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Why is the sum or product of two rational numbers rational?
Suppose p/q and r/s are rational numbers where p, q, r and s are integers and q, s are non-zero.Then p/q + r/s = ps/qs + qr/qs = (ps + qr)/qs.Since p, q, r, s are integers, then ps and qr are integers, and therefore (ps + qr) is an integer.q and s are non-zero integers and so qs is a non-zero integer.Consequently, (ps + qr)/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.Also p/q * r/s = pr/qs.Since p, q, r, s are integers, then pr and qs are integers.q and s are non-zero integers so qs is a non-zero integer.Consequently, pr/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.
What is the perpendicular bisector equation of the chord y equals x plus 5 within the circle x2 plus 4x -18y plus 59 equals 0?
x² + 4x - 18y + 59 = 0 is not a circle; it can be rearranged into: y = (x² + 4x + 59)/18 which is a parabola. You have missed out a y² term. ------------------------------------------------------------ Assuming you meant: x² + 4x + y² - 18y + 59 = 0, then: The perpendicular bisector of a chord passes through the centre of the circle. The slope m' of a line perpendicular to another line with slope m is given by m' = -1/m The chord y = x + 5 has slope m = 1 → the perpendicular bisector has slope m' = -1/1 = -1 A circle with centre Xc, Yc and radius r has an equation in the form: (x - Xc)² + (y - Yc)² = r² The equation given for the circle can be rearrange into this form by completing the square in x and y: x² + 4x + y² - 18y + 59 = 0 → (x + (4/2))² - (4/2)² + (y - (18/2))² - (18/2)² + 59 = 0 → (x + 2)² +(y - 9)² - 2² - 9² + 59 = 0 → (x + 2)² + (y - 9)² = 4 + 81 - 59 → the circle has centre (-2, 9) (The radius, if wanted, is given by r² = 4 + 81 - 59 = 36 = 6²) The equation of a line with slope m' through a point (Xc, Yc) has equation: y - Yc = m'(x - Xc) → y - 9 = -1(x - -2) → y - 9 = -x - 2 → y + x = 7 The perpendicular bisector of the chord y = x + 5 within the circle x² + 4x + y² - 18y + 59 = 0 is y + x = 7
How do you subtract rational numbers with the same or different signs?
Suppose x and y are two rational number.Then x = p/q and y = r/s where p, q, r and s are integers, with q and s being non-zero. Then x - y = p/q - r/s = pq/qs - qr/qs = (pq - rs)/qs. The signs of x and y do not matter, in so far as their signs will be used to determine the signs of p,q, r and s.
Why is the difference between two rational numbers always a rational number?
Suppose x and y are two rational numbers. Therefore x = p/q and y = r/s where p, q, r and s are integers and q and s are not zero.Then x - y = p/q - r/s = ps/qs - qr/qs = (ps - qr)/qsBy the closure of the set of integers under multiplication, ps, qr and qs are all integers,by the closure of the set of integers under subtraction, (ps - qr) is an integer,and by the multiplicative properties of 0, qs is non zero.Therefore (ps - qr)/qs satisfies the requirements of a rational number.
If line m is perpendicular to line n and line n is perpendicular to line r how is line m relatred to line r?
Lines r and m are parallel or line r is line m continued
If R and S are two points in the plane the perpendicular bisector of RS is the set of all points equidistant from R and S?
True
What is the expression for the electric field in the perpendicular bisector plane of a dipole?
In the perpendicular bisector plane of a dipole, the electric field expression is given by: E = (kqd)/(r^3), where E is the electric field, k is Coulomb's constant, q is the magnitude of the charge at each end of the dipole, d is the separation distance between the charges, and r is the distance from the midpoint of the dipole.
Construct a circle with a radius r. mark a point on the circle. construct a tangent thourgh this point.?
1) draw the circle with a radius r and the center at O. 2) mark a point, A, on the circle 3) draw a line from O to A and beyond to point B, a little longer than the radius 4) draw a perpendicular bisector at point A using line OB 5) the perpendicular bisector is the tangent at point A In case, you forgot about drawing the perpendicular bisector. Here is the procedure: a) use your compass and mark equidistant points C and D from point A on line OB (make the length slightly less than half the radius); one point should be outside the circle and the other within. b) use your compass and draw an arc from point C and then from point D, with the arc radii being identical and about as long as the circle radius; the two arcs should intercept at two locations, E and F, one on each side of line OA. c) join points E and F to form the perpendicular bisector of line CD ===============================
Is it true or false If R And S Are Two Points In The Plane The Perpendicular Bisector Of RS Is The Set Of All Points Equidistant From R And S?
