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Tutor profile: Ankit P.

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Ankit P.
Student worker at university's Student Succes Center
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Questions

Subject: Basic Math

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Question:

30 bottles of orange juice costs $$ $45$$. What will one bottle cost and how many bottles will you get with $$ $10$$.

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Ankit P.
Answer:

To being with lets find out how much one bottle will cost. We do this by dividing 45 by 30. $$ \frac{30}{45}$$ = $$0.66$$. This answers the first part of the question for the second part how many bottles can we get with $$ $10 $$ we know that one bottle costs $$ $0.66 $$. We multiply $$ $10 * $0.66$$. We get $$ 6.6$$ You can not have 6.6 bottles so we will round down to 6 bottle. If the question as for the maximum amount we could round up

Subject: Calculus

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Question:

Integrate the following integral by using integration by parts only. Be sure to include C for the constant when you done integrating, Integral: $$\int x^5ln(x)dx $$.

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Ankit P.
Answer:

To being with we write down the formula for integration by parts. $$\int udv = uv - \int vdu$$ Let $$ u = ln(x) $$ and $$dv = x^5dx$$ $$du$$ is simply $$ u $$ differentiated and $$v$$ is the integral of $$dv$$ so $$du = \frac{1}{x}dx $$ and $$v= \frac{x^6}{6}$$ So now we can say: Let $$ u = ln(x) $$ and $$dv = x^5dx$$ and Let $$ du = \frac{1}{x}dx $$ and $$v = \frac{x^6}{6}$$ Now we can rewrite our formular using the values for $$u ,v ,dv ,du$$ that we found $$\int x^5ln(x) dx = ln(x)*\frac{x^6}{6}-\int \frac{x^6}{6}*\frac{1}{x}dx$$ We can see the first part of the equation is just our original equation. So lets work on the right half. Lets split it up to make it easier for us to work on. One half will be $$ln(x)*\frac{x^6}{6}$$ and the other will be $$\int \frac{x^6}{6}*\frac{1}{x}dx$$ Lets work with the first half: $$ln(x)*\frac{x^6}{6}$$ $$ln(x)$$ is the same as $$\frac{ln(x)}{1}$$ so we can multiply the two fraction $$\frac{ln(x)}{1} * \frac{x^6}{6}$$ and get $$\frac{x^6ln(x)}{6}$$ Now we will tackle the second half: $$\int \frac{x^6}{6}*\frac{1}{x}dx$$ We will rewrite the equation in a simplifed form. The $$ x's$$ cancel and we can take out a $$ \frac{1}{6}$$ we are left with $$ \frac{1}{6}\int x^5 dx$$ Lets integrate the $$\int x^5$$ and we get $$\frac{1}{6}*x^6$$ Let us combine the two halves now $$\frac{x^6ln(x)}{6}$$ -$$\frac{1}{6}*$$ $$\frac{1}{6}*x^6$$ Simplify it by multiplying $$\frac{1}{6}*$$ $$\frac{1}{6}$$ Our final answer is $$\frac{x^6ln(x)}{6}$$ -$$\frac{x^6}{36}$$ + C

Subject: Algebra

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Question:

Simplify the following rational expressions. $$ \frac{x^2−7x+10}{x+2} $$ ÷ $$ \frac{x^2−3x-40}{x+2} $$

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Ankit P.
Answer:

Firstly we will start by attempting to factor the numerator of the two equations. $$ x^2−7x+10$$ and $$ x^2−3x-40 $$. Due to the coefficient of $$x^2$$ being one we can easily factor it out. For the first equation the factor is $$(x-2)(x-5)$$ The second is $$(x+8)(x-5)$$ To check you can multiply the parenthesis out using the foil method. Now we subsititute the factors back into the original equation $$ \frac{(x-2)(x-5)}{x+2} $$ ÷ $$ \frac{(x+8)(x-5)}{x+2} $$ Dividing a fraction is the same as multiplying by its reciprocal. Our equation becomes $$ \frac{(x-2)(x-5)}{x+2} $$ * $$ \frac{x+2}{(x+8)(x-5)} $$ Now we can cancel like terms $$(x-5)$$ $$(x+2)$$ We are left with $$ \frac{x-2}{x+8} $$ This is our final answer simplified

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