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2 years ago

Write 10 places in India and write their population in Indian and international system

Mathematics

2 answers:

Delvig [45]2 years ago

6 0

Answer:

Mumbai 12,442,373
Delhi 11,007,835
Bangalore 8,436,675
Hyderabad 6,809,970
Ahmadabad 5,570,585
Chennai 4,681,087
Kolkata 4,486,678
Surat 4,467,797
Pune 3,124,458
Jaipur 3,046,163.

Step-by-step explanation:

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gulaghasi [49]2 years ago

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Solution :-

Following are the population of five major cities of India according to the Census of 2011.

1) Mumbai

a) Indian System :

1,24,42,373 - One Crore Twenty Four Lakh Forty Two Thousand Three Hundred Seventy Three.

b) International system :

12,442,373 - Twelve Million Four Hundred Forty Two Thousand Three Hundred Seventy Three.

2) Delhi

a) Indian System :

1,10,34,555 - One Crore Ten Lakh Thirty Four Thousand Five Hundred Fifty Five.

b) International System :

11,034,555 - Eleven Million Thirty Four Thousand Five Hundred Fifty Five.

3) Bangalore

a) Indian System :

84,43,675 - Eighty Four Lakh Forty Three Thousand Six Hundred Seventy Five.

b) International system :

8,443,675 - Eight Million Four Hundred Forty Three Thousand Six Hundred Seventy Five.

4) Hyderabad

a) Indian System :

67,31,790 - Sixty Seven Lakh Thirty One Thousand Seven Hundred Ninety.

b) International System :

6,731,790 - Six Million Seven Hundred Thirty One Thousand Seven Hundred Ninety.

5) Ahmedabad

a) Indian System :

55,77,940 - Fifty Five Lakh Seventy Seven Thousand Nine Hundred Forty.

b) International System :

5,577,940 - Five Million Five Hundred Seventy Seven Thousand Nine Hundred Forty.

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dusya [7]

7. Law of cosines

a² = b² + c² - 2bc cos A

where a is the side opposite to ∠A and b and c are adjacent sides to ∠A.

Here, a=BC= 3.8

b=CA =4.8

c = AB = 3.2

Substituting the values in the formula,

3.8² = 4.8² + 3.2² - 2(4.8)(3.2) cos A

30.72 cos A =23.04 + 10.24 -14.44=18.84

cos A = 18.84/30.72 = 0.6133

∴ cos A =0.6133

8. It is a right angled triangle.

Pythagoras theorem

hypotenuse² = Perpendicular² + base²

Here, hypotenuse = 6.7, base = 5.4

Perpendicular² = 6.7² -5.4²=44.89 - 29.16 =15.73

Perpendicular = √15.73 = 3.966 = 3.97

∴ The length of the missing side is 3.97.

9. Given: Side of new cube = 5 * Side of original cube

Let the side of the original cube measure 1 unit.

So, volume of original cube = side³ = 1

The side of new cube = 5 * 1 =5 units

So, volume of new cube = side³ = 5³ =125

The volume of the bigger cube is 125 times larger than the original cube.

10. Given : Height of the cone = 7 in

Radius of the base of the cone = 2 in

Volume of cone = (1/3) πr²h

=(1/3)π(2²)(7)

=29.32 cubic inches

∴ Volume of the cone is 29.32 cubic inches.

11. The given equation is:

x² -18x +10 =0

(x² -18x + 9²) +10 -9² =0

(x-9)²-71 =0

(x-9)² =71

x-9 =√71

x = 9 + √71

∴ x = 9+√71

7 0

3 years ago

What are the degree and leading coefficient of the polynomial? <br><br> - 2u + 12u ^ 2 + 8 - u ^ 9

USPshnik [31]

Answer:

Step-by-step explanation:

hello :

- 2u + 12u ² + 8 - u ^ 9 = -1u ^ 9+12u² -2u+8...standard form

the degree is 9 and leading coefficient is -1

7 0

3 years ago

Misha and Dontavious were asked to solve the following problem using linear programming. A local school district can operate a l

bearhunter [10]

Answer:

did you get the answer?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Simplify each exponential expression using the properties of exponents and match it to the correct answer.

saveliy_v [14]

1) (2 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^2 = 2

2) (3^3) (4^0)^2 (3 $\times$ 2)^-3 (2^2) = 1/2

3) (3^7 $\times$ 4^7) (2 $\times$ 5)^-3 (5)^2 / (12^7) (5^-1) (2^-4) = 2

4) (2.3)^-1 (2^0) / (2.3)^-1 = 1

<u>Step-by-step explanation</u>:

Step 1 :

(2 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^2

⇒ (2^3) (3^-6) (5^2) (3^4) / (3^-2) (10^2)

⇒ (2.2^2.5^2) (3^-6.3^4) / (3^-2) (10^2)

⇒ (2)(10^2) (3^-2) / (3^-2) (10^2)

⇒ 2

Step 2 :

(3^3) (4^0)^2 (3 $\times$ 2)^-3 (2^2)

Any number with power 'zero' is 1.

⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)

⇒ (2^(-3+2))

⇒ 2^-1 = 1/2

Step 3 :

(3^7 $\times$ 4^7) (2 $\times$ 5)^-3 (5)^2 / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (2^-3) (5^(-3+2)) / (5^-1) (2^-4)

⇒ (2^-3) (5^-1) / (5^-1) (2^-4)

⇒ 1 / (2^-1)

⇒ 2

Step 4 :

(2.3)^-1 (2^0) / (2.3)^-1

⇒ (2^0)

⇒ Any number with power 'zero' is 1

⇒ 1

8 0

3 years ago

I really need help please (links will be reported)

mrs_skeptik [129]

Answer: all the ones that do not have a negative sign in front easy.

7 0

3 years ago

Write 10 places in India and write their population in Indian and international system​

Write 10 places in India and write their population in Indian and international system