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Ask QuestionPosted by Ravindra Bhasme 4 years, 4 months ago
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Posted by Ajay Ray 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
Let us assume that √5 is a rational number.
we know that the rational numbers are in the form of p/q form where p,q are integers.
so, √5 = p/q
p = √5q
we know that 'p' is a rational number. so √5 q must be rational since it equals to p
but it doesn't occurs with √5 since its not an integer
therefore, p =/= √5q
this contradicts the fact that √5 is an irrational number
hence our assumption is wrong and √5 is an irrational number.
Posted by Tarleen Kaur 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
Let construct a rhombus with diagonal of 16 cm and d2
Area = 1/2 × d1 × d2
96 cm2 = 1/2 × 16 cm × d2
96 cm2 = 8 cm × d2
96 cm2 / 8 cm = d2
12 cm = d2
Now half the diagonals to make a small triangle AOB
16/2 = 8 cm
12/2 = 6 cm
{tex}{h}^{2} = {b}^{2} + {p}^{2} \\ {h}^{2} = {8}^{2} + {6}^{2} \\ {h}^{2} = 64 + 36 \\ {h}^{2} = 100 \: {cm}^{2} \\ h = \sqrt{100} \\ h = 10 \: cm{/tex}
Hence side are length of 10 cm
Perimeter = 4 (10) = 40 cm
Posted by Priya Maurya 4 years, 4 months ago
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Rajat Singh 4 years, 4 months ago
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Meet.B. Dattani 4 years, 4 months ago
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Posted by ꧁ঔৣ☬✞J₳₮Ŀ₦✞☬ঔৣ꧂ Gamer 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
3y = ax + 7
Put x = 3 and y = 4, we get
3(4) = a(3) + 7
12 = 3a + 7
3a = 12 - 7
3a =5
a = 5/3
Posted by Smriti Garg 4 years, 4 months ago
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Posted by Ketul Jadav 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
Equation : y = x
Keeping x = 1 , 2 and 3 once
y = 1 , y = 2 and y = 3.
Coordinates are : ( 1, 1 );( 2, 2 );( 3, 3 )
Equation, y = - x
Keeping y = 1 , 2 and 3
1 = - x , 2 = - x and 3 = - x
Coordinates are : ( - 1, 1 );( - 2, 2 );( - 3, 3 )
On plotting both equations we get,
That two equation intersect at coordinate (0,0) i.e. origin.
Posted by Priyansh Diwan 4 years, 4 months ago
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Posted by Himanshi Singh 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
3/5 and 4/5
(3×10)/(5×10) = 30/50
(4×10)/(5×10)= 40/50
30/50 < 32/50 < 34/50 < 36/50 < 38/50 <39/50 < 40/50
Answer :- 32/50 , 34/50 , 36/50 , 38/50 , 39/50
Posted by Himanshi Singh 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
3/5 and 4/5
(3×10)/(5×10) = 30/50
(4×10)/(5×10)= 40/50
30/50 < 32/50 < 34/50 < 36/50 < 38/50 <39/50 < 40/50
Answer :- 32/50 , 34/50 , 36/50 , 38/50 , 39/50
Posted by Mritunjay Rai 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
Theorem functions on an actual case that a polynomial is comprehensively dividable, at least one time by its factor in order to get a smaller polynomial and ‘a’ remainder of zero. This acts as one of the simplest ways to determine whether the value ‘a’ is a root of the polynomial P(x).
That is when we divide p(x) by x-a we obtain
p(x) = (x-a)·q(x) + r(x),
as we know that Dividend = (Divisor × Quotient) + Remainder
But if r(x) is simply the constant r (remember when we divide by (x-a) the remainder is a constant)…. so we obtain the following solution, i.e
p(x) = (x-a)·q(x) + r
Observe what happens when we have x equal to a:
p(a) = (a-a)·q(a) + r
p(a) = (0)·q(a) + r
p(a) = r
Hence, proved.
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