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i) A intersection B is B as B is subset of Aii) A intersection C is C as C is subset of A.iii) A intersection D is D as D is subset of A

(-3/5) * (-10/-9) * ( 21/-4) * (-6)

= (-3/5 * - 10/-9) * (21/-4 * - 6)

= (-2/3) * (21*3/2)

= - 21

when vectors are perpendicular then dot product is 0 so here

12 +2× lamda - 22k² = 0so lamda = (22k² -12)/2 = 11k² -6

3+6=216×3+3=218+11=? 8×3+11=35

tq

We have R = {(x,y) :x ∈ N, y ∈ N, 2x + y = 24} y = 24 - 2xAs y ∈ NSo, 24 - 2x >= 1-2x >= - 24x =< 12Domain = {1,2,3,.....,12} So Domain is first 12 natural numbers 2x + y = 242x = 24 - yx = (24 - y) /2 As x ∈ N(24 - y) /2 >= 124 - y >= 2y =< 22So, range is first 22 natural numbers

4/9*(-3/7) +(3/14) +(-3/7)*(2/9)= -4/21 +3/14 -2/21= 3/14-6/21= 9/42 -12/42....... ( L.C.M of 14 and 21 is 42)= -3/42 = -1/14 ✓

X+7 -8x/3 = 17/6 – 5x/2

X+7 -8x/3 +5x/2= 17/6

[ Transposing 5x/2 to L.H.S]

LCM of the denominators 2, 3 is 6.

[6(x+7)- 2(8x)+3(5x)]/6=17/6

6x+42-16x+15x=6× (17/6)

6x-16x+15x+42= 17

-10x+15x=17-42

[Transposing42 to R.H.S]

5x=-25

x= -25/5

X= -5

2x - 6y = 12..... (1)5x + 8y = 12....(2)

Multiply eq(1) by 5 and eq(2) by 210x - 30y = 6010x + 16y = 24

Subtract these equations-46y = 36y = - 18/23

Put value of y in eq(1)2x - 6*(-18)/23 = 12x = 6 - 54/23x = 138-54/23 = 84/23

Ramu can plant more trees

1 and 3 are in inverse proportion

the answer of the question is -24

the answer of the question is-24

The answer of the question is _ 24

Ratio of the sides of the triangle = 12 : 17 : 25

Let the common ratio be x then sidesare 12x, 17x and 25x

Perimeter of the triangle = 540cm

12x + 17x + 25x = 540 cm

⇒ 54x = 540cm

⇒ x = 10Sides of triangle are,

12 × 10 = 120cm

17x = 17 × 10 = 170cm

25x = 25 × 10 = 250cm

Semi perimeter of triangle(s) = 540/2 = 270cm

Using heron's formula,Area of the triangle = √s (s-a) (s-b) (s-c) = √270(270 - 120) (270 - 170) (270 - 250)cm2

= √270 × 150 × 100 × 20 cm² = 9000 cm²

Ratio of the sides of the triangle = 12 : 17 : 25Let the common ratio be x then sides are 12x, 17x and 25xPerimeter of the triangle = 540cm12x + 17x + 25x = 540 cm⇒ 54x = 540cm⇒ x = 10Sides of triangle are,12x = 12 × 10 = 120cm,17x = 17 × 10 = 170cm,25x = 25 × 10 = 250cmSemi perimeter of triangle(s) = 540/2 = 270cmUsing heron's formula,Area of the triangle = √s (s-a) (s-b) (s-c) = √270(270 - 120) (270 - 170) (270 - 250)cm^2 = √270 × 150 × 100 × 20 cm2 = 9000 cm^2.Ans

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12x+17x +25x =540cm 54x =540cm x =540/54x =10

1) Merits:1. It is easiest to calculate and simplest to understand even for a beginner.2. It is one of those measures which are rigidity defined.2) Demerits:1. Range is not based on all the terms. Only extreme items reflect its size. Hence range cannot be completely representative of the data as all other middle values are ignored.2. Due to above reason range is not a reliable measure of dispersion.3) a quantity expressing by how much the members of a group differ from the mean value for the group.

what is green revolution?

let the common ratio be xthen sides are 12x 17x and 25x perimeter of the triangle =540cm 12x+17x+25x=540cm =54x =540cm=x=10 sides of triangle are 12 x =12'×10=120 17×10=170 25×10 =250 semi perimeter of triangle 540/2= 270 area of the triangle =square s(s-a)(s_b) (s_c) square 270(270_120)(270_170) and (270_250)×150×100×20cm2=9000cm2

if we put x=0 it will be in 0/0 form so using L'hospital rulelim x->0 (e^x-e^(-x))/(-3sin3x+5sin5x)if we put x=0 it will be in 0/0 form so using L'hospital rulelim x->0 (e^x+e^(-x))/(-9cos3x+25cos5x)=2/(-9+25)=2/16=1/8

Given : x^2 + 1/x^2 = 98 ---- ( 1 )

We take :(x + 1/x)^2 = x^2 + 1/x^2 + 2( x )(1/x)⇒( x + 1/x)^2 = x^2 + 1/x^2 + 2,

Substitute value from equation 1 and get ⇒( x + 1/x)^2 = 98 + 2 ⇒( x + 1/x)^2 = 100⇒( x + 1/x) = 10

Taking whole cube on both hand side and get ⇒( x + 1/x)^3 = 10^3⇒x^3 + 1/x^3 + 3 (x)^2 (1/x) + 3(x)(1/x)^2 = 1000⇒x^3 + 1/x^3 + 3x + 3/x = 1000⇒x^3 + 1/x^3 + 3(x + 1/x) = 1000⇒x^3 + 1/x^3 + 3 × 10 = 1000⇒x^3 + 1/x^3 + 30 = 1000⇒x^3 + 1/x^3 = 970

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