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attashe74 [19]
1 year ago
14

Determine the slope and y-intercept of the line represented by the linear equation:y + 2 = 0. Graph the line using the slope and

y-intercept. Describe, in complete sentences, what type of linear relationship the solutions (points) of the line have with each other. Choose two points on the line and prove that they are solutions of the linear equation.
Mathematics
1 answer:
BigorU [14]1 year ago
3 0
  1. The slope and y-intercept of this line represented by the linear equation are 0 and -2 respectively.
  2. The linear relationship between the solutions (points) of the line values is that the points on the y-axis are constant while the points on the x-axis increased from left to right.

<h3>How to graph the line?</h3>

Mathematically, the standard form of the equation of a straight line is given by;

y = mx + c

Where:

  • x and y are the points.
  • m is the slope.
  • c is the intercept.

In order to graph this line using the slope and y-intercept, we would have to express the given equation in the standard form as follows:

y + 2 = 0

y = -2

Therefore, the graph of the given equation is an horizontal line with y-intercept at -2

By critically observing the graph of the given equation, we can logically deduce that the linear relationship between the solutions (points) of the line values is that the points on the y-axis are constant while the points on the x-axis increased from left to right.

<h3>How to prove that the points are solutions of the linear equation?</h3>

Slope, m = Δy/Δx

Slope, m = (-2 + 2)/(6 - 2) = 0.

At point (2, -2), we have:

y = mx + c

y = 0(2) - 2

y = 0 - 2

y = -2 (Proved).

At point (6, -2), we have:

y = mx + c

y = 0(6) - 2

y = 0 - 2

y = -2 (Proved).

Read more on slope and y-intercept here: brainly.com/question/17920127

#SPJ1

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Compare 1/7 to consecutive multiples of 1/9. This is easily done by converting the fractions to a common denominator of LCM(7, 9) = 63:

1/9 = 7/63

2/9 = 14/63

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Then 1/7 falls between 1/9 and 2/9, so 1/7 = 1/9 plus some remainder. In particular,

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We do the same sort of comparison with the remainder 2/63 and multiples of 1/9² = 1/81. We have LCM(63, 9²) = 567, and

1/9² = 7/567

2/9² = 14/567

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2/63 = 18/567

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2/63 = 2/9² + 4/567

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Compare 4/567 with multiples of 1/9³ = 1/729. LCM(567, 9³) = 5103, and

1/9³ = 7/5103

2/9³ = 14/5103

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1/9⁴ = 7/45927

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1/5103 = 9/45927

Then

1/5103 = 1/9⁴ + 2/45927

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1/7 = 1/9¹ + 2/9² + 5/9³ + 1/9⁴ + 2/45927

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