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2 years ago

Using Cramer’s rule, what is the value of x in the solution to the system of linear equations below?

Mathematics

1 answer:

Aloiza [94]2 years ago

6 0

9514 1404 393

Answer:

d 2

Step-by-step explanation:

The system of equations can be written in augmented matrix form as ...

$\displaystyle\left[\begin{array}{cc|c}-\frac{1}{2}&3&-4\\-1&-1&-1\end{array}\right]$

The value of x found by Cramer's rule is the ratio of two determinants. The numerator is the coefficient matrix with the x-column replaced by the constant column. The denominator is the coefficient matrix.

$\displaystyle x=\frac{\left|\begin{array}{cc}-4&3\\-1&-1\end{array}\right| }{\left|\begin{array}{cc}-\frac{1}{2}&3\\-1&-1\end{array}\right|}=\frac{(-4)(-1)-(-1)(3)}{(-\frac{1}{2})(-1)-(-1)(3)}=\frac{4+3}{\frac{1}{2}+3}\\\\\boxed{x=2}$

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Triangle ABC has vertices A(-2, 3), B(0, 3), and C(-1,-1). Find the coordinates of the image after a reflection over the

motikmotik

Given:

Given that the triangle ABC has vertices A(-2,3), B(0,3) and C(-1,-1).

We need to determine the coordinates of the image after a reflection over the x - axis.

Let A'B'C' denote the coordinates of the triangle after a reflection over the x - axis.

Coordinates of the point A':

The general rule to reflect the coordinate across the x - axis is given by

$(x,y)\rightarrow (x,-y)$

Substituting the point A(-2,3), we get;

$(-2,3)\rightarrow (-2,-3)$

Thus, the coordinates of the point A' is (-2,-3)

Coordinates of the point B':

The general rule to reflect the coordinate across the x - axis is given by

$(x,y)\rightarrow (x,-y)$

Substituting the point B(0,3), we get;

$(0,3)\rightarrow (0,-3)$

Thus, the coordinates of the point B' is (0,-3)

Coordinates of the point C':

The general rule to reflect the coordinate across the x - axis is given by

$(x,y)\rightarrow (x,-y)$

Substituting the point C(-1,-1), we get;

$(-1,-1)\rightarrow (-1,1)$

Thus, the coordinates of the point C' is (-1,1)

Hence, the coordinates of the image after a reflection over the x - axis is A'(-2,-3), B(0,-3) and C(-1,1)

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3 years ago

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Answer:

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

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3 years ago

Solve the following system of equations x^2+2y^2=59 2x^2+y^2=43

aivan3 [116]

Answer: x = (-3,3), y = (-5,5)

Step-by-step explanation:

Add both of the equations together, for $3x^{2}+3y^{2} = 102$ . Now we can divide both sides by 3, getting $x^2+y^2=34$ . we subtract the first equation from our equation we just got, getting $y^2=25$ y= (5,-5). once we plug that in, we get $50+x^2 = 59$ , $x^2 = 9$ x = (-3,3)

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3 years ago