In the figure, find teh perimeter of the shaded region where. ADC, AEB and BFC are semi-circles on diameters AC, AB and BC respectively. -Maths 10th

In the figure, find teh perimeter of the shaded region where. ADC, AEB and BFC are semi-circles on diameters AC, AB and BC respectively. -Maths 10th
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Given: Diameter of semicircle AEB= 2.8 cm RADIUS OF SEMICIRCLE(AEB)= 2.8/2= 1.4 cm Diameter of semicircle BFC=1.4 cm RADIUS OF SEMICIRCLE(BFC)= 1.4/2= 0.7 cm Diameter of semicircle ADC= 2.8+1.4 = 4.2 cm RADIUS OF SEMICIRCLE(ADC) = 4.2/2= 2.1 cm. Perimeter of the shaded region= sum of the arc of semicircle AEB, BFC and ADC= [π×1.4) +(π×0.7)+(π×2.1)] cm = π(1.4+0.7+2.1) cm =( 22/7) ×4.2 = 22× .6 = 13.2 cm Hence, the perimeter of the shaded region is 13.2 cm HOPE THIS WILL HELP YOU...
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In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Description : If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Last Answer : (c2 – ab) x2 + 2(bc - a2 ) x+ (b2 – ac) = 0 Comparing with Ax2 + Bx + C = 0 A = (c2 – ab), B = 2(bc - a2 ) and C = b2 – ac According to the question, B2 - 4AC = 0 Put the values in the above equation we get 4a(a3 + b3 + c3 -3abc) = 0 hence, a = 0 or a3 + b3 + c3 = 3ab

In the given figure, AB is the diameter where AP = 12 cm and PB = 16 cm. If the value of π is taken 3, what is the perimeter of the shaded region? (a) 58 cm (b) 116 cm (c) 29 cm (d) 156 cm

Description : In the given figure, AB is the diameter where AP = 12 cm and PB = 16 cm. If the value of π is taken 3, what is the perimeter of the shaded region? (a) 58 cm (b) 116 cm (c) 29 cm (d) 156 cm

Last Answer : (a) 58 cm

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Description : In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th

Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Last Answer : answer:

Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b) -Maths 10th

Description : Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b) -Maths 10th

Last Answer : Distance between two points (x1 ,y1 ) and (x2 ,y2 ) is (x1 −x2 )2+(y1 −y2 )2 (i) Distance between(2,3) and (4,1) is =(2−4)2+(3−1)2 =(−2)2+(2)2 =4+4 =8 =22 (ii) Distance between (−5,7) and (−1,3) ... 42 (iii )Distance between (a,b) and(−a,−b) is =(a−(−a))2+(b−(−b))2 =(2a)2+(2b)2 =4a2+4b2 =2a2+b2

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. -Maths 10th

Description : Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. -Maths 10th

Last Answer : Then , seven years ago, Aftab's age = x − 7 His daughter's age = y − 7 According to the question, x − 7 = 7 ( y − 7 ) x − 7 = 7 y − 4 9 x − 7 y = − 4 9 + 7 x − 7 y = − 4 2 ( ... these two equations x − 7 y = − 4 2 ⟹ x = 7 y − 4 2 x = 7 y − 4 2 4 2 3 5 4 9 y 1 2 1

The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. -Maths 10th

Description : The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. -Maths 10th

Last Answer : Step-by-step explanation: Let x be the cost of 1 bat Let y be the cost of 1 ball Cost of 3 bats = 3x Cost of 6 balls = 6y Cost of 3 balls = 3y The coach of a cricket team buys 3 ... 6 balls for ₹ 3900 She buys another bat and 3 more balls of the same kind for ₹1300. So, situation algebraically :

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. -Maths 10th

Description : The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. -Maths 10th

