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Preeti Dabral 20 hours ago
Let p be any positive integer and b = 3. Applying division Lemma with p and b =3 ,
we have
p = 3q + r, where 0 {tex}\leq{/tex} r < 3 and q is some integer
So r=0,1,2
If r=0 , p=3q
If r=1, p=3q+1
If r=2, p=3q+2
Therefore any positive integer is of form 3q,3q+1,3q+2 for some integer q.
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Preeti Dabral 1 day, 23 hours ago
Suppose the numerator of the fraction be x
Denominator of the fraction be y
{tex}\therefore{/tex} the fraction is {tex}\frac{x}{y}{/tex}
According to the question,
The sum of the numerator and denominator of the fraction is 12.
{tex}\Rightarrow x + y = 12{/tex}
{tex}\Rightarrow{/tex} {tex}x + y - 12 = 0{/tex}
If the denominator is increased by 3, the fraction becomes {tex}\frac{1}{2}{/tex}.
{tex}\Rightarrow \frac{x}{{y + 3}} = \frac{1}{2}{/tex}
{tex}\Rightarrow{/tex} {tex}2x = (y + 3){/tex}
{tex}\Rightarrow{/tex} {tex}2x - y - 3 = 0{/tex}
So, we have two equations
{tex}x + y - 12 = 0{/tex}
{tex}2x - y - 3 = 0{/tex}
Here x and y are unknowns.
We have to solve the above equations for x and y.
By using cross-multiplication,
{tex}\Rightarrow \frac{x}{{(1) \times ( - 3) - ( - 1) \times - 12}}{/tex} {tex} = \frac{{ - y}}{{1 \times ( - 3) - 2 \times - 12}}{/tex} {tex}= \frac{1}{{1 \times ( - 1) - 2 \times (1)}}{/tex}
{tex}\Rightarrow \frac{x}{{ - 3 - 12}}{/tex} {tex}= \frac{{ - y}}{{ - 3 + 24}} = \frac{1}{{ - 1 - 2}}{/tex}
{tex}\Rightarrow \frac{x}{{ - 15}} = \frac{{ - y}}{{21}} = \frac{1}{{ - 3}}{/tex}
{tex}\Rightarrow \frac{x}{{15}} = \frac{y}{{21}} = \frac{1}{3}{/tex}
{tex}\Rightarrow x = \frac{{15}}{3},\;y = \frac{{21}}{3}{/tex}
{tex}\Rightarrow{/tex} {tex}x = 5, y = 7{/tex}
The fraction is {tex}\frac{5}{7}{/tex}
Posted by Armaan Hizkiel 1 day, 23 hours ago
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Preeti Dabral 1 day, 23 hours ago
Given :
To Find : solve it by cross multiplication method
Solution:
Cross multiplication method
Equations :
Comparing the given equations
So, x = 4 and y = 1
Posted by Harsh Kushwah 1 day, 23 hours ago
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Preeti Dabral 1 day, 23 hours ago
Consider quadratic polynomial
P(x) = 2x2 – 16x + 30.
Now, 2x2 – 16x + 30 = (2x – 6) (x – 3)
= 2 (x – 3) (x – 5)
The zeros of P(x) are 3 and 5.
Sum of the zeros = 3 + 5 = 8 = −(−16)2 = -[coefficient of xcoefficient of x2]
Product of the zeros = 3 × 5 = 15 = 302 = [constant term coefficient of x2]
So if ax2 + bx + c, a ≠ 0 is a quadratic polynomial and α, β are two zeros of polynomial then
α+β=−ba
αβ=ca
In general, it can be proved that if α, β, γ are the zeros of a cubic polynomial ax3 + bx2 + cx + d, then
α+β+γ=−ba
αβ+βγ+γα=ca
αβγ=−da
Note: ba, ca and da are meaningful because a ≠ 0.
Armaan Armaan 1 day, 22 hours ago
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