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lozanna [386]
3 years ago
7

Examine this system of linear equations. y – x = 2, x + y = 4 Which is a solution of the system of equations? (1, 3) (2, 2) (3,

1) (4, 2)
Mathematics
2 answers:
Illusion [34]3 years ago
5 0

Answer:

A.

Step-by-step explanation:

A. (1,3)

Blizzard [7]3 years ago
4 0

Answer:

(1, 3)

Step-by-step explanation:

Solve the following system:

{y - x = 2 | (equation 1)

{x + y = 4 | (equation 2)

Add equation 1 to equation 2:

{-x + y = 2 | (equation 1)

{0 x+2 y = 6 | (equation 2)

Divide equation 2 by 2:

{-x + y = 2 | (equation 1)

{0 x+y = 3 | (equation 2)

Subtract equation 2 from equation 1:

{-x+0 y = -1 | (equation 1)

{0 x+y = 3 | (equation 2)

Multiply equation 1 by -1:

{x+0 y = 1 | (equation 1)

{0 x+y = 3 | (equation 2)

Collect results:

Answer: x = 1, y = 3

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4 years ago
(3.2x10^3) + (5.78 x 10^5) ?
meriva
Simple....

you have: (3.2* 10^{3} )+(5.78* 10^{5} )

Simplify....

(3.2* 10^{3} )=3200

and

(5.78* 10^{5} )=578,000

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=581,200

Thus, your answer.
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3 years ago
What is the answer to this equation? (1.03×10to the 9)-(4.7×10to the 7)
STALIN [3.7K]

Answer:

                x10000000 • (103x990000000 - 470)

 —————————————————————————————————

                                               100    

7 0
3 years ago
What are the points on y-axis which are at a distance of 4 units from 4x+3y=12
marissa [1.9K]

Given:

The equation of line is

4x+3y=12

To find:

The points on y-axis which are at a distance of 4 units from the given line.

Solution:

The point lie on the y-axis. So, their x-coordinate must be zero.

Let the points are in the form of (0,k).

The distance between a point (x_0,y_0) and a line ax+by+c=0 is

d=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}

The given equation can be written as

4x+3y-12=0

The distance between (0,k) and 4x+3y-12=0 is 4 units.

4=\dfrac{|4(0)+3(k)-12|}{\sqrt{4^2+3^2}}

4=\dfrac{|0+3k-12|}{\sqrt{16+9}}

4=\dfrac{|3k-12|}{\sqrt{25}}

4=\dfrac{|3k-12|}{5}

Multiply both sides by 5.

20=|3k-12|

\pm 20=3k-12

12\pm 20=3k

\dfrac{12\pm 20}{3}=k

Now,

k=\dfrac{12-20}{3}\text{ and }k=\dfrac{12+20}{3}

k=\dfrac{-8}{3}\text{ and }k=\dfrac{32}{3}

Therefore, the two points are \left(0,\dfrac{-8}{3}\right)\text{ and }\left(0,\dfrac{32}{3}\right).

7 0
3 years ago
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