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  • 4 answers

Himanshi Sharma 4 years, 4 months ago

0

Shanvi Sneh 4 years, 4 months ago

Answer will be 0.

T Lalitha Rani Saraswathi 4 years, 4 months ago

Sorry it's 1

T Lalitha Rani Saraswathi 4 years, 4 months ago

2
  • 1 answers

Himanshi Gautam 4 years, 4 months ago

4-5√3= a/b -5√3= a/b-4 √3=4/5-a/b RHS is not equal to LHS HENCE IT IS AN IRRATIONAL NUMBER.
  • 0 answers
  • 1 answers

Preeti Dabral 4 years, 4 months 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}

  • 1 answers

Preeti Dabral 4 years, 4 months 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

  • 3 answers

Preeti Dabral 4 years, 4 months ago

<header>

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.

</header>

Armaan Armaan 4 years, 4 months ago

Like in 3x^2 + 5 In this 3 is a coefficient

Armaan Armaan 4 years, 4 months ago

Zeroes are number like a and b. Coefficient are numbers side the x and y. ?
  • 3 answers

Armaan Armaan 4 years, 4 months ago

19/30,20/30,21/30,22/30,23/30

Vinayak Rathod Vishal 4 years, 4 months ago

Find the 6th term from the end of the A. P 17,14 , 11...-40

Akarshit Sharma 4 years, 4 months ago

3/5=0.6 4/5=0.8 So numbers between between 0.6 and 0.8 are 0.65 , 0.70 , 0.71 , 0.75 , 0.61
  • 1 answers

Tejas Bramhe 4 years, 4 months ago

They can not be added
  • 3 answers

Bishnu Gond 4 years, 4 months ago

X coordinate 0 and y coordinate 3

Sarita Panda 4 years, 4 months ago

(5,-2), (6,-3), (0,3), (3,0) and infinitely many more.

Vipin Patel 4 years, 4 months ago

2,1
4-1
  • 1 answers

Armaan Armaan 4 years, 4 months ago

3
  • 4 answers

Armaan Armaan 4 years, 4 months ago

79

Vipin Patel 4 years, 4 months ago

79

Miswah Anwar 4 years, 4 months ago

79

Sanjay Yadav 4 years, 4 months ago

79
  • 2 answers

Tejas Bramhe 4 years, 4 months ago

It cannot be solved by factorisation method

Armaan Armaan 4 years, 4 months ago

5/2,7/2
  • 2 answers

Vinod Khandelwal 4 years, 4 months ago

2x²-5x+7 2x²-(2+7)x+7 2x²-2x-7x+7 2x(x-1)-7(x-1) (2x-7)(x-1) HOPE IT HELPS U

Kashish Goyal 4 years, 4 months ago

2x²-5x+7 2x²-(2+7)x+7 2x²-2x-7x+7 2x(x-1)-7(x-1) (2x-7)(x-1) answer
  • 0 answers
  • 1 answers

Preeti Dabral 4 years, 4 months ago

The method of substitution involves three steps: Solve one equation for one of the variables. Substitute (plug-in) this expression into the other equation and solve. Resubstitute the value into the original equation to find the corresponding variable.

  • 0 answers

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