:

Umnica [9.8K]3 years ago

6 0

Answer:

Step-by-step explanation:

blagie [28]3 years ago

5 0

Answer:

function

Step-by-step explanation:

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Using the properties of equality, find the value of x in the equation below.

Natali5045456 [20]

Answer:

-1/5 or - .2

Step-by-step explanation:

4x(2+8)= -8

(4*-.2)(2+8)= -8

-.8(2+8)= -8

-.8(10)= -8

-8=-8

8 0

3 years ago

Divide.<br><br> (5/6)÷(−2/3)

SSSSS [86.1K]

Hii

The answer is - 1 1/4

Sorry if it incorrect

8 0

2 years ago

Find the constant of proportionality on the graph.

Goryan [66]

Answer : 2.5

Step - by - step explanation :

X / Y = COP

5 / 2 = 2.5 or 2 1/2

Y = 2.5x

7 0

2 years ago

solve the following systems of equations. Express your answer as an ordered pair in the format(a,b), with no spaces between the

kodGreya [7K]

If you multiply the top equation by -2 you get:

-4x -14y = 14 => -4x = 14+14y

now you can plug in the -4x into the second equation:

14+14y - 3y = -19 => 11y = -33 => y=-3

Fill in y=-3 in either equation to find x:

2x - 21 = -7 => 2x = 14 => x=7

So your ordered pair is (7,-3)

7 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