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a) y + 120 = 180 linear pair y = 180-120 = 60 x + y + 50 = 180 x + 60 + 50 = 180 x = 180-110 x = 70

b) y = 80 vertically opposite angles

x + y + 50= 180 x + 80 + 50 = 180-130 x = 50

c) y + 50 + 60 = 180 y = 180-110 y = 70

x + y = 180 linear pair x = 180 - 70 x = 60

1)area=h×b/2when b=4.8m then h=6marea=4.8×6/2=14.4m^22)now area=14.4m^2b=6.4marea=h×b/214.4=h×6.4/2h=4.5m

Answer is wrong correct answer is 72

since XA||BCso ang(a)=ang(XAB)....... {alternate anglesa=50°similarly ang(b)=130°

Draw the diagram I describe and follow the calculation as set out below. Let O1, O2 be the two centers distance 'd' apart, and let r1, r2 be the radii of the circles with r2>r1. Join O1, O2 and draw the radii from the two centers to the points of contact with the common tangent of length 11 units. From O1 draw a perpendicular to the radius r2 (just drawn) to meet this radius at point S. Now we have a right angled triangle O1 S O2, with the shorter sides of length 11 and (r2-r1) and the hypotenuse equal to 'd'. Then by Pythagoras: d^2 = 11^2 + (r2-r1)^2 ......(1) We now turn to the transverse tangent of length 7 units. Draw the lines from the centers O1, O2 to meet this tangent. Now from the foot of the r2 radius where it meets the tangent draw a line parallel to 'd' (line joining centers) until it meets the line of r1. This will require you to extend r1 backwards to the other side of O1. Let T be the point where the lines meet. We now have another right angled triangle, with vertices at T and the two points of contact of the tangent with the circles. The shorter sides are of length 7 and (r2+r1), and the hypotenuse is 'd'. Again applying Pythagoras we get: d^2 = 7^2 + (r2+r1)^2 .......(2) Now subtract (2) from (1) and we have: 0 = 11^2 - 7^2 + (r2-r1)^2 - (r2+r1)^2 0 = 121 - 49 + r2^2 - 2r2.r1 +r1^2 - r2^2 - 2r2.r1 - r1^2 0 = 72 - 4r2.r1 4r2.r1 = 72 r2.r1 = 18 And so the product of the radii is 18.

Weknowthatlengthofthetangentsdrawnfromanexternalpointtoacirclearealwaysequal.

So,PQ=PTandPR=PT.

Hence,PQ=PT=PR

Now,PQ=3.8cm

So,PT=3.8cm

Now,PR=3.8cmn

ow,QR=PQ+PR=3.8+3.8=7.6cm

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LHS = Tan^3A / ( 1+ Tan^2A) + Cot^3A / (1 + Cot^2a)

= Tan^3A / Sec^2A + Cot^3A / Cosec^2A

= (sin^3A/cos^3A) / (1 / Cos^2A) + (Cos^3A/Sin^3A) / (1 / Sin^2A)

= Sin^3A/CosA + Cos^3A/SinA

= (Sin^4A + Cos^4A) / SinA.CosA

= [ (Sin^2A + Cos^2A)^2 - 2Sin^2A.Cos^2A] / SinA.CosA

= ( 1- 2Sin^A.Cos^A)/ SinA.CosA

RHS = SecA CosecA - 2sinAcosA

= 1/CosA . 1/SinA - 2SinACosA

= (1 - Sin^2A.Cos^2A) / sinAcosA

Hence LHS = RHS (PROVED)

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tq

y=50°(corresponding angle)x+y+30=180°(sum of angles of triangle)then x=100°x=z=100°(corresponding angle)

thanks 😊

(16 sqrt2+1)/8 rationalizing it gives (16×16×2-1)/8(16 sqrt2+1)511/8(16 sqrt2+1)

7x-10y=4...eq(1)12x+18y=1...eq(2)

Multiply eq.1 by 18Multiply eq.2 by 10

126x-180y=72....eq(3)120x+180y=10......eq(4)

add eq.(3)&(4)

246x=82x=3

y=17/10

To find the value of 4x+6y

substitute value of x and y

12+10.222.2

To find the value of 8y-x

13.6+316.6

4x+6y=22.28y-x=16.6

value of 4x+6y=22.28y-x=16.6

3 will be the answer

4x + 6y=22.2 8y-x =16.6

4x +6y=22.28y-x=16.6

Length of latus rectum is = 4a

=> (7,5) -(7,3) = 4a=> 4a = √(7-7)²+(5-3)² = 2=> a = 2/4 = 1/2

now focus will be have y coordinate as mid point of (7,5) and (7,3) = (7,4)

and x coordinate of vertex is "a" distance from focus

so, the x coordinates are = 7±1/2 = 13/2 and 15/2

so, possible points are (13/2, 4) and (15/2,4)

option A and D

Length of latus rectum is = 4a

=> (7,5) -(7,3) = 4a=> 4a = √(7-7)²+(5-3)² = 2=> a = 2/4 = 1/2

now focus will be have y coordinate as mid point of (7,5) and (7,3) = (7,4)

and x coordinate of vertex is "a" distance from focus

so, the x coordinates are = 7±1/2 = 13/2 and 15/2

so, possible points are (13/2, 4) and (15/2,4)

option A and D

l0

Given : 5y² -7y + 1

a and b are zeroes so,

a + b = -(-7)/5 = 7/5

ab = 1/5

So, 1/a + 1/b

(a + b)/ab

(7 × 5)/5 = 7

Ans: 7

no this is wrong

can u try again

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8m³ + 9m² + 5m - 2 - (5m³ -3m² + 7m -8)8m³ - 5m³ + 9m² + 3m² + 5m - 7m -2 + 83m³ + 12m² -2m + 6

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3x^2+7x+4alpha+ beeta= -7/3= -b/aalpha * beeta= 4/3= c/a

Observe :

7 y × 6 ------- yyy

7 4 × 6 ------- 444

So y value is 4 option (1).

The correct answer is 9,801

9801 is the correct answer of the question

9801 is the correct answer

1

find the equation of circle whose diameter is join of the point (x₁,y₁) and (x₂, y₂) . Let A ≡ (x₁,y₁) and B ≡ (x₂,y₂)Then, equation of circle is given by (x - x₁)(x - x₂) +(y - y₁)(y - y₂) = 0 [ by formula ] x² - (x₁ + x₂)x + x₁x₂ + y² - (y₁ + y₂)y + y₁y₂ = 0x² + y² -(x₁ + x₂)x - (y₁ + y₂)y + x₁x₂ + y₁y₂ = 0

Hence, the equation of circle is x² + y² -(x₁ + x₂)x - (y₁ + y₂)y + x₁x₂ + y₁y₂ = 0

eska radius kya hoga bhai

8-7y=1-7y=1-8-7y=-7y=1hence y=1 is the root

Given : Opposite sides are equal so opposite angles are also equal therefore,x = y .sum of angle in a triangle is 180 degree so,

x + x + 30° = 180°

2x = 180° - 30°

2x = 150°

x = 150°/2

x = 75° so, y = 75°