True
Why is the sum or product of two rational numbers rational?
Suppose p/q and r/s are rational numbers where p, q, r and s are integers and q, s are non-zero.Then p/q + r/s = ps/qs + qr/qs = (ps + qr)/qs.Since p, q, r, s are integers, then ps and qr are integers, and therefore (ps + qr) is an integer.q and s are non-zero integers and so qs is a non-zero integer.Consequently, (ps + qr)/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.Also p/q * r/s = pr/qs.Since p, q, r, s are integers, then pr and qs are integers.q and s are non-zero integers so qs is a non-zero integer.Consequently, pr/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.
Why is the sum of two rational numbers always rational numbers?
Suppose p/q and r/s are rational numbers where p, q, r and s are integers and q, s are non-zero.Then p/q + r/s = ps/qs + qr/qs = (ps + qr)/qs. Since p, q, r, s are integers, then ps and qr are integers, and therefore (ps + qr) is an integer. q and s are non-zero integers and so qs is a non-zero integer. Consequently, (ps + qr)/qs is a ratio of two integers in which the denominator is non-zero. That is, the sum is rational.
What is the perpendicular bisector equation of the chord y equals x plus 5 within the circle x2 plus 4x -18y plus 59 equals 0?
x² + 4x - 18y + 59 = 0 is not a circle; it can be rearranged into: y = (x² + 4x + 59)/18 which is a parabola. You have missed out a y² term. ------------------------------------------------------------ Assuming you meant: x² + 4x + y² - 18y + 59 = 0, then: The perpendicular bisector of a chord passes through the centre of the circle. The slope m' of a line perpendicular to another line with slope m is given by m' = -1/m The chord y = x + 5 has slope m = 1 → the perpendicular bisector has slope m' = -1/1 = -1 A circle with centre Xc, Yc and radius r has an equation in the form: (x - Xc)² + (y - Yc)² = r² The equation given for the circle can be rearrange into this form by completing the square in x and y: x² + 4x + y² - 18y + 59 = 0 → (x + (4/2))² - (4/2)² + (y - (18/2))² - (18/2)² + 59 = 0 → (x + 2)² +(y - 9)² - 2² - 9² + 59 = 0 → (x + 2)² + (y - 9)² = 4 + 81 - 59 → the circle has centre (-2, 9) (The radius, if wanted, is given by r² = 4 + 81 - 59 = 36 = 6²) The equation of a line with slope m' through a point (Xc, Yc) has equation: y - Yc = m'(x - Xc) → y - 9 = -1(x - -2) → y - 9 = -x - 2 → y + x = 7 The perpendicular bisector of the chord y = x + 5 within the circle x² + 4x + y² - 18y + 59 = 0 is y + x = 7
What letters in the English language have both parallel and perpendicular lines?
There are letters in the alphabet with both parallel and perpendicular lines. In alphabetical order, they are E, F, and H. If the joining point can be considered perpendicular and parallel, then B, D, P, and R also match the criterion.
How do you subtract rational numbers with the same or different signs?
Suppose x and y are two rational number.Then x = p/q and y = r/s where p, q, r and s are integers, with q and s being non-zero. Then x - y = p/q - r/s = pq/qs - qr/qs = (pq - rs)/qs. The signs of x and y do not matter, in so far as their signs will be used to determine the signs of p,q, r and s.
Why is the difference between two rational numbers always a rational number?
Suppose x and y are two rational numbers. Therefore x = p/q and y = r/s where p, q, r and s are integers and q and s are not zero.Then x - y = p/q - r/s = ps/qs - qr/qs = (ps - qr)/qsBy the closure of the set of integers under multiplication, ps, qr and qs are all integers,by the closure of the set of integers under subtraction, (ps - qr) is an integer,and by the multiplicative properties of 0, qs is non zero.Therefore (ps - qr)/qs satisfies the requirements of a rational number.
Why is the difference of two rational numbers are rational numbers?
Suppose A and B are two rational numbers. So A = p/q where p and q are integers and q > 0 and B = r/s where r and s are integers and s > 0. Then A - B = p/q - r/s = ps/qs - qr/qs = (ps - qr)/qs Now, p,q,r,s are integers so ps and qr are integers and so x = ps-qr is an integer and y = qs is an integer which is > 0 Thus A-B can be written as a ratio of two integers, x/y where y>0. Therefore, A-B is rational.
If line m is perpendicular to line n and line n is perpendicular to line r how is line m relatred to line r?
Lines r and m are parallel or line r is line m continued
