Rebhav B.

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Pre-Calculus

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Question:

Give equations for the following lines in both point-slope and slope-intercept form. (a) The line which passes through the point (1, 2) having slope 4. (b) The line which passes through the points (−1, 1) and (2, −1). (c) The line parallel to y =1/2x + 2, with y-intercept (0, −1). (d) The line perpendicular to y = −3x + 1 which passes through the origin.

Rebhav B.

Answer:

(a) The point-slope form is y − 2 = 4(x − 1). Solving for y, y = 4(x − 1) + 2 = 4x − 4 + 2 = 4x − 2, yields the slope-intercept form, y = 4x − 2. (b) First, we compute the slope using the familiar “rise-over-run” formula, m =(−1 − 1)/2 − (−1) = −2/3. The point-slope form (using the first point) is, y − 1 = −2/(x + 1)/3, and solving for y yields the slope-intercept form, y = −2/3x +1/3. (c) The slope of our desired line is 1/2, since parallel lines must have the same slope. The point-slope form is, y − (−1) = 1/2 (x − 0), and the slope-intercept form is y =1/2 x − 1. (d) The slope of our desired line is 1/3, since it must be the negative reciprocal of the slope any line to which it is perpendicular. The point-slope form is, y − 0 =1/3 (x − 0), and the slope-intercept form is, y =1/3x.

Trigonometry

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Question:

Prove the identity (1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x)

Rebhav B.

Answer:

Use the identities cos(2x) = 2 cos2(x) - 1 and sin(2x) = 2 sin(x) cos(x) in the left hand side of the given identity. [ 1 + cos(x) + cos(2x) ] / [ sin(x) + sin(2x) ] = [ 1 + cos(x) + 2 cos2(x) - 1 ] / [ sin(x) + 2 sin(x) cos(x) ] = [ cos(x) + 2 cos2(x) ] / [ sin(x) + 2 sin(x) cos(x) ] = cos(x) [1 + 2 cos(x)] / [ sin(x)( 1 + 2 cos(x) ) ] = cot(x)

Algebra

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Question:

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Rebhav B.

Answer:

Multiply factors. -15x - 10 - x + 3 = -16x - 20 +13 Group Like Terms -16x - 7 = -16x - 7 Add 16x + 7 to both sides and write the equation as follows 0 = 0 The above statement is true for all values of x and therefore all real numbers are solutions to the given equation

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