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What is 4q 5r 7p 7r 2q 3p?

Updated: 4/28/2022
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11y ago

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It is a string of algebraic terms.

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Q: What is 4q 5r 7p 7r 2q 3p?
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What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q on the Cartesian plane?

Points: (p, q) and (7p, 3q) Midpoint: (4p, 2q) Slope: q/3p Perpendicular slope: -3p/q Perpendicular bisector equation:- => y-2q = -3p/q(x-4p) => qy-2q^2 = -3p(x-4p) => qy-2q^2 = -3px+12p^2 => qy = -3px+12p^2+2q^2 In its general form: 3px+qy-12p^2-2q^2 = 0


What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?

A line with slope m has a perpendicular with slope m' such that:mm' = -1→ m' = -1/mThe line segment with endpoints (p, q) and (7p, 3q) has slope:slope = change in y / change in x→ m = (3q - q)/(7p - p) = 2q/6p = q/3p→ m' = -1/m = -1/(q/3p) = -3p/qThe perpendicular bisector goes through the midpoint of the line segment which is at the mean average of the endpoints:midpoint = ((p + 7p)/2, (q + 3q)/2) = (8p/2, 4q/2) = (4p, 2q)A line through a point (X, Y) with slope M has equation:y - Y = M(x - x)→ perpendicular bisector of line segment (p, q) to (7p, 3q) has equation:y - 2q = -3p/q(x - 4p)→ y = -3px/q + 12p² + 2q→ qy = 12p²q + 2q² - 3pxAnother Answer: qy =-3px +12p^2 +2q^2


What is the answer to 7p-12 equals 3p?

3p +12= 7p 7p-3p=12 12?3=4 p=4


How do you form an equation for the perpendicular bisector of the line segment joining the points of p q and 7p 3q showing all details of your work?

First find the midpoint the slope and the perpendicular slope of the points of (p, q) and (7p, 3q) Midpoint = (7p+p)/2 and (3q+q)/2 = (4p, 2q) Slope = (3q-q)/(7p-p) = 2q/6p = q/3p Slope of the perpendicular is the negative reciprocal of q/3p which is -3p/q From the above information form an equation for the perpendicular bisector using the straight line formula of y-y1 = m(x-x1) y-2q = -3p/q(x-4p) y-2q = -3px/q+12p2/q y = -3px/q+12p2/q+2q Multiply all terms by q and the perpendicular bisector equation can then be expressed in the form of:- 3px+qy-12p2-2q2 = 0


-7p plus 5r-6p plus 3r?

-7p + 5r - 6p + 3r = -7p - 6p + 5r + 3r = -(7 + 6)p + (5 + 3)r = -13p + 8r


What is p and q when 2p add 4q is 16 and 7p add 12q is 52?

2p+4q=16 (now divide the equation by two) p+2q=8 (now subtract 2q) p=8-2q 7p+12q=52 (substitue the answer you got for p in the previous equation) 7(8-2q)+12q=52 (multiply the first equation by 7) 56-14q+12q=52 (subtract 14q from 12q) 56-2q=52 (subtract the 56 from 52) -2q=-4 (multiply by -1) 2q=4 (divide by 2) q=2 p=8-2q (substitute the value of q) p=8-2(2) (multiply) p=8-4 (subtract) p=4 2p+4q=16 (check your answers with the new values of p and q) 2(4)+4(2)=16 8+8=16 true 7p+12q=52 7(4)+12(2)=52 28+24=52 true


What is the perpendicular bisector equation of the line segment of p q and 7p 3q?

In its general form it works out as: 3px+qy-12p2-2q2 = 0Improved Answer:-Points:(p, q) and (7p, 3q)Slope: q/3pPerpendicular slope: -3p/qMidpoint: (4p, 2q)Equation: y-2q = -3p/q(x-4p) => yq = -3px+12p2+2q2Perpendicular bisector equation in its general form: 3px+qy-12x2-2q2 = 0


How do you work out and find the perpendicular bisector equation meeting the straight line segment of p q and 7p 3q?

First find the mid-point of the line segment which will be the point of intersection of the perpendicular bisector. Then find the slope or gradient of the line segment whose negative reciprocal will be the perpendicular bisector's slope or gradient. Then use y -y1 = m(x -x1) to find the equation of the perpendicular bisector. Mid-point: (7p+p)/2 and (3q+q)/2 = (4p, 2q) Slope or gradient: 3q-q/7p-p = 2q/6p = q/3p Slope of perpendicular bisector: -3p/q Equation: y -2q = -3p/q(x -4p) y = -3px/q+12p2/q+2q Multiply all terms by q to eliminate the fractions: qy = -3px+12p2+2q2 Which can be expressed in the form of: 3px+qy-12p2-2q2 = 0


What is 5p-15p plus 7p in algebra?

-3p


What is 5p 15p plus 7p in algebra?

5p + 15p + 7p = 27p 5p - 15p + 7p = -3p 5p x 15p + 7p = 75p2 + 7p 5p/15p + 7p = 7p + 1/3


What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q showing work and final answer in its general form?

The perpendicular bisector of a line has a gradient (m') which is the negative reciprocal of the gradient (m) of the line, and passes through the mid-point of the line.The equation of a line with gradient m through a point (x0, y0) has an equation of the form:y - y0 = m(x - x0)The gradient m of a line between two points (x0, y0) and (x1, y1) is given by:m = change_in_y / change_in_x = (y1 - y0) / (x1 - x0)Thus the line through (p, q) and (7p, 3q) has gradient:m = (3q - q) / (7p - p) = 2q / 6p = q/3pand the perpendicular bisector has gradient:m' = -1 / m = -1 / (q/3p) = -3p/qThe midpoint of the line through (p, q) and (3p, 3q) is:midpoint = ((p + 7p)/2), (q + 3q)/2) = (4p, 2q)Thus the perpendicular bisector of the line between (p, q) and (7p, 3q) has equation:y - 2q = -3p/q (x - 4p)→ qy - 2q² = -3px + 12p²→ qy + 3px = 12p² + 2q²Additional Information:-Final answer in its general form: 3px+qy-12p^2-2q^2 = 0


What is 7p more than 3p?

10