Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th

Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th
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Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th

Description : Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th

Last Answer : Solution :-

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

Evaluate using identities -Maths 9th

Description : Evaluate using identities -Maths 9th

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

Evaluate using the identities -Maths 9th

Description : Evaluate using the identities -Maths 9th

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

Description : 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

Last Answer : Rationalisation of denominator

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

Using suitable identity, evaluate the following. -Maths 9th

Description : Using suitable identity, evaluate the following. -Maths 9th

Last Answer : Evaluate the following

Using suitable identity, evaluate the following -Maths 9th

Description : Using suitable identity, evaluate the following -Maths 9th

Last Answer : Evaluation of following

Evaluate using identities -Maths 9th

Description : Evaluate using identities -Maths 9th

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

Evaluate using the identities -Maths 9th

Description : Evaluate using the identities -Maths 9th

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

Description : 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

Last Answer : Rationalisation of denominator

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

Using suitable identity, evaluate the following. -Maths 9th

Description : Using suitable identity, evaluate the following. -Maths 9th

Last Answer : Evaluate the following

Using suitable identity, evaluate the following -Maths 9th

Description : Using suitable identity, evaluate the following -Maths 9th

Last Answer : Evaluation of following

Evaluate 185 x 185 - 15 x 15 -Maths 9th

Description : Evaluate 185 x 185 - 15 x 15 -Maths 9th

Last Answer : Solution :-

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

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

Last Answer : Solution :-

Evaluate x if log3 (3 + x) + log3 (8 – x) – log3 (9x – 8) = 2 – log39 -Maths 9th

Description : Evaluate x if log3 (3 + x) + log3 (8 – x) – log3 (9x – 8) = 2 – log39 -Maths 9th

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.

Evaluate each of the following: (i) 1113 – 893 (ii) 463 + 343 (iii) 1043 + 963 (iv) 933 – 1073 -Maths 9th

Description : Evaluate each of the following: (i) 1113 – 893 (ii) 463 + 343 (iii) 1043 + 963 (iv) 933 – 1073 -Maths 9th

Last Answer : answer:

Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Description : Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Last Answer : answer:

Evaluate each of the following using identities: (i) (399)2 (ii) (0.98)2 (iii) 991 x 1009 (iv) 117 x 83 -Maths 9th

Description : Evaluate each of the following using identities: (i) (399)2 (ii) (0.98)2 (iii) 991 x 1009 (iv) 117 x 83 -Maths 9th

Last Answer : answer:

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Last Answer : Prove that a3 +b3 +c3 – 3abc = -25

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Last Answer : Prove that a3 +b3 +c3 – 3abc = -25

If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Description : If 5^(3x^2 log10 2) = 2^((x+1/2)log10 25), then the value of x is: -Maths 9th

Last Answer : (d) \(-rac{1}{3}\)\(5^{{3x^2}log_{10}2}\) = 2\(\big(x+rac{1}{2}\big)\)log10 25⇒ \(5^{{3x^2}log_{10}2}\) = 2\(\big(rac{2x+1}{2}\big)\) x log10 5 = 2(2x+1)log10 5⇒ \(5^{{3x^2}log_{10}2}\) = 2(2x+1)log2 5. ... aloga x = x]⇒ 3x2 = 2x + 1 ⇒ 3x2 - 2x - 1 = 0 ⇒ (x - 1) (3x + 1) = 0⇒ x = 1, \(-rac{1}{3}\)

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Description : A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. -Maths 9th

Last Answer : Inner radius of hemispherical bowl = 5cm Thickness of the bowl = 0.25 cm Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of ... 22/7) (5.25)2 = 173.25 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25 cm wide and 25 cm high. -Maths 9th

Description : A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25 cm wide and 25 cm high. -Maths 9th

Last Answer : Length of greenhouse, say l = 30cm Breadth of greenhouse, say b = 25 cm Height of greenhouse, say h = 25 cm (i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh] = [2(30 ... (30+25+25)] (after substituting the values) = 320 Therefore, 320 cm tape is required for all the 12 edges.

x^4+x^2+25 factorise this -Maths 9th

Description : x^4+x^2+25 factorise this -Maths 9th

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

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Description : Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Last Answer : 14 repeat maximum number of times (4 times) in the given data. Mode = 14

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

x^4+x^2+25 factorise this -Maths 9th

Description : x^4+x^2+25 factorise this -Maths 9th

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

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Description : Find the median of the values 37, 31, 42, 43, 46, 25, 39, 45, 32. -Maths 9th

Last Answer : Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46 Here, number of observations = 9 (odd) Median = value of (9+1 / 2)th Observation = Value of 5th Observation = 39

Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Description : Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Last Answer : 14 repeat maximum number of times (4 times) in the given data. Mode = 14

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Last Answer : According to question find the lengths QC and PD.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : NEED ANSWER

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : NEED ANSWER

factorise the following 4x^2+20x+25 -Maths 9th

Description : factorise the following 4x^2+20x+25 -Maths 9th

Last Answer : NEED ANSWER

The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Last Answer : According to question find the lengths QC and PD.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : (b) Since, the difference between mid value is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option (c) and (d) are wrong. Since, the mid value is 20 which can get only, if we take option (b)

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : (b) Given, mean of 25 observations = 36 ∴ Sum of 25 observations = 36 x 25 = 900 Now, the mean of first 13 observations = 32 ∴ Sum of first 13 observations = 13 x 32 = 416 and the mean of last 13 ... - (Sum of 25 observations) = (520 + 416)-900 = 936 - 900 = 36 Hence, the 13th observation is 36.