Explore topic-wise InterviewSolutions in Current Affairs.

The sum of the deviations from the mean is zero.

This will always be the case as it is a property of the sample mean, i.e., the sum of the deviations below the mean will always equal the sum of the deviations above the mean

Thesumof thedeviationsfrom themeanis zero. This will always be the case as it is a property of the samplemean, i.e., thesumof thedeviationsbelow themeanwill always equal thesumof thedeviationsabove the mean.

It is by means of chain rule that we divide or disintegrate the derivative into two partial derivatives.

2. 10 + y = 11 y = 11 - 10 = 1

5. d/7 = 11 d = 77

10+y=11

y=11-10

y=1

8. 12 - 9t = 3 9t = 12 - 3 = 9 t = 1

d/7=11

d=7×11

d=77

77/48 because when we devide one of them has to be reciprocaled

A triangular pyramid has 4 faces, including the base, are triangles, 4 veritces and 6 edges.

Faces = 4

Vertices = 4

Edges = 6

F + V = E + 2

F + V - E = 2

This relationship is called Euler's Formula.

⇒ 4 + 4 - 6 = 2

⇒ 8 - 6 = 2

⇒ 2 = 2

L.H.S. = R.H.S.

Hence, The Euler's Formula is verified for a Triangular Pyramid.

A triangular pyramid has 4 faces, including the base, are triangles, 4 veritces and 6 edges.

Faces = 4

Vertices = 4

Edges = 6

F + V = E + 2

F + V - E = 2

This relationship is called Euler's Formula.

⇒ 4 + 4 - 6 = 2

⇒ 8 - 6 = 2

⇒ 2 = 2

L.H.S. = R.H.S.

Hence, The Euler's Formula is verified for a Triangular Pyramid.

we have to check how many times an equilateral triangle fits on to itself during a full rotation of 360 degrees.

Please look at the images of the equilateral triangle in the order A,B and C. A is the original image. The images B and C are generated by rotating the original image A.

When we look at the below images of equilateral triangle, it fits on to itself 3 times during a full rotation of 360 degrees.

Hence, an equilateral triangle has rotational symmetry of order 3.

sphere

triangular pyramid

cube

cuboid

cylinder

triangular prism

Let the triangle and parallelogram have common base b,let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), thenArea of triangle =12∗b∗h1Area of rectangle =b∗h2As per question12∗b∗h1=b∗h212∗b∗h1=b∗100h1=100∗2=200m

in square pyramid Number of faces:5

Number of edges:8

Number of vertices:

in triangular prismNumber of faces:5

Number of edges:9

Number of vertices:6

Thanks

The number of faces in triangular prism are 5 (b)

Square Pyramid : In geometry, a square pyramid is a pyramid having a square base. If the apex is perpendicularly above the center of the square, it is a right square pyramid, and has C₄ᵥ symmetry.

It has 5 vertices and 8 edges.

Triangular prism :

In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. A right triangular prism has rectangular sides, otherwise it is oblique.

It has 9 edges and 6 vertices.

thank u sir

In the figure above ∠1, ∠4, ∠5, and ∠7 are all acute angles. They are all congruent to each other. ∠1 ≅ ∠4 are vertical angles. ∠4 ≅ ∠5 are alternate interior angles, and ∠5 ≅ ∠7 are vertical angles. The same reasoning applies to the obtuse angles in the figure: ∠2, ∠3, ∠6, and ∠8 are all congruent to each other

qrs is 120qps is 120q is 60

given,<QRX=25°<SRY=35°a.)<QRX+<QRS+<SRY=180° [angles on st. line]putting the values;25°+<QRS+35°=180°<QRS=180°-25°-35°<QRS=180°-60°<QRS=120°b.)For a kite opposite angles are equalso,<QPS=<QRS<QPS=120°c.) given, <S=60°therefore <Q=60° [opposite angles of akite are equal]

If you find this solution helpful, Please like it.

∠AEC =∠DEB (Vertically Opposite Angle)join OA and OCso in triangle OAE and triangle OCE we have,OA=OC (radii of same circle)OE=OE (common)∠OAE=∠OCE (90° as the tangent is always perpendicular to the radius at the point of contact)∴ΔOAE ≡ΔOCEso∠AEO =∠CEO (CPCT)similarly for other circle we have∠DEO' =∠BEO' (CPCT)Now∠AEC=∠DEB⇒1/2(∠AEC) = 1/2(∠DEB)⇒∠AEO =∠CEO =∠DEO'=∠BEO'so all 4∠'s are equal and bisected by OE and OE'∴ O,E,O' are collinear

Thank you but I was hoping to see this shown using Intermediate Value Theorem and Rolle's Theorem

That i have used without mentioning it as it quite noticeable fact as f(x) is continuous and differentiable, there it is negitive for some value and positive fo some and as it is strictly increasing because f'(x) > 0so it cuts the x axis one time. that is exactly one zero.

We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0<r<3i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)n = 3q is divisible by 3or n +2 = 3q +1+2 = 3q +3 also divisible by 3or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3Hence n, n+2 , n+4 are divisible by 3.

40 च्यक्तीयों की औसत 4200 है तो योग 4200×40 = 168000 35 च्यक्तीयों की औसत 4000 है तो योग 4000×35 = 140000 तो 40+35 = 75 च्यक्तीयों का योग 168000 + 140000 = 308000 औसत = 308000/75 = 4106.67

This can be solved by remainder theorem,that is by finding out the unit digit of every num.1^99 will always end with 1.5^99 will always end with 5Now 2^99 can be written as (2^4)^24 × 2^3 as we know 2^4 will always end with 6 hence (2^4)^24 will also end with 6 and × (2^3=8) the last digit will be 8.Similarly 3^99 can be written (3^4)^24 × 3^3 which will end with 7.And 4^99 can be written as (4^2)^49 × 4 which will end with 4.Hence adding all unit digits 1+8+7+4+5=25 Last digit is 5 means the num is divisible by 5 -

thank you bhai

Solution:-

Let ABC be the triangular field with sides Ac = 15 m, AB = 16 m and BC = 17 mAnd,Let the place where the cow, the buffalo and thehorse are tied, are three sectors i.e. sector ADE, sector BFG and sector CHIThe area of triangular field by Heron's formula =√s(s-a)(s-b)(s-c)s = (a+b+c)/2s = (15+16+17)/2s = 48/2s = 24 m√24(24-15)(24-16)(24-17)√24*9*8*7√12096Area of triangular field = 109.98 sq mArea of the grazed part = Area of the triangular field = Area of the sector ADE + Area of sector BFG + Area of sector CHI= π*7²*∠A/360 + π*7²*∠B/360 +π*7²*∠C/360=π*7²(∠ A + ∠ B + ∠ C)/360= 22/7*7*7*180/360= 154/2Area of the grazed part = 77 sq mNow, the area of the field which cannot be grazed by these animals= 109.98 - 77= 32.98 sq mAnswer.

Height (h) = 5cmarea of its base (πr²) = 36π cm²Where, r is the radius of it's base.

πr² = 36π

r² = 36

r = 6cm

Now,we have r = 6cm ,h = 5cm.

and C.s.a of cylinder = 2πrh

2πrh = 2 * π * 6 * 5

= 60π cm²

Hence, the curved surface area of the cylinder = 60π cm²