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xcos=1ysin=1

ysin/xcos=1ytan/x=1tan=x/y

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3 2/3+1 5/6 +2=9+2/3 +6+5/6+2=11/3+11/6+2=11*2/3+11/6+2=22/6+11/6+2=33/6+2=11/2+215/2

We know that tangent is always perpendicular to the radius at the point of contact.

So, ∠OAP = 90

We know that if 2 tangents are drawn from an external point, then they are equally inclined to the line segment joining the centre to that point.

So, ∠OPA = 12∠APB = 12×60° = 30°

According to the angle sum property of triangle-

In ∆AOP,∠AOP + ∠OAP + ∠OPA = 180°⇒∠AOP + 90° + 30° = 180°⇒∠AOP = 60°

So, in triangle AOP

tan angle AOP = AP/ OA

√ 3= AP/a

therefore, AP = √ 3a

hence, proved

First 10 prime numbers are: 2,3,5,7,11,13,17,19,23,29

Since all the numbers occur only once, there is no mode of the first 10 prime numbers.

5x^2y - 15xy^25xy(x-3y)

(2x+y)(2x-y) - (2x+y)^2= (2x+y)(2x - y - 2x - y)= (2x+y)(-2y)= - 4xy - 2y

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As you know that √2 & √3 are irrational number, so we can represent them in non repeating decimal form i.e

√2=1.4142135623730950488…[Square root of 2]

√3=1.7320508075688772935…[Square root of 3]

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(x+1)(x+1) = 2(x-3) x*x + 2x + 1 = 2x - 6 x*x = -7 x = √7i

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D is the right answer

(d) option is a correct answer

Answer : (d) option is right

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range=highest value-lowest value

first 10 prime numbers:

2,3,5,7,11,13,17,19,23,29

=29-2

=27

theefore, the range of first ten prime no.s is 27

the range of first prime number is 27

56 = 2 × ( 4x + 3x)56 = 2 × 7x14x = 56x = 56/14 = 4x = 4

ilu

56 × 2 = 112 is the answer of the question

52 * 2 = 112 is the answer

56×2=112......................

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Solution :

1 ^ 3 + 1 = 1 * 1 * 1 + 1 = 2.

2 ^ 3 + 2 = 2 * 2 * 2 + 2 = 10.

3 ^ 3 + 3 = 3 * 3 * 3 + 3 = 30.

4 ^ 3 + 4 = 4 * 4 * 4 + 4 = 68.

5 ^ 3 + 5 = 5 * 5 * 5 + 5 = 130.

Answer = 130.

please do last two questions

thx

A) I² = I , is the correct option as unit matrix is Identity therefore I × I = I

Answer of this question is x=11

Scale factor for the sides of these triangles

k=3/2

Therefore the ratio of area will be:

k^2 =Area Triangle A/Area triangle B

k^2=(3/2)^2

=9/4

Hence if sides of two similar triangles are in the ratioa:b, their areas are in the proportiona^2:b^2

let the no..s be 3x and 2x

2x+3x=1805x=180x=180/5x=36

3x=3×36=1082x=2×36=72

Given⇒Area of the two similar triangles are 225 cm² and 81 cm².

By using the theorem,When the two triangles are similar, then the ratio of there areas is equal to the ratio of the square of there corresponding sides.

∴ Area of theΔABC/Area of theΔPQR = AB²/PQ²∴ 225/81 = AB²/12²⇒ AB² = 225× 144/81⇒ AB = √400∴ AB = 20 cm.

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Answer: speed of the passenger train is 40km/h

Speed of the express train is 60km/h

Step-by-step explanation: Let the speed of the passenger train is x km/h

Speed of the express train is y km/h

A/c to question,

Passenger train takes 2 hours more than an express train to travel a distance of 240km. Also the speed of the express train is more than that of the passenger train by 20km/h

e.g., x + 20 = y .............(1)

240/x = 240/y + 2 .........(2) [ time = distance/speed ]

⇒ 240/x - 240/(x +20) = 2 [ putting equation (1) in equation (2); ]

⇒ 240 [ (x + 20 - x)/x(x +20) ] = 2

⇒ 120 [ 20/(x² + 20x) ] = 1

⇒ 120 × 20 = x² + 20x

⇒ x² + 20x - 2400 = 0

⇒ x² + 60x - 40x -2400 = 0

⇒ x(x + 60) -40(x + 60) = 0

⇒ (x + 60)(x - 40) = 0

⇒ x = 40 and -60 but speed can’t be negative so, x ≠ -60 km/h

Hence, speed of the passenger train is 40km/h

And speed of the express train is 60km/h

The two points lying in the second quadrant are

(-1,+1),(-4,+5), (-2,1)

Thus, if the abscissa of a point (x coordinate) of a point is zero, the point lies on the y axis. When the abscissa is 0, it means that it is at the origin. On the other hand, the abscissae on the right and left sides of the x axis have positive and negative values respectively.

let the point be (a,b)and the slope of function y = x³ is given by dy/dx = 3x²

now dy/dx at (a,b) = 3a²so from equation slope "3a²" = b

=> 3a² = b => a² = b/3 ; a = ±√b/3.

therefore points are (+√b/√3, b) and (-√b/√3,b)

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In diagram pic