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igomit [66]
3 years ago
9

Which statement correctly explains how Dana can solve the following system of linear equations for x using the elimination metho

d?
4x+8y=20
-4x+2y= -30
Add the two equations, solve for y, and then substitute –1 for y to find x.
Multiply the bottom equation by –4, solve for y, and then substitute –1 for y to find x.
Multiply the bottom equation by –1, combine the two equations, solve for y, and then substitute 1 for y to find x.
Multiply the two equations, solve for y, and then substitute 1 for y to find x.
Mathematics
2 answers:
Vanyuwa [196]3 years ago
6 0
1. The correct statement is the first one, which is: <span>Add the two equations, solve for y, and then substitute –1 for y to find x.
</span>
 2. Therefore, you have the following system of equations:
<span>
 4x+8y=20
 -4x+2y= -30

 3. When you a</span>dd the two equations, and clear the variable "y", you obtain:

 10y=-10
 y=-1

 4. Now, you must substitute -1 for y (y=-1) to find x, as below:

 4x+8y=20
 4x+8(-1)=20
 4x-8=20
 4x=20+8
 4x=28
 x=28/4
 x=7
patriot [66]3 years ago
4 0

Answer:

A

Step-by-step explanation:

3dg 2020 answer

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A<br> 9x - 40<br> B<br> 3x + 20<br> x = [?]
Mrrafil [7]

Answer:

x = 10

Step-by-step explanation:

9x - 40 = 3x + 20

<u>9</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>x</u> - 40 = <u>3x - 3x</u> + 20

6x - 40 = 20

6x <u>-</u><u> </u><u>40</u><u> </u><u>+</u><u> </u><u>40</u> = <u>20</u><u> </u><u>+</u><u> </u><u>40</u>

6x = 60

<u>6x</u><u> </u><u>/</u><u> </u><u>6</u> = <u>60</u><u> </u><u>/</u><u> </u><u>6</u>

x = 10

Now plug the x value in the equation to make the statement true that A is parallel to B.

9x - 40

<u>9</u><u>(</u><u>10</u><u>)</u> - 40

<u>90</u><u> </u><u>-</u><u> </u><u>40</u>

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3x + 20

<u>3</u><u>(</u><u>10</u><u>)</u> + 20

<u>30</u><u> </u><u>+</u><u> </u><u>20</u>

50

Therefore, x = 10 making the statement true that A is parallel to B. Hope this helps and stay safe, happy, and healthy, thank you :) !!

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Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.
alina1380 [7]

Answer:

v_1=(\frac{1}{10},-\frac{3}{10})

v_2=(-\frac{1}{10},\frac{3}{10})

Step-by-step explanation:

First we define two generic vectors in our \mathbb{R}^2 space:

  1. v_1 = (x_1,y_1)
  2. v_2 = (x_2,y_2)

By definition we know that Euclidean norm on an 2-dimensional Euclidean space \mathbb{R}^2 is:

\left \| v \right \|= \sqrt{x^2+y^2}

Also we know that the inner product in \mathbb{R}^2 space is defined as:

v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

\left \| v_1 \right \|= \sqrt{x^2+y^2}=1

and

\left \| v_2 \right \|= \sqrt{x^2+y^2}=1

As second condition we have that:

v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0

v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0

Which is the same:

y_1=-3x_1\\y_2=-3x_2

Replacing the second condition on the first condition we have:

\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}

Since x_1^2= \frac{1}{10} we have two posible solutions, x_1=\frac{1}{10} or x_1=-\frac{1}{10}. If we choose x_1=\frac{1}{10}, we can choose next the other solution for x_2.

Remembering,

y_1=-3x_1\\y_2=-3x_2

The two vectors we are looking for are:

v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})

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