Enable contrast version

# Tutor profile: Suman G.

Inactive
Suman G.
Online Math tutor for 12 years.
Tutor Satisfaction Guarantee

## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Find an equation of a line in slope intercept form which is perpendicular to line 8x+6y=11 and passing through point (1,2 )

Inactive
Suman G.

We know that slopes of two perpendicular lines must be negative reciprocal to each other. So we first find slope of given line. For that we isolate y and then the coefficient of x will be the slope. 8x + 6y =11 6y = -8x +11 y = -8/6(x) +11/6 Here coefficient of x is -8/6 which can be simplified to -4/3. That means slope of given line = -4/3 So the slope of perpendicular line = -1/(-4/3) =3/4 And we can write the equation of perpendicular line as y= (3/4)x +b Now we plugin given point (1,2) to find value of b. Plugin x=1 and y=2 into y=(3/4)x+b 2=(3/4)*1 +b 2-(3/4) = b 5/4 = b Using values of m and b we get slope intercept form y=mx+b as y = (3/4)x + (5/4) This is the equation of perpendicular line in slope intercept form. Enjoy Math!!

### Subject:Calculus

TutorMe
Question:

The volume of a cube is increasing at a constant rate. Prove that increase in surface area of cube varies inversely as the length of the edge of cube.

Inactive
Suman G.

We know that surface area S, and volume V of a cube is given as... S= 6x^2 and V= x^3 where x is the length of edge of cube. It is given that dV/dt= K(constant) Now V= x^3 Taking derivative of both sides with respect to time(t)..... dV/dt = 3x^2 * dx/dt k =3x^2 * dx/dt [ plugged in given value of dV/dt as k ] k/3x^2 = dx/dt ---------(1) And S = 6x^2 Taking derivative of both sides with respect to t, dS/dt = 12x * dx/dt Now replace dx/dt with k/3x^2 as we get from equation (1) above dS/dt = 12x * k/3x^2 dS/dt = 12xk/3x^2 dS/dt = 4k/x dS/dt inversely proportional to 1/x This shows that rate of increase in surface area varies inversely as the length of edge of cube. Enjoy Math!!

### Subject:Algebra

TutorMe
Question:

Solve the following absolute value equation for x. -5 |-x+2| + 8 = -12

Inactive
Suman G.

Here we try to reach out to expression inside absolute signs. For that we need to get rid of first 8 and then -5. So first step is : subtract 8 from both sides. We get.. -5 |-x+2 | +8 = -12 -8 -8 ____________________ -5 |-x+2 | = -20 Second step: divide both sides by -5 to get rid of -5 -5 |-x+2 | / -5 =- 20 /-5 |-x+2 | =4 Third step : Now we solve the absolute expression using two cases, one with positive sign and other with negative sign. -x+2 = 4 -x+ 2 = -4 -2 -2 -2 -2 ------------------------ ----------------------- -x = 2 -x = -6 x = -2 x = 6 So this equation has two solutions x = -2 ,6 Don't forget to plugin back these x values into original equation to check the correct solution. Here both solutions satisfy the given equation so both are solutions.

## Contact tutor

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

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