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2 years ago

Use the question In the image to put the symbols in place.

Mathematics

1 answer:

AnnZ [28]2 years ago

5 0

The true equation of 24 ÷ 3 + 9 * 5 - 2 = 6 is 24 ÷ (3 + 9) * (5 - 2) = 6

<h3>How to make the equation true?</h3>

The equation is given as:

24 ÷ 3 + 9 * 5 - 2 = 6

The above equation is false because

24 ÷ 3 + 9 * 5 - 2 = 51

And 51 and 6 do not have equal values

Next, we add grouping symbols in the expression on the left-hand side to make the equation true

So, we have:

24 ÷ (3 + 9) * (5 - 2) = 6

The above equation is true because

24 ÷ (3 + 9) * (5 - 2) = 6

Hence, the true equation of 24 ÷ 3 + 9 * 5 - 2 = 6 is 24 ÷ (3 + 9) * (5 - 2) = 6

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Which matrix represents the system of equations shown below? 3x-5y=12 4x-2y=15

tatuchka [14]

Answer:

$\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]$

Step-by-step explanation:

When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:

a column for the values of x in each equation
a column for the values of y in each equation
a column for the independent values of each equation

since our system of equations is:

$3x-5y=12\\ 4x-2y=15$

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:

$\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]$

Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:

$\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]$

Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:

$\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]$

usually there is a line separating the columns for the values of x and y, and the independent values:

$\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]$

this is the matrix of the system of equations

6 0

3 years ago

Calcular en cuanto se convertira, un capital de $15,500 en 8 meses al 6.5%

dybincka [34]

En algunos bancos, cuentas de ahorro o incluso en prestamos y empeños podemos observar que se cobra un interés mensual. Esto no significa que todos los intereses deban ser así.

El periodo, es el tiempo en el que se aplica el interés, este puede ser anual, mensual, semanal, o incluso por día.

8 0

3 years ago

What is the prime factorization of 96? 8 · 12 2 · 2 · 2 · 2 · 3 25 · 3 25 · 32

olganol [36]

96
/ \
8 • 12
/ \ / \
4 • 2 • 4 • 3
/ \ / \
2 2 2 2
96 is 2•2•2•2•2•3 or 2^5•3

8 0

3 years ago

If you know about functions pls help

katrin [286]

Answer:

f(x) is exponential g(x) is quadratic

Step-by-step explanation:

c) the values of function g will exceed function f

5 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