Enable contrast version

Inactive
Former math teacher w/ 17 yrs tutoring exp
Tutor Satisfaction Guarantee

## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Find lim (x->3) (x^2 - 9)/(x-3)

Inactive

Direct substitution in this case yields 0/0, which is undefined. We therefore try a different approach. To solve this problem, we first observe that x^2 - 9 is a difference of squares, and can be factored as (x-3)(x+3), so we rewrite the problem as lim (x->3) [(x-3)(x+3)]/(x-3). We have a common factor of (x-3) in the numerator and denominator. This cancels out, leaving lim (x->3) (x+3)/1 = lim (x->3) (x+3). Since we no longer have a denominator, we can now use direct substitution because the possibility of dividing by zero has been eliminated: lim (x->3) (x+3) = 3+3 = 6.

### Subject:Calculus

TutorMe
Question:

Differentiate with respect to x. Assume y is a function of x. 3y^2 - 16x^3 + x^2 - 4 = 0

Inactive

According to the chain rule, to find the derivative of terms involving y, we differentiate the given terms, and multiply the result by the derivative of y with respect to x. We also differentiate other terms with respect to x: 6yy' - 48x^2 + 2x = 0 Next, we divide both sides by the Greatest common factor of 2: 3yy' - 24x^2 + x = 0 To isolate the term that contains y as a derivative of x, we first gather other terms that are strictly defined in terms of x to the other side of the equation: 3yy' = 24x^2 - x Finally, we divide both sides by 3y: y' = (24x^2 - x)/3y

### Subject:Algebra

TutorMe
Question:

Solve for x: x^2 - 10x - 75 = 0

Inactive

We factor this quadratic equation by seeking two factors of our constant term, -75, whose sum is the linear coefficient, -10. The only two numbers that work are -15 and 5. We now factor by writing these numbers in two pairs of parentheses, each following a linear term of x to get a product of binomials: (x - 15)(x + 5) = 0. Since the only way two numbers could multiply to a product of zero is if one of them is zero, we have two cases. Either x - 15 = 0 or x + 5 =0. We solve both cases and have: x - 15 = 0. Add 15 to both sides: x = 15, and x + 5 = 0 Subtract 5 from both sides: x = -5. The solutions are 15 and -5. Check: x = 15: 15^2 - 10*15 -75 = 225 - 150 - 75 = 75 -75 = 0. x = -5: (-5)^2 -10*(-5) - 75 = 25 + 50 - 75 = 75 - 75 = 0.

## Contact tutor

Send a message explaining your

Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage