Enable contrast version

# Tutor profile: Aakash S.

Inactive
Aakash S.
I have been teaching for 5 years.
Tutor Satisfaction Guarantee

## Questions

### Subject:Basic Math

TutorMe
Question:

Solve the follwoing system of equations -x/2 + y/3 = 0 x + 6y = 16

Inactive
Aakash S.

We first multiply all terms of the first equation by the LCM of 2 and 3 which is 6. 6(-x/2 + y/3) = 6(0) x + 6y = 16 We then solve the following equivalent system of equations. -3x + 2y = 0 x + 6y = 16 which gives the solution x = 8/5 and y = 12/5

### Subject:Algebra

TutorMe
Question:

How many real solutions does each quadratic equation shown below have? a) x 2 + (4 / 5) x = - 1/4 b) x 2 - 7x + 10 = 0 c) x 2 - (2/3) x + 1/9= 0

Inactive
Aakash S.

Find the discriminant of each equation. a) (4/5) 2 - 4(1)(1/4) = 16/25 - 1 < 0 , no real solutions. b) (-7) 2 - 4(1)(10) = 49 - 40 = 9 > 0 , 2 real solutions. c) (-2/3) 2 - 4(1)(1/9) = 4/9 - 4/9 = 0 , 1 real solutions.

### Subject:Physics

TutorMe
Question:

Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels?

Inactive
Aakash S.

Given: vi = 0 m/s vf = 444 m/s t = 1.83 s Find: a = ?? d = ?? a = (Delta v)/t a = (444 m/s - 0 m/s)/(1.83 s) a = 243 m/s2 d = vi*t + 0.5*a*t2 d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s2)*(1.83 s)2 d = 0 m + 406 m d = 406 m (Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)

## Contact tutor

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

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