Question

Find (2x + 3y)2 using algebraic identities.
2. Using suitable identities find (1092)2.
3. Using the identity (a-b)2 = a2 � 2ab + b2, find (5a � 7b)2.
4. Find 194 * 206 using suitable identity.
5. Use a suitable identity to find the product of (3a � 1/3)(3a � 1/3).
6. The length and breadth of a rectangle are 3x2 � 2 and 2x+ 5 respectively. Find its area.
Can anyone pls help me to solve these questions?

Answer 1

(2x+3y)^2 =2x^2+3y^2+2(2x)(3y)
=4x^2+9y^2+12xy

2)(1092)^2=(1100-8)^2
(1100-8)^2=(1100)^2+8^2-2(1100)(8)
=(1210000+64)-17600
=1210064-17600
=1192400

3)(5a-7b)^2=5a^2+7b^2-2(5a)(7b)
=25a^2+49b^2-70ab

4)194×206=(200-6)(200+6)

(200-6)(200+6)=(200)^2-(6)^2
=40000-36
=39964
5)(3a-1/3)(3a+1/3)=3a^2-(1/3)^2
=9a^2-1/9

6)(3x^2-2)(2x+5)

Area of rectangle=l×b
=(3x^2-2)(2x+5)
=3x^2(2x+5)-2(2x+5)
=6x^2+15x^2-4x-10
=21x^2-4x-10

Answer 2

We know that according to Ohm’s law

V = IR

where

V= potential difference

I= Current

R = Resistance

We can also modify the equation as

I=V/R ——- (i)

Now given that the potential difference across the two ends of the component decreases to half

∴ So let the new potential difference be Vʹ=V/2

Resistance remains constant across the electrical component

So the new current is drawn through the electrical component is Iʹ = Vʹ/R

= (V/2)/R {Substituting Vʹ=V/2 in the above equation}

= (1/2) (V/R)

= (1/2) I = I/2

Therefore, the amount of current flowing through the electrical component is reduced by half.