Evaluate 105 x 108 without multiplying directly. -Maths 9th

Evaluate 105 x 108 without multiplying directly. -Maths 9th
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A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : NEED ANSWER

A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.

If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

Last Answer : a+b+c=9 and a2+b2+c2=35 Using formula, (a+b+c)2=a2+b2+c2+2(ab+bc+ca) 92=35+2(ab+bc+ca) 2(ab+bc+ca)=81−35=46 (ab+bc+ca)=23 using formula, (a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c) a3+b3+c3−3abc=(35−23)×9=9×12=108

One joule is equal to a.105 ergs b.108 ergs c.104 ergs d.107 ergs e.106 ergs

Description : One joule is equal to a.105 ergs b.108 ergs c.104 ergs d.107 ergs e.106 ergs

Last Answer : d. 107 ergs

Given E = 40xyi + 20x 2 j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3). a) 105 b) 106 c) 107 d) 108

Description : Given E = 40xyi + 20x 2 j + 2k. Calculate the potential between two points (1,-1,0) and (2,1,3). a) 105 b) 106 c) 107 d) 108

Last Answer : b) 106

The open loop gain of the amplifier is ______ (a) 105 (b) 103 (c) 106 (d) 108

Description : The open loop gain of the amplifier is ______ (a) 105 (b) 103 (c) 106 (d) 108

Last Answer : (a) 105

A charge of 10μC and 14.4 μC are 12 cm apart, the force between them is: (a) 9 × 105 N (b) 9 × 107 N (c) 90 N (d) 108 × 107 N

Description : A charge of 10μC and 14.4 μC are 12 cm apart, the force between them is: (a) 9 × 105 N (b) 9 × 107 N (c) 90 N (d) 108 × 107 N

Last Answer : (c) 90 N

A cantilever of length 3m carries a uniformly distributed load of 15KN/m over a length of 2m from the free end.If I= 108 mm4 and E= 2×105 N/mm2,find the slope at the free end? a.0.00326 rad b.0.00578 rad c.0.00677 rad d.0.00786 rad

Description : A cantilever of length 3m carries a uniformly distributed load of 15KN/m over a length of 2m from the free end.If I= 108 mm4 and E= 2×105 N/mm2,find the slope at the free end? a.0.00326 rad b.0.00578 rad c.0.00677 rad d.0.00786 rad

Last Answer : a.0.00326 rad

A cantilever of length 3 m carries two point loads of 2 KN at the free end and 4KN at a distance of 1m from the free end .What is the deflection at the free end? Take E= 2×105 N/mm2and I= 108 mm4. a.2.56 mm b.3.84 mm c.1.84 mm d.5.26mm

Description : A cantilever of length 3 m carries two point loads of 2 KN at the free end and 4KN at a distance of 1m from the free end .What is the deflection at the free end? Take E= 2×105 N/mm2and I= 108 mm4. a.2.56 mm b.3.84 mm c.1.84 mm d.5.26mm

Last Answer : c.1.84 mm

A cantilever of length 3m carries a point load of 60 KN at a distance of 2m from the fixed end.If E= 2×105 and I=108, what is the deflection at the free end?. a.7 mm b.14 mm c.26 mm d.52 mm.

Description : A cantilever of length 3m carries a point load of 60 KN at a distance of 2m from the fixed end.If E= 2×105 and I=108, what is the deflection at the free end?. a.7 mm b.14 mm c.26 mm d.52 mm.

Last Answer : b.14 mm

A cantilever of length 2m carries a point load of 30KN at the free end.If I = 108 mm4 and E= 2×105 N/mm2. What is the slope of the cantilever at the free end? a.0.503 rad b.0.677 rad c. 0.003 rad d.0.008

Description : A cantilever of length 2m carries a point load of 30KN at the free end.If I = 108 mm4 and E= 2×105 N/mm2. What is the slope of the cantilever at the free end? a.0.503 rad b.0.677 rad c. 0.003 rad d.0.008

Last Answer : c. 0.003 rad

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Last Answer : (c) 4log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) = 2 - log39 ⇒ log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) + log39 = 2⇒ log3 \(\bigg[rac{(3+x)(8-x)(9)}{(9x-8)}\bigg]=2\)⇒ \(rac{9(24+8x-3x-x^2)}{(9x- ... = 9x - 8 ⇒ x2 + 4x - 32 = 0 ⇒ (x + 8) (x - 4) = 0 ⇒ x = - 8, 4. Taking the positive value x = 4.

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Last Answer : (i) 1023 = (100 + 2)3 = (100)3 + 3(100)2 (2) + 3(100) (2)2 + 23 = 10,00,000 + 60,000 + 1200 + 8 = 10,61,208 (ii) 993 = (100 - 1)3 = 1003 - 3(100)2 (1) + 3(100) (1)2 - 13 = 10,00,000 - 30,000 + 300 - 1 = 9,70,299

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Last Answer : (i) 505 × 503 = (500 + 5) (500 + 3) = 5002 + (5 + 3) (500) + 5 × 3 = 250000 + 4000 + 15 = 254015 (ii) 37 × 16 = (30 + 7) (30 - 4) = 302 + (7 - 4) (30) - 7 × 4 = 900 + 90 - 28 = 962

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

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Using suitable identity, evaluate the following -Maths 9th

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Last Answer : (i) 1023 = (100 + 2)3 = (100)3 + 3(100)2 (2) + 3(100) (2)2 + 23 = 10,00,000 + 60,000 + 1200 + 8 = 10,61,208 (ii) 993 = (100 - 1)3 = 1003 - 3(100)2 (1) + 3(100) (1)2 - 13 = 10,00,000 - 30,000 + 300 - 1 = 9,70,299

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Last Answer : (i) 505 × 503 = (500 + 5) (500 + 3) = 5002 + (5 + 3) (500) + 5 × 3 = 250000 + 4000 + 15 = 254015 (ii) 37 × 16 = (30 + 7) (30 - 4) = 302 + (7 - 4) (30) - 7 × 4 = 900 + 90 - 28 = 962

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Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Last Answer : (i) (2x - 1/x)2 [Use identity: (a - b)2 = a2 + b2 - 2ab ] (2x - 1/x)2 = (2x) 2 + (1/x)2 - 2 (2x)(1/x) = 4x2 + 1/x2 - 4 (ii) (2x + y) (2x - y) [Use identity: (a - b)(a + b) = a2 - b 2 ] (2x + y) (2x - ... ) = a2 - b 2 ](1.5 x 2 - 0.3y2 ) (1.5x2 + 0.3y2 ) = (1.5 x 2 ) 2 - (0.3y2 ) 2 = 2.25 x4 - 0.09y4

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

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Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Last Answer : Linear equation

x varies directly as the cube of y, x is 32 when y is 4, then what is the value of y when x is 108 ?

Description : x varies directly as the cube of y, x is 32 when y is 4, then what is the value of y when x is 108 ?

Last Answer : x varies directly as the cube of y, x is 32 when y is 4, then what is the value of y when x is 108 ?

Force applied on a body of mass 5 kg is directly proportional to the acceleration produced in the body. Represent this solution as a linear equation in two variables. -Maths 9th

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Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

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