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# Tutor profile: Olakunle A.

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Olakunle A.
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## Questions

### Subject:Geometry

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Question:

Find the distance between the plane x + 2y - 3z + 1 = 0 and the point (1,3,5)

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Olakunle A.

The equation can be written in the form n . (x,y,z) = -1 where n = (1,2,-3). We then find the foot of the perpendicular from the point (1,3,5) to the plane. Suppose the foot is a point (x,y,z). Then (1,3, 5) - (x, y, z) = \lambda (1,2,-3) for some constant \lambda. Now we take the dot product of both sides with the vector n, to have (1,3,5).(1,2,-3) + 1 = \lambda(1,2,-3).(1,2,-3. ). So, \lambda = -7 / 14 = -1/2. Giving 1 - x = -1/2 (x = 3/2), y = 4, z = 7/2. so (3/2, 4, 7/2) is the foot of the perpendicular and so the square of the distance is (1 - 3/2)^2 + (3 - 4)^2 + (5 - 7/2)^2 = 1/4 + 1 + 9/4 = 14/4 . So the distance is sqrt(14)/2.

### Subject:Discrete Math

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Question:

A family of seven consisting the father , mother and five children are to be seated on seven seats, what is the number of ways to arrange them so that the husband is not seated next to his wife?

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Olakunle A.

There are 5! ways of arranging the 5 children. In any of this arrangements, there are six possible places for one of them and 5 possible places left for the other. So the total number of ways of arranging them is 5! X 6 X 5 = 120 X 30 = 3600 ways

### Subject:Calculus

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Question:

What is the minimum value of the function f(x) = x^2 + y^2 subject to the condition that x + y= 1 ?

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Olakunle A.

Let P = x^2 + y^2 be the objective function . From the given condition, we have that y = 1 - x substitute into the objective function to have P(x) = x^2 + (1 - x)^2. To find a critical point, we take the first derivative P'(x) = 2x -2(1-x) = 4x - 2. Set P'(x) = 0, to have 4x -2 = 0, so x = 2/4 = 1/2. We now check that this actually corresponds to a minimum point. We take second derivative P''(x) = 4, so P''(1/2) > 0. Hence (x,y) = (1/2, 1/2) gives a minimum point . The minimum value of the function is (1/2)^2 + (1/2)^2 = 1/2.

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