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# Tutor profile: Ariel O.

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Ariel O.
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## Questions

### Subject:Basic Math

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Question:

A basketball team played 40 games and won 70 percent of them. How many games did it win?

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Ariel O.

First lets right the 70 percent in the correct form which is $$\frac{70}{100}$$. This can be simplified to $$\frac{7}{10}$$ because we can divide both terms by 10 as that is the common factor. Now in order to find the number of games won we must multiply $$40 * \frac{7}{10}$$ $$4* \frac{7}{1}$$ We get rid of the 10. Now we just multiply 7 * 4 $$28$$ is the amount of game the team won

### Subject:Calculus

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Question:

Given the following rational function. $$\frac{x-4}{(x^2+x+9)^3}$$. What would the correct form of decomposition look like?

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Ariel O.

To being with lets look at the ratios of the two polynomial functions $$p(x)=x-4$$ and $$q(x)=(x^2+x+9)^3$$. The degree or exponent of the numerator is less than the degree or exponent of the denominator. This means we have a proper rational function. First step in decomposition is to make sure the $$q(x)$$ is factorized completely into a product form $$(ax+b)^m$$ and $$(ax^2+bx+c)$$. Remember m and n are positive integers and $$a b$$ and $$c$$ are real numbers. Each factor of the form $$(ax^2+bx+c)^n$$ cannot be factored further over real numbers. In order to insure it can not be factored further over real numbers we can check the discriminant. If the discriminant is less than 0 it cannot be factorized further. $$b^2-4ac<0$$ this is the discriminant formula. $$x^2+x+9$$ lets plug in our values for a=1, b=1 and c =9 $$(1)^2-4(1)(9)<0$$ $$1-36<0$$ $$-35<0$$ this function is full factorized. now lets put it into the partial fraction form. $$\frac{B_1x+C_1}{ax^2+bx+c} + \frac{B_2x+C_2}{(ax^2+bx+c)^2}+...+\frac{B_nx+C_n}{(ax^2+bx+c)^n}$$ this is the form we will put it in. $$\frac{x-4}{(x^2+x+9)^3} = \frac{B_1x+C_1}{x^2+x+9}+\frac{B_2x+C_2}{(x^2+x+9)^2}+\frac{B_3x+C_3}{(x^2+x+9)^3}$$ This is the correct form for the partial function.

### Subject:Algebra

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Question:

Solve the following radical equation $$\sqrt{5x-3}=4$$?

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Ariel O.

Since the radical equation is already isolated on the left we need to eliminate the radical. In order to eliminate the radical we must square both sides of the function $$(\sqrt{5x-3})^2=(4)^2$$ $$5x-3=16$$ Next lets combine like terms We can add 3 to both sides $$5x-3+3=16+3$$ $$5x=19$$ Now we can divide both sides by 5 to isolate the x by itself $$x=\frac{19}{5}$$ That is our final answer

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