Last Answer : y 20 40 60 80Let the cost of 1 kg of apples be x and that of 1 kg be y So the algebraic representation can be as follows: 2x+y=160 4x+2y=300⇒2x+y=150 The situation can be represented ... that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x – 2x = (- 2) (3-x) (iii) (x – 2) (x + 1) = (x – 1) (x + 3) (iv) (x – 3) (2x + 1) = x (x + 5) (v) (2x – 1) (x – 3) = (x + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Description : Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x - 2x = (- 2) (3-x) (iii) (x - 2) (x + 1) = (x - 1) (x + 3) (iv) (x - 3) (2x + 1) = x (x + 5) (v) (2x - 1) (x - 3) = (x ... vi) x2 + 3x + 1 = (x - 2)2 (vii) (x + 2)3 = 2x(x2 - 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Last Answer : this is the correct answer!

What are trigonometric ratios? -Maths 10th

Description : What are trigonometric ratios? -Maths 10th

Last Answer : The trigonometric ratios of the angle A in the right triangle ABC are defined as follows: sine of ∠A = side opposite to angle Ahypotenuseside opposite to angle Ahypotenuse = BCACBCAC cosine of ∠A = side ... of angle A = ACABACAB cotangent of ∠A = 1tangent of angle A1tangent of angle A = ABBCABBC

If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y). -Maths 10th

Description : If ax + by = a2 – b2 and bx + ay = 0, find the value of (x + y). -Maths 10th

Last Answer : Given, ax+by=a2−b2......(1) bx+ay=0.......(2). Now adding (1) and (2) we get, x(a+b)x+(a+b)y=a2−b2 or, (a+b)(x+y)=a2−b2 or, x+y=a−b.

How many solutions does the pair of equations y = 0 and y = -5 have? -Maths 10th

Description : How many solutions does the pair of equations y = 0 and y = -5 have? -Maths 10th

Last Answer : y = 0 and y = -5 are Parallel lines, hence no solution.

Find the radius of the largest right circular cone that can be cut out from a cube of edge 4.2 cm. -Maths 10th

Description : Find the radius of the largest right circular cone that can be cut out from a cube of edge 4.2 cm. -Maths 10th

Last Answer : Radius of the largest right circular cone 1/2 (Edge of the Square) =4.2/2 = 2.1 cm

If 2k + 1, 6, 3& + 1 are in AP, then find the value of k. -Maths 10th

Description : If 2k + 1, 6, 3& + 1 are in AP, then find the value of k. -Maths 10th

Last Answer : For an AP a, b, c; 2b = a + c ⇒ (2k + 1) + (3k + 1) = 2×6 5k+2=12 ⇒ 5k = 10 ⇒ k = 2

Find the zeroes of the of the polynomials p(x) = 4x2 – 12x + 9 -Maths 10th

Description : Find the zeroes of the of the polynomials p(x) = 4x2 – 12x + 9 -Maths 10th

Last Answer : P(x)= 4x² - 12x + 9 4x²- 12x + 9 = 0 4x²- 6x - 6x +9 = 0 2x(2x - 3) - 3(2x - 3) = 0 (2x - 3) (2x - 3) x = 3/2 x = 3/2 x = 3/2 ; 3/2 are the zeros of the given polynomial.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? -Maths 10th

Description : 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? -Maths 10th

Last Answer : Given, Number of army contingent members=616 Number of army band members = 32 If the two groups have to march in the same column, we have to find out the highest common factor between the two groups. HCF(616, ... HCF (616, 32) = 8. Hence, the maximum number of columns in which they can march is 8.

