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elena55 [62]
3 years ago
13

Which point below is not on the graph of h(x) ^3√x+64 = ? (65, 5) (-37, 3) (-72, -2) (-64, 0)

Mathematics
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0

Answer:

(-37, 3)

Step-by-step explanation:

Natalija [7]3 years ago
5 0
(-37,3) is correct answer
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MrRissso [65]

Answer:

5.3

Step-by-step explanation:

Cos(68) = adjacent/hypotenuse

Cos(68) = 2/AC

AC = 2/cos(68)

AC = 5.3389343253

AC = 5.3

4 0
3 years ago
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M=p/n+4 make n the subject​
Kipish [7]

Answer:

n = \frac {p - 4m}{m}

Step-by-step explanation:

Given the algebraic expression;

m = \frac {p}{n + 4}

To make n the subject of formula.

Cross-multiplying, we have;

m*(n+4) = p

mn + 4m = p

Rearranging the equation, we have;

mn = p - 4m

n = \frac {p - 4m}{m}

6 0
3 years ago
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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
2 years ago
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sveta [45]

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Let XY = x

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360x = 10173.60

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Answer is C. 28.3 cm

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alukav5142 [94]

Answer:

what are the expressions?

Step-by-step explanation:

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2 years ago
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