TutorMe homepage
Subjects
PRICING
COURSES
SIGN IN
Start Free Trial
Suman G.
Online Math tutor for 9 years.
Tutor Satisfaction Guarantee
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 )

Suman G.
Answer:

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!!

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.

Suman G.
Answer:

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!!

Algebra
TutorMe
Question:

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

Suman G.
Answer:

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.

Send a message explaining your
needs and Suman will reply soon.
Contact Suman
Ready now? Request a lesson.
Start Session
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 Session" 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.