The vertices of a quadrilateral are, (0,2), (4,2), (-3,0), (5,0) .

Mathematics

1 answer:

sasho [114]3 years ago

3 0

By plotting, (please see the image attached) we can clearly see that neither side is equal nor perpendicular to each other. Same condition of we use the distance formula to compare the sides. The type of quadrilateral that has meets the condition is a trapezoid

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x=-1 x=8
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Determine whether Rolle's Theorem can be applied to f on the closed interval [a, b]. (Select all that apply.)f(x) = 6 cos πx, [0

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Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

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2 years ago

At what point does the curve have maximum curvature? y = 9 ln(x) (x, y) =

Andrews [41]

y = 9ln(x)
y' = 9x^-1 =9/x
y'' = -9x^-2 =-9/x^2

curvature k = |y''| / (1 + (y')^2)^(3/2)

= |-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,

we find where k' = 0. 
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.

x=9/√2=6.36

y=9 ln(x)=9ln(6.36)=16.66

the answer is
(x,y)=(6.36,16.66)

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