The questions are in the link. SOS PLEASE HELP

P is the point on the line 2x+y-10=0 such that the length of OP, the line segment from the origin O to P, is a minimum. Find the

nirvana33 [79]

The minimum distance is the perpendicular distance. So establish the distance from the origin to the line using the distance formula.
The distance here is: d2=(x−0)^2+(y−0)^2
 =x^2+y^2

To minimize this function d^2 subject to the constraint, 2x+y−10=0
If we substitute, the y-values the distance function can take will be related to the x-values by the line:y=10−2x
You can substitute this in for y in the distance function and take the derivative:
d=sqrt [x2+(10−2x)^2]

d′=1/2 (5x2−40x+100)^(−1/2) (10x−40)
Setting the derivative to zero to find optimal x,
d′=0→10x−40=0→x=4

This will be the x-value on the line such that the distance between the origin and line will be EITHER a maximum or minimum (technically, it should be checked afterward).
For x = 4, the corresponding y-value is found from the equation of the line (since we need the corresponding y-value on the line for this x-value).

Then y = 10 - 2(4) = 2.
So the point, P, is (4,2).

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4 years ago

What is the volume of the box pictured below?

SpyIntel [72]

Answer:

The volume of the box is 3/4 cubic meters.

Step-by-step explanation:

We know that volume = length * width * height. We have length, width, and height already, so all we have to do is multiply.

5/6 * 1 1/8 * 4/5. First we convert 1 1/8 from a mixed fraction, to give us 9/8, as 8/8 = 1.

5/6 * 9/8 * 4/5. 5 * 9 = 45, and 6 * 8 = 48. 45/48 reduces to 15/16.

15/16 * 4/5. 15 * 4 = 60, and 16 * 5 = 80. 60/80 reduces to 6/8, which reduces to 3/4.

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