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Kulraj Singh?? 5 years, 4 months ago
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Neeraja Vashisht 5 years, 4 months ago
Yogita Ingle 5 years, 4 months ago
tan 48° tan 23° tan 42° tan 67°
= tan (90° - 42°) tan (90° - 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°)
= 1×1
= 1
Posted by Kulraj Singh?? 5 years, 4 months ago
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? Pranali.A.P ? 5 years, 4 months ago
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Ramadevi.K? K 5 years, 4 months ago
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Yogita Ingle 5 years, 4 months ago
x2+2x-8=0
x2+4x-2x-8=0
x(x+4)-2(x+4)=0
(x-2) (x+4)=0
x=2 or x= -4
Posted by Dinesh Singh 5 years, 4 months ago
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? Pranali.A.P ? 5 years, 4 months ago
Yogita Ingle 5 years, 4 months ago
lets assume that √p is rational,
⇒ √p = a/b ( where 'a' and 'b' are co primes, meaning they don't have any common factors except for 1)
From squaring both sides,
p = a²/b²
⇒pb² = a²
⇒ b² = a²/p
Since 'p' divides a², it also divides 'a' meaning 'a' has a factor of p
Let 'a' = pm (where m is a positive integer) ⇒ a² = p²m²
Now, pb² = a²
pb² = p²m²
pb²/p²= m²
b²/p =m²
∴ 'p' divides 'b' ⇒ 'b' also has a factor 'p'
∴ 'a' and 'b' are not co primes and our assumption was wrong
⇒ √p is irrational
Similarly √q is irrational
∴⇒ √p + √q is irrational
Sumesh ☺️☺️☺️ 5 years, 4 months ago
Posted by Jatin Kumar Nangaliya 5 years, 4 months ago
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? Pranali.A.P ? 5 years, 4 months ago
Gaurav Seth 5 years, 4 months ago
To Find :- The quadratic polynomial.
Quadratic Equation Should be in the form of : ax² + bx + c.
For Finding the quadratic polynomial we use this formula :- k = (x² - (α + β)x + αβ ).
It's Given that :-
Sum of roots ( α + β ) = -√2
Product of roots ( αβ ) = 3/2
Now, Putting this in the formula.
k = (x² - (α + β)x + αβ)
k= [ x² - (-√2)x +[tex]\huge\sf\dfrac{3}{2}]
k = x² + √2x + 3/2
k = 2x² + 2√2x + 3 / 2
Let k = 2.
2 = 2x² + 2√2x + 3 / 2
Hence, the required quadratic polynomial is 2x² + 2√2x + 3
Posted by Kulraj Singh?? 5 years, 4 months ago
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Kulraj Singh?? 5 years, 4 months ago
? Pranali.A.P ? 5 years, 4 months ago
Posted by Harsh Ydv 5 years, 4 months ago
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Posted by Deepak Naagar 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
The decimal expansion of a rational number 
will terminate after 3 decimal places.
Step-by-step explanation:
Given : The decimal expansion of rational number 327 upon 2 raise to power 3 into 5.
To find : The decimal expansion will terminate after ?
Solution :
Rational number is
Multiply and divide by
The decimal expansion of a rational number will terminate after 3 decimal places.
Posted by Shaan Singh 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
Let radius = r
Perimeter = radius + radius + length of arc
Perimeter of quadrant = r + r + angle/360 * 2 * pi * r = 25
2r + [1/4 * 2 * 22/7 * r] = 25
25/7 * r = 25
r = 7
Area = angle/ 360 * pi * r * r
1/4 * 22/7 * 7 * 7 = 38.5 sq.cm
Posted by Jatin Khanna 5 years, 4 months ago
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Neha Kakkar ?? 5 years, 4 months ago
Simranpreet Kaur 5 years, 4 months ago
Posted by Anuj Kumar 5 years, 4 months ago
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Yogita Ingle 5 years, 4 months ago
935/10500
-> 935 /10500= 187 / 2250
-> 187 / (5)3 *(3)2* 2
so , it can't be divide properly
Posted by Suryansh Singh 5 years, 4 months ago
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Posted by Prajnasree Behera 5 years, 4 months ago
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Yogita Ingle 5 years, 4 months ago
2x2+7x -4=0
2x2+8x -x -4=0
2x(x+4)-1(x+4)=0
(X+4)(2x -1)=0
X+4=0
X=-4
2x-1=0
X=1/2
The zeroes of the polynomial 2x²+7x-4 are -4 , and 1/2.
Posted by Sahil Narang 5 years, 4 months ago
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Yogita Ingle 5 years, 4 months ago
For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such that
a=bq+ra=bq+r, where 0≤r<b0≤r<b
Explanation:
Thus, for any pair of two positive integers a and b; the relation
a=bq+ra=bq+r, where 0≤r<b0≤r<b
will be true where q is some integer.
Posted by Md Tamimul Haque 5 years, 4 months ago
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? Pranali.A.P ? 5 years, 4 months ago
Gaurav Seth 5 years, 4 months ago
Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0)
Cubing both sides : 6=a^3/b^3
a^3 = 6b^3
a^3 = 2(3b^3)
Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number
divides the product of two integers then it must divide one of the two integers
Since all the terms here are the same we conclude that 2 divides a
.
Now there exists an integer k such that a=2k
Substituting 2k in the above equation
8k^3 = 6b^3
b^3 = 2{(2k^3) / 3)}
Therefore, 2 divides b^3. Using the same logic as above. 2 divides b.
Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime. Therefore, the initial assumption is wrong.
cube root 6 is irrational.
Posted by Harman Dhaliwal 5 years, 4 months ago
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? Pranali.A.P ? 5 years, 4 months ago
Gaurav Seth 5 years, 4 months ago
π (Pi) is irrational number.
We cannot write down a simple fraction that equals Pi. The popular approximation of 22/7 = 3.1428571428571... is close but not accurate. Another clue is that the decimal goes on forever without repeating.
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Kulraj Singh?? 5 years, 4 months ago
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Gaurav Seth 5 years, 4 months ago
The system of linear equations with variables x and y are given to be
⇒ b³x - a³y = a²b²(a + b) .......... (1)
⇒ b³x - a³y = a³b² + a²b³ ........... (2)
And, the other equation is
b²x - a²y = 2a²b² ............ (3)
⇒ b³x - a²by = 2a²b³ ............ (4)
Now, solving equations (2) and (4) we get,
a²by - a³y = a³b² + a²b³ - 2a²b³ = a³b² - a²b³
⇒ a²(b - a)y = a²b²(a - b)
⇒ y = - b²
Now, from equation (3) we get
b²x - (-b²)a² = 2a²b²
⇒ b²x = a²b²
⇒ x = a²
Therefore, the solution for x is x = a². (Answer)
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