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MariettaO [177]

2 years ago

14

I need help solving this

1 answer:

ladessa [460]2 years ago

5 0

I’m not that sure but
B is (-2,2)
A is (-4,-2)
C is (2,2)
Hope it helps!!

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Can someone help me figure out what is 5/12 of 1 ⅓ =

lorasvet [3.4K]

Answer:

Gotta multiply them

Step-by-step explanation:

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2 years ago

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Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu

Alona [7]

Answer:

We have the equation

$c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]$

Then, the augmented matrix of the system is

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

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

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

This matrix is in echelon form. Then, now we apply backward substitution:

1.

$8c_4=0\\c_4=0$

2.

$4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0$

3.

$3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0$

4.

$c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0$

Then the system has unique solution that is $(c_1,c_2c_3,c_4)=(0,0,0,0)$ and this imply that the vectors $v_1,v_2,v_3,v_4$ are linear independent.

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3 years ago

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2 years ago

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alukav5142 [94]

The answer to that question is b

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2 years ago

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brilliants [131]

Answer:

1 Answer->3

Step-by-step explanation:

To factor this out find the factors of 9 which add up to 6.

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1 year ago