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In Δ ABC

AB² + BC² = 8² + 6² = 64 + 36 = 100 = AC² 

Δ ABC is a right angled triangle.

If we draw a circle taken in a AC as diameter, ∠B = 90°, Therefore the point B on the circle.

Δ ABC is a right-angled triangle. 

If we draw a circle taking BC as diameter,

∠A < 90°, Therefore the position of point A will be outside the circle.

If we draw a circle taking AB as diameter, 

∠C < 90°, Therefore the position of point C will be outside the circle.

a. On the circle

∠B = 90°

b. Outside the circle Position of the vertex of an triangle with opposite side as the diameter is outside the circle, because. ∠A <9o°.

C inside the circle ∠C > 90°.

D on the circle ∠D = 90°.

E outside the circle ∠E <90°.

QR = BC

∠A = ∠P

PR Diameter of the circumcircle of Δ ABC Diameter = PR= \(\sqrt{12^2+5^2}\) = \(\sqrt{169}=\) 13cm.

The correct answer is option 4

All the Carry flag (CY), Auxiliary carry flag (AC) and Zero flag (Z) are affected when an arithmetic operation is performed.

Auxiliary carry flag (AC):

In the addition of two 8 bit hexadecimal numbers, if a carry occurs when the LS hex digits of the two numbers are added, then such a carry is called Auxiliary carry and AC flag is set.

Carry flag (CY):

Carry flag is set when two 8 bit numbers are added and the result has a carry.

Zero flag (Z):

Zero flag is set when any arithmetic operation is performed and the result is 00H.

Assembly language: It is a low-level programming language that uses symbols, variables, and functions that work directly with CPU.

High-level language: It is a human-friendly language that uses variables and functions, independent of the computer architecture.

Let radius of sphere be r

Since, after immersing sphere into a cylindrical container the water level upto the tip of clinder form 10 m height.

\(\therefore\) Volume of sphere = Raised volume

⇒ \(\frac43\pi r^3\) = πr²(14 - 10)

⇒ r = \(\frac34\times4\) = 3

Hence, radius of solid sphere is 3m.

h = height of cylinder  = 2 m

r = base radius of cylinder  = 2m

\(\therefore\) volume of cylinder = πr²h

 = π.2².2

= 8π m³

Volume of metal tablet whose diameter is 2cm

 = \(\frac43\pi r^3\) = \(\frac43\pi(\frac22)^3\) = \(\frac43\pi cm^3\) (\(\because\) r = \(\frac d2\))

Volume of metal tablet whose diameter is 6 cm

 = \(\frac43\pi R^3\) = \(\frac43\pi (\frac62)^3\) = \(\frac43\pi\times27cm^3\)

\(\because\) The price of 2cm diameter metal tablet  = Rs. 5

\(\therefore\) The price of metal quantity of \(\frac43\pi cm^3\) = Rs. 5

\(\therefore\) The price of metal quantity of \(\frac43\pi\times27 cm^3\) = Rs. (5 x 27) = Rs. 135

Volume of the cone = 1/3πr³ = 1/3×(9)²×36

Volume of a Sphere = 4/3πr³

Volume of the old solid = Volume of the new solid formed

1/3π×(9)² × 36 = 4/3πr³

= (9)²×9×4 = 4×r³

= r³ = (9)³ 

⟹ radius of sphere formed = 9

Radius of cylinder = 3cm

Height of cylinder = 3r = 3 × 3 = 9cm

Volume = πr²h

= 22/7 × 3 × 3 × 9 

= 1782/7

= 254.57cm³

Radius of hemisphere is r = 3

(By comparing x² + y² + z² = r²)

\(\therefore\) Volume of hemisphere = \(\frac23\pi r^3\) 

 = \(\frac23\pi\times3^3\) = 18\(\pi\)

Let side of cube be x cm.

\(\therefore\) volume of cube is x³

\(\therefore\) x³ = 3375 = 15³

\(\therefore\) x = 15

Hence, side of cube is 15 cm.

Now total surface area of cube

= 6x²

= 6 x 15²

= 6 x 225

= 1350 cm²

(a) NH₃ + 3Cl₂(Excess) → NCl₃ + 3HCl

(b) Na₂SO₃ + 2HCl → 2NaCl + H₂O + SO₂

(c) Br₂ + 3F₂ → 2BrF₃

Digits are 1, 2, 3, 4, 5, 7, 9 

Multiple of 11 → Difference of sum at even & odd place is divisible by 11. 

