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Tanveera Alam 1 year ago

Rupee 20 as 5%of rupee 100 is rupee 5 So,if we double it will become rupee (5×2)=rupee 10
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Tanveera Alam 1 year ago

Yes
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We can prove that 2/5√3 is irrational by proof by contradiction. Assume for the sake of contradiction that 2/5√3 is rational. This means it can be expressed as a fraction in its simplest form, where the numerator and denominator have no common factors other than 1. Let's call this fraction a/b, where a and b are integers. Multiplying both sides of the equation by 5√3, we get: 2 = (a/b) * 5√3 2√3 = a/b This equation implies that a is a multiple of √3. However, if a is a multiple of √3, then it cannot be in its simplest form with a denominator of b, contradicting our initial assumption. Therefore, our initial assumption that 2/5√3 is rational must be false. Hence, 2/5√3 is irrational.
Prove that 2/5√3 is irrational
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Ashish Kumar 1 year ago

Hlo baat kro gi mujhse
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Tanveera Alam 1 year ago

2×2×2×
Factors of 2048 and 960 are · 2048 = 2¹¹×1 960 = 2⁶×3×5 H.C.F (2048,960) = 2⁶ =64 OR 2×2×2×2×2×2=64 64 is the answer.
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Tejas Swami 1 year ago

X=6 and y=36
Mat
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Mental ability test
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Tanveera Alam 1 year ago

2.12132034356

Khushi P 1 year ago

2 ×1/root 2 +1/root 2
1
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Mudit Bhadada 1 year ago

Given equations are, 37x+43y=123; 43x+37y=117 . Multiply equation 37x+43y=123 by 43, ⇒1591x+1849y=5289 ……….(1) Equation 43x+37y=117 by 37 , ⇒1591x+1369y=4329 ………..(2) Subtract equation (1) by equation (2), ⇒1591x+1849y−1591x−1369y=5289−4329 ⇒480y=960 ⇒y=2 Substitute,y=2 for x in the equation 43x+37y=117 . ⇒43x+37(2)=117 ⇒43x+74=117 ⇒x=1 Thus, the value of x and y is x=1 and y=2 .
X=2, y=1

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