Visualize the representation of 5.3777... on the number line upto 5 decimal places, that is, up to 5.37777. -Maths 9th

Visualize the representation of 5.3777... on the number line upto 5 decimal places, that is, up to 5.37777. -Maths 9th
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show the following numbers on the number line. (a) 0.2 (b) 1.9 (c) 1.1 (d) 2.5 -Maths 9th

Description : show the following numbers on the number line. (a) 0.2 (b) 1.9 (c) 1.1 (d) 2.5 -Maths 9th

Last Answer : (a) 0.2 lies between the points 0 and 1 on the number line. The space between 0 and 1 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 0.2 is the second point ... parts. Therefore each equal part will be equal to one-tenth. 2.5 is the fifth point between 2 and 3

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Last Answer : (a) 0.8 lies between 0 and 1 0.8 is nearer to 1 (b) 5.1 lies between 5 and 6 5.1 is nearer to 5 (c) 2.6 lies between 2 and 3 2.6 is nearer to 3 (d) 6.4 lies between 6 and 7 6.4 is nearer to 6 (e) 9.1 lies between 9 and 10 9.1 is nearer to 9 (f) 4.9 lies between 4 and 5 4.9 is nearer to 5

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Last Answer : 1.Draw a line segment QR = 5 cm. 2.With Q as centre, construct an angle of 90° and let this line through Q is QX. 3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Description : Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. -Maths 9th

Last Answer : 1.Draw a line segment QR = 5 cm. 2.With Q as centre, construct an angle of 90° and let this line through Q is QX. 3. With R as centre, construct an angle of 90° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.

A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Last Answer : NEED ANSWER

A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Last Answer : This answer was deleted by our moderators...

Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Description : Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Last Answer : Solution :- x= -4

Find the value of k if the line on 2x + y = k passes through the point (3,5). -Maths 9th

Description : Find the value of k if the line on 2x + y = k passes through the point (3,5). -Maths 9th

Last Answer : Solution :-

Two points with coordinates (3, 4) and (–5, 4) lie on a line parallel to which axis? Justify your answer. -Maths 9th

Description : Two points with coordinates (3, 4) and (–5, 4) lie on a line parallel to which axis? Justify your answer. -Maths 9th

Last Answer : Solution :- y-coordinate of both the points is 4. So, both points lie on the line y = 4 which is parallel to x-axis.

Find the ratio in which the x-axes divides the line joining the points (–2, 5) and (1, –9) ? -Maths 9th

Description : Find the ratio in which the x-axes divides the line joining the points (–2, 5) and (1, –9) ? -Maths 9th

Last Answer : Let the co-ordinates of the point of internal division A be (x, y). Then,\(x\) = \(rac{2 imes(-7)+3 imes8}{2+3}\) = \(rac{-14+24}{5}\) = \(rac{10}{5}\) = 2y = \(rac{2 imes4+3 imes9}{2+3}\) = \(rac{8+27}{5}\) = \(rac{35}{5}\) = 7∴ Co-ordinates of the point for internal division are (2, 7).

Find the slope and inclination of the line which passes through the points (1, 2) and (5, 6) ? -Maths 9th

Description : Find the slope and inclination of the line which passes through the points (1, 2) and (5, 6) ? -Maths 9th

Last Answer : Let A ≡ (x, y), B ≡ (2, 1), C ≡ (3, -2) Area of ΔABC = \(rac{1}{2}\) |{x1 (y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}|= \(rac{1}{2}\) | \(x\)(1 + 2) + 2(-2 - y) + 3(y - 1) | = \(rac{1}{2 ... \(rac{3}{2}\)∴ Co-ordinates of A are \(\bigg(rac{7}{2},rac{13}{2}\bigg)\) or \(\bigg(rac{-3}{2},rac{3}{2}\bigg)\)

Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Description : Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Last Answer : Let C(x, y) be the centre of the circle passing through the points P(6, -6), Q(3, -7) and R(3, 3) Then, PC = QC = RC(Being radius of the same circle) PC2 = QC2 ⇒ (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 ⇒ x2 - ... i), we get 3\(x\) + (-2) - 7 = 0 ⇒ 3\(x\) = 9 ⇒ \(x\) = 3 ∴ The centre is (3, -2).