Enable contrast version

# Tutor profile: Tavina O.

Inactive
Tavina O.
PhD Candidate in Chemistry
Tutor Satisfaction Guarantee

## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Find the vertex of the parabola \$\$ y = x^{2} + 6x - 2 \$\$.

Inactive
Tavina O.

For a quadratic equation (\$\$ y = ax^{2} + bx + c\$\$), the x-coordinate of the vertex is calculated by solving the following: \$\$ x = -b/2a \$\$. The value of x can then be substituted into the quadratic equation to solve for y. \$\$ x = -b/2a; b = 6\$\$ \$\$ x= -6/2(1) \$\$ \$\$ x = -6/2 \$\$ \$\$ x = -3 \$\$ Now, plug in the value for x into the quadratic equation to solve for y. \$\$ y = x^{2} + 6x - 2 \$\$ \$\$ y = (-3)^{2} + 6(-3) -2 \$\$ \$\$ y = 9 - 18 - 2 \$\$ \$\$ y = -11 \$\$

### Subject:Basic Chemistry

TutorMe
Question:

Balance the following chemical equation: \$\$ NaBr + Cl_{2} --------> NaCl + Br_{2}\$\$

Inactive
Tavina O.

When balancing chemical equations, the number of atoms should be the same on both the reactant side and the product side since matter is neither created nor destroyed (Law of Conservation). The following balanced chemical equation is: \$\$ 2 NaBr + 1Cl_{2} --------> 2NaCl + 1Br_{2}\$\$ Start with Br, which in the product side, we see there are 2. We must add a 2 in front of the NaBr on the reactant side to balance the \$\$Br_{2}\$\$ on the product side. We then must add a 2 in front of the NaCl on the product side to balance the 2 Na atoms on the reactant side; this also balances the Cl atoms as well. The order in which you go about balancing each atom is entirely up to you.

### Subject:Algebra

TutorMe
Question:

Is (2,10) a solution to this system of equations? \$\$ y = x + 8 \$\$ \$\$ y = 4x + 2 \$\$

Inactive
Tavina O.

There are a few different ways to answer this question: 1) The first way is to plug in the values for x and y into BOTH equations and see if they hold true: \$\$ y = x + 8 \$\$ \$\$ 10 = 2 + 8 \$\$ Since this is true (2 + 8 does equal 10), then (2,10) is a solution to the equation, \$\$ y = x + 8 \$\$. \$\$ y = 4x + 8 \$\$ \$\$ 10 = 4(2) + 2 \$\$ \$\$ 10 = 8 + 2 \$\$ Since this is true, (2,10) is also a solution to the equation, \$\$ y = 4x + 8 \$\$. Therefore, (2,10) is a solution to the system of equations since it is a solution to BOTH equations. 2) The second way to solve this question is to use the Substitution method, where we solve for one variable in one equation. Substitute what you get for that variable in the other equation. In this example, both equations have already been solved for y. Lets plug in what y equals for the first equation into the second equation: \$\$ y = x + 8 \$\$ \$\$ y = 4x + 2 \$\$ \$\$ x + 8 = 4x + 2 \$\$ Now, we can solve for x. \$\$ x = 4x + 2 - 8 \$\$ \$\$ x = 4x - 6 \$\$ \$\$ -3x = -6 \$\$ \$\$ x = 2 \$\$ Now, we can plug in 2 for x, and solve for y in both equations. \$\$ y = x + 8 \$\$ \$\$ y = 2 + 8 \$\$ \$\$ y = 10 \$\$ \$\$ y = 4x + 2 \$\$ \$\$ y = 4(2) + 2 \$\$ \$\$ y = 10 \$\$ Using the substitution method to solve for x and y, we see that (2,10) is in fact a solution for this system of equations. 3) The third way to solve this question is to use the Elimination method, where we multiply one of the equations by a certain number that will eliminate x or y. Lets multiply the second equation by -1 in order to eliminate y: \$\$ y = 4x + 2 \$\$ -----> multiply by -1 \$\$ -y = -4x -2 \$\$ Now we can add this equation to the first equation to eliminate y. \$\$ y = x + 8 \$\$ \$\$-y = -4x -2 \$\$ __________ \$\$ 0y = -3x + 6 \$\$ \$\$ 0 = -3x + 6 \$\$ \$\$ -6 = -3x \$\$ \$\$ 2 = x \$\$ Now we can substitute 2 in for x into both equations like we did above to solve for y. 4) The fourth way to solve this equation is to graph both equations on the same coordinate system, and find the point where both lines intersect. You can graph this by solving for y at various values of x (0, 1, 2) for the first equation, y = x + 8. You will get the following: x y 0 8 1 9 2 10 Do the same for the second equation, y = 4x + 2, and you should get the following: x y 0 2 1 6 2 10 When you graph both equations on the same coordinate system, you will see that they intersect at (2,10). This is also apparent by looking at the tabulated values of x and y for both equations.

## Contact tutor

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