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Posted by Pratibha Kumari 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
Let us assume √2 is rational number.
a rational number can be written into he form of p/q
√2=p/q
p=√2q
Squaring on both sides
p²=2q²__________(1)
.·.2 divides p² then 2 also divides p
.·.p is an even number
Let p=2a (definition of even number,'a' is positive integer)
Put p=2a in eq (1)
p²=2q²
(2a)²=2q²
4a²=2q²
q²=2a²
.·.2 divides q² then 2 also divides q
Both p and q have 2 as common factor.
But this contradicts the fact that p and q are co primes or integers.
Our supposition is false
.·.√2 is an irrational number
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Yogita Ingle 6 years, 2 months ago
{tex} Value \: of \:\frac{cos45}{(sec30+cosec30)}{/tex}
{tex}=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}{/tex}
{tex}=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2\sqrt{3}}{\sqrt{3}}}{/tex}
{tex}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2(1+\sqrt{3})}{/tex}
{tex}=\frac{\sqrt{2}}{\sqrt{2}\times \sqrt{2}} \times \frac{\sqrt{3}\times (\sqrt{3}-1)}{2(\sqrt{3}+1)(\sqrt{3}-1)}{/tex}
{tex}=\frac{\sqrt{6}(\sqrt{3}-1)}{4\left(\big(\sqrt{3}\big)^{2}-1^{2}\right)}{/tex}
{tex}=\frac{\sqrt{6}(\sqrt{3}-1)}{4(3-1)}{/tex}
{tex}=\frac{\sqrt{6}(\sqrt{3}-1)}{8}{/tex}
Therefore,
Value{tex} \: of \:\frac{cos45}{(sec30+cosec30)}\\=\frac{\sqrt{6}(\sqrt{3}-1)}{8}{/tex}
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Yogita Ingle 6 years, 2 months ago
LHS = √(1 + sin∅)/√(1 - sin∅)
= √(1 + sin∅) × √(1 + sin∅)/√(1 -sin∅)×√(1 +sin∅)
= √(1 + sin∅)²/√(1 -sin²∅)
= (1 + sin∅)/√cos²∅
= (1 + sin∅)/cos∅
= 1/cos∅ + sin∅/cos∅
= sec∅ + tan∅ = RHS
Posted by Gursumeet Judge11 6 years, 2 months ago
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Posted by Ang Solanki 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
Given pth term = 1/q
a + (p - 1)d = 1/q
aq + (pq - q)d = 1 --- (1)
Similarly,
ap + (pq - p)d = 1 --- (2)
From (1) and (2), we get
aq + (pq - q)d = ap + (pq - p)d
aq - ap = d[pq - p - pq + q]
a(q - p) = d(q - p)
Therefore, a = d
Equation (1) becomes,
dq + pqd - dq = 1
d = 1/pq
Hence a = 1/pq
Consider, Spq = (pq/2)[2a + (pq - 1)d]
= (pq/2)[2(1/pq) + (pq - 1)(1/pq)]
= (1/2)[2 + pq - 1]
= (1/2)[pq + 1]
Posted by Kashyap Anurag 6 years, 2 months ago
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Posted by Kashyap Anurag 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
second term=7
common difference =4
second term=a+d
a+4=7
a=7-4
a=3
So, tenth term=a+9d
= 3+9(4)
= 3+36
= 39
Step-by-step explanation:
second term=first term+common difference
nth term =a+(n-1)d
Here
n=10
n-1= 10-1
=9
so tenth term= a+9d
Posted by Zeeshan Jameel 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
Area of square = 1600 m²
length of side, l = √1600 = 40m
so diameter of semicircles = length of side of square = 40m
radius, r = 40/2 = 20m
Area of one semicircle = {tex}\frac{ \pi r^2}{2} = \frac{22*20^2}{7*2} =628.57\ m^2{/tex}
Area of all 4 semicircles = 4×628.57 = 2514.28 m²
total area = 2514.28+1600 = 4114.28
rate = Rs. 1.25 per m²
total cost = 1.25×4114.28 = Rs. 5142.85
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