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volume of a coboid=l*b*hhere sine l,b and h r doubled,we can write it as(2l) (2b) (2h) which is equal to 8 (l)(b)(h)therefore ans=12*8=96cm^3

According to the figure

3x+15+2x+30 = 180=> 5x+45 = 180=> 5x = 180-45. = 135=> x = 27°

27kaisa aya hai

5x=135so x= 135/5 = ??

Volume=64 cm^3Let the edge of the cube be a cm.Hence, by formula,a^3=64=>a=4 cmThus,total surface area of the cube=6a^2=6*4*4 cm^2=96 cm^2

These are 3 isometric cuboids . Each one should have different dimensions of the front face.

These are 3 isometric cuboids . Each one should have different dimensions of the front face.

7x + 6 + 6x + 4 + 5x + 8 = 180°18x + 18° = 180°18x = 180° - 18°18x = 162°x = 162°/18x = 9

I don't know answer friend sorry for that come and enjoy the party in Jammu and Kashmir near the border of pakistan

power power will be +so answer is x to the power 6 and 1

xy+7z is the correct answer of the given question

(4xy-9z);(3xy-16z)=(4xy-9z)-(3xy-16z)=4xy-9z-3xy+16z=xy+7z ans

(4xy-9z);(3xy-16z)=(4xy-9z)-(3xy-16Z)=4xy-9z-3xy+16z=xy+7z is correct answer.

xy + 7z is the correct answer

i) 6abc -9a²c = 3ac(2b-3a)

9z = 81

z = 81/9

z = 9

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Given, AB=AC

= whole root of (0-3)^2 + (2-p)^2 = whole root of (0-p)^2 + (2-5)^2

= 9 + (2-p)^2 = p^2 + 9

= 4 + p^2 - 4p = p^2

4p =4

p = 1.

Substitute of p-value in AB

Length of AB = whole root of (0-3)^2 + (2-1)^2

= whole root of 9 + 1

= root 10.

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sorry but I didn't understand

tan x = 8/15; (H)^2 = (8)^2 + (15)^2= 64 + 225 = 289 = (17); (1- sinx)(2+2sinx) _____________________ (1+ cosx)( 2-2 sinx); = = (1- sinx)(2+2 sinx)/ (1+ cosx)(2-2sinx) = (1 + 8/17)(2+2 (8/17) / (1+ 15/17)(2+2(15/17)) = (17+8/17)(34+16/17) / (17+15/17)(34+30/17) = (25)(50)/(32)(64) = = 1250 / 2048 = 625/1024 = = 25/32

Consider cosec theta - sin theta = a³⇒ !/sin theta - sin theta = a³⇒ 1 - sin² theta/sin theta = a³cos² theta/ sin theta = a³→ (1)⇒ (cos² theta/sin theta)²/³ = (a³)²/³⇒ cos⁴/³ theta/sin²/³ theta = a²→ (2)Now consider, sec theta - cos theta = b³⇒ 1/cos theta - cos theta = b³⇒ 1 - cos²theta/cos theta = b³⇒ sin² theta/cos theta = b³→ (3)⇒ (sin² theta/cos theta)²/³ = (b³)²/³⇒ sin⁴/³ theta/cos²/³ theta = b²→ (4)Multiply (2) and (4), we get(cos⁴/³ theta/sin²/³ theta)× (sin⁴/³ theta/cos²/³ theta) = a²b²→ (5)a² + b² =(cos⁴/³ theta/sin²/³ theta) + (sin⁴/³ theta/cos²/³ theta)(cos² theta + sin² theta)/(sin²/³ theta cos²/³ theta)= 1/sin²/³ thetacos²/³ thetaConsider, a²b²(a²+b²) =(sin²/³ theta cos²/³ theta)× 1/sin²/³ theta cos²/³ theta= 1 Hence proved.

replacing m by a and n by b

Given that Sin A + 2 cos A = 1

Squaring on both sides, we get

(sin A + 2 cos A)^2 = 1

We know that (a+b)^2 = a^2 + b^2 + 2ab.

(sin^2 A + 4 cos^2 A + 4 sin A cos A) = 1

4 cos^2 A + 4 sin A cos A = 1 - sin^2 A

4 cos^2 A + 4 sin A cos A = cos^2 A

3 cos^2 A + 4 sin A cos A = 0

3 cos^2 A = - 4 sin A cos A ---- (1).

Given 2 sin A - cos A

Squaring on both sides, we get

(2 sin A - cos A)^2 = 4 sin^2 A + cos^2 A - 4 sin A cos A

= 4 sin^2 A + cos^2 A + 3 cos^2 A

= 4 sin^2 A + 4 cos^2 A

= 4(sin^2 A + cos^2 A)

= 4.

2 sin A - cos A = 2.LHS = RHS.

main ek lekhak hon in hindi

The point equidistant from the vertices is called it's circumcentre

circumcenter is the correct answer

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(i) DC, EF, HG(ii) FG, BC, AD(iii) GC, BF, DH(iv) No, they are perpendicular (v) No, they are not parallel

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if possible please solve it too

Option ADegree of differential equation is 2

By trigonometric sum identity,

cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B) andcos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)

ii) Adding both above, cos(A+B) + cos(A-B) = 2cos(A)*cos(B)

iii) Now, let A = (x + y)/2 and B = (x - y)/2

==> A + B = (x + y + x - y)/2 = 2x/2 = x

and A - B = (x + y - x + y)/2 = 2y/2 = y

iv) Thus replacing these in (ii) above,

cos(x) + cos(y) = 2cos[(x + y)/2]*cos[(x - y)/2] [Proved]

16 t + 20S12 t + 4S----------------28 t + 24S