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Simplify 75/455 When u will divide 75/455 = 15/91 you'll get this. So deniminatior is 91 And it's factors are 91= 7×13 75/455 now, see 455 it contains its prime factors 7 and 13 but not 2 and 5 So it's an non terminating decimal expansion.

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value of sin x equal to 4 by 5 find the value of 10 x 90 - x

Posted by Sushant Raj 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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If value of sin x equal to 4 by 5 find the value of 10 x 90 - x

Posted by Sushant Raj 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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Find the quadratic polynomial whose sum and products of zeroes are -2/3 , 3/2

Posted by Aditya Pratap Singh Parihar Singh 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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1.To find the HCF of two numbers experimentally based on Euclid's division lemma. 2. To find zeroes of a quadratic polynomials graphically. 3. To verify the conditions for consistency of a system of linear equations in two variables by graphical representation. 4. To find the solutions of a quadratic equation by completing square method by cutting and pasting. 5. To verify that the given sequence is an arithmetic progression by paper cutting and pasting method. 6. To verify the basic proportionality theorem. 7. To verify that equal chords subtend equal angles at the centre of a circle 8. To understand the meaning of different trigonometric ratios using the circular board. 9. To understand the concept of surface area and volume of solids by deriving the formulas for all the six shapes taught. 10. To set the idea of probability of an event through any experiment of your choice.

Posted by Misbah 10Th 'C' 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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Factorise - 4x^2-115x+375

Posted by ??Mr_Siddhant?? . 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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If 1/5 and -2 are respectively product and sum of the zeroes of a quadratic polynomial , find the polynomial.

Posted by Priya Tomar 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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Which constant should be added and subtracted to solve the quadratic equation 4x²-root3x- 5=0 by the method of completing the square

Posted by Milina Limboo 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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The traffic lights at three different road Crossings change after every 36 seconds 60 seconds and 72 seconds if they change simultaneously at 8 a.m. after what time will they change again simultaneously

Posted by Sachin Thakur 4 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 5 months ago

We know that in order to find the time when the three lights will change simultaneously again after 7 a.m., we need to find the LCM of 48, 72, and 108.

Finding the LCM using the prime factorization method, we get the prime factors of LCM as :

Factors = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 1 × 3 = 432

Hence, converting 432 seconds into minutes and seconds, we get:

432 seconds = 7 minutes and 12 seconds.

Thus, the three lights will change simultaneously again after 7 a.m. at 7:07:12 a.m.

Thus, The three lights will change simultaneously again after 7 a.m. at 7 a.m. + 7 minutes and 12 seconds = 7:07:12 a.m.

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the product of three consecutive positive integers is divisible by 6 is this statement true or false

Posted by Lakshya Garg 4 years, 5 months ago

CBSE > Class 10 > Mathematics

3 answers

Sia ? 4 years, 5 months ago

The statement is true that the product of any three consecutive positive numbers can be divisible by 6.

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Jiya Gujjar 4 years, 5 months ago

Yes this is true

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Vanshika Sharma 4 years, 5 months ago

Yes, For example Let's assume the 3 consecutive integers be 2,3,4 2×3×4/6 = 24/6 = 4 See its divisible by 6

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If alpha and beta are zeroes of x^2-2x-1, form a quadratic polynomial whose zeroes are 2alpha -1 and 2beta -1

Posted by Tani Gupta 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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Prove that one of every 3 consecutive positive integers is divisible by 3.

