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# Tutor profile: Michael R.

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Michael R.
Tutor at my local college
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## Questions

### Subject:Trigonometry

TutorMe
Question:

convert the line that passes through (-4,0) and (0,5) to polar form

Inactive
Michael R.

First we will find the slope. This is $$\frac{5-0}{0-(-4)}=\frac{5}{4}$$ Then we use that to find the equation of the line with $$y-y_1=m(x-x_1)$$ $$y=\frac{5}{4}(x-0)+5$$ $$y=\frac{5}{4}x+5$$ now we multiply both sides by 4 so that we have whole number coefficients in front of the variables to get $$4y=5x+20$$ and then we move both variables to one side by subtracting 5x from each side to get $$4y-5x=20$$, and then substitute $$rsin\theta$$ for y and $$rcos\theta$$ for x, to get $$4rsin\theta-5rcos\theta=20$$, and then we factor out the r to get $$r(4sin\theta-5cos\theta)=20$$, and then solve for r $$r=\frac{20}{4sin\theta-5cos\theta}$$ to get our solution.

### Subject:Chemistry

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Question:

If a 356 mL solution of .406 molar hydrofluoric acid is titrated with 1.00 molar NaOH until completion, what is the pH of the solution? the Ka of hydrofluoric acid is 7.20E-4

Inactive
Michael R.

first, we will assume that the NaOH reacts with the hydrofluoric acid 100%, in the reaction $$NaOH + HF -> Na^++ F^- + H_{2}O$$ because $$Na^+$$ is the conjugate acid of a strong base, it will play no role in the pH of the solution, so all we are concerned with is the $$F^-$$ We now need to know the concentration of $$F^-$$, so we will need to know how many solutions were added with the NaOH we can figure this out with dimensional analysis. $$356mL * \frac{1L}{1000mL} * \frac{.406molsF^-}{1L}*\frac{1molNaOH}{1molF^-} * \frac{1L}{1molNaOH} * \frac{1000mL}{1L} = 144.536mL$$ we add this to the original amount of 356mL and find that the final volume is 500.536mL from the earlier dimensional analysis, we can find that we began with 0.144536 mols of $$F^-$$, which diluted to 500.5mL gives us a molarity of $$\frac{.144536mols}{.5005L}$$ or $$.2888M$$ because $$F^-$$ is a conjugate base of a weak acid, we know that some of it will turn back into HF, forming $$OH^-$$ in the process. We can use the chemical equation $$F^-+H_2O->HF+OH^-$$ we know that $$[F^-]$$ starts at .2888M, but will lose some concentration x during the reaction. $$[HF]$$ starts at 0 but gains x and the same for hydroxide. because we are now using a formula with hydroxide rather than hydronium, it would be better to use a Kb value, rather than the given Ka value, so we will find Kb using the relationship $$Kb=\frac{1E-14}{Ka}$$ to find that $$Kb=1.389E-11$$ now we know that $$\frac{x^2}{.2888-x}=1.389E-11$$, this looks like we may need the quadratic formula, but because x is so small, we can approximate the change in concentration to 0, and say that $$\frac{x^2}{.2888}=1.389E-11$$, and $$x_1=2.00E-6$$ we can now substitute $$x_1$$ for x, and find that $$x_2=2.00E-6$$, this is accurate enough. Because x is equal to the concentration of hydroxide in solution, we can take the negative log to find pOH=5.70 we can then use the relationship $$pH=14-pOH$$ to find that pH=8.30

### Subject:Algebra

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Question:

solve the equation $$ax^2+bx+c=0$$ for x, where a, b, and c are constants.

Inactive
Michael R.

