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Let x = 0.33333..... ..1 Multiplying eq. 1 with 10 So, 10x = 3.33..... ..2 Subtracting eq.2 from 1 So, 10x - x = 3 9x = 3 x = 3/9 x = 1/3 Thus 0.333...=1/3

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The site of food made and him what does he suggested

Posted by Tanisha Gupta 3 years, 5 months ago

CBSE > Class 10 > Mathematics

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(sin A + cosec A ) ^ 2 * (cos A + sec A ) ^ 2 = 7 + tan 2 A + cot 2 A

Posted by Dheeraj Manuja 3 years, 5 months ago

CBSE > Class 10 > Mathematics

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Find the zeroes of the following quadratic polynomial and verify the relation between its zeroes and coefficient x^2-2

Posted by Priyanshu Baghel 3 years, 5 months ago

CBSE > Class 10 > Mathematics

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Rishabh Sharma 3 years, 5 months ago

X^2-2=0 X^2=2 X=√2

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Root 3x²+10x+7root3=0

Posted by Uiojvff Huiohgg 3 years, 6 months ago

CBSE > Class 10 > Mathematics

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A boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream in 6 hours and 30 minutes. The speed of the boat in still water is

Posted by Sujal Sujal 3 years, 6 months ago

CBSE > Class 10 > Mathematics

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A gulab jamun contains sugar syrup up to about 30%of its volume find approximately how much syrup would be found in 45 gulab jamun each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8cm

Posted by Aditey Srivastava 3 years, 6 months ago

CBSE > Class 10 > Mathematics

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Tashvi Sud 3 years, 6 months ago

Volume of 1 gulab jamun = volume of cylinder + 2* vol. Of hemisphere. πr²h + 2*2/3πr³ = πr²(h+4/3r) = 6.16 * (12.2÷3) Vol. Of 45 gulab jamuns = 45 * 6.16 * 12.2/3 = 6.16 * 12.2 * 15. As gulab jamun contains 30% of sugar syrup of its volume so 6.16 *12.2*15*30%. On solving we will get the answer 338.184cm³

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Tashvi Sud 3 years, 6 months ago

height of cylinder = total height of gulab jamun - 2 * radius of hemisphere. 5cm - (2*1.4) = 5cm - 2.8cm. So height of cylinder = 2.2cm

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Tashvi Sud 3 years, 6 months ago

Radius of cylinder = Radius of hemisphere

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2-√3 prove that it is irrational number

Posted by Aryan Dhiman 3 years, 6 months ago

CBSE > Class 10 > Mathematics

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show that only one of the numbers n , n+2 , n+4 is divisible by 3

Posted by Bsbsb Kenen 3 years, 6 months ago

CBSE > Class 10 > Mathematics

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Khushi Sharma 3 years, 6 months ago

Case-I: When n = 3q In this case, we have n= 3q, which is divisible by 3 Now, n = 3q n+2 = 3q+2 n+2 leaves remainder 2 when divided by 3 Again, n = 3q n+4 = 3q+4=3(q+1)+1 n+4 leaves 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 remainder 1 when divided by 3. n is divisible by 3 Now, n = 3q+1 n+2 = (3q+1)+2=3(q+1) n+2 is divisible by 3. Again, n = 3q+1 n+4 = 3q+1+4=3q+5=3(q+1)+2 n+4 leaves 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. Now, 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 divisible by 3. Again, 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 3.

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Show that 21n cannot end with the digits 0, 2, 4, 6 and 8 for any natural number n.

Posted by Bsbsb Kenen 3 years, 6 months ago

CBSE > Class 10 > Mathematics