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Let diameter of earth = D

diameter of moon =D/4

=) radius of earth = D/2

=) radius of moon = D/8

earth and moon are like spheres.

so volume 0f earth = 4/3×π(D/2)^3

and volume of moon = 4/3 × π(D/8)^3

so ratio of their volumes

= 512/8 = 64:1

curved surface area of earth = 4π(D/2)^2

curved surface area of moon = 4π(D/8)^2

so ratio of their surface area

= 64/4 = 16:1

So distance between their centres is OC+OC'=28 cm

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Angle AOB =360°minus angle OAC minus angle OBC minus angle ACB.=360°-90°-90°-65°.=180°-65°=115°

115 degree is the best answer

Let,∠AOB=2x and∠BOC=3x∠AOC=∠AOB+∠BOC=75°or, 2x+3x=75°or, 5x=75°or, x=15°∴,∠AOB=2×15°=30° and∠BOC=3×15°=45°

the answer of the question is (c)

c) is the right answer of the following

a) is correct answer

The mode ofnutrition in amoebais Holozoic.Nutritionin anamoebaoccurs through a process called phagocytosis where the entire organism pretty much engulfs the food it plans on eating up. Asamoebais a unicellular organism, it does not have any specialized organ for the mechanism ofnutrition.

ΔABC is a right angled triangle.∠ABC = 90°.A circle is drawn with AB as diameter intersecting AC in P, PQ is the tangent to the circle which intersects BC at Q.

Join BP.

PQ and BQ are tangents drawn from an external point Q.

∴ PQ = BQ -------------- (1) (Length of tangents drawn from an external point to the circle are equal)

⇒ ∠PBQ = ∠BPQ (In a triangle, angles opposit to equal sides are equal)

Given that, AB is the diameter of the circle.

∴ ∠APB = 90° (Angle in a semi-circle is a right angle)

∠APB + ∠BPC = 180° (Linear pair)

∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°

Consider ΔBPC,

∠BPC + ∠PBC + ∠PCB = 180° (Angle sum property of a triangle)

∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ----------- (2)

∠BPC = 90°

∴ ∠BPQ + ∠CPQ = 90° ...(3)

From equations (2) and (3), we get

∠PBC + ∠PCB = ∠BPQ + ∠CPQ

⇒ ∠PCQ = ∠CPQ (Since,∠BPQ = ∠PBQ)

Consider ΔPQC,

∠PCQ = ∠CPQ

∴ PQ = QC ----------- (4)

From equations (1) and (4), we get

BQ = QC

Therefore, tangent at P bisects the side BC.

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I am not able to understand that how you wrote PT^2=PB×PA.

Here is your answer:

In ΔAOB and ΔCOB,

AB = BC (we know that the sides of the square are equal)

BO = BO (Common)

OA = OC (The diagonals of square will bisect with each other)

ΔAOB ≅ ΔCOB (SSS)

∠AOB = ∠COB [CPCT]

∠AOB + ∠COB = 180° (this is the linear pair)

∠AOB + ∠AOB = 180°

∠AOB = 90°

No 7 is most bigest that question Bro

hiii and good night my

No 7 is most biggest that question Bro.

thank you for this question

According to the question,

AB = 8 cm, BC = 15 cm, AC = 17 cm

According to the diagram,

AB = AdjacentSide, BC = Opposite Side, AC = Hypotenuse.

Now we need to find Sin A, Cos A, and Tan A.

Let us calculate Sin A first,

Sin A = Opposite / Hypotenuse

=> Sin A = BC / AC

=> Sin A = 15 / 17

Cos A = Adjacent / Hypotenuse

=> Cos A = AB / AC

=> Cos A = 8 / 17

Tan A = Opposite / Adjacent

=> Tan A = BC / AB

=> Tan A = 15 / 8

infinite number of points

if a combination reaction how many products are formed?

ΔBCM & ΔEDM are similar. MD=CM => BC = DE, as AD = BC, AE = 2 BCΔACL & ΔAEL are similar. EL / BL = AE / BC = 2So EL = 2 BL

a=5-ba+b=5-------------(1)b=6÷aab=6---------------(2)a³+b³=(a+b)(a²+b²-ab) =(a+b)(a²+b²+2ab-3ab) =(a+b)((a+b)²-3ab)put values from eq 1 and 2a³+b³=5*(5²-3*6) =5*(25-18)=35

a=5-ba+b=5.........(1)ab=6............(2)eq(1) or (2)......a³+b³=5*(5²-3*6)=5*(25-18)35 ans

your answer is wrong

f(a)= a^3-a^3+2a+a-13a-1will be the answer

Option B is the correct answer.

take sushi hdhgdh hehehe bdhd

Abe ans chahiye pagleti nhi

sorry galti se bheja Gaya

it's ok

thanks you bhai it's Shreyansh

Kon shreyansh

sorry bhaiya ..... ..

discount will be 7800*10/100780 rupees

no bad maths

no bad maths

As AB = AC so angle B = angle CA + B + C = 180So 2B = 100, B =50

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D. Imposter

An imposter is a person who pretends to be someone else in order to deceive others.

Rogue is a dishonest person.