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# Tutor profile: Jennifer Z.

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Jennifer Z.
Senior at University of Chicago
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## Questions

### Subject:Geometry

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Question:

Parallelogram ABCD has an perimeter of $$\text{44 }cm$$ and area of $$\text{64 }cm^2$$. The length of side AB is $$3x+2$$ and the length of side AD is $$5x+4$$. Find the supplemental angles (angles A and B). $$\textit{*This problem requires trigonometry.}$$

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Jennifer Z.

$$\textbf{Step 1:}$$ This problem is a little difficult without a diagram, so you should draw out a diagram. Things to keep in mind when drawing the diagram: 1) Opposite sides are congruent (AB = DC). 2) Opposite angels are congruent (A = C). 3) Consecutive angles are supplementary (A + D = 180°). $$\textbf{Step 2:}$$ Solve for $$x$$ using the perimeter $(44=2(3x+2)+2(5x+4)$) $(44=6x^2+4+10x+8$) $(44=16x+12$) $(32=16x$) $(x=2$) From this, we know that length of side AB is 8cm and side AD is equal to 14cm. $$\textbf{Step 3:}$$ Calculate the height. Use the area formula (Area = Base * Height) $(64=14*Height$) $(Height=64\div14$) $(Height=\frac{32}{7}$) $$\textbf{Step 4:}$$ Solve for the measurement of angle A. $(\sin{A}=\frac{\frac{32}{7}}{8}$) $(A=\arcsin{\frac{\frac{32}{7}}{8}}$) $(A=\arcsin{\frac{4}{7}}$) $(A=34.8^{\circ}$)

### Subject:Pre-Algebra

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Question:

There are 212 pages in a book. Harrison started reading the book a week ago and still has 46 pages left. How many pages has Harrison read?

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Jennifer Z.

$$\textbf{Step 1:}$$ As with all word problems, we approach this problem by first writing out an equation. Assign a variable to the unknown. In this case, the unknown is the number of pages Harrison has already read. $(\text{let x = pages read}$) $(212-x=46$) $(-x=46-212$) $(-x=-166$) $((-1)(-x)=(-166)(-1)$) $(x=166$) $$\textbf{Solution:}$$ Harrison has already read 166 pages. $$\textit{*As a reminder, don't forget that word problems should be answered with a sentence.}$$

### Subject:Algebra

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Question:

Solve using the elimination method. $(6x - 5y = 8$) $(-12x + 2y = 0$)

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Jennifer Z.

$$\textbf{Step 1:}$$ Modify one or both equations such that one of the variables has the same coefficient but with opposite signs. $(6x - 5y = 8 \overset{x2}{\rightarrow} 12x - 10y = 16$) $( -12x + 2y = 0 \overset{no change}{\rightarrow} -12x - 2y = 0$) In this step, we multiply the first equation by 2 so that the coefficient of $$x$$ becomes 12. This makes the coefficient the opposite sign from the $$-12$$ coefficient on $$x$$ in the second equation. $$\textbf{Step 2:}$$ Add the modified equations together. $(12x - 10y = 16$) $(\underline{+ -12x + 2y =0}$) $(-8y = 16$) $$\textbf{Step 3:}$$ Solve for $$y$$. $(-8y = 16$) $(y = 16\div8$) $(y=-2$) $$\textbf{Step 4:}$$ Plug in the value found for $$y$$ and solve for $$x$$. $(6x-5y)=8$) $(6x-5(-2)=8$) $(6x+10=8$) $(6x=8-10$) $(6x=2$) $(x=2\div6$) $(x=-\frac{1}{3}$) $$\textbf{Solution:}$$ $$x=-\frac{1}{3}$$, $$y=-2$$

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