Maths Teachers in Bazar Road, 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   

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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