Posted by Niharika Das 4 years, 5 months ago

CBSE > Class 10 > Mathematics

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Solve 2x-y=4 , 3y-x=3 graphically and also , find the coordinates of the points where these lines intersect the 2 axis

Posted by Harshita Chaudhary 4 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Jiya Gujjar 4 years, 5 months ago

Nice

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Sia ? 4 years, 5 months ago

2x - y = 4

3y - x = 3

2x - y = 4

x = 0 => y = -4 (0, -4)

y = 0 => x = 2 (2 , 0)

3y - x = 3

x = 0 => y = 1 (0, 1)

y = 0 => x = -3 (-3 , 0)

2x - y = 4

3y - x = 3 => 6y - 2x = 6

=> 5y = 10

=> y = 2

& x = 3

( 3 ,2 )

lines intersect at ( 3, 2)

x 2x - y = 4 3y -x = 3

-6 -16 -1

-3 -10 0

0 -4 1

3 2 2

6 8 3

9 14 4

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Prove that 2+√3 is an irrational number , given that √3 is irrational number.

Posted by Anand Kumar 4 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Shrishti Singh ? 4 years, 5 months ago

Let's suppose 2+√3 be rational Since 2+√3 - 2 also be rational ( difference of two rational number is rational ) Now 2+√3 - 2 = √3 which is rational But this contradicts the fact that √3 is irrational The contradiction arises by assuming √3 rational Therefore; 2+√3 is irrational Hence proved .

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Solve *x4 6i*π€cx

Posted by Adicherla Sathvik 4 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 5 months ago

Please ask question with complete information.

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Show that one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer.

Posted by Stuti Singla 4 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Adicherla Sathvik 4 years, 5 months ago

We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur Case I : When n = 3q In this case, we have, n = 3q, which is divisible by 3 n = 3q = adding 2 on both sides n + 2 = 3q + 2 n + 2 leaves a remainder 2 when divided by 3 Therefore, n + 2 is not divisible by 3 n = 3q n + 4 = 3q + 4 = 3(q + 1) + 1 n + 4 leaves a remainder 1 when divided by 3 n + 4 is not divisible by 3 Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3 Case II : When n = 3q + 1 In this case, we have n = 3q +1 n leaves a reaminder 1 when divided by 3 n is not divisible by 3 n = 3q + 1 n + 2 = (3q + 1) + 2 = 3(q + 1) n + 2 is divisible by 3 n = 3q + 1 n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 n + 4 leaves a remainder 2 when divided by 3 n + 4 is not divisible by 3 Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3 Case III : When n = 3q + 2 In this case, we have n = 3q + 2 n leaves remainder 2 when divided by 3 n is not divisible by 3 n = 3q + 2 n + 2 = 3q + 2 + 2 = 3(q + 1) + 1 n + 2 leaves remainder 1 when divided by 3 n + 2 is not divsible by 3 n = 3q + 2 n + 4 = 3q + 2 + 4 = 3(q + 2) n + 4 is divisible by 3 Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3 . Thanks I HOPE YOU UNDERSTAND?

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Sia ? 4 years, 5 months ago

We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur
Case I : When n = 3q
In this case, we have,
n = 3q, which is divisible by 3
n = 3q
= adding 2 on both sides
n + 2 = 3q + 2
n + 2 leaves a remainder 2 when divided by 3
Therefore, n + 2 is not divisible by 3
n = 3q
n + 4 = 3q + 4 = 3(q + 1) + 1
n + 4 leaves a remainder 1 when divided by 3
n + 4 is not divisible by 3
Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3
Case II : When n = 3q + 1
In this case, we have
n = 3q +1
n leaves a reaminder 1 when divided by 3
n is not divisible by 3
n = 3q + 1
n + 2 = (3q + 1) + 2 = 3(q + 1)
n + 2 is divisible by 3
n = 3q + 1
n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2
n + 4 leaves a remainder 2 when divided by 3
n + 4 is not divisible by 3
Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3
Case III : When n = 3q + 2
In this case, we have
n = 3q + 2
n leaves remainder 2 when divided by 3
n is not divisible by 3
n = 3q + 2
n + 2 = 3q + 2 + 2 = 3(q + 1) + 1
n + 2 leaves remainder 1 when divided by 3
n + 2 is not divsible by 3
n = 3q + 2
n + 4 = 3q + 2 + 4 = 3(q + 2)
n + 4 is divisible by 3
Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3 .