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ANSWER:

COZ

I THINK your so UPSET or depressed by some thing you have hated or not WILLING to do

Answer:

⇒ Classmark=

2

Lowerlimit+Upperlimit

⇒ Given CLASS INTERVAL is 50−60

⇒ Lower LIMIT =50 and upper limit =60

∴ Classmark=

2

50+60

=

2

110

=55

∴ The class mark of class onterval 50−60 is 55.

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HTTPS://learnodo-newtonic.com/van-gogh-famous-paintings

Answer:

x=9

Total MONEY=29.50 RUPEES

Step-by-step EXPLANATION:

ATQ

(2x+5)+(3x-2)=48

   2x+5+3x-2=48

   2x+3x+5-2=48

            5x+3=48

                5x=48-3

                5x=45

                  x=45÷5

                 x=9

Total money = 50(2x+5) + 75(3x-2)

                     =50[(2×9)+5] +75[(3×9)-2]

                     =50(18+5) + 75(27-2)

                     =50(23) + 75(25)

                     =1150 + 1800

                     =2950 paise

                     =29.50 rupees

-22/-77

=-2/-7

= 2/7

so UR ANS is 2/7

mark as BRAINLIST

Answer:

Counting RELATIONS. Since any subset of A × B is a relation from A to B, it follows that if A and B are finite SETS then the number of relations from A to B is 2|A×B| = 2|A|·|B|. ONE way to SEE this is as the number of subsets of A × B.

Here MATE here is the ANSWER.....

ANSWER:

AREA of Circle=π×radius×radius

=π×6×6=36π

This is the BEST POSSIBLE answer

Hope you have understood

Mark it as

ANSWER:

MARK me as brainliest

follow me on BRAINLY

PLS pls pls pls pls pls

the quadrilateral OBTAINED by joining MIDPOINTS of the sides of any quadrilateral ABCD is RECTANGLE or SQUARE

CONVERT into fraction

7.05 = 2×1(× =1) (× + 2)

Answer:

TOTAL rope= 120+8=128m

Step-by-step explanation:

PERIMETER of rope required= 4×30= 120m

Extra rope for tying= 4×2=8m

Hope it HELPS!

PLEASE mark as brainliest.

Good luck for exams.

Answer:

1.sachin INVESTED 70000 RS in 12yrs

Step-by-step EXPLANATION:

PLEASE POST the QUESTION PROPERLY

ANSWER:

SQUARE

Step-by-step EXPLANATION:

is formed by joining these points on the GRAPH

ANSWER:

hiiiiiiiiiiiiiiiiiiiiiiiiii

ANSWER:

Step-by-step EXPLANATION:

HEY MATE here is your ANSWER:

ANSWER:

you should ADD x^2 + 2XY - y^2 to MAKE it 2x^2 + 3XY

Answer:

1. a

2. b

Step-by-step EXPLANATION:

Step-by-step explanation:

perimeter of triangle = sum of all side

Perimeter of ABC = AB + BC + CA

= 4cm + 5cm + 1cm

= 10 cm

ANSWER:

The VALUE of K is 4 not 7.

ANSWER:

what is this jdjrjdjeejdsj

ANSWER:

Don't KNOW SORRY BRO...

ANSWER:

LET numbers be = x, y

Equation (i):

x/y = 2/3

=> x = 2y/3

Equation (ii):

x+5/y+5 = 5/7

=> 7x + 35 = 5Y + 25

=> 7(2y/3) + 35 = 5y + 25

=> 14y + 105 = 15y + 75

=> y = 105 - 75

=> y = 30

x = 2y/3 = 2 × 30/3 = 2 × 10

x = 20

_______________

as PROVED in IMAGE that bac=90-1/2a

then

boc=90-1/2*70

boc=90-35

boc=55 DEGREE

Answer:

We KNOW that,

VOLUME of cube=a^3=1000 cubic cm

a=10 cm

Total SURFACE area of Cube=6a^2=6×(10)^2

=6×10×10=600 SQ cm

This is the best possible answer

Hope you have understood

Please mark it as

HELLO mate here's your AnsWer for your Question is

hello mate here's your AnsWer for your Question isfor 1 right ANGLE is formed at angle E.

hello mate here's your AnsWer for your Question isfor 1 right angle is formed at angle E.other answer I. am giving with pictures.

