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Sia ? 6 years, 6 months ago
f(x) = x4 + 4x2 + 5
= (x2)2 + 4x2 + 5
Let x2 =n,
Then, f(x) = n2 + 4n + 5,
Here a=1,b=4,c=5
The discriminant(D) = {tex}\text{b}^2-4\mathrm{ac}=\;(4)^2-4\times1\times5=16-20=-4{/tex}
Since the discriminant is negative so this polynomial has no zeros
Hence, f(x) = x4 + 4x2 + 5 has no zero.
Posted by Dileep Kumar 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let p(x) ={tex} x^3 + 2x^2 + kx + 3{/tex}
Now, x - 3 = 0
{tex}\Rightarrow{/tex} x = 3
By the remainder theorem, we know that when p(x) is divided by (x - 3), the remainder is p(3).
Now, p(3) = (3)3 + 2(3)2 + k(3) + 3
= 27 + 2(9) + 3k + 3
= 30 + 18 + 3k
= 48 + 3k
But, remainder = 21
{tex}\Rightarrow{/tex} 48 + 3k = 21
{tex}\Rightarrow{/tex} 3k = 21 - 48
{tex}\Rightarrow{/tex} 3k = -27
{tex}\Rightarrow{/tex} k = {tex}\frac{-27}3{/tex}
{tex}\Rightarrow{/tex} k = -9
So, the value of k is -9.
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Sia ? 6 years, 6 months ago
In statistics, an ogive is a graph showing the curve of a cumulative distribution function. The points plotted are the upper class limit and the corresponding cumulative frequency. The ogive for the normal distribution, resembles one side of an Arabesque or ogival arch.
Posted by Manika Priyadarshini 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We have,
{tex}x + 2y - 4=0{/tex}
Putting {tex}y = 0{/tex}, we get
{tex}x + 0 - 4 = 0{/tex}
{tex} \Rightarrow {/tex} {tex}x = 4{/tex}
Putting x = 0, we get
{tex}0 + 2y - 4 = 0{/tex}
{tex} \Rightarrow {/tex} {tex}y = 2{/tex}
Thus, two solutions of equation {tex}x + 2y - 4 = 0{/tex} are:
| x | 4 | 0 |
| y | 0 | 2 |
We have,
{tex}2x + 4y - 12 = 0{/tex}
Putting {tex}x = 0{/tex}, we get
{tex}0 + 4y - 12 = 0{/tex}
{tex} \Rightarrow {/tex} {tex}y = 3{/tex}
Putting {tex}y = 0{/tex}, we get
{tex}2x + 0(12) = 0{/tex}
{tex} \Rightarrow {/tex} x = 6
Thus, two solutions of equation {tex}2x + 4y - 12 = 0{/tex} are:
| x | 0 | 6 |
| y | 3 | 0 |
Now, we plot the points A (4, 0) and B (0, 2) and draw a line passing through these two points to get the graph of the line represented by the equations (i).
We also plot the points P (0, 3) and Q (6, 0) and draw a line passing through these two points to get the graph of the line represented by the equation (ii).
We observe that the lines are parallel and they do not intersect any where.
REMARK The graphical representation of the above pair of linear equations provides us a pair of parallel lines.
Let us write the pair of linear equations,
{tex}x + 2y - 4 = 0{/tex}
{tex}2x + 4y -12 = 0{/tex}
as {tex}a_1x + b_1y + c_1=0{/tex}
{tex}a_2x + b_2y +c_2 =0{/tex}
where {tex}a_1=1, b_1= 2, c_1 = -4{/tex},
{tex}a_2 = 2, b_2 = 4\ and \ c_2 = -12{/tex}
{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { 1 } { 2 } , \frac { b _ { 1 } } { b _ { 2 } } = \frac { 2 } { 4 } = \frac { 1 } { 2 } \text { and } \frac { c _ { 1 } } { c _ { 2 } } = \frac { - 4 } { - 12 } = \frac { 1 } { 3 }{/tex}
{tex}\therefore \quad \frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }{/tex}
will represent parallel lines, if
{tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }{/tex}
The converse is also true for any pair of linear equations.
It follows from the above examples that the pair of linear equations
{tex}a_1x + b_1y + c_1 = 0{/tex}
{tex}a_2x + b_2y +c_2=0{/tex}
will represent:
- intersecting lines, if {tex}\frac { a _ { 1 } } { a _ { 2 } } \neq \frac { b _ { 1 } } { b _ { 2 } }{/tex}
- coincident lines, if {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex}
- parallel lines, if {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } \neq \frac { c _ { 1 } } { c _ { 2 } }{/tex}
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Sia ? 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
A.P. is 17,14,11,..., - 40
We have,
l = Last term = -40 , a = 17 and, d = Common difference= 14 - 17 = - 3
{tex}\therefore{/tex} 6th term from the end = l - (n -1)d
= l - (6-1) d
= -40 - 5 {tex}\times{/tex} (-3 )
= -40 + 15
= -25
So, 6th term of given A.P. is -25.
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Sia ? 6 years, 6 months ago
Prime Factorisation method-
Express each one of the given numbers as the product of prime factors.
The product of least powers/index of common prime factors gives H.C.F.
Find the H.C.F. of 8 and 14 by Prime Factorisation method.
Solution: 8 = 2 x 2 x 2 and 14 = 2 x 7
Common factor of 8 and 14 = 2.
Thus, Highest Common Factor (H.C.F.) of 8 and 14 = 2.
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