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-2/3*3/5 + 5/2 - 3/5*1/6

= - 2/5 + 5/2 - 1/10

Take LCM

= (-4 + 25 - 1)/10

= 20/10

= 2

(3x+2y)(4x+3y) = 3x*4x + 3x*3y + 2y*4x + 2y*3y = 12x*x + 9xy + 8xy + 6y*y = 12x*x + 17xy + 6y*y

Proofs in maths generally have three methods

1. Contradiction

2. Contrapositive

3. Mathematical induction

First we should remember that percent means per hundred.

So , as per the problem given ,

A student scored 180 out of 900 . So his percentage will be :-

900 --- 180100 --- ?=( 180 ÷ 900 ) × 100= 0.2 × 100 = 20 •/•

Hence his percentage will be 20 .

15/8 is the correct answer

15/8 is correct answer

15/8 is the right answer

15/8 is the right answer.

We know that, the lengths of tangents drawn from an external point to a circle are equal.∴ TP = TQIn ΔTPQ,TP = TQ⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)We know that, a tangent to a circle is perpendicular to the radius through the point of contact.OP ⊥ PT,∴ ∠OPT = 90º⇒ ∠OPQ + ∠TPQ = 90º⇒ ∠OPQ = 90º – ∠TPQ⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)From (1) and (2), we get∠PTQ = 2∠OPQ

In ABC and ABD AC= ADDAB= BAC AB = ABby SAS rule ABC and ABD are congruent so BC = BD by CPCT.

(2/9)^n + (2/9)^-6 = 1(2/9)^n-6 = (2/9)^0n-6=0n=6

so we can useL'Hospital's rule.

d/dx(numerator ) = sec ^2 x -1 = tan ^2 x d/dx(denominator = (x^2 sec ^2 x+ 2x tan x)

so tan ^2 x/((x^2 sec ^2 x+ 2x tan x)

= (tan x/x) (tan x/( x sec^2 x+ 2 tan x)

tan x/x -> 1 and

tan x/( x sec ^2 x + 2 tan x)

is reciprocal of

( x sec ^2 x + 2 tan x)/ tan x = ( x (1/cos^2x tan x) + 2 = x/(cos x sin x) + 2 = (x/sin x)(1/cos x) + 2 = 1 .1 + 2 = 3

so tan x/( x sec ^2 x + 2 tan x) = 1/3

so tan ^2 x/((x^2 sec ^2 x+ 2x tan x) or given expression = 1/3

the answer is not same. the answer is 72

1cm=0.01m 25cm=0.25m required tiles = (16×12)/(0.25×0.25) = 3072

thanx

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thnku

54 is the correct answer

6+49-6+5=55-6+5=49+5=54

6+49-6+5=54.......hope this helps

10C7 = 10×9×8/3×2×1 = 720/6 = 120

Let the number of mangoes be x.

when x ÷ 2 leaves remainder as 1.when x ÷ 3 leaves remainder as 2.when x ÷ 4 leaves remainder as 3.when x ÷ 5 leaves remainder as 4.when x ÷ 6 leaves remainder as 5.when x ÷ 7 leaves remainder as 0.⇒ x is divisible by 7.

The remainder in each case is 1 less than the divisor.⇒ (x + 1) is the LCM of 2, 3, 4, 5 and 6.

LCM of 2, 3, 4, 5 and 6 = 60.

If x + 1 = 60, then x = 59.But 59 is not divisible by 7.

If x + 1 = 120, then x = 119.

119 is divisible by 7 also it satisfies all the conditions.

Hence the number of mangoes = 119.

excellent thanks

conjugate of i) 3+i = 3-i ii) 3-i = 3+i iii) -√5-√7i = -√5+√7iiv) -√-5 or -√5i has conjugate of √5iv) 5i has conjugate of -5i

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solve no 3 and 4

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first solve

On 1st Jan,She saved 10 Rs. Next Day 11 Rs. Next Day 12 rs

So money saved by her= 10 + 11 + 12 + 13 + .........................+ (till 31st Dec)

Now, Above Sum is the sum of a Arithmetic Progression witha = 10d = 1n = 366 ( since year 2016 is a Leap Year)

=>Sum of an A.P. = (n/2){ 2a + (n-1)d} = (366/2){ 20 + 365 } = 70455

So, she saved Rs. 70455 till the end of the year

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3-i3+I-√5+√7i is the correct answer of the given question

(3-i)+(3+i)3^2-i^29-1=8

[3-i]+[3+i]3^2-i^29-1=8

will you plzz find the amplitude of the 5th que

(x+5)(x-7) + 35x² -7x + 5 x -35 + 35x² -2xx(x-2)

let the number be x ,so ,x × 1/4 = 100x/4 = 100x = 400

First of all know the definition of complex numbers. They consist of real, imaginary and real+imaginary.

1st case….Real number

Let z1= -5

Absolute value is 5

And let z2 = 5

Again absolute value is five. So its not necessary.

Similarly Its applicable to iota, real + imaginary numbers also

No it is not necessary tgat z1=z2

As if z1 value is +-1 and z2=+-2

then for some case +1is not equal to -1

no it may or may not equalbecause if we take minus 5 as Z1 and 5 asZ2 then the modulus of minus 5 is equal to the modulus of 5