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# Tutor profile: Jayne H.

Jayne H.
Aspiring Math Teacher; Pre-Calculus & Calculus Tutor

## Questions

### Subject:Pre-Calculus

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Question:

a. Write an equation in slope-intercept form for the following situation. Assume a linear relationship between $$x$$ and $$y$$. In 2007 Sandra’s bank account balance was $500. In 2010, it was$1,400.  b. Assuming she saves the same amount each year, how many years will it take for her balance to become $3,500? Jayne H. Answer: a. Let $$y$$ represent the balance in Sandra’s bank account after $$x$$ years. We must first find the linear change in her bank account balance. The question states that in 2007, Sandra’s account balance was$500 and it was $1,400 in 2010. So, the linear change in her bank account balance is: $$\frac{1400-500}{2010-2007} = \frac{900}{3} = 300$$ Since Sandra started out with$500, the y-intercept in the equation will be 500. So, the linear relationship can be given as: $$y=300x+500$$ b. To determine when Sandra’s balance will be $3,500, we can plug in 3,500 for $$y$$ (which represents the account balance) and solve for $$x$$ (which represents the number of years it takes to reach that balance). $$y=300x+500$$ $$3,500=300x+500$$ $$3,000=300x$$ $$10=x$$ So, in 10 years Sandra’s balance will be$3,500.

### Subject:Calculus

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Question:

Using the definition of the derivative, find the derivative of: $$f(x) = x^3 – 2x^2 + x – 1$$

Jayne H.

The definition of the derivative is: $$f'(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$$ Plugging in the function to the definition of the derivative, $$f'(x) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h} = \lim\limits_{h \to 0}\frac{[(x+h)^3 – 2(x+h)^2 + (x+h) – 1]-[x^3 – 2x^2 + x – 1]}{h} = \lim\limits_{h \to 0}\frac{[(x+h)^3 – 2(x+h)^2 + (x+h) – 1]-x^3 + 2x^2 - x + 1]}{h}$$ (Make sure to distribute the minus sign in front of the second equation!) Now, use algebra to combine like terms: $$f'(x) = \lim\limits_{h \to 0}\frac{[(x+h)^3 – 2(x+h)^2 + (x+h) – 1]-x^3 + 2x^2 - x + 1]}{h} = \lim\limits_{h \to 0}\frac{x^3+3x^2h+3xh^2+h^3 – 2(x^2+2xh+h^2) + x+h – 1-x^3 + 2x^2 - x + 1}{h} = \lim\limits_{h \to 0}\frac{x^3+3x^2h+3xh^2+h^3 – 2x^2-4xh-2h^2 + x+h – 1-x^3 + 2x^2 - x + 1}{h} = \lim\limits_{h \to 0}\frac{h(3x^2+3xh+h^2 -4x-2h + 1)}{h} = \lim\limits_{h \to 0}3x^2+3xh+h^2 -4x-2h + 1 = 3x^2+3x(0)+(0)^2 -4x-2(0) + 1 = 3x^2 -4x + 1$$ So, the derivative of f(x) is: $$f'(x) = 3x^2 -4x + 1$$

### Subject:Algebra

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Question:

Solve for $$y$$: $$12y + d = −19y + t$$

Jayne H.

To solve for $$y$$, we want to get all of the terms with $$y$$ onto one side; so, we can add $$19y$$ to both sides to cancel out the $$-19y$$ on the right side of the equation. $$12y (+19y) + d = -19y (+19y) + t$$ $$31y + d = 0y + t$$ $$31y + d = t$$ Now we want the $$y$$ term to be isolated, without any other terms added to or subtracted from it. So, we can subtract $$d$$ from both sides to cancel out the $$d$$ on the left side of the equation. $$31y + d (–d) = t (–d)$$ $$31y + 0d = t – d$$ $$31y = t – d$$ Finally, we want the coefficient of the $$y$$ term to be $$1$$. To do that, we can divide $$31y$$ by $$31$$, because $$31y ÷ 31 = 1y = y$$. $$\frac{31y}{31} = \frac{t – d}{31}$$ $$1y = \frac{t – d}{31}$$ $$y = \frac{t – d}{31}$$

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