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Dating Contained in this Triangles Maintaining Analytical Competence
Explanation: The slope of the given line is \(\frac < 1> < 3>\). Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line. The slope of the perpendicular line = -3 Substitute the values in y = mx + c 1 = -3(3) + c 1 = -9 + c c = 1 + 9 c = 10 use the slope intercept form of a linear equation again substitute m, c y = -3x + 10
Explanation: The slope of the given line is -3. Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line. The slope of the perpendicular line = \(\frac < 1> < 3>\) Substitute the values in y = mx + c -3 = –\(\frac < 1> < 3>\)(4) + c c = -3 + \(\frac < 4> < 3>\) = \(\frac < -9> < 3>\) = \(\frac < -5> < 3>\) use the slope intercept form of a linear equation again substitute m, c y = \(\frac < 1> < 3>\)x + \(\frac < -5> < 3>\) y = \(\frac < 1> < 3>\)x – \(\frac < 5> < 3>\)
Explanation: The slope of the given line is -4. Since the slope of the perpendicular line must be the negative reciprocal of the slope of the given line. The slope of the perpendicular line = \(\frac < 1> < 4>\) Substitute the values in y = mx + c -2 = \(\frac < 1> < 4>\)(-1) + c c = -2 + \(\frac < 1> < 4>\) = \(\frac < -8> < 4>\) = \(\frac < -7> < 4>\) use the slope intercept form of a linear equation again substitute m, c y = \(\frac < 1> < 4>\)x + \(\frac < -7> < 4>\) y = \(\frac < 1> < 4>\)x – \(\frac < 7> < 4>\)
Explanation: At least means ? and no more than means < w ? -3 and w < 8 -3 ? w < 8
Explanation: more than means > and less than means < m > 0 and m < 11 0 < m < 11
Explanation: less than or equal to means ? and greater than means > s ? 5 or s > 2 2 < s ? 5