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Description : 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Last Answer : Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r

Use Euclid’s division algorithm to find the HCF of: i. 135 and 225 ii. 196 and 38220 iii. 867 and 225 -Maths 10th

Description : Use Euclid’s division algorithm to find the HCF of: i. 135 and 225 ii. 196 and 38220 iii. 867 and 225 -Maths 10th

Last Answer : 135 and 225 As you can see, from the question 225 is greater than 135. Therefore, by Euclid's division algorithm, we have, 225 = 135 1 + 90 Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get, ... (867,225) = HCF(225,102) = HCF(102,51) = 51. Hence, the HCF of 867 and 225 is 51

Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8 -Maths 10th

Description : Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8 -Maths 10th

Last Answer : Let us consider a and b where a be any positive number and b is equal to 3. According to Euclid's Division Lemma a = bq + r where r is greater than or equal to zero and less than b (0 ≤ r < b) a = 3q + r so ... 8 Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Description : Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. -Maths 10th

Last Answer : Let a' be any positive integer and b = 6. ∴ By Euclid's division algorithm, we have a = bq + r, 0 ≤ r ≤ b a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4, ... numbers. Therefore, any odd integer can be expressed is of the form 6q + 1, or 6q + 3, or 6q + 5 where q is some integer

Use Euclid’s division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 -Maths 10th

Description : Use Euclid’s division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 -Maths 10th

Last Answer : (i) Given numbers are 135 and 225. On applying Euclid's division algorithm, we have 225 = 135 x 1 + 90 Since the remainder 90 ≠ 0, so again we apply Euclid's division algorithm to 135 and 90, to get 135 = 90 x ... division lemma, to a =867 and b=255 to find q and r such that 867 = 255q + r, 0 ≤ r

Is polynomial y4 + 4y2 + 5 have zeroes or not? -Maths 10th

Description : Is polynomial y4 + 4y2 + 5 have zeroes or not? -Maths 10th

Last Answer : Y4+4y2+5=y4+4y2+4 +1 =(y2+2)2+1 The first term being square,it will be equal to zero or greater than zero. Therefore (y2+2)2+1 will not be zero for any value of y. Therefore given polynomial have no zeroes.

Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Description : Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Last Answer : Here, p(x) = x4 + 2x3 - 4x2 + 6x - 3, g(x) = x2 - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).

If the product of two zeroes of polynomial 2x3 + 3x2 – 5x – 6 is 3, then find its third zero. -Maths 10th

Description : If the product of two zeroes of polynomial 2x3 + 3x2 – 5x – 6 is 3, then find its third zero. -Maths 10th

Last Answer : The third zero is 1. Step-by-step explanation: Given the polynomial The product of two zeroes is 3 we have to find the third zero As product of two zeroes is 3 Let ab=3 Therefore, 3c=3 ⇒ c=1 hence, the third zero is 1.

If one zero of the polynomial 5z2 + 13z – p is reciprocal of the other, then find p. -Maths 10th

Description : If one zero of the polynomial 5z2 + 13z – p is reciprocal of the other, then find p. -Maths 10th

Last Answer : The value of p is -5 Step-by-step explanation: Given that if one zero of polynomial is reciprocal of the other then we have to find the value of p. a=5, b=13 and c=-p since the zeroes are reciprocal to each other. therefore their product must be 1 ⇒ ⇒ Hence, the value of p is -5

The sum of two children is ‘a’. The age of the father is twice the ‘a’. After twenty years, his age will be equal to the addition of the ages of his children. Find the age of father. -Maths 10th

Description : The sum of two children is ‘a’. The age of the father is twice the ‘a’. After twenty years, his age will be equal to the addition of the ages of his children. Find the age of father. -Maths 10th

Last Answer : Let the sum of the ages of two children will be x and age of father will be y. A.T.Q. , y = 2x ----------(i) and , After 20 years, x+40=y+20 ==> ... i), x-2x = -20 ==> x = 20 Now, put x= 20 in (i), y = 2 x 20 = 40 Hence age of father will be 40 years.

