Answer:
One equation would be y= -1.5x +11
Step-by-step explanation:
Here's a table to show the coordinates
x: y:
8 -1
-------------------
-4 17
the x coor. 8 goes down by 12 to get to the next x coor. -4.
8-12= -4
the y coor. -1 goes up by 18 to get to the next y coor. 17.
-1+18= 17
therefore 18/-12 is -1.5 which can be your k in y=kx+b.
Check your answer.
-1.5*8= -12 but when you add 11, you get the y coor. -1
Do the same with the other x coor.
-4*-1.5= 6 + 11 =17
11 can be your b in the slope-intercept form equation.
One equation may be y = -1.5x +11
y = 9ln(x)
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2
curvature k = |y''| / (1 + (y')^2)^(3/2)
<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2)
= (9/x^2) / (1 + 81/x^2)^(3/2)
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)]
= 9x(x^2 + 81)^(-3/2).
To maximize the curvature, </span>
we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2)
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2]
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2)
Setting k' = 0 yields x = ±9/√2.
Since k' < 0 for x < -9/√2 and k' > 0 for x >
-9/√2 (and less than 9/√2),
we have a minimum at x = -9/√2.
Since k' > 0 for x < 9/√2 (and greater than 9/√2) and
k' < 0 for x > 9/√2,
we have a maximum at x = 9/√2. </span>
x=9/√2=6.36
<span>y=9 ln(x)=9ln(6.36)=16.66</span>
the
answer is
(x,y)=(6.36,16.66)
X=120
This is the answer because 120-21=99
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