A civil engineer found that the durability dd of the road, she is laying depends on two functions y1y1 and y2y2 as follows: d=y22+4y1−21d=y22+4y1−21. Functions y1y1 and y2y2 depend on the amount of plastic (xx) mixed in bitumen, and their variations are linear functions of xx. Let y1=8y1=8 and y2=3y2=3 for x=0x=0 and y1=0y1=0 and y2=7y2=7 for x=7x=7. Find the durability of the road (upto 2 decimal points), if the amount of plastic is such that both the functions are equal.
Answers
Answer:
19.36
Step-by-step explanation:
y1 = -8x/7 +8
y2 = 4x/7 +3
y1=y2 then find x
x= 2.92
Sub x in given equation to find durability
d=19.36
Given : d = y₂² + 4y₁ - 21
durability d of the road depends on two functions y₁ and y₂
functions y₁ and y₂ depend on the amount of plastic ( x ) mixed in bitumen,
variations are linear functions of x
y 1 = 8 and y 2 = 3 for x = 0 and y 1 = 0 and y 2 = 7 for x = 7
To Find : the durability of the road (upto 2 decimal points), if the amount of plastic is such that both the functions are equal.
Solution:
y₁ = 8 for x = 0
=0 for x = 7
=> ( y₁ - 0) = ( 0 - 8)/(7 - 0) ( x - 7)
=> y₁ = (-8/7)(x - 7)
=> y₁ = -8x/7 + 8
y₂ = 3 for x = 0
7 for x = 7
( y₂ - 7) = (7 - 3)/(7 - 0) ( x - 7)
=> y₂ = (4/7)(x - 7) + 7
=> y₂ = 4x/7 - 4 + 7
=> y₂ = 4x/7 + 3
both the functions are equal.
4x/7 + 3 = -8x/7 + 8
=> 12x/7 = 5
=> x = 35/12
y₁ = -8x/7 + 8
x = 35/12
=> y₁ = -8 (35/12)/7 + 8 = - 10/3 + 8 = 14/3
y₂ = 4x/7 + 3 = 14/3 for x = 35/12
d = y₂² + 4y₁ - 21
d = (14/3)² + 4(14/3) - 21
d = 196/8 + 168/9 - 189/9
d = 175/9
d = 19.44
durability of the road = 19.44
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