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3 years ago

What is the value of f(-4) ?

Mathematics

1 answer:

poizon [28]3 years ago

6 0

Answer:

-64

Step-by-step explanation:

Since this is a piecewise defined function

We need to find where x=- 4 is defined

f(-4) = x^3 -5<=x<=2

= (-4)^3

= -64

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Amy is 5 years older than Ben. Three times Amy's age added to six times Ben's age is 42. How old are Amy and Ben?

I am Lyosha [343]

Answer:

Amy: 8, Ben: 3

Step-by-step explanation:

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3 years ago

The sum of five consecutive multiples of 4 is 20.Find the smallest integer.

Aloiza [94]

Answer:

The smallest integer is -4.

Step-by-step explanation:

Let the smallest integer be x.

Since the integers are all multiples of 4, they are

=> x, x+4, x+8, x+12 and x+16

=> x + x+4 + x+8 + x+12 + x+16 = 20

=> 5x + 40 = 20

Subtract 40 from both sides of the equation

5x +40 - 40 = 20 - 40

5x = -20

Divide both sides by the coefficient of x(which is 5)

$\frac{5x}{5} = \frac{-20}{5}$

x = -4

∴ the smallest integer is -4.

Hope this helps!!!

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3 years ago

{x|x is an odd whole number less than 8}

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Answer:

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Step-by-step explanation:

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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:

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