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The question is incomplete in the sense that there is no +/- sign between q2 and 2q.
Considering plus '+' sign.
q3 - q2 + 2q - 2
q2(q - 1) + 2(q - 1)
(q2 + 2)(q - 1)
Considering minus '-' sign.
q3 - q2 - 2q - 2
The expression can't be factored.
However if we consider the question complete then it is written as:
q3 - q2(2q) - 2
q3 - 2q3 - 2
-q3 - 2
-(q + 21/3)(q2 - 21/3 + 22/3)
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Is the square root of 18 rational and why?
No. √18 cannot be expressed as a fraction of the form p/q. 18 = 2 x 9 = 2 x 32 ⇒ √18 = √(2 x 32) = (√2) x 3. So if √2 is rational then √18 is rational. Assume √2 is rational. Then p and q can be found such that √2 = p/q is in its simplest form, that is p and q have no common factor. Consider: (√2)2 = (p/q)2 ⇒ 2 = p2/q2 ⇒ p2 = 2q2 Thus p2 is even, and so p must be even. Let p = 2r. Then: p2 = (2r)2 = 2q2 ⇒ 4r2 = 2q2 ⇒ 2r2 = q2 Thus q2 is even, and so q must be even. Let q = 2s. Thus p = 2r, q = 2s and so p and q have a common factor of 2. But p and q are such that they have no common factor. Contradiction. Thus the assumption that √2 is rational is false, that is √2 is not rational, so √18 is not rational.
Numbers that have 12 factors?
There are infinitely many such numbers.If p is a prime then p^11 has 12 factors.If p and q are any two primes, then p*q^5 has 12 factors.If p and q are any two primes, then p^2*q^3 has 12 factors.If p, q and r are any three primes, then p*q*r^2 has 12 factors.
If p is 50 of q then what percent of p is q?
If p = 50 of q then q is 2% of p.
How do you write a -2 in rational form?
If you mean, (by rational form), in the form "p/q", let p= -2 and q = 1
What do you do if the base number is not a fraction but the exponent is a fraction?
Consider have x^(p/q) where the base, x, is a whole number. p and q are also whole numbers (q is not 0) so that the exponent, p/q, is a fraction. Then x^(p/q) = (x^p)^(1/q), that is, the qth root of x^p or equivalently, x^(p/q) = [x^(1/q)]^p, that is, the pth power of the qth root of x. For example, 64^(2/3) = 3rd root of 64^2 = 3rd [cube] root of 4096 = 16 or (cube root of 64)^2 = 4^2 = 16. If p/q is negative, the answer is the reciprocal of the answer obtained with positive p/q.
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
Simplify 4(3q-q)?
steps to solve 4(3q-q): =(4)(3q+-q) =(4)(3q)+(4)(-q) =12q-4 answer: 8q
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
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
How do you solve 3q - 5 13 for q.?
3q = 18 q = 6
A number q tripled plus z doubled in algebraic expression?
3q + 2z, although it could be 2*(3q+z)
Solve -3q plus 4 equals 6q equals 13?
If -3q + 4 = 13 then 3q = -9 : q = -3 If 6q = 13 then q = 13/6 = 2 1/6 The question contains an anomaly and is invalid.
What is the quadratic 30q3 plus 14q2-4q equals 0?
30q3 + 14q2 - 4q = 0 can be factored as q(30q2 + 14q - 4) = 0 The bracketed term can be factored (30q2 + 14q - 4) = (10q - 2)(3q + 2) The equation can now be written : q(10q - 2)(3q + 2) = 0 The equation = 0 when either q = 0 or one of the bracketed terms = 0 When 10q - 2 = 0 then q = 2/10 = 1/5 : and when 3q + 2 = 0 then q = -2/3.
What is the expression for three times the quantity q minus eleven?
(3)(q-11) = 3q-33
3(2 plus q) plus 15?
2q + 21.
Is the square root of 18 rational and why?
No. √18 cannot be expressed as a fraction of the form p/q. 18 = 2 x 9 = 2 x 32 ⇒ √18 = √(2 x 32) = (√2) x 3. So if √2 is rational then √18 is rational. Assume √2 is rational. Then p and q can be found such that √2 = p/q is in its simplest form, that is p and q have no common factor. Consider: (√2)2 = (p/q)2 ⇒ 2 = p2/q2 ⇒ p2 = 2q2 Thus p2 is even, and so p must be even. Let p = 2r. Then: p2 = (2r)2 = 2q2 ⇒ 4r2 = 2q2 ⇒ 2r2 = q2 Thus q2 is even, and so q must be even. Let q = 2s. Thus p = 2r, q = 2s and so p and q have a common factor of 2. But p and q are such that they have no common factor. Contradiction. Thus the assumption that √2 is rational is false, that is √2 is not rational, so √18 is not rational.
What are the factors of q3-q2 plus 2q-2?
Factoring the left and right parts separately gives: q3 - q2 + 2q - 2 = q(q2 - 1) + 2(q - 1) = q(q + 1)(q - 1) + 2(q - 1) Now we have a common factor (q - 1) that we can take out. I'll combine the remaining terms again: (q - 1)[q(q + 1) + 2(q - 1)] = (q - 1)[q2 + q + 2q - 2] = (q - 1)(q2 + 3q - 2) The right part has no obvious factorization; but you can use the quadratic formula to get the factors, which will probably include square roots.
