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alexandr1967 [171]
1 year ago
13

HELP HELP HELP Just 5 quick algebra 1 questions for 100 points!

Mathematics
1 answer:
serious [3.7K]1 year ago
8 0

1. x = a

2. y = b

3. The slope of a vertical line is undefined..

4. Change in y/change in x (m) or rise/run

5. Rewrite the equation in slope-intercept form to find the value of the y-intercept. The value of x will always be 0.

<h3>What is the Equation of a Line?</h3>

The equation, y = mx + b, in slope-intercept form, defines a line, where m is the slope and the y-intercept = b.

1. Slopes of vertical lines are always undefined, therefore, for a vertical line that contains (a, b), the equation would be: x = a.

2. Horizontal lines have a slope of 0, therefore, the equation of a horizontal line that contains the point, (a, b), will  be: y = b.

3. The slope of a vertical line is undefined.

4. Two ways of finding the slope if we are given two points on a line are:

Change in y/change in x (m) = y2 - y1 / x2 - x1

or

rise/run

5. If we have an equation in point-slope form, for example, y - 2 = 2(x - 4), rewrite the equation in slope-intercept form and find the value of the y-intercept:

y - 2 = 2x - 8

y = 2x - 8 + 2

y = 2x - 6

The y-intercept = -6, therefore, the coordinates of the y-intercept would be: (0, -6).

Learn more about the equation of a line on:

brainly.com/question/19417700

#SPJ1

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A home purchased for $120,000 in 2012 is sold for $156,000 in 2015. What is the percent increase in the price?
almond37 [142]

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HELP PELASE ILL MARK BRAINLIEST AND POINTS!
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\qquad\qquad\huge\underline{{\sf Answer}}

Both the given lines in graph intersect at point (5 , -2), the point must satisfy both the equations.

and if we check the options, we will observe that D. option has both the equations that satisfy the given conditions.

that is ~

\qquad \sf  \dashrightarrow \: x + y = 3

put y = - 2,we will get :

\qquad \sf  \dashrightarrow \: x - 2 = 3

\qquad \sf  \dashrightarrow \: x = 5

so, 1st equation satifys the condition. let's check out 2nd one.

\qquad \sf  \dashrightarrow \: x + 2y = 1

put y = -2

\qquad \sf  \dashrightarrow \: x + (2 \times  - 2) = 1

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