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We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur

Case I : When n = 3q

In this case, we have,n=3q, which is divisible by 3n=3q= adding 2 on both sidesn + 2 = 3q + 2n + 2 leaves a remainder 2 when divided by 3Therefore, n + 2 is not divisible by 3n = 3qn + 4 = 3q + 4 = 3(q + 1) + 1n + 4 leaves a remainder 1 when divided by 3n + 4 is not divisible by 3Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3

Case II : When n = 3q + 1

In this case, we haven = 3q +1n leaves a reaminder 1 when divided by 3n is not divisible by 3n = 3q + 1n + 2 = (3q + 1) + 2 = 3(q + 1)n + 2 is divisible by 3n = 3q + 1n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2n + 4 leaves a remainder 2 when divided by 3n + 4 is not divisible by 3Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3

Case III : When n = 3q + 2

In this case, we haven = 3q + 2n leaves remainder 2 when divided by 3n is not divisible by 3n = 3q + 2n + 2 = 3q + 2 + 2 = 3(q + 1) + 1n + 2 leaves remainder 1 when divided by 3n + 2 is not divsible by 3n = 3q + 2n + 4 = 3q + 2 + 4 = 3(q + 2)n + 4 is divisible by 3Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3 .

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I have posted one question please tell me the solution

I have posted one question please tell me the solution

tq very much

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To prove that a given number is a factor of the polynomial, substitute the value for the variable and check if the result is zero. If it is zero then the given number is a factor of the polynomial.

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We know √2 is irrational.

Let 5 + 3 √2 be rational so,

5 + 3 √2 = r, where r is rational.

√2 = (r - 5)/3

So, this is an absurd result because RHS is rational and RHS is irrational therefore a contradiction to our assumption.

So, 5 + 3 √2 is irrational.

20%=0.20Kg = 1000g1000×0.2=200g0.2kg

(-5) × 19 × (-60)

First we group -5 and -60 as it is commuative

So, (-5) × (-60) = 300

So, 19 × 300 = 5700

2×65×52×5×6510×65=650

A. Cylinder example is pipe

B. Triangle example is Arrow, cake

Perimeter will be sum of all sides.P= 8+10+6+2+4+(10-4-2)=30+4 =34 cm

80

Suppose length us l and breadth is b now 5 is subtracted from length and breadth so l-5 and b-5 Perimeter is 26 2(l-5)+2(b-5) = 26 2l + 2b - 10 - 10 = 26 2l + 2b = 46 l + b = 23

For equation px^2 - 14x + 8 = 0Let a and b are roots

Given,a = 6b

Sum of roots = - (-14)/pa + b = 14/p6b + b = 14/p7b = 14/pb = 2/p.............. (1)

Product of roots = 8/pa*b = 8/p6b*b = 8/p6b^2 = 8/p6*(2/p)^2 = 8/pp^2/6*4 = p/8p^2 = 3pp^2 - 3p = 0p(p - 3) = 0p = 0 or p = 3

Correct format

((10×10)÷10+50

(100÷10)+50

10+50

Answer is 60

10% of ₹50 = 10/100 × 50 = ₹5So, ₹50 - ₹5 = ₹45

1)Theadvantages of classifying organismsare as follows: (i)Classificationfacilitates the identification oforganisms. (ii) helps to establish the relationship among various groups oforganisms. (iii) helps to study the phylogeny and evolutionary history oforganisms.

2)Before developing a hierarchy in classification, we need to decide which characteristics should be used as the bass for making the broadest divisions. Then we should pick up next set of characteristics for making sub-groups. This process must continue and each time new characteristics should be used. The characteristics that decide the broadest divisions among living organisms should be independent of any other characteristics.

For example, nature of cells and form of the body is considered to classify organisms into broad divisions.

The characteristics in the next level should be dependent on the previous one that will decide the subsequent divisions of the groups.

3)The five kingdom classification was proposed by R.H. Whittaker in 1969. It includes Kingdom Monera, Kingdom Protista, Kingdom Fungi, Kingdom Plantae, and Kingdom Animalia.

The five kingdoms were formed on the basis of following characteristics.*cell structure*mode of nutrition*source of nutrition*body organisation

No we can't

as

There is a property with triangles that sum of two sides of triangle should always be greater than the 3rd side.

but here3+5= 8that is equal to third side so it is not possibleplease like the solution 👍 ✔️

(a+b+c)^2=a^2+b^2+c^2+2(ab+BC+CA)14^2=96+2(ab+bc+CA)196-96=2(ab+bc+ac)100/2=ab+bc+acab+bc+ac=50hence it will be 50

32-17=15.............

Volume of prism =area of base * height

Volume of prism = 6cm² * 10 cm

Volume of prism =60 cm³

Thanks anna

are of triangle=1/2 x3x4x5 =30volume of prism=base area x height =30x10 300 cm

Total volume of 3 cubes = (33+ 43+ 53) cm3

= (27 + 64 + 125)cm3

= 216 cm3

Now volume of new cube = 216 cm3

So edge of new cube = cube root of 216 cm3

= 6 cm

how many cubes of side 12 mm are contained in a cube of side 24 cm.

For a rectangle,

CASE 1 AB=CD

(2-14)^2+ (-2-10)^2= (11+1)^2+ (13-1)^2

288 = 288

case 1 verified

CASE 2 AD=BC

(2+1)^2+ (-2-1)^2= (14-11)^2+ (10-13)^2

18=18

case 2 verified

CASE 3 AC=BD

(2-11)^2+ (-2-13)^2= (14+1)^2+ (10-1)^2

306=306

case 3 verified

Therefore it is a rectangle.

axb =a.bsin∅ = 8 and |a| = 2 and |b| = 5.so, axb = 2.5.sin∅ = 8=> sin∅ = 8/10 = 4/5. so cos∅ is √5²-4²/5 = 3/5

now , a.b = |a|.|b|cos∅ = 2.5.3/5 = 6.

axb=a.bsinø but we don't know the value of a.b so u cannot use a.b as 2.5=10

A4+a8=24a+3d+a+7d=242a+10d=24. .........1

a6+a10=34a+5d+a+9d=442a+14d=44. .,.......2Subtracting 2 from 12a+14d=442a+10d=24- - -4d=20d=4/20or 1/5Put d= 1/5 in 12a+10×1/5=242a+2=242a=24-22a=22a=11

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Given,a = b + 5a - b = 5 a^2 + b^2 = 41

We know, (a - b)^2 = a^2 + b^2 - 2ab5*5 = 41 - 2ab2ab = 41 - 252ab = 16ab = 16/2 = 8

Value of ab = 8

(B) is correct option