Subjects
PRICING
COURSES
SIGN IN
Start Free Trial
Madison R.
Lab Technician and Avid Tutor
Tutor Satisfaction Guarantee
Basic Chemistry
TutorMe
Question:

Balance the following equation: C6H12O6 + O2 ---> CO2 + H2O

Madison R.
Answer:

Balancing a chemical equation involves adjusting the coefficients on molecules so that the total number of each type of atom is the same on both sides of the equation. We will begin with carbon. There are 6 carbons on the left side of the equation, and 1 on the right. To balance the carbon atoms, a coefficient of 6 must be added to CO2. This give us: C6H12O6 + O2 ---> 6CO2 + H2O. Next, we will balance the hydrogens. There are 12 hydrogen atoms on the left and 2 on the right. To balance the hydrogens, a coefficient of 6 must be added to H2O, giving us C6H12O6 + O2 ---> 6CO2 + 6H2O. Finally, we will balance the oxygens. This is the trickiest step, since each molecule in the equation contains oxygen. There are 8 total on the left, 6 from C6H12O6 and 2 from O2. There are 18 on the right, 12 from 6CO2 and 6 from 6H2O. We need an additional 10 oxygens on the left to balance the equation. We cannot change the coefficient on C6H12O6, because this would change the number of carbon and hydrogen atoms. Therefore, we will add a coefficient of 6 to O2. The final equation is C6H12O6 + 6O2 ---> 6CO2 + 6H2O

Calculus
TutorMe
Question:

Find the derivative of y=x^2sinx

Madison R.
Answer:

Since this equation is the product of two x variables, we will use the product rule which states that d/dx(ab)=ab'+a'b, where a' and b' represent the derivatives of a and b. In this question, a=x^2 and b=sinx. The derivative of x^2 is 2x, and the derivative of sin x is cosx. Therefore, a'=2x and b'=cosx. By plugging these values into the product rule, we get y'=x^2cosx+2xsinx.

Algebra
TutorMe
Question:

Simplify the following equation: x^4 -25x^3 -23x^2 -50x -50

Madison R.
Answer:

Solving this problem involves several factoring steps. We see that there are 5 obvious factors available: x^3, x^2, x, 50, and 25. We will begin by factoring out x^2 from x^4, -25x^3, and -23x^2. This gives us: x^2(x^2-25x-23)-50x-50. Next we need to factor x^2-25x-23. By adding and subtracting 2, we can create the difference of two squares, or x^2-25x-25+2. This factors to (x-5)(x+5)+2. The equation is now x^2[(x-5)(x+5)+2]-50x-50. x^2 can be distributed again, giving us x^2(x-5)(x+5)+2x^2-50x-50. Factor a 2 from 2x^2-50x-50, giving 2(x^2-25x-25), then use the difference of two squares to get 2(x-5)(x+5). Our equation is now x^2(x-5)(x+5)+2(x-5)(x+5). (x-5)(x+5) is now a common factor, so x^2 and 2 can be combined. The equation becomes (x^2+2)(x-5)(x+5). This is the simplified equation.

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