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3 years ago

5

Help with these pls!!

1 answer:

evablogger [386]3 years ago

8 0

#17, x=24
#19, ×=31 are the answers

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24 to 16 simplest form

faltersainse [42]

24:16 in simplest form, is.... 3:2 3 to 2

6 0

3 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.

4 0

3 years ago

Simplify 28 over 35

maria [59]

4 / 5
use the 7 to reduce each numer of the fraction

8 0

3 years ago

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Fill in the blank with a number to make the expression a perfect square.

Sati [7]

Answer:

16

Step-by-step explanation:

Substituting 16 will make the expression valid and correct since the square of -4 is 16 and the addition of it is -8.

$v^2-8v+16=(v-4)^2$

3 0

3 years ago

(11x-3) (4x+39) find x<br>

patriot [66]

Answer:

x=6

Step-by-step explanation:

3 0

3 years ago