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Tutor profile: Clayton G.

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Clayton G.
College graduate of Mathematics. Served as a supplemental instruction leader in college.
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Questions

Subject: Pre-Calculus

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Question:

Let $$f(x)=(x+2)^2$$ Calculate $$\frac{f(a+h)-f(a)}{h}.$$

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Clayton G.
Answer:

Note that $$f(x)=x^2+4x+4.$$ Then $$f(a)=a^2+4a+4$$ and $$f(a+h)=(a+h)^2+4(a+h)+4=a^2+2ah+h^2+4a+4h+4.$$ It follows that $$\frac{f(a+h)-f(a)}{h}=\frac{a^2+2ah+h^2+4a+4h+4-(a^2+4a+4)}{h}=\frac{a^2+2ah+h^2+4a+4h+4-(a^2+4a+4)}{h}=\frac{2ah+h^2+4h}{h}=\frac{h(2a+h+4)}{h}=2a+h+4.$$. Therefore $$\frac{f(a+h)-f(a)}{h}=2a+h+4.$$

Subject: Calculus

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Question:

The position $$s$$ (measured in meters) of a moving object at time $$t\geq 0$$ is given by the function $$s(t)= \frac{t^3-1}{t+2}$$ . Find the acceleration of the object at t=2 seconds.

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Clayton G.
Answer:

Note that the derivative of the position function $$s(t)$$ yields the velocity function $$v(t)$$, and the derivative of the velocity function $$v(t)$$ yields the acceleration function $$a(t)$$. Thus, we must find the second derivative of $$s(t)$$; that is, $$s''(t)=a(t)$$. Taking the first derivative of $$s(t)$$ by using the quotient rule for differentiation, we get $$s'(t)=v(t)=\frac{2t^3+6t^2-1}{(t+2)^2}.$$ We then use the quotient rule for differentiation once again to find the second derivative, which we find to be $$s''(t)=v'(t)=a(t)=\frac{2t^3+12t^2+24t-2}{(t+2)^3}.$$ Now that we have the function for acceleration $$a(t)$$, we may let $$t=3$$ and solve for $$a(2)$$. Solving, we find that $$a(2)=\frac{55}{32}$$ meters per second. Thus, the acceleration of the object at $$t=2$$ is $$a(2)==\frac{55}{32}\frac{m}{s}.$$

Subject: Algebra

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Question:

Jimmy sells apples and oranges at his fruit stand. He sells apples for $1.25 and oranges for $1.50. Today, he sold a total of 43 fruits and earned a total of $59.75. How many of each fruit did Jimmy sell today?

Inactive
Clayton G.
Answer:

Let $$x$$ and $$y$$ represent the number of apples and oranges sold today, respectively. By the given, a total of 43 fruits have been sold today. It follows that $$x+y=43$$ (we will call this equation A). Furthermore, we know by the given that apples and oranges are sold for $1.25 and $1.50, respectively, and that the total earnings of today are $59.75. It follows that $$ 1.25x + 1.50y = 59.75 $$ (we will call this equation B). Since we now have two equations, we may solve for the two unknowns $$x$$ and $$y$$. Multiplying both sides of equation A by 1.25, we find that $$1.25x+1.25y=53.75$$. Subtracting equation A from equation B, we find that $$0.25y=6$$. It follows that $y=24.$ Plugging this value for $$y$$ into equation A, we find that $$x+24=43$$ and therefore $$x=19$$. Thus, Jimmy sold 19 apples and 24 oranges today.

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