hello mate here's your AnsWer for your Question isfor 1 right angle is formed at angle E.other answer I. am giving with pictures.by the way, SORRY I mistakenly WROTE angle C inside triangle

hello mate here's your AnsWer for your Question isfor 1 right angle is formed at angle E.other answer I. am giving with pictures.by the way, sorry I mistakenly wrote angle C inside trianglebut

hello mate here's your AnsWer for your Question isfor 1 right angle is formed at angle E.other answer I. am giving with pictures.by the way, sorry I mistakenly wrote angle C inside trianglebut If you like it Mark this as BRAINLIEST Answer

hello mate here's your AnsWer for your Question isfor 1 right angle is formed at angle E.other answer I. am giving with pictures.by the way, sorry I mistakenly wrote angle C inside trianglebut If you like it Mark this as brainliest Answerthank you

ANSWER:

Answer:Three consecutive INTEGERS are 7,8 and 9

Solution:

Let the numbers are x, x+1 and x+2

(Since these are consecutive integers)

they are TAKEN in INCREASING order and multiplied by 2,3 and 4 respectively

So x BECOMES 2x

(x+1) becomes 3(x+1)

(x+2) becomes 4(x+2)

they add up to 74,so ATQ

So,

Integers are 7,8,9

Verification:

Hence the calculate integers are correct.

Hope it helps you.

Step-by-step explanation:

MRK AS BRAINLIST

Hey mate here is your ANSWER in Photo..

PLZE give me FIRST BRAINLIEST TAG pls pls pls pls......

Answer.

LET the SIDES of the TRIANGLE be AB, BC and AC

AB =2x

BC= 3x

AC =4x

AC = 20 + AB

=20 + 4x

AB^2+BC^2=AC^2

2x^2 + 3x^2 = (20+4x)^2

=20^2 +2×20×4x +4x^2 + 4x^2

ANSWER:

K= -4

Step-by-step explanation:

x+kx+3=0

Since, it has EQUAL roots, x=1,

Hence,

1+k×1+3=0

1+k+3=0

4+k=0

Therefore , k= -4

Glad to HELP

ANSWER:

in TRIANGLE UM of ANGLE is 180

Step-by-step explanation:

180-100=80

then answer is AC

HELLO MATE here's your ANSWER for your QUESTION is

3,552 is the NUMBER

ANSWER:

what is your QUESTIONS

Answer:

2.5%

Step-by-step explanation:

1DAY = 24hr

1hr = 60MINS

1day =24multiply by 60

= 1440

36/1440multiply by 100

=2.5%

Answer:

Step-by-step explanation:

PROOF OF CAUCHY’S THEOREM

KEITH CONRAD

The converse of Lagrange’s theorem is false in general: if G is a finite group and d | |G|

then G doesn’t have to contain a subgroup of order d. (For example,|A4| = 12 and A4

has no subgroup of order 6). We will show the converse is true when d is prime. This is

Cauchy’s theorem.

Theorem. (Cauchy 1845) Let G be a finite group and p be a prime factor of |G|. Then

G contains an element of order p. Equivalently, G contains a subgroup of order p.

The equivalence of the existence of an element of order p and a subgroup of order p is

easy: an element of order p generates a subgroup of order p, while conversely any nonidentity

element of a subgroup of order p has order p because p is prime.

Before treating Cauchy’s theorem, let’s SEE that the special case for p = 2 can be proved

in a simple WAY. If |G| is even, consider the set of pairs {g, g−1}, where g 6= g

−1

. This

takes into account an even number of elements of G. Those g’s that are not part of such

a pair are the ONES satisfying g = g

−1

, i.e., g

2 = e. Therefore if we count |G| mod 2, we

can ignore the pairs {g, g−1} where g 6= g

−1 and we obtain |G| ≡ |{g ∈ G : g

2 = e}| mod 2.

One solution to g

2 = e is e. If it were the only solution, then |G| ≡ 1 mod 2, which is false.

Therefore some g0 6= e satisfies g

2

0 = e, which gives us an element of order 2.

Now we prove Cauchy’s theorem.

Proof. We will use induction on |G|.

1 Let n = |G|. Since p | n, n ≥ p. The base case is

n = p. When |G| = p, any nonidentity element of G has order p because p is prime. Now

suppose n > p, p | n, and the theorem is true for all groups having order less than n that is

divisible by p. We will treat separately abelian G (using homomorphisms) and nonabelian

G (using conjugacy classes).