This question is tricky because we cannot simply factor out the x, or solve by rearranging, so we will have to use a different technique called completing the square. In this technique we will replace the original equation with an equivalent one in which it is easier to solve for x, by finding a factor whose square is equivalent to $$ax^2+bx$$ plus or minus a constant. to do this we will first say that that factor is equal to $$(yx+z)$$ through f.o.i.l.ing, we know that $$(yx)^2=ax^2$$, and so to solve for y we can take the square root of both sides, and divide by x, telling us that $$y=\sqrt{a}$$ also through f.o.i.l.ing, we know that $$bx=zyx+zyx=2zyx$$, and by dividing both sides by $$2yx$$, we see that $$z=\frac{bx}{2yx}=\frac{b}{2y}$$, and then by substituting $$\sqrt{a}$$ for $$y$$ we see that $$z=\frac{b}{2\sqrt{a}}$$ so our factor is $$(\sqrt{a}x+\frac{b}{2\sqrt{a}})$$, if we square this we will get $$\sqrt{a}x*\sqrt{a}x+\sqrt{a}x*\frac{b}{2\sqrt{a}}+\frac{b}{2\sqrt{a}}*\frac{b}{2\sqrt{a}}$$ through foiling, or $$ax^2+bx+\frac{b^2}{4a}$$ because we have unbalanced the left hand side by adding $$\frac{b^2}{4a}$$ to the left hand side, we must rebalance the equation by adding it to the right side, so when we bring this factor back to the original equation, we will say that $$(\sqrt{a}x+\frac{b}{2\sqrt{a}})^2+c=\frac{b^2}{4a}$$, and now we have the equation in a form in which we can solve for x by simple rearrangement. First we will subtract c from both sides, giving us $$(\sqrt{a}x+\frac{b}{2\sqrt{a}})^2=\frac{b^2}{4a}-c$$ then we will get a common denominator on the right hand side by multiplying the $$c$$ by $$4a$$, giving us $$(\sqrt{a}x+\frac{b}{2\sqrt{a}})^2=\frac{b^2-4ac}{4a}$$ we will then take the square root of both sides, giving us $$\sqrt{a}x+\frac{b}{2\sqrt{a}}=\sqrt{\frac{b^2-4ac}{4a}}$$ next we will isolate the x term by subtracting $$\frac{b}{2\sqrt{a}}$$ from both sides, giving us $$\sqrt{a}x=\sqrt{\frac{b^2-4ac}{4a}}-\frac{b}{2\sqrt{a}}$$ and then we will completely isolate x by dividing both sides by $$\sqrt{a}$$, giving us $$x=\frac{\sqrt{\frac{b^2-4ac}{4a}}-\frac{b}{2\sqrt{a}}}{\sqrt{a}}$$, and now that we have solved for x, all we have left to do is simplify the right hand side. We will do this by separating the numerator into two terms, and simplifying each individually. First the right term, $$\frac{\frac{-b}{2\sqrt{a}}}{\sqrt{a}}$$, when we have a compound fraction like this, we simply multiply the numerator by the reciprocal of the denominator, which in this case equates to $$\frac{-b}{2\sqrt{a}}*\frac{1}{\sqrt{a}}$$, which simplifies to $$\frac{-b}{2a}$$, in the final equation this will be our leading term. The left term is a bit trickier, but we will start by getting a common denominator with the term we just simplified, to do this we will divide the term we want by the term we wish to alter, so $$\frac{2a}{\sqrt{a}}$$, or $$2\sqrt{a}$$, so we will multiply the left term by $$\frac{2\sqrt{a}}{2\sqrt{a}}$$, giving us $$\frac{2\sqrt{a}*\sqrt{\frac{b^2-4ac}{4a}}}{2a}$$ we will then move the $$2\sqrt{a}$$ inside the radical by multiplying the inside of the radical by the square of the factor to be brought inside, or in this case, $$4a$$ the $$4a$$ inside the radical cancels with the $$4a$$ we just brought into the numerator, leaving us with $$\frac{\sqrt{b^2-4ac}}{2a}$$, but because at this point we have squared a number and then taken its square root, we must be careful that we have added an additional solution to the equation, and so must take the positive or negative value for the radical, giving us $$\frac{±\sqrt{b^2-4ac}}{2a}$$ We can then combine the two terms, and finally see that $$x=\frac{-b±\sqrt{b^2-4ac}}{2a}$$ This gives us a very important formula in mathematics, the quadratic formula. Once you know it, you may never have to complete the square again for a problem of this type, as long as you can get a polynomial into the form $$ax^2+bx+c=0$$, you can just plug the numbers into the formula, and get your answer(s) (side note, I accidentally exited the page when adding the plus or minus symbol near the end, and had to write this whole thing twice, just something to think about)

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