What is the value of x? x= -8 2x + 5 – įx = 3 O x = 0.x = -2 0 x = 1/2 x = 8

Mathematics

1 answer:

Dafna11 [192]2 years ago

4 0

Dude doesn’t x basically equal 8 X=8 ? Or am I just wrong

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The Frostburg-Truth bus travels on a straight road from Frostburg Mall to Sojourner Truth Park. The mall is 3 miles west and 2 m

Verizon [17]

To start, write your locations as points, with north and south being positive and negative y respectively and east and west being positive and negative x respectively. Doing this gives us the mall at (-3,-2) and the park at (4,5). Now, we use our distance formula $( \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 } )$ to solve for the unknown distance. Plugging in with the park values as our second values and our mall values as our first values (as well as with our unknown distance as d), we get $d= \sqrt{(4-(-3))^2+(5-(-2))^2} = \sqrt{(7)^2+(7)^2} = \sqrt{49+49} = \sqrt{98}$ . This square root can be rounded to 9.9 miles.

4 0

3 years ago

[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation

irina1246 [14]

Answer:

$y=-\frac{158}{579}$

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

$A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]$

And the vector B is formed with the solution of each equation of the system: $b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]$

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called $A_{2}$ .

$A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]$

The value of y using Cramer's rule is:

$y=\frac{det(A_{2}) }{det(A)}$

Find the value of the determinant of each matrix, and divide:

$y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}$