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B*B - 2*2 using formula a*a - b*b = (a+b)(a-b) we can write following = (B+2)(B-2)

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tan(A+ B)=V3=tan 60°\•, A+B=60°, tan(A-B)=1/V3= Tan 30°= A-B=30°, A+B=60 A-B=30/2A=90;A=45°,B=15°

A45;B15 by using trigonometry

tan(A+B)=√3= tan 60°A+B= 60-----(1)tan(A-B)= 1/√3= tan 30°A-B= 30---(2)A+B= 60--(1)by adding 2A= 90A= 90/2= 45°A+B= 60°45+B= 60B= 60-45= 15°

Let present age of man be x

Then,As per given condition(x + 40) = (x - 32)^2x + 40 = x^2 + 1024 - 64xx^2 - 65x + 984 = 0x^2 - 41x - 24x + 984 = 0x(x - 41) - 24(x - 41) = 0(x - 24)(x - 41) = 0x = 24, 41

Therefore,Present age of man is 41 years

3x^2 - 2x - 8 = 0

We know, sum of zeros for equation ax^2 + bx + c = 0 is equal to - b/a

Therefore,Sum of zeroes for equation 3x^2 - 2x - 8 = 0 is equal to-(-2)/3 = 2/3

For equal rootsD= 0x^2-2x-k= 0(-2)^2= 4k4= 4kk= 1

x + 1 is factor of p(x) = 3x² - kx Then by reaminder theorem p(-1) = 0So,p(-1) = 3(-1)² - k (-1) = 03 + k = 0k = -3

√{(8/15 + 15/8) ÷ (5/6)}

= √{64+225/120) ÷ (5/6)}

= √{289*6/120*5)

= √(1734/600)

= √(2.89)

= 1.7

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First, we'll assume that √p + √q is rational, where p and q are distinct primes√p + √q = x, where x is rational

Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.

(√p + √q)² = x²p + 2√(pq) + q = x²2√(pq) = x² - p - q

√(pq) = (x² - p - q) / 2

Now x, x², p, q and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - p - q) / 2 is rational.

But since p and q are both primes, then pq is not a perfect square and therefore √(pq) is not rational. But this is a contradiction. Original assumption must be wrong.

So √p + √q is irrational, where p and q are distinct primes

5/7=0.7142 9/11=0.818181first we have to change the the number into decimal or rationaldecimal form of 5/7 is 0.71428571 and the decimal form of 9/11 is 81818181hence the rational number are between 0.71428571 and 0.81818181

rational numbers between 5/7 and 9/11 1) 0.724285712)0.734285713)0.74428571 . . . .upto 0.80818181

let the two numbers are 4 and 3 then sum is 7 and difference is 1 that is 7times of 1

if the two numbers are 5and 2 then the sum is 7 and difference is 1that is 7 times of 1

let the function be √x²-9

so the given function can't have a value as -ve under its root.. because it will be not be real solution then.

so the minimum value of this function is = 0 , that is at x = ±3

and maximum value is ∞

so range =[0,∞)

^^^^^ above person..

your solution is completely wrong.. how can it be possible that the value under root is negetive...

your range is (-∞, -3) so , that means the function √(x²-9) = - 3... for some value of X

now please solve it.. to get that value on which.. you are getting the range as -3. it will be imaginary number..

you have solved the domain actually... domain is (-∞,-3] U [ 3,∞) and not the range.

check your answer again..

ssryy I was in a little bit confused between domain and range......but my answer is appropriate for domain...

Jai mata di you very much for

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Let the number of boys = xnumber of girls = ywe need to find= x/y

The average score of boys in the examination is 71total score of boys = 71×x = 71xThe average score of girls in the examinationis 73total score of girls = 73× y = 73ySo total mark of school = 71x + 73y------(1)

average score of school in examination 71.8so total mark = 71.8× (x+y) -----(2)

From equations (1) and (2);71x + 73y = 71.8(x+y)⇒ 71x + 73y = 71.8x + 71.8y⇒ 73y - 71.8y = 71.8x - 71x⇒ 1.2y = 0.8x⇒ 0.8x = 1.2y⇒ x/y = 1.2/0.8 = 3/2

Ratio of boys and girls is 3:2.

please answer rest of the questions

answer iscosine30°=√3/2

Cos 30= root(3)/2

1/√3 is the correct answer of the given question

x²-3x-10=x²-5x+2x-10=x(x-5)+2(x-5)=(x+2)(x-5)

x= {-3+2√2}/2√2 ORx= {-3-2√2}/2√2

X^2 - 3X - 10=0X^2 - 5X + 2X - 10=0X(X-5)+2(X-5)=0(X+2)(X-5)=0X=-2,5

Let the cost of a chair be x.

Let the cost of a table be y.

Given that 3 chairs and 2 tables cost 700 rupees

3x + 2y = 700 ------ (1)

Given that 5 chairs and 3 tables cost 1100 rupees.

5x + 3y = 1100 ------ (2)

On solving (1)*3 and (2)*2, we get

10x + 6y = 2200

9x + 6y = 2100----------------------

x = 100.

Substitute x = 100 in (1), we get

3x + 2y = 700

3(100) + 2y = 700

300 + 2y = 700

2y = 700 - 300

2y = 400

y = 400/2

y = 200.

Therefore the cost of 2 chairs = 2 * 100

= 200 rupees.

Therefore the cost of 2 tables = 2 * 200

= 400 rupees.

Verification:

3x + 2y = 700

3(100) + 2(200) = 700

300 + 400 = 700

700 = 700.

5x + 3y = 1100

5(100) + 3(200) = 1100

500 + 600 = 1100

1100 = 1100.

Hope this helps!

thanks broooo

We know a, b, c are in AP then 2b = a + c

So, here

2(2k - 1) = k + 9 + 2k + 7

4k - 2 = 3k + 16

4k - 3k = 16 + 2

k = 18

= By dividing by 6 if remainder is 4, the number is 2 less than the multiple of 6.

= Means the number maybe 10, 16, 22, 28, 34

= By dividing these numbers by 9 the remainder might be 1 or 4.

We knowa^3 + b^3 = (a + b) (a^2 - ab + b^2 ).

Given, a^2 + b^2 = ab

Then we get, a^3 + b^3 = (a + b) (ab - ab) = (a + b)*0 = 0

points A , B , C be collinear when ,slope of AB = slope of BC = slope of CA

slope of AB = (b² - a²)/(b - a) = (b-a)(b + a)/(b-a)= a + b similarly,slope of BC = (c² - b²)/(c - b) = b + c slope of CA = c + a

here it is clear that , slope of AB ≠ slope of BC ≠ slope of CA so, points A, B , C are never be collinear.

hlo

Replace p and q by a and b.

thanks