Let the number be p and q.
According to question,
(p + q)2 = 4pq
? p2 + q2 + 2pq ? 4pq = 0
? (p ? q)2 = 0
? p = q
Hence , required ratio is 1 : 1 .
Given that :- a2 + b2 + c2 = 2 (a ? b ? c) ? 3
? a2 + b2 + c2 - 2 (a ? b ? c) + 3 = 0
? a2 + b2 + c2 - 2a + 2b + 2c + 3 = 0
? ( a2 + 1 - 2a ) + ( b2 + 1 + 2b ) + ( c2 + 1 + 2c ) = 0
? ( a - 1 )2 + ( b + 1 )2 + ( c + 1 )2 = 0
This is possible when ( a - 1 )2 = 0, ( b + 1 )2 = 0, and ( c + 1 )2 = 0
? a = 1, b = ?1, c = ?1
Thus, 2a ? 3b + 4c = 2 (1) ? 3 (?1) + 4 (?1)
2a ? 3b + 4c = 2 + 3 ? 4 = 1.
We know that , ?6 = 2.44, ?5 = 2.23, ?3 = 1.73
a = ?6 - ?5 = 2.44 - 2.23 = 0.21
b = ?5 - 2 = 2.23 - 2 = 0.23
c = 2 - ?3 = 2 - 1.73 = 0.27
Let the numbers are p and q.
p ? q = 3 ? (1)
p2 ? q2 = 39
(p ? q) (p + q) = 39 ? x + y = 13 ? (2)
Adding eqn (1) and (2)
p + q + p ? q = 16 ? p = 8
? q = 3
Hence, 8 is the larger number.
As indicated in the graph, the line passing through the points cuts Y-axis only.
Since two digit number = 10p + q
According to question , ? q = 2p ? 1.. (i)
When digits are interchanged then new number = 10q + p
then original number ? [new number ? original number] = 20
? 10p + q ? [10q + p ? (10p + q)] = 20
? 10p + q ? 10q ? p + 10p + q = 20
19p ? 8q = 20
19p ? 8 (2p ? 1) = 20 (Using eq. (i))
19p ? 16p + 8 = 20
3p = 12 ? p = 4
From (i) q = 2 × 4 ? 1 ? q = 7
? original number = 10p + q = 10 × 4 + 7 = 47
The linear equation is y = 4x,
When, x = 1, y = 4
Let the number of 25 paise coins and 50 paise cons be x and y,respectively.
According to the question,
x + y = 38
Also, 25x + 50y = 1350
On multiplying Eq.(i) by 25 and substracting from Eq.(ii), we get y =16
On substituting the value of y in Eq.(i),we get y=22
So,the number of 25 paise coins in 22.
When f(x) = x4 - 6x3 + 2x2 + 3x + 1 is divided by (x - 2), then remainder can be calculated by putting x - 2 = 0
i.e. x = 2
Now, put x = 2 in f(x)
f(2) = (2)4 - 6(2)3 + 2(2)2 + 3 x 2 + 1
= 16 - 48 + 8 + 6 + 1 = -17
? The remainder will be - 17.
Since, p(x) = ax3 + 4x2 + 3x - 4 and q(x) = x3 -4x + a divided by x - 3, hence p(3) = q(3)
? a x 33 + 4 x 32 + 3 x 3 - 4 = 33 - 4 x 3 + a
? a = -1
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