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In ∆ OAB,

angle B < angle A

So, AO < BO ....(1)

In ∆ ODC,

angle C < angle D

OD < OC ....(2)

Adding (1) and (2)

AO + OD < BO + OC

AD < BC

c) is correct

f-¹(1) , means the value of x , for which the function will have the value of 1.

=> x²-5x+7 = 1 => x²-5x+7-1 = 0 => x²-5x+6 = 0 => x²-3x-2x+6 = 0 => x(x-3) -2(x-3) = 0 => (x-3)(x-2) = 0 ==> x = 3,2.

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1)80+650=730 litresthe maximum capacity of container which can measure the oil of both tankers when used on exact number of times is 730 litres

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please mention the question which should be answered

Sara ka sara galat hai

Z=40Y=50+z=50+40=90

k= -10 the correct answer of the given question

if x-2=0; x=2, 3x^2+4x+2k 3(2)^2+4(2)+2k=3(4)+8+2k=12+8+2k; 20+2k; 2k=-20; k=-20/2=-10

thank you for solving my querie

I do it with a different method

by substracting the loss percentage to 100 percentage

This is an another method in which you have to just substitute the value in formulae

I dont think this answer is correct....

since opposite angles are equal angle CED is also 36°

3x + 120 = 360

3x = 360-120

x = 240/3

x = 80

one angle is 80 second angle is 80 third angle is 80 fourth angle is 120 total 360

I) is the correct answer

I) is the right answer of the following

If two lines intersect each other, then the vertically opposite angles are equal.

Proof :In the statement above, it is given that ‘two lines intersect each other’. So, let AB and CD be two lines intersecting at O as shown in Fig. 6.8. They lead to two pairs of vertically opposite angles, namely,

(i) ∠ AOC and ∠ BOD (ii) ∠ AOD and ∠ BOC.

We need to prove that ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC.

Now, ray OA stands on line CD.

Therefore, ∠ AOC + ∠ AOD = 180° (Linear pair axiom) ………..(1)

Can we write ∠ AOD + ∠ BOD = 180°? (Linear pair axiom)……………(2)

From (1) and (2), we can write

∠ AOC + ∠ AOD = ∠ AOD + ∠ BOD

This implies that ∠ AOC = ∠ BOD

Similarly, it can be proved that ∠AOD = ∠BOC

Solution:-Slant height of the conical cup, 'l' = radius of the semi-circular sheet, R = 14 cmLet radius and height of the conical cup be 'r' and 'h' respectively.Circumference of the base of the cone = Length of arc of the semi-circleOr, 2πr = (1/2)2πROr, 2πr = (1/2)(2π)(14)Or, r = 7 cmNow, we know that l² = h² + r²(14)² = (h)² + (7)²h² = 196 - 49h =√147Height or depth of the conical cup = 12.124 cmNow, capacity of the conical cup = 1/3πr²h= 1/3*22/7*7*7*12.124= 13069.672/21Capacity of the conical cup = 622.365 cu cm

CP = a

SP = a x 125/100 = 1.25a

LP = 1.25a x 125/100 = 1.5625a

If he had sold it in list price,

Profit = 1.5625a - a = 0.5625a

in percent = 0.5625a/a x 100 = 56.25%

|1 -1||2 3|=3-(-2)=3+2=5

Thanks

Given expression is the expansion of

[3p-(1/6)]³

=(3p)³ - (1/6)³ - 3×(3p)²×1/6 +3×3p×(1/6)²=(3p - 1/6)³=(3p-1/6)(3p-1/6)(3p-1/6)

27p^3- 1/216 - 9/2p^2+1/4p

(3p)^3-(1/6)^3- 3(3p)^2(1/6) + 3(3p)(1/6)^2

(3p - 1/6)^3= (3p - 1/6)(3p - 1/6)(3p - 1/6)

thanks Yar , but I was just testing the app

As AB || FE, FA becomes the tranversal line cutting parallel lines. So angle A + angle F = 180°.

sum of other four angles = 720° - 180°= 540°.So 6 x + 4 x + 2 x + 3 x = 15 x = 540°.x = 36°.

Angles B C D E are: 216°, 144°, 72°, 108°

a)f ,b)e ,c)e ,d)d ,e)f

Area of bigger semi circle=πr^2/2=(22/7)*14*14/2=308cm^2Area of smaller semi circle=πr^2/2=(22/7)*7*7/2=77cm^2Area of shaded region=308-2(77)=308-154=154cm^2

the answer of the following is 14

28 is the answer please like

14 is the correct answer for you

14 is right answer of this question

14 is the correct answer

πr² = 616

r² = 616*7/22

r = √196

r = 14m

diagonal² = 14²+14²

diagonal = √392

diagonal = 14√2 cm

plz send me details

first consecutive number =30,second consecutive number =31