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Jurgen X.
Tutor for 3 years
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Pre-Calculus
TutorMe
Question:

\newline a) Find the domain of $f(x)=\sqrt{2x-4}+ln(3-x)$

Jurgen X.
Answer:

\newline a) For $\sqrt{2x-4}$, the condition is $2x-4\geq0$ which leads to $x\geq2$. For $ln(3-x)$, the condition is $(3-x)>0$ which leads to $x<3$. Therefore, the domain is $[2,3)$

Geometry
TutorMe
Question:

\newline a) What is the area of rhombus $ABCD$ with \angle ABC being 90 $^{\circ}$ and AB being 5cm long?

Jurgen X.
Answer:

\newline a) Since one of the rhombus' angles is a right angle, then the angle facing it is a right angle too. Since two angles next to each other add up to 180 $^{\circ}$ in a rhombus, then the other two angles are 90 $^{\circ}$ each. That means $ABCD$ is actually a square. Therefore we can easily find the area $5\times 5 = 25cm^2$

Algebra
TutorMe
Question:

\newline This is a sample question for Algebra: \newline a) Does the line $y= \frac{2}{3}x-6$ intersect with line $2y+3x=7$ ? If so, find the angle. If not, explain why in a sentence or two. \newline b) What are the coordinates of the common point (s) for these two lines? If there is no common point, what can we change to make them intersect?

Jurgen X.
Answer:

\newline a) Slope for the first line is $m_{1}=\frac{2}{3}$ and slope to second line is $m_2=\frac{-3}{2}$. Then we calculate the product $m_{1}\times m_{2}=-1$. Since the product of two slopes is $-1$, then the lines do intersect and they are perpendicular, therefore the angle is 90$^{\circ}$. \newline b) We start by plugging in $\frac{2}{3}x-6$ for $y$ in the second line's equation and we get $2\times [\frac{2}{3}x-6] +3x=7$. Solving for $x$, we get $x=\frac{57}{13}$. Substituting $\frac{57}{13}$ in the equation of first (or second) line, we find $y=\frac{-40}{13}$. Finally, the solution is ($\frac{57}{13}$;$\frac{-40}{13}$).

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