Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never be negative. So, x = 40.
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1
Answers
Answer:
Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never be negative. So, x = 40.
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1 negative. So, x = 40.Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never be negative. So, x = 40.
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never be negative. So, x = 40.
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never be negative. So, x = 40.
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never be negative. So, x = 40.
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1Then the increased speed of the train = (x + 5) km/hr.
Distance = 360 km
According to question,
360
360
= 1
(x + 5)
1800
= 1
x(x + 5)
x2 + 5x – 1800 = 0
x(x + 45) - 40(x + 45) = 0 =>
= 1
360(x + 5) - 360x
x(x + 5)
x(x + 5) = 1800
x2 + 45x - 40x - 1800 = 0
(x + 45) (x - 40) = 0
x = -45 or 40
Speed of the train can never be negative. So, x = 40.
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1
Hence, normal speed of the train = 40 km/hr
ercise 4.5
d on number and fraction
Find the numbers.
im of two numbers is 16 and sum of their reciprocal is
1
Answer:
please make me brainlist answer I will follow you