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3 years ago

50 POINTSSSS

Mathematics

1 answer:

defon3 years ago

6 0

Answer:

<h3>Given</h3>

m∠REG = 78°
mAR = 46°
ER ≅ GA

<h3>Solution</h3>

m∠GAR = 180° - m∠REG = 180° - 78° = 102° (supplementary angles sum to 180°)
m∠TAR = 1/2mAR = 1/2(46°) = 23° (tangent chord angle is half the size of intercepted arc)
m∠GAN = 180° - (m∠TAR + m∠GAR) = 180° - (23° + 102°) = 55° (straight angle is 180°)
mAG = 2m∠GAN = 2(55°) = 110°
mRE = mAG = 110° (as ER ≅ GA)
mGE = 360° - (mAG + mAR + mRE) = 360° - (110° + 46° + 110°) = 94° (full circle is 360°)

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3 years ago

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The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

$\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 1 by 1/3:

$\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

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4 years ago

2x + y - z = -8

Andreyy89

Answer:

Multiply row 1 by $\frac{1}{2}$ .

Step-by-step explanation:

The augmented matrix of the system of linear equation is described below:

$\left[\begin{array}{cccc}2&1&-1&-8\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]$

Where $a_{11} = 2$ , if we need to create $a_{11} = 1$ , we need to multiply row 1 by $\frac{1}{2}$ , that is to say:

$\left[\begin{array}{cccc}1&\frac{1}{2} &-\frac{1}{2} &-4\\0&2&3&-6\\-\frac{1}{2} &1&1&-4\end{array}\right]$

Hence, the correct answer is: Multiply row 1 by $\frac{1}{2}$ .

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3 years ago

*****50 POINTSSSS*****

50 POINTSSSS