Case 1: G is abelian.

Assume no element of G has order p and we will get a contradiction.

No element has order divisible by p: if g ∈ G has order r and p | r then g

r/p would have

order p.

Let G = {g1, g2, . . . , gn} and let gi have order mi

, so each mi

is not divisible by p. Let m

be the least common multiple of the mi

’s, so m is not divisible by p and g

m

i = e for all i.

Because G is abelian, the function f : (Z/(m))n → G given by f(a1, . . . , an) = g

a1

1

· · · g

an

n

is

a homomorphism:

2

f(a1, . . . , an)f(b1, . . . , bn) = f(a1 + b1, . . . , an + bn).

That is,

g

a1

1

· · · g

an

n

g

b1

1

· · · g

bn

n = g

a1

1

g

b1

1

· · · g

an

n

g

bn

n = g

a1+b1

1

· · · g

an+bn

n

1Proving a theorem on groups by induction on the order of the group is a very fruitful idea in group

theory.

2This function is well-defined because g

m

i = e for all i, so g

a+mk

i = g

a

i

for any k ∈ Z.

1

2 KEITH CONRAD

from COMMUTATIVITY of the gi

’s. This homomorphism is surjective (each element of G is a

gi

, and if ai = 1 and other aj ’s are 0 then f(a1, . . . , an) = gi), so by the first isomorphism

theorem (Z/(m))n/ ker f ∼= G. Therefore

|G| =

|(Z/(m))n

|

| ker f|

=

mn

| ker f|

,

so |G|| ker f| = mn

. Thus |G| is a factor of mn

, but p divides |G| and mn

is not divisible by

p, so we have a contradiction.

Case 2: G is nonabelian.

Assume no element of G has order p and we will get a contradiction.

In every proper subgroup H of G there is no element of order p (H may be abelian or

nonabelian), so by induction no proper subgroup of G has order divisible by p. For each

proper subgroup H, |G| = |H|[G : H] and |H| is not divisible by p while |G| is divisible by

p, so p | [G : H] for every proper subgroup H of G.

Since G is nonabelian it has some conjugacy classes with size greater than 1. Let these

be represented by g1, g2, . . . , gk. Conjugacy classes in G of size 1 are the elements in Z(G).

Since the conjugacy classes in G form a partition of G, computing |G| by adding the sizes

of its conjugacy classes implies

Check the attachment.

In the sum i have not done it SEPARATELY like first finding INTEREST n then maturitu VALUE. i have DIRECTLY substituted in maturity value so that it makes the sum easier & small.

Answer:

55

Step-by-step EXPLANATION:

Sum of ANGLES in QUADRILATERAL is 360

so,

The smallest ANGLE is

Step-by-step EXPLANATION:

very good RAM, u not studied WELL and QUESTION is incomplete

HEY MATE ,

2,4,8,12 ETC

ANSWER:

PLEASE WRITE QUESTION PROPERLY

ANSWER:

ANSWER:

i don't KNOW SIS................

Given :

The rate of interest = 12% compounded monthly

Future value of amount = Rs 55,000

Number of payments = 12

To Find :

The Amount invested each month

Solution :

Future value = PRESENT value ×

Or, Rs 55,000 = P  ×

Or, Rs 55,000 = P  ×

Or, Rs 55,000 = P  × 1.126

∴                    P =

i.e Present value = P = Rs 48845.4

Thus, Amount invested every month =

                                                             = Rs 4070.45

Hence, The Amount invested invested by person every month is Rs 4070.45    ANSWER

HEY MATE ,

ANSWER is ,

(0,0) and Y AXIS

Answer:

Sides of the FLOOR = 5M and 3 cm ( length and BREADTH)

                             = 500cm and 300cm            [ 1m = 100cm]

Area of the floor = lb

                           = 500 × 300 = 150000cm².

Square tile : Side = 25cm

   ∴ Area of a tile= 25 × 25

                            = 625cm².

∴No. of tiles required = Area of the floor ÷ Area of a tile

                                   = 150000 ÷ 625

                                  =240.

∴ 240 square tiles are required to cover the room.

PLEASE MARK ME AS BRAINLIEST

ANSWER:

area of a rhombus= 1/2 × d1 × D2 where d1 and D2 are the DIAGONALS of the rhombus

KINDLY give an APPROPRIATE question,with complete information,?

Step-by-step EXPLANATION:

each ANGLE is ACUTE angle