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1 year ago

How many decimals in multiplication.

Mathematics

1 answer:

kari74 [83]1 year ago

8 0

Answer:

In multiplication of a decimal by a decimal number we will learn how to find the product of a decimal by another decimal.

To multiply a decimal number by a decimal number, we first multiply the two numbers ignoring the decimal points and then place the decimal point in the product in such a way that decimal places in the product is equal to the sum of the decimal places in the given numbers.

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Translate the phrase into a numerical expression of One hundred divided by the quantity three times five

dimulka [17.4K]

$\frac{100}{3*5}$

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2 years ago

Whats the answer and how

Jet001 [13]

Answer: its c or 35

Step-by-step explanation: because at 10 salespeople its 40, so 9 salespeople is 35

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2 years ago

2(x-5)=3(x+7) Show the process, please.

chubhunter [2.5K]

Answer:

-31

Step-by-step explanation:

(2×x) - (2×5) = (3×x) + (3×7)

2x-10 = 3x+21

2x-3x = 21+10

-x = 31

therefore, x = -31

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2 years ago

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Refer to your answer to the previous question: A digital media company uses the function f(p), where p is the number of people w

Lyrx [107]

Answer:

i think the answer is A

Step-by-step explanation:

again "i think". so i could be wrong

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2 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

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

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2 years ago