The larger of two supplementary angles exceeds thrice the smaller by 20 degrees. Find them. -Maths 10th

Description : The larger of two supplementary angles exceeds thrice the smaller by 20 degrees. Find them. -Maths 10th

Last Answer : Two angles are supplementary. If one angle is x, then other angle is (180-x) let x > (180-x) x = 3(180-x)+20 = 560 - 3x hence 4x = 560 or x = 140 hence angles are 140° and 40°

A diver rowing at the rate of 5 km/h in still water takes double the time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. -Maths 10th

Description : A diver rowing at the rate of 5 km/h in still water takes double the time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. -Maths 10th

Last Answer : 5/3 km/h Step-by-step explanation: Speed in still water = 5km/h Define x: Let the speed of the steam be x Upstream speed = (5 - x) km/h Downstream speed = (5 + x) km/h Time taken going upstream: Time taken = ... 40x = 400 - 80x 120x = 200 x = 200 120 = 5/3 The speed of the stream is 5/3 km/h

There are two points on a highway a,b. They are 70 km apart. An auto starts from A and another auto starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they meet in 1 hour. Find how fast the two autos are -Maths 10th

Description : There are two points on a highway a,b. They are 70 km apart. An auto starts from A and another auto starts from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they meet in 1 hour. Find how fast the two autos are -Maths 10th

Last Answer : When they travel in same direction, suppose they meet when B travels for x m, then A will have travelled 70+x m in the same time Since, Speed =TimeDistance Speed of auto at A =7x+70 And Speed of auto at B =7x ... So, Speed of auto at A =7x+70 =7280 =40km/hr Speed of auto at B =7x =7210 =30km/hr

Solve graphically 4x-3y+4=0, 4x+3y-20=0 -Maths 10th

Description : Solve graphically 4x-3y+4=0, 4x+3y-20=0 -Maths 10th

Last Answer : 4x−3y+4=0 ⟹x=4−4+3y Let, y=4, then x=4−4+3(4) =2 Similarly, when y=0,x=−1 4x+3y−20=0 ⟹x=420−3y Let, y=4, then x=420−3(4) =2 Similarly, when y=0,x=5 Plotting the above points on the graph, ... point i.e(2,4) Area of the region bounded by these lines and x-axis =21 6units 4 units =12 sq. units.

A number say z is exactly the four times the sum of its digits and twice the product of the digits. Find the numbers -Maths 10th

Description : A number say z is exactly the four times the sum of its digits and twice the product of the digits. Find the numbers -Maths 10th

Last Answer : This is the Answer

When you add two numbers and the number obtained by reversing the order of its digits is 165. If the both numbers differ by three, find the number. -Maths 10th

Description : When you add two numbers and the number obtained by reversing the order of its digits is 165. If the both numbers differ by three, find the number. -Maths 10th

Last Answer : Let the No. Be xy ATGC , xy + yx = 165 10x + y + 10y+x = 165 11x + 11y = 165 Divide Both sides By 11 x + y = 15 __________(1) Also,x ... By 1 and 2 we get 2x = 18 x = 9 y = 6 so the number is So the number is 96 or 69 ______________________

Find the distance between the points (-8/5, 2) and (2/5, 2). -Maths 10th

Description : Find the distance between the points (-8/5, 2) and (2/5, 2). -Maths 10th

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Find a point on the y-axis equidistant from (-5, 2) and (9, -2). -Maths 10th

Description : Find a point on the y-axis equidistant from (-5, 2) and (9, -2). -Maths 10th

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Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

Description : Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

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A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. -Maths 10th

Description : A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. -Maths 10th

Last Answer : The given speed of the motor boat in the still water is equal to 18 km/ hr. The given distance travelled by motor boat is equal to 24 km. The time taken to travel 24 km downstream by motor boat = 1 hour. x=122=6 km/ hr.

In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Description : In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Last Answer : Let P has obtained x in mathematics and y in science. Then by the problem, x+y=28.......(1). If P would have got 3 marks more in mathematics, then P would have got (x+3) in mathematics ... P obtained 12 marks in mathematics and 16 in science. or, P obtained 9 marks in mathematics and 19 in science.

At �t� minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than minutes. Find ‘t’. -Maths 10th

Description : At �t� minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than minutes. Find ‘t’. -Maths 10th

Last Answer : answer: