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3 years ago

AB lies on a number line. The coordinate of point A is -6. Given that AB= 20, what are the two possible coordinates for point B?

1 answer:

7 0

The fact that you must remember is that these points lie on a plane that have a a distance that is ALWAYS an absolute value. If point A lies on -6 you can have two points that are of a distance 20 points in either direction. If moving the left, point B can have a value of -26 and moving to the right, point B will have a value of 14. Hope this is clear!

Good luck!

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3 years ago

R+r+2r-r+4r combine like terms and solve

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3 years ago

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3 years ago

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Match each system of equations to its graph. y = 2x + 1 y = x + 2 y = 3x y = x + 3 y = 2x − 2 y = x − 2 y = 2x + 3 y = x + 5 y =

GalinKa [24]

The matchup are:

(1st picture): y=2x+3 and y=x+5
(2nd picture): y=4x+2 and y =3x+2
(3rd picture): y=2x+1 and y =x+2
(4th picture): 3x and x+3

<h3>What does the graph of an equation shows?</h3>

The graph of the linear equation is known to be one that often brings or set the points that can be found on the coordinate plane and it is one that shows all the solutions to the equation.

Note that when all variables stands for real numbers, a person can be able to use graph to show the equation and this is often done through plotting the points to show a pattern and then link up the points to have all the points.

From the above, the pictures that have all the points as shown on the graph are:

(1st picture): y=2x+3 and y=x+5
(2nd picture): y=4x+2 and y =3x+2
(3rd picture): y=2x+1 and y =x+2
(4th picture): 3x and x+3

Learn more about equations from

brainly.com/question/2972832

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4 0

2 years ago

Find two numbers whose sum is 25 and the sum of whose reciprocals is 1/6

Katen [24]

Answer:

10 and 15

Step-by-step explanation:

Let 'x' and 'y' are the numbers we need to find.

x + y = 25 (two numbers whose sum is 25)

(1/x) + (1/y) = 1/6 (the sum of whose reciprocals is 1/6)

The solutions of the this system of equations are the numbers we need to find.

x = 25 - y

1/(25 - y) + 1/y = 1/6 multiply both sides by 6(25-y)y

6y + 6(25-y) = (25-y)y

6y + 150 - 6y = 25y - (y^2)

y^2 - 25y + 150 = 0 quadratic equation has 2 solutions

y1 = 15

y2 = 10

Thus we have :

First solution: for y = 15, x = 25 - 15 = 10

Second solution: for y = 10, x = 25 - 10 = 15

The first and the second solution are in fact the same one solution we are looking for: the two numbers are 10 and 15 (since the combination 10 and 15 is the same as 15 and 10).

5 0

3 years ago