Enable contrast version

# Tutor profile: Ross E.

Inactive
Ross E.
Third-year computer science student, math tutor for 1 year
Tutor Satisfaction Guarantee

## Questions

### Subject:Java Programming

TutorMe
Question:

What is the purpose of implementing the Comparable interface in Java?

Inactive
Ross E.

The purpose is to be able to compare Objects of the same type to one another which don't have the typical ordering scheme as for example integers (1 < 2 < 3, etc).

### Subject:Calculus

TutorMe
Question:

Given: $(f(x)=14x^2+8x$) Find the first and second derivatives of f(x)

Inactive
Ross E.

To find both the first and second derivatives, we must use the power rule: $(\frac{d}{dx}[x^n]=nx^{n-1}$) Thus: $(f'(x)=28x+8$) $(f''(x)=28$)

### Subject:Algebra

TutorMe
Question:

Find the x-intercepts of the function: $(f(x)=-3x^2+4x+5$)

Inactive
Ross E.

Finding the x-intercepts means we are finding where: $(f(x)=0$) Setting the given function equal to zero gives us: $(0=-3x^2+4x+5$) Now that we have just one variable, we can solve for x using the quadratic formula: $(\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}$) From the given function we can determine the following: $(\begin{array}{*{20}c} {x = \frac{{ - 4 \pm \sqrt {4^2 - 4(-3)(5)} }}{{2(-3)}}} \\ \end{array}$) $(\begin{array}{*{20}c} {x = \frac{{ - 4 \pm \sqrt {76} }}{{-6}}} \\ \end{array}$) Now it's just arithmetic to determine the two x-intercepts:$(x_1\approx-0.79$) $(x_2\approx2.12$)

## Contact tutor

Send a message explaining your
needs and Ross will reply soon.
Contact Ross