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Tutor profile: Victor A.

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Victor A.
Private tutor for five years, I am a physicist, I master calculus and physics.
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Questions

Subject:Pre-Calculus

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Question:

In the $$xy$$-plane, the graph of the function $$h$$ is a line. If $$h(-1)=4$$ and $$h(5) = 1$$, what is the value of $$h(0)$$?

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Victor A.

The variation of a linear function depends on the value of the slope, so we must determine the value of the slope of the line. Note that the searched value corresponds to the intersection of the line with the y-axis. The slope is: $( m=\frac{h(5)-h(1)}{5-(-1)}= \frac{1-4}{5+1}=\frac{-3}{6}=-\frac{1}{2}$) The negative sign of the slope shows us that the line is decreasing. The value of the slope indicates that every two units in which the independent variable $$(x)$$ varies, there is one unit of variation of $$h(x)$$. Therefore, if the independent variable changes from -1 to 0, that is, one unit. The value of h is reduced from 4 to 3.5. Result: $( h(0)=3.5$)

Subject:Calculus

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Question:

Determine the concavity of the function: $( y=-\frac{x^2}{4x+4}$)

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Victor A.

To analyze any function from its derivatives, we must obtain the two derivatives (Remember the derivative of a quotient): First: $(y'=-\frac{x(x+2)}{4(x+1)^2}$) Second: $( y'' = - \frac{1}{2(x+1)^3}$) In order to analyze concavity, we analyze the domain of the second derivate. Therefore, the value $$x = -1$$ is where the second derivative does not continue. Therefore, just take a value greater than -1, such as zero, and substitute in the second derivative: $( y''(0)=- \frac{1}{2}$) As the value is negative and according to the criterion of the second derivative, that says: If $$f''(x)>0$$, then graph of $$f$$ is concave up. If $$f''(x)<0$$, then graph of $$f$$ is concave down. We can conclude that the function is concave upward at $$(-\infty, -1)$$ and concave downward at $$(-1, \infty)$$.

Subject:Physics

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Question:

A typical question: A block of mass $$m$$ sliding down an incline at constant speed is initially at a height $$h$$ above the ground. The coefficient of kinetic friction between the mass and the incline is $$\mu$$. If the mass continues to slide down the incline at a constant speed, how much energy is dissipated by friction by the time the mass reaches the bottom of the incline? (A) $$\frac{mgh}{\mu}$$ (B) $$mgh$$ (C) $$\frac{\mu mgh}{\sin \theta}$$ (D) $$mgh \sin \theta$$

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Victor A.

Although, the typical way to solve this problem is by using a free-body diagram and considering the forces on the body. A more efficient solution is to use the principle of conservation of energy. So if the block has a constant speed, the kinetic energy is the same at the top and bottom of the slope, so the only energy available to dissipate is gravitational potential energy. Finally, the answer is $$mgh$$.

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