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Let the width of the room be 'b' m

Length of the room = 15 m

Area of the room = 15 × b = 15b m2

Total cost of carpeting = Rs 19200

Cost of carpet per metre = Rs 80

Length of the carpet = 19200 / 80 = 240 m

Width of the carpet = 75 cm = 0.75 m

Area of the carpet = 240 x 0.75

Area of the room = Area of the carpet

⇒15b = 240 × 0.75

⇒ b = (240 × 0.75) / 15 = 12

Therefore, width of the room is 12 m.

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(a)^-2/3*b * (b)^-2/3*c * (c)^-2/3*a

= (a)^[-2/3 + 1] * (b)^[-2/3 + 1] * (c)^[-2/3 + 1]

= a^1/3 * b^1/3 * c^1/3

= (abc)^1/3

a = 115° + 100° + 70° + a = 360° a = 275°- 360°a = 85°

115°+100°+70°+a=360°275°+a = 360°a=360°-275°a=85°

sum of this figure is 360 ,115+100+70+a=360 ,a =360-275 ,a =275

the total no. of sum of angles in a polygon is 360.

115+100+70+X= 360X= 85.

a+115°+70°+100°=360° ( A.S.P of parallelogram)a+285°=360°a=360°- 285°a=75°

115+100+70+a=360285+a=360a=360-285a=75

root(2):(1+root(3)::root(6):x

root(2)*x = root(6)*(1+root(3))

x = root(2)*root(3)(1+root(3))/ root(2)

x = root(3)(1 + root(3))

x = root(3) + 3

a) it has four lines of symmetryb) it has only one line of symmetryc) it has four lines of symmetry

First term = a , common difference = d

Sm = n

→m/2(2a + (m-1)d) =n …….(1)

Sn = m

→n/2(2a + (n-1)d) =m ……..(2)

Eq. (2) – (1)

2a(n-m) + d(n2-n-m2+m)=2m-n

n-m{2a + d(n+m-1)}=-2(n-m......(3)

2a + d(n+m-1) = -2

Now, Sm+n = m+n/2{2a+(m+n-1)d}

=-(m+n)(using eq. 3)

First term = a , common difference = d

Sm = n

→m/2(2a + (m-1)d) =n …….(1)

Sn = m

→n/2(2a + (n-1)d) =m ……..(2)

Eq. (2) – (1)

2a(n-m) + d(n2-n-m2+m)=2m-n

n-m{2a + d(n+m-1)}=-2(n-m) …….(3)

2a + d(n+m-1) = -2

Now, Sm+n = m+n/2{2a+(m+n-1)d}

=-(m+n) (usingeq. 3)

thanks

Given sum of m terms of and A.P. is the same as the sum of its n terms

We know sum of n terms of AP= n/2(2a+(n-1)d)Therefore sum of m terms =m/2(2a+(m-1)d)----(1)

sum of n terms=n/2(2a+(n-1)d)-----(2)

According to the problem (1)=(2)

m/2(2a+(m-1)d)=n/2(2a+(n-1)d)

2ma+(m2-m)d=2na+(n2-n)d

(2m-2n)a=(n2-n)d-(m2-m)d

2(m-n)a=n2d-nd-m2d+md

2a(m-n)=-d(m-n)(m+n-1)

a=-d/2(m+n-1)----(3)

From (3) we got a,sum of m+n terms = m+n/2(2(-d/2(m+n-1)+(m+n-1)d)which is 0

Hence sum of its(m+n) terms is Zero

I want to know the answer

bahi photo sahi se le lo

apap

(a+b)(a+c)= a^2+ac+bc+abx^2+2x+16x+32x^2+18x+32 Answer

48ac+68bcwill be the answerplease like the solution 👍 ✔️

2A = 3B = 4C = k A = k/2 B = k/3 C = k/4 A : B : C = k/2 : k/3 : k/4 = 6 : 4 : 3

Opposite angle in cyclic have sum 180°So,angle A + angle C = 180°4x + 25 + 5x + 20 = 180°9x + 45 = 180°9x = 180° - 45°9x = 135°x = 135°/ 9x = 15°So, angle C = 5x + 20 = 75 + 20 = 95°(4) option is correct

Factorise:1.x^2-36. =x^-6^2=(x+6)(x-6). (using a^2-b^2)

x^2-36x^2-6^2=(x+6)(x-6)

2.4a^2-9=(2a)^2-(3)^2=(2a+3)(2a-3)

3a(2b+5c) +3c(2a+2b) = 6ab+15ac+6ac+6bc = 6ab+21ac+6bc = 6(ab+3.3ac+bc)

plz any other reply

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Can you please specify the question ! :)

It's written in fraction form

like this

Thank you so much!

we take all value in a formsso, b/4=a/3b=4a/3similarly, c=5a/3

given, a-2b+3c/x = a/3so put the value of 'b' and 'c' in 'a' form, (a-8a/3+15a/3) / x =a/3on solving it, we get, 3a-8a+15a / 3x = a/310a/x=ax=10a/ax=10(Answer) _____________________

(a+3b+4c+4a-b+9c-2b+3c-a)-(2a-3b+4c)=(4a+4c)-(2a-3b+4c)=4a+4c-2a+3b-4c=2a+3b

waste not the correct a ms pulka

x = 2 is an equation in one variablex+y = 6 is an equation in two variablewhere x and y are two variable

Thankyou 😊😊

x - 2 = 7Given, x – 2 = 7By adding 2 to both sides, we getx – 2 + 2 = 7 + 2⇒ x = 7 + 2⇒ x = 9y + 3 = 10Given,y + 3 = 10By subtracting 3 from both sides, we gety + 3 – 3 = 10 – 3⇒ y = 10 – 3⇒ y = 7