let a and b positive integers such that 90 less than a+b less than 99 and 0.9 less than a/b less than 0.91. Find ab/46 please answer fast About the author Maya

Given :– Let a and b are positive integers such that 90 < a + b < 99 and 0.9 < a/b < 0.91 . Find (ab/46) ? Answer :– → 90 < a + b < 99 ——— Eqn.(1) → 0.9 < a/b < 0.91 ——– Eqn.(2) taking Eqn.(2), → 0.9 < a/b < 0.91 → 0.9b < a < 0.91b adding b , → 0.9b + b < a + b < 0.91b + b using Eqn.(1) now , we get, → 0.9b + b < 99 => 1.9b < 99 => b < 52.1 → 0.91b + b > 90 => 1.91 b > 99 => b > 51.8 since it is given that, b is a positive integer , → b = 52 . putting this value we get, → 0.9 * 52 < a < 0.91 * 52 → 46.8 < a < 47.3 → a = 47 . but, → a + b < 99 → 52 + 47 < 99 → 99 < 99 therefore, b must be 51 and a must be 46 . hence, → (ab)/46 → (51 * 46)/46 → 51 (Ans.) Note :- if first condition was 90 ≤ a + b ≤ 99 , a will be 52 and b will be 47 . Learn more :- if the positive square root of (√190 +√ 80) i multiplied by (√2-1) and the product is raised to the power of four the re… https://brainly.in/question/26618255 Reply

Given : a and b be positive integers such that 90<a+b<99 and 0.9<a/b<0.91 To Find : ab/46 Solution: 90<a+b<99 0.9<a/b<0.91 0.9b < a a < 0.91b 0.9b < a a+b<99 => a < 99-b => 0.9b < 99-b => 1.9b < 99 => b < 52.1 a < 0.91b 90<a+b => 90 – b < a => 90 – b < 0.91b => 90 < 1.91b => 47.12 < b 47.12 < b < 52.1 => b = 48 , 49 , 50 , 51 , 52 0.9b < a < 0.91b b = 48 => 43.2 < a < 43.68 no integer values of a b = 49 => 44.1 < a < 44.59 no integer values of a b = 50 => 45 < a < 45.5 no integer values of a b = 51 => 45.9 < a < 46.41 so a = 46 b = 52 => 46.8 < a < 47.32 so a = 47 but 52 + 47 not less than 99. b = 51 and a = 46 only satisfy ab /46 = 46 * 51/46 = 51 Learn More Find ab/46 https://brainly.in/question/40052768 Reply

Given:–Answer:–→ 90 < a + b < 99 ——— Eqn.(1)

→ 0.9 < a/b < 0.91 ——– Eqn.(2)

taking Eqn.(2),

→ 0.9 < a/b < 0.91

→ 0.9b < a < 0.91b

adding b ,

→ 0.9b + b < a + b < 0.91b + b

using Eqn.(1) now , we get,

→ 0.9b + b < 99 => 1.9b < 99 => b < 52.1

→ 0.91b + b > 90 => 1.91 b > 99 => b > 51.8

since it is given that, b is a positive integer ,

→ b = 52 .

putting this value we get,

→ 0.9 * 52 < a < 0.91 * 52

→ 46.8 < a < 47.3

→ a = 47 .

but,

→ a + b < 99

→ 52 + 47 < 99

→ 99 < 99

therefore, b must be 51 and a must be 46 .

hence,

→ (ab)/46

→ (51 * 46)/46

→

51(Ans.)Note:- if first condition was 90 ≤ a + b ≤ 99 , a will be 52 and b will be 47 .Learn more:-if the positive square root of (√190 +√ 80) i multiplied by (√2-1) and the

product is raised to the power of four the re…

https://brainly.in/question/26618255

Given :a and b be positive integers such that 90<a+b<99 and 0.9<a/b<0.91To Find :ab/46Solution:90<a+b<99

0.9<a/b<0.91

0.9b < a

a < 0.91b

0.9b < a

a+b<99

=> a < 99-b

=> 0.9b < 99-b

=> 1.9b < 99

=> b < 52.1

a < 0.91b

90<a+b

=> 90 – b < a

=> 90 – b < 0.91b

=> 90 < 1.91b

=> 47.12 < b

47.12 < b < 52.1

=> b = 48 , 49 , 50 , 51 , 52

0.9b < a < 0.91b

b = 48

=> 43.2 < a < 43.68 no integer values of a

b = 49

=> 44.1 < a < 44.59 no integer values of a

b = 50

=> 45 < a < 45.5 no integer values of a

b = 51

=> 45.9 < a < 46.41 so a = 46

b = 52

=> 46.8 < a < 47.32 so a = 47

but 52 + 47 not less than 99.

b = 51 and a = 46 only satisfy

ab /46 = 46 * 51/46 =

51Learn More

Find ab/46

https://brainly.in/question/40052768