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  • 1 answers

Shivani Lama 7 years, 5 months ago

kon sai
  • 2 answers

Nachiketh Neelaraddi 7 years, 5 months ago

We can solve quadratic equation by taking our own value of X and Y

Shivani Lama 7 years, 5 months ago

quadratic equation solve then put a graph
  • 1 answers

Sia ? 6 years, 6 months ago

Let digit at unit’s place = x and digit at hundred’s place = y
{tex}\therefore{/tex} Middle digit ={tex} x + y + 1{/tex}
Number = {tex}100y + 10(x + y + 1) + x{/tex}
Number obtained by reversing the digits = {tex}100x + 10(x + y + 1) + y{/tex}
ATQ.,
{tex}x + y + (x + y + 1) = 17{/tex}
{tex}\Rightarrow{/tex} {tex}2x + 2y = 16{/tex} ..(i)
and {tex}100x + 10(x + y + 1) = x - 396{/tex} ..(ii)
{tex}\Rightarrow{/tex} {tex}99x - 99y = -396{/tex}
{tex}\Rightarrow{/tex} {tex}x - y = -4{/tex}
By Solving equation (i) and (ii), we get
x = 2, y = 6 and Number = 692

  • 3 answers

Alok Alok 7 years, 5 months ago

Rational Number

Monu Monu 7 years, 5 months ago

Rational no.

Parteek Parteek 7 years, 5 months ago

Rational
  • 0 answers
  • 2 answers

..... ...... 7 years, 5 months ago

24√5

Gandham Cherishma 7 years, 5 months ago

6√5+7
  • 3 answers

Anchal Singh 7 years, 5 months ago

Let Us assume that root 2 is a rational no. Than, exist of Co-prime a and b. B is not equal to zero Root 2 = a/b Since a/b is rational no. But root 2 is irrational no so root 2 is irrational no.

Amar Nagargoje 7 years, 5 months ago

See NCERT pg no-12

Goutam Kumar Kushwaha 7 years, 5 months ago

Lat wrong assumption
  • 1 answers

Ayush Dalmia 7 years, 5 months ago

sin(90-17)+sin17 cos17+sin17
  • 1 answers

Sia ? 6 years, 6 months ago

We have,
{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}
In order to factorize {tex}(a + b)^2x^2 - 4abx - (a - b)^2{/tex}, we have to find two numbers 'l' and 'm' such that.
l + m = -4ab and lm = -(a + b)2(a - b)2
Clearly, (a - b)2 + [-(a + b)2] = -4ab and lm = -(a + b)2(a - b)2
l = (a - b)2 and m = -(a + b)2
Now,
{tex}(a + b)^2x^2 - 4abx - (a - b)^2 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}(a + b)^2x^2 - (a + b)^2x + (a - b)^2x - (a - b)^2 = 0{/tex}
{tex}\Rightarrow{/tex} (a + b)2x [x - 1] + (a - b)2[x - 1] = 0
{tex}\Rightarrow{/tex} (x - 1)[(a + b)2x + (a - b)2] = 0
{tex}\Rightarrow{/tex} x - 1 = 0 or (a + b)2x + (a - b)2 = 0
{tex}\Rightarrow{/tex} x = 1 or {tex}x = - \frac{{{{(a - b)}^2}}}{{{{(a + b)}^2}}}{/tex}

  • 4 answers

Anchal Singh 7 years, 5 months ago

P/ h

Amar Nagargoje 7 years, 5 months ago

P/h

Nikki Kushwah 7 years, 5 months ago

Perpendicular /Hypotenus P/H

Arpit Singh 7 years, 5 months ago

perpendicular/hypotenus
  • 1 answers

Kushagra Kumar Ghai 7 years, 5 months ago

6x+10y-4=0..............eq1
3x+5y+11=0.............eq2
now subtract eq2 from eq1
gives 3x+5y=-14.............eq 3
hat is not possible so i recommend u to recheck the question

  • 1 answers

Sanjana Yadav 5 years, 8 months ago

Hello....
  • 0 answers
  • 2 answers

Mh Kamali 7 years, 5 months ago

Let the fixed charge for 1st two days is Rs.x and additional charge for each day is Rs.y For Latika, A/q x+5y=22 ------(1) For Anand, Also, x+3y=16 ------(2) On subtracting equation (2) from equation(1) We get, (x+5y)-(x+3y)=22-16 => x+5y-x-3y=6 => 2y=6 => y=6/2 =>y=3 Now, put value of y in equation (2) x+3×3=16 => x+9=16 => x=16-9 => x=7 Hence, fixed charge for 1st two days is Rs.7 and charge for each day is Rs.3

Sohan Pavan 7 years, 5 months ago

Let the fixed charge be x and let the additional charge be y
  • 1 answers

Sia ? 6 years, 6 months ago

Thales is credited with the following five theorems of geometry:

  • A circle is bisected by its diameter.
  • Angles at the base of any isosceles triangle are equal.
  • If two straight lines intersect, the opposite angles formed are equal.
  • If one triangle has two angles and one side equal to another triangle, the two triangles are equal in all respects. (See Congruence)
  • Any angle inscribed in a semicircle is a right angle. This is known as Thales' Theorem.
  • 0 answers
  • 0 answers

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