Y = x + 5A linear equation (in slope-intercept form) for a line perpendicular to y = -x + 12 with a y-intercept of 5.y = 1/2x - 5Convert the equation 4x - 8y = 40 into slope-intercept form.y = -1/2x + 5A linear equation (in slope-intercept form) which is parallel to x + 2y = 12 and has a y-intercept of 5.3x - y = -5A linear equation (in standard form) which is parallel to the line containing (3, 5) and (7, 17) and has a y-intercept of 5.y = -3x + 1A linear equation (in slope-intercept form) which contains the points (10, 29) and (-2, -7).y = -5A linear equation which goes through (6, -5) and (-12, -5).x = -5A linear equation which is perpendicular to y = 12 and goes through (-5, 5).y = 5A linear equation which is parallel to y = 12 and goes through (-5, 5).y = -x + 5A linear equation (in slope-intercept form) which is perpendicular to y = x and goes through (3, 2).y = -5xA linear equation (in slope-intercept form) which goes through the origin and (1, -5).x = 2A linear equation which has undefined slope and goes through (2, 3).y = 3A linear equation which has a slope of 0 and goes through (2, 3).2x + y = -9A linear equation (in standard form) for a line with slope of -2 and goes through point (-1, -7).3x +2y = 1A linear equation (in standard form) for a line which is parallel to 3x + 2y = 10 and goes through (3, -4).y + 4 = 3/2 (x - 3)A linear equation (in point-slope form) for a line which is perpendicular to y = -2/3 x + 9 and goes through (3, -4).y - 8 = -0.2(x + 10)<span>The table represents a linear equation. Which equation shows how (-10, 8) can be used to write the equation of this line in point-slope form?</span>
We base ourselves in the one piece of crucial evidence: She has 2 quarters. -from the rest of the problem, we know that she has 2 more NICKEL'S than QUARTERS,
