All categories

2 years ago

Jason knows that AK←→⊥KB←→ , AK←→∥EF←→ , and AK←→∥CB←→ .

Mathematics

2 answers:

netineya [11]2 years ago

4 0

I think the answer is KB//DM because they are on different lines and will never touch, they go sideways for eternity.

Lemur [1.5K]2 years ago

4 0

I took the test it's

m∠EJM=90°

KB←→⊥CB←→

You might be interested in

Find c1 and c2 such that M2+c1M+c2I2=0, where I2 is the identity 2×2 matrix and 0 is the zero matrix of appropriate dimension.

Katyanochek1 [597]

The question is missing parts. Here is the complete question.

Let M = $\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$ . Find $c_{1}$ and $c_{2}$ such that $M^{2}+c_{1}M+c_{2}I_{2}=0$ , where $I_{2}$ is the identity 2x2 matrix and 0 is the zero matrix of appropriate dimension.

Answer: $c_{1} = \frac{-16}{10}$

$c_{2}=\frac{-214}{10}$

Step-by-step explanation: Identity matrix is a sqaure matrix that has 1's along the main diagonal and 0 everywhere else. So, a 2x2 identity matrix is:

$\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

$M^{2} = \left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right]$

$M^{2}=\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]$

Solving equation:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+c_{1}\left[\begin{array}{cc}6&5\\-1&-4\end{array}\right] +c_{2}\left[\begin{array}{cc}1&0\\0&1\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

Multiplying a matrix and a scalar results in all the terms of the matrix multiplied by the scalar. You can only add matrices of the same dimensions.

So, the equation is:

$\left[\begin{array}{cc}31&10\\-2&15\end{array}\right]+\left[\begin{array}{cc}6c_{1}&5c_{1}\\-1c_{1}&-4c_{1}\end{array}\right] +\left[\begin{array}{cc}c_{2}&0\\0&c_{2}\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

And the system of equations is:

$6c_{1}+c_{2} = -31\\-4c_{1}+c_{2} = -15$

There are several methods to solve this system. One of them is to multiply the second equation to -1 and add both equations:

$6c_{1}+c_{2} = -31\\(-1)*-4c_{1}+c_{2} = -15*(-1)$

$6c_{1}+c_{2} = -31\\4c_{1}-c_{2} = 15$

$10c_{1} = -16$

$c_{1} = \frac{-16}{10}$

With $c_{1}$ , substitute in one of the equations and find $c_{2}$ :

$6c_{1}+c_{2}=-31$

$c_{2}=-31-6(\frac{-16}{10} )$

$c_{2}=-31+(\frac{96}{10} )$

$c_{2}=\frac{-310+96}{10}$

$c_{2}=\frac{-214}{10}$

For the equation, $c_{1} = \frac{-16}{10}$ and $c_{2}=\frac{-214}{10}$

6 0

2 years ago

What is the quotient (3x4 – 4x2 + 8x – 1) ÷ (x – 2) brainly

Karolina [17]

To find our quotient we are going to perform long division.

Long division step by step:

Step 1. Write the polynomial (dividend) in descending order (from highest to least exponent). If you encounter a missing term, use zero to fill the space of the term in the descending sequence.

Notice that even tough our polynomial is written in descending form, $x^3$ is missing, so we are going to use $0x^3$ to fill the gap:

$3x^4-4x^2+8x-1$

$3x^4+0x^3-4x^2+8x-1$

Step 2. Divide the first term of the polynomial (dividend) by the first term of the divisor.

In our case the first term of the dividend is $3x^4$ and the first term of the divisor is $x$ , so $\frac{3x^4}{x} =3x^3$ . Notice that $3x^3$ is the first term of the quotient.

Step 3. Multiply the first term of the quotient by the terms of the divisor and subtract them from the respective term of the dividend.

The first term of our quotient (from the previous step) is $3x^3$ and the divisor is $(x-1)$ , so $3x^3(x-2)=3x^4-6x^3$ . Now, we are going to subtract them from the respective term (the term with the same power) of the dividend. The respective terms of the dividend are $3x^4$ and $0x^3$ , so $3x^4-3x^4=0$ and $0x^3-(-6x^3)=0x^3+6x^3=6x^3$

Step 4. Bring down the next term in the dividend and repeat the process for the remaining terms.

After finish the process (check the attached picture), we can conclude that the quotient of $3x^4-4x^2+8x-1$ ÷ $(x-2)$ is $3x^3+6x^2+8x+24$ with a remainder of $47$ , or in a different notation: $3x^3+6x^2+8x+24+\frac{47}{x-2}$