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player 1 need 7 runs player 2 need 3 runs player 3 need 11 runsplayer 4 need 1 run so all answers are correct if you think the answer is correct please like

21b - 32 + 7b - 20b

28b - 20b - 32

8b - 32

-1032-878= -154-154 + (-34) = -188

38+ (-87)= -49-134-(-49) =-86

Prime numbers are : 37,41,43,73,89.

Non prime numbers are :35,39,51,111

composite no. are 49 59 57 65prime 41 37 89 43 73non primev 111 39 51 35

prime - 41 37 89 43 73composite - 49 59 57 65

Answer => 2

Given : x = 8ab / (a + b) => x = 4a * 2b / (a + b)=> x / 4a = 2b / (a + b) - - - (1)

Similarly, x / 4b = 2a / (a + b) - - - -(2)

Now, using Componendo & Dividendo in (1) & (2)and adding left to left and right to right we get,(x + 4a) / (x - 4a) + (x + 4b) / (x - 4b) =(2a + a + b) / (2a - a - b) + (2b + a + b) / (2b - a - b)

=> (x + 4a) / (x - 4a) + (x + 4b) / (x - 4b) = (3a + b) / (a - b) + (3b + a) / (b - a )=> (x + 4a) / (x - 4a) + (x + 4b) / (x - 4b) = (3a + b) / (a - b) - (3b + a) / (a - b )=> (x + 4a) / (x - 4a) + (x + 4b) / (x - 4b) = (3a + b - 3b - a) / (a - b)=> (x + 4a) / (x - 4a) + (x + 4b) / (x - 4b) = (2a - 2b) / (a - b) = 2(a - b) / (a - b)

=> (x + 4a) / (x - 4a) + (x + 4b) / (x - 4b) = 2 . . . .(Answer)

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Area= theeta/360* π r^2

= 60/360* 22/7* 14*14102.62 cm^2

Perimeteras 2π r consist = 360°1° will consist= 2πr/360now 60° will consist= 2πr/360*60= 2π*14/628π/16cm

step by step

Solution:

Given;

Length of wall(l)= 15 m,Breadth of wall(b)= 10 m Height of wall(h)= 7 m

Total area to be painted= area of 4 walls + area of ceiling

= 2(l+b)h + lb

=2(15+10)7 + 15×10

= 2×25×7+ 150

=50× 7+150= 350+ 150=500

Total area to be painted=500m²

Given 100m² area can be painted from each can.

Number of cans Required=

Area of hall/ area of 1 can

= 500/100= 5

Hence, 5 cans are required to paint the room.

Given that AB and CD are two chords of a circle, with centre O intersecting at a point E.XY is a diameter passing through E, such that ∠ AEY = ∠ DEYDraw OP⊥ AB and OQ ⊥ CD.In right angled DOPE∠POE + 90° + ∠ PEO =180° (Angle sum property of a triangle)∴∠POE = 90° – ∠PEO = 90° – ∠AEY = 90° – ∠DEY= 90° – ∠QEO = ∠QOEIn triangles OPE and OQE,∠PEO = ∠QEO ∠POE = ∠QOE(Proved)OE = OE (Common side)∴ ΔOPE ≅ ΔOQE ⇒ OP = OQ (CPCT)Thus,AB = CD

1) 15x-8×= 7x2) 14a^2b-6a^2b= 8a^2b3) 10ab-(-6ab)= 16ab4) -14x^2y-8x^2y= -22×^2y

34696/16= 2 1 6 8 .3

option (iii) is correct gave likes to each other

34696/16 =21683 is correct answer

1

2

So the answer is y<x<z

18 is the correct answer

18 is the correct answer

answer is x=21 and y =24

answer is x=21 and y =24

answer is x=21 and y =24

Let a be any positive integer and b = 3a = 3q + r, where q ≥ 0 and 0 ≤ r < 3Therefore, every number can be represented as these three forms. There are three cases.Case 1: When a = 3q,Where m is an integer such that m = Case 2: When a = 3q + 1,a^3= (3q +1)^3a^3= 27q^3+ 27q^2+ 9q + 1a^3= 9(3q^3+ 3q^2+ q) + 1a^3= 9m + 1Where m is an integer such that m = (3q^3+ 3q^2+ q)Case 3: When a = 3q + 2,a^3= (3q +2)^3a^3= 27q^3+ 54q^2+ 36q + 8a^3= 9(3q^3+ 6q^2+ 4q) + 8a3= 9m + 8Where m is an integer such that m = (3q^3+ 6q^2+ 4q)Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m+8

a+c=bput the values b/2+b/2=bhence proved.