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The expression "p + 2q" represents the sum of a variable p and twice the value of another variable q. This can also be written as p + 2 * q, where the asterisk denotes multiplication. In algebraic terms, this expression cannot be simplified further unless specific values are assigned to the variables p and q.

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10y ago

3q + 2p

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Q: P q p plus 2q
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Related questions

What is 1p plus q-p plus 2q equals?

3


What is the answer to this question 7p plus 2q equals 46 and 5p plus 3q equals 36?

7p + 2q = 46 . . . . (A) 5p + 3q = 36 . . . . (B) 3*(A): 21p + 6q = 138 2*(B): 10p + 6q = 72 Subtracting gives 11p = 66 so that p = 6 Substitute for p in (A): 7*6 + 2q = 46 or 42 + 2q = 46 which gives 2q = 4 so that q = 2 Solution: (p, q) = (6,2)


If angle P is three less than twice the measure angle Q angle P and angle Q are supplementary angles what are their measurements?

Assuming the angles are expressed in degrees: P = 2Q -3° (because "angle P is three less than twice the measure angle Q") P + Q = 180° (because they are supplementary angles) P+Q = 2Q - 3° + Q = 3Q -3° = 180° 3Q = 183° Q = 61° P = 2∙61° -3° = 122° - 3° = 119° If the angles are expressed in radians, the math is similar except you start with P = 2Q - 3 and P+Q = π yielding P = 2π/3 -1 and Q = π/3 +1


What equals twice the difference of p and q?

The difference of p and q can be written : p - q Twice the difference is therefore 2 x (p - q) which can also be written as 2(p - q) OR 2p - 2q. Consequently you can create another variable (say) y and make this equal to twice the difference of p and q by simply writing, y = 2(p -q)


Why is sqrt 2 irrational?

First we will assume that sqrt(2) is rational, meaning that it can be written as a ratio of two integers say (p/q) p and q must have no common factors sqrt(2)=p/q, square both sides 2=p^2/q^2, multiply both sides by q^2 2q^2=p^2, since 2 divides by the LHS, so does the RHS, meaning that p^2 is evenand because p^2 is even, so is p itselfLet p=2r with r being an integer so that p^2=2q^2=(2r)^2=4r^ 2Since 2q^2=4r^2, q^2=2r^2Because q^2 is 2r^2, then q^2 is even, meaning q itself is evenSince p and q are even, they have a common factor of 2THEREFORE, sqrt(2) cannot be rational


What is the sum of -p plus q?

q - p


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 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


How do you put q p plus 5 in a word phrase?

Q(p+5) or q(p)+5


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


What is p plus r plus q plus 113 plus 80 plus 54?

p + r + q + 175


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