Math, asked by madhura8039, 2 months ago

Simplify and write in exponential form: a

10 × a

22 × a

11​

Answers

Answered by freef3497
0

Step-by-step explanation:

Find the value of:

(i) 26 (ii) 93 (iii) 112 (iv) 54

Answer:

(i) 26 = 2 * 2 * 2 * 2 * 2 * 2 = 64

(ii) 93 = 9 * 9 * 9 = 729

(iii) 112 = 11 * 11 = 121

(iv) 54 = 5 * 5 * 5 * 5 = 625

Question 2:

Express the following in exponential form:

(i) 6 * 6 * 6 * 6 (ii) t * t (iii) b * b * b * b (iv) 5 * 5 * 7 * 7 * 7 (v) 2 * 2 * a * a

(vi) a * a * a * c * c * c * c * d

Answer:

(i) 6 * 6 * 6 * 6 = 64

(ii) t * t = t2

(iii) b * b * b * b = b4

(iv) 5 * 5 * 7 * 7 * 7 = 52 * 73

(v) 2 * 2 * a * a = 22 * a2

(vi) a * a * a * c * c * c * c * d = a3 * c4 * d

Question 3:

Express each of the following numbers using exponential notations:

(i) 512 (ii) 343 (iii) 729 (iv) 3125

Answer:

(i) 512 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 29

Class_7_Exponents_&_Powers_Primefactorization_of_512

(ii) 343 = 7 * 7 * 7 = 73

Class_7_Exponents_&_Powers_Primefactorization_of_343

(iii) 729 = 3 * 3 * 3 * 3 * 3 * 3 = 36

Class_7_Exponents_&_Powers_Primefactorization_of_729

(iv) 3125 = 5 * 5 * 5 * 5 * 5 = 55

Class_7_Exponents_&_Powers_Primefactorization_of_3125

Question 4:

Identify the greater number, wherever possible, in each of the following:

(i) 43 and 34 (ii) 53 or 35 (iii) 28 or 82 (iv) 1002 or 2100 (v) 210 or 102

Answer:

(i) 43 = 4 * 4 * 4 = 64

34 = 3 * 3 * 3 * 3 = 81

Since 64 < 81

So, 34 is greater than 43

(ii) 53 = 5 * 5 * 5 = 125

35 = 3 * 3 * 3 * 3 * 3 = 243

Since 125 < 243

So, 35 is greater than 53

(iii) 28 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256

82 = 8 * 8 = 64

Since, 256 > 64

Thus, 28 is greater than 82

(iv) 1002 = 100 * 100 = 10,000

2100 = 2 * 2 * 2 * 2 * 2 * …..14 times * ……… * 2 = 16,384 * ….. * 2

Since, 10,000 < 16,384 * ……. * 2

Thus, 2100 is greater than 1002.

(v) 210 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 1,024

102 = 10 * 10 = 100

Since, 1,024 > 100

Thus, 210 is greater than 102

Question 5:

Express each of the following as product of powers of their prime factors:

(i) 648 (ii) 405 (iii) 540 (iv) 3,600

Answer:

(i) 648 = 23 * 34

Class_7_Exponents_&_Powers_Primefactorization_of_648

(ii) 405 = 5 * 34

Class_7_Exponents_&_Powers_Primefactorization_of_405

(iii) 540 = 22 * 33 * 5

Class_7_Exponents_&_Powers_Primefactorization_of_540

(iv) 3,600 = 24 * 32 * 52

Class_7_Exponents_&_Powers_Primefactorization_of_3600

Question 6:

Simplify:

(i) 2 * 103 (ii) 72 * 22 (iii) 23 * 5 (iv) 3 * 44 (v) 0 * 102

(vi) 52 * 33 (vii) 24 * 32 (viii) 32 * 104

Answer:

(i) 2 * 103 = 2 * 10 * 10 * 10 = 2,000

(ii) 72 * 22 = 7 * 7 * 2 * 2 = 196

(iii) 23 * 5 = 2 * 2 * 2 * 5 = 40

(iv) 3 * 44 = 3 * 4 * 4 * 4 * 4 = 768

(v) 0 * 102 = 0 * 10 * 10 = 0

(vi) 52 * 33 = 5 * 5 * 3 * 3 * 3 = 675

(vii) 24 * 32 = 2 * 2 * 2 * 2 * 3 * 3 = 144

(viii) 32 * 104 = 3 * 3 * 10 * 10 * 10 * 10 = 90,000

Question 7:

Simplify:

(i) (-4)3 (ii) (-3) * (-2)3 (iii) (-3)2 * (-5)2 (iv) (-2)3 * (-10)3

Answer:

(i) (-4)3 = (-4) * (-4) * (-4) = -64

Answered by tanishanegi249
0

Answer:

10×a=10a

22×a=22a

11×a=11a

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