Please help me with this question fast!!!!

Answer!!!!!! Please!!!!!!

shusha [124]

Answer:

C

Step-by-step explanation:

Using calculator to evaluate

$2^{\sqrt{19} }$ ≈ 20.5191 → C

6 0

3 years ago

Read 2 more answers

solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations. x-2y+

Alekssandra [29.7K]

Answer:

x = -5, y = -6, z = -3

Step-by-step explanation:

Given the system of three equations:

$\left\{\begin{array}{l}x-2y+3z=-2\\6x+2y+2z=-48\\x+4y+3z=-38\end{array}\right.$

Write the augmented matrix for the system of equations

$\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)$

Find the reduced row-echelon form of the augmented matrix for the system of equations:

$\left(\begin{array}{ccccc}1&-2&3&|&-2\\6&2&2&|&-48\\1&4&3&|&-38\end{array}\right)\sim \left(\begin{array}{ccccc}1&-2&3&|&-2\\0&-14&16&|&36\\0&-6&0&|&36\end{array}\right)\sim \left(\begin{array}{ccccc}1&3&-2&|&-2\\0&16&-14&|&36\\0&0&-6&|&36\end{array}\right)$

Thus, the system of three equations is

$\left\{\begin{array}{r}x+3z-2y=-2\\16z-14y=36\\-6y=36\end{array}\right.$

From the last equation:

$y=-6$

Substitute it into the second equation:

$16z-14\cdot (-6)=36\\ \\16z=36-84\\ \\16z=-48\\ \\z=-3$

Substitute y = -6 and z = -3 into the first equation:

$x+3\cdot (-3)-2\cdot (-6)=-2\\ \\x=-2+9-12\\ \\x=-5$

7 0

3 years ago

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Maurinko [17]

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)\times(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&1\\4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$\dfrac{\mathbf v}{\|\mathbf v\|}=\dfrac{\vec i+3\,\vec j-4\,\vec k}{\sqrt{1^2+3^2+(-4)^2}}=\dfrac1{\sqrt{26}}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .