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

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:

$8c_4=0\\c_4=0$

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

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

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

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

$13,957 is invested, part at 7% and the rest at 6%. If the interest earned from the amount invested at 7% exceeds the interest e

ch4aika [34]

Answer:

The Amount invested at 7% interest is $12,855

The Amount invested at 6% interest = $1,102

Step-by-step explanation:

Given as :

The Total money invested = $13,957

Let The money invested at 7% = $p_1$ = $A

And The money invested at 6% = $p_2$ = $13957 - $A

Let The interest earn at 7% = $I_1$

And The interest earn at 6% = $I_2$

$I_1$ - $I_2$ = $833.73

Let The time period = 1 year

Now,<u> From Simple Interest method</u>

Simple Interest = $\dfrac{\textrm principal\times \textrm rate\times \textrm time}{100}$

Or, $I_1$ = $\dfrac{\textrm p_1\times \textrm 7\times \textrm 1}{100}$

Or, $I_1$ = $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$

And

$I_2$ = $\dfrac{\textrm p_2\times \textrm 6\times \textrm 1}{100}$

Or, $I_2$ = $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$

∵ $I_1$ - $I_2$ = $833.73

So, $\dfrac{\textrm A\times \textrm 7\times \textrm 1}{100}$ - $\dfrac{\textrm (13,957 - A)\times \textrm 6\times \textrm 1}{100}$ = $833.73

Or, 7 A - 6 (13,957 - A) = $833.73 × 100

Or, 7 A - $83,742 + 6 A = $83373

Or, 13 A = $83373 + $83742

Or, 13 A = $167,115

∴ A = $\dfrac{167115}{13}$

i.e A = $12,855

So, The Amount invested at 7% interest = A = $12,855

And The Amount invested at 6% interest = ($13,957 - A) = $13,957 - $12,855

I.e The Amount invested at 6% interest = $1,102

Hence,The Amount invested at 7% interest is $12,855

And The Amount invested at 6% interest = $1,102 . Answer

8 0

3 years ago

PLEASEE HELPP<br><br>3. Evaluate the following equation for x. 2 X = -20 5

KengaRu [80]

Answer:

x=50

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

In circle R with QRS=90 and QR= 3 units find the length of arc QS round to the nearest hundredth

luda_lava [24]

Hello, I Am BrotherEye

Answer:

27.92 units

Step-by-step explanation:

Radius (qr) = r = 16 units

<QŘS = θ = 82°

Length of an arc = (Θ / 360) * 2πr

L = (θ / 360) *2πr

L = (82 / 360) * 2π * 16

L = 0.2777 * 100.53

L = 27.917 units

L = 27.92 unit

Best Of Luck

BrotherEye

6 0

3 years ago