Enable contrast version

Inactive
Application Engineer and Avid Tutor
Tutor Satisfaction Guarantee

## Questions

### Subject:Basic Chemistry

TutorMe
Question:

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

Inactive

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

### Subject:Calculus

TutorMe
Question:

Find the derivative of y=x^2sinx

Inactive

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.

### Subject:Algebra

TutorMe
Question:

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

Inactive

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.

## Contact tutor

Send a message explaining your