Let number of the form abcdefg 

\(\therefore\) (a + c + e + g) – (b + d + f) = 11x 

a + b + c + d + e + f = 31

\(\therefore\) either a + c + e + g = 21 or 10

 \(\therefore\) b + d + f = 10 or 21

Case- 1

a + c + e + g = 21

b + d + f = 10

(b, d, f) ∈ {(1, 2, 7) (2, 3, 5) (1, 4, 5)} 

(a, c, e, g) ∈ {(1, 4, 7, 9), (3, 4, 5, 9), (2, 3, 7, 9)} 

\(\therefore\) Total number in case-1 = (3! × 3) (4!) = 432

Case- 2 

a + c + e + g = 10

b + d + f = 21

(a, b, e, g) ∈ {1, 2, 3, 4)} (b, d, f) & {(5, 7, 9)} 

\(\therefore\) Total number in case 2 = 3! × 4! = 144 

\(\therefore\) Total numbers = 144 + 432 = 576

Correct option is (D) 72

To make a no. divisible by 3 we can use the digits 1,2,5,6,7 or 1,2,3,5,7. 

Using 1,2,5,6,7, number of even numbers is 

= 4 x 3 x 2 x 1 x 2 = 48 

Using 1,2,3,5,7, number of even numbers is 

= 4 x 3 x 2 x 1 x 1 = 24 

Required answer is 72.

Given circle x² + y² + 4x + 6y - 5 = 0

Differentiate w.r.t x, we get

\(2x + 2y\frac{dy}{dx} + 4 + 6\frac {dy}{dx} = 0\)

⇒ \((2y + 6) \frac{dy}{dx} = - 2x - 4 \)

⇒ \(\frac{dy}{dx} = \frac{-2x - 4}{2y + 6} = \frac{-x -2}{y + 3}\)

Slope of line x + y = -13 is -1.

\(\therefore\) For point where slope of circle is -1, we have

\(\frac {-x_1 - 2}{y_1 + 3} = -1\)

⇒ \(-x_1 - 2 = -y_1 - 3\)

⇒ \(1 -x_1 = -y_1\)

⇒ \(y_1 = x_1 - 1\)

\(\therefore\) From equation of circle, we get

\({x_1}^2 + (x_1 - 2)^2 + 4x_1 + 6(x_1 - 1) - 5 = 0\)

⇒ \(2{x_1}^2 - 2x_1 + 1+ 4x_1 + 6x_1 - 6-5 =0\)

⇒ \(2{x_1}^2 + 8x_1 - 10 = 0\)

⇒ \({x_1}^2 + 4x_1 - 5 = 0\)

⇒ \((x_1 + 5)(x_1 - 1) = 0\)

⇒ \(x_1 = -5 \;or\; x_1 = 1\)

\(\therefore x_1 = 5\)

\(\therefore\) \(y_1 = -5 - 1 = -6\)

\(\therefore\) Point which on circle which is closest to line is (-5, -6).

The shortest distance between line & circle 

= Distance of point (-5, -6) from line x + y = -13

= \(\left|\frac{-5 + (-6) + 13}{\sqrt{1 + 1}}\right| = \frac 2{\sqrt 2} =\sqrt 2 \, units\)

Distance between (-5, -6) & (-6, -7)

\(= \sqrt{(-6 + 5)^2 + (-7 + 6)^2} = \sqrt{1 + 1} = \sqrt 2 \, units\)

& (-6, -7) lies on line x + y = -13.

\(\therefore\) Point on line x + y +13 = 0 which is closet to given circle is (-6, - 7).

By this process we can easily solve this problem.please follow the picture given below

Correct option is (2) 1120

Total no. of way to select 3 boys and 3 girls = ¹⁰C₃ - ⁵C₃ = 1200

When particular boys B₁ and B₂ both are in team = ⁸C₁.⁵C₃ = 80

So required number of ways = 1200 - 80 = 1120

(B) If a ≠ – 3, then the system has a unique solution for all values of λ and μ 

(C) If λ + μ = 0, then the system has infinitely many solutions for a = – 3 

(D) If λ + μ ≠ 0, then the system has no solution for a = – 3

System has unique solution for a/3 ≠ 2/-2

system has infinitely many solutions for a/3 = 2/-2 = λ/μ

and no solution for a/3 = 2/-2 ≠ λ/μ

LHS = sin2A + sin 2B -sin 20 

= 2sin (A+ B). Cos (A – B) – 2 sin C Cos C[ using transformation formula] 

= 2sin C. Cos (A – B) – 2sin C. Cos C 

= 2sin C[Cos (A – B) – Cos C] 

= 2 sin C [Cos (A – B)+ Cos (A + B)] 

∵ Cos (A + B) = – Cos C 

= 2 sin C. 2.cos A. Cos B 

= 4 Cos A. sin C. Cos B [∵ CosC + CosD = 2cos\(\frac{C + D}{2}cos\frac{C - D}{2}]\) 

= 4 cos A. Cos B. sin C = RHS 

∴ sin 2A + sin 2 B sin 2 C = 4 Cos A. Cos B. sin C.