Maths Teachers in Gandhi Nagar bazar, Delhi

Don't waste time in taking common term in denominator. It is of no use. Use the same time for other useful purpose.

Answer:

We know that the diagonals of a parallelogram bisect each other. Given BD= 7cms, AC=4.8cms and AB= 5 cms. Let the diagonals AC and BD bisect each other at O. Because of bisection we have AO= AC/2 =  4.8/2 = 2.4 cms. Now proceed as follows

1.Draw line segment BD= 7cms.

2.Bisect BD at O.

3.With  O as centre and radius AO= 2.4 cms draw an arc.

4.With B as centre and radius AB = 5cms draw another arc intersecting the above arc at A.

5. Join  AB. Join AD

6.Produce AO to C such that AO=OC=2.4 cms.

7.Join BC and  CD..

ABCD is the required Parallelogram   

This doesn't contain any question!

This can be done by converting the limit to definite integral as shown in the pic given below

The given function Y is neither differentiable nor continuous at X=3. Because , the greatest int. func. Isn't continuous at any integer.and any discontinuous function is non-differentiable at the same point.

All the numbers from 9990 to 9999.

Option C is correct because ✓5 is irrational number and 3 is rational number then the sum of these numbers is irrational

here is the answer

Solving in the Substution Method we get:

5x+2y+13=0

∴5x+2y=-13

and 7x-5y+26=0

∴7x-5y=-26 Suppose: Consider; 5x+2y=-13→(1) and 7x-5y=-26→(2) Taking equation (1), we have 5x+2y=-13 ⇒5x=-2y-13 ⇒x=-2y-135→(3) Putting x=-2y-135 in equation (2), we get  7x-5y=-26 7(-2y-135)-5y=-26 ⇒-14y-91-25y=-130 ⇒-39y-91=-130 ⇒-39y=-130+91 ⇒-39y=-39 ⇒y=1→(4) Now, Putting y=1 in equation (3), we get x=-2y-13 x=-2(1)-135 ⇒x=-2-135 ⇒x=-155 ⇒x=-3 ∴x=-3 and y=1.

Solving in the Elimination Method we get:

5x+2y+13=0 ∴5x+2y=-13 and 7x-5y+26=0 ∴7x-5y=-26 5x+2y=-13→(1) 7x-5y=-26→(2) equation(1)×5⇒25x+10y=-65 equation(2)×2⇒14x-10y=-52  Adding ⇒39x=-117 ⇒x=-11739 ⇒x=-3 Putting x=-3 in equation (1), we have 5(-3)+2y=-13 ⇒2y=-13+15 ⇒2y=2 ⇒y=1 ∴x=-3 and y=1.

Solving in the Cross Multiplication Method we get:

5x+2y+13=0 ∴5x+2y=-13 and 7x-5y+26=0 ∴7x-5y=-26 5x+2y+13=0→(1) 7x-5y+26=0→(2) Here, a1=5,b1=2,c1=13 a2=7,b2=-5,c2=26  x=(b1⋅c2-b2⋅c1)/(a1⋅b2-a2⋅b1) =(2)(26)-(-5)(13)(5)(-5)-(7)(2) =(52)-(-65)(-25)-(14) =117-39 x=-3 y=(c1⋅a2-c2⋅a1)/(a1⋅b2-a2⋅b1) =(13)(7)-(26)(5)(5)(-5)-(7)(2) =(91)-(130)(-25)-(14) =-39-39 y=1 ∴x=-3 and y=1.

It has two answers either add x-2 or x²+3x-5