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# Tutor profile: Owen K.

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Owen K.
Student at University of Oregon
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## Questions

### Subject:Physics (Newtonian Mechanics)

TutorMe
Question:

Determine the net force required to accelerate a 18099.45N car from 0 to 27 m/s (60 mph) in 10.0 seconds.

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Owen K.

F=ma m=18099.45/9.81 m=1845kg a=dv/dt a=(27m/s)/10s a=2.7m/s/s F=1845(2.7) F=4981.5N

### Subject:Geometry

TutorMe
Question:

Jon wants to make a cabbage garden but the pesky Lemurs keep steeling his cabbages. He has 200 meters of fencing and wants to make the largest enclosure possible to keep his cabbages safe. What length and width should Jon use to maximize the area of his garden?

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Owen K.

The perimeter of the fence is 200=2L+2W the area is A=LxW I solve for one of the variables using the perimeter equation. (200=2L+2W)/2 100=L+W L=100-W I substitute the L in the area equation for (100-W) A=(100-W)(W) distribute A=100W-W^2 or A= -W^2+100W <<<Parabola! The maximum area is the vertex of the parabola at the point (h,k) h= -b/2a h= -(100)/2(-1) = 50 k= -h^2+100h k= -50^2+100(50)=2500 Therefore the max area is 2500 square meters when the width is 50meters. To find L just plug in 50 meters into either the area or the perimeter equation. Let's use Area. 2500=Lx50 2500/50=L L=50 Well that turned out nice, didn't it :) To wrap up: A=2500m^2 W=50m L=50m

### Subject:Calculus

TutorMe
Question:

Find d/dx of (x^2 + 4x)^(1/3)

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Owen K.

d/dx of (x^2 + 4x)^(1/3) = 1/3 (x^2 + 4x)^(-2/3) (2x + 4)

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