10 Most Common TSI Math Questions

10 Most Common TSI Math Questions

B. \(\frac{4}{5}\)

C. \(\frac{4}{3}\)

D. \(\frac{3}{4}\)

7- Find the slope-intercept form of the graph \(6x – 7y = – 12\)

A. \(y = –\frac{7}{6} x – \frac{12}{7}\)

B. \(y = –\frac{6}{7} x – \frac{12}{7}\)

C. \(y = \frac{6}{7} x +\frac{12}{7}\)

D. \(y = \frac{7}{6} x +\frac{12}{7}\)

8- Ella (E) is 4 years older than her friend Ava (A), who is 3 years younger than her sister Sofia (S). If E, A, and S denote their ages, which one of the following represents the given information?

A.
\(E=A+4\)
\(S=A-3\)

B.
\(E=A+4\)
\(A=S+3\)

C.
\(A=E+4\)
\(S=A-3\)

D.
\(E=A+4\)
\(A=S-3\)

9- Which of the following points is the solution of the system of equations?
\(5x + y = 9 \)
\(10x-7y= -18\)

A. \((2, 4)\)

B. \((2, 2)\)

C. \((1, 4)\)

D. \((0, 4)\)

10- Find the Center and Radius of the graph \((x – 3)^2 + (y + 6)^2 = 12\)

A. \((3, 6), \sqrt{3}\)

B. \((3, -6), 2\sqrt{3}\)

C. \((-3, 6), 2\sqrt{3}\)

D. \((3, -6), \sqrt{3}\)

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Answers:

1- D
\( (g – f)(x) = g(x) – f(x) = (– x^2 – 1 – 2x) – (5 + x)
– x^2 – 1 – 2x – 5 – x = – x^2 – 3x – 6\)

2- B
\(\frac{|3+x|}{7}≤ 5⇒|3+x|≤35⇒-35≤3+x≤35⇒-35-3≤x≤35-3⇒
-38≤x≤32\)

3- D
\(tan\space (-\frac{π}{6}) = -\frac{\sqrt3}{3}\)

4- A
\(\frac{\sqrt{32a^5 b^3} }{\sqrt{2ab^2} }=\frac{4a^2 b\sqrt{2ab}}{b\sqrt{2a}}=4a^2 \sqrt b\)

5- D
\(c(3)=(3)^2+10(3)+30=9+30+30=69\)
\(4×3=12⇒12-69=-57⇒57,000 \space loss\)

6- B
\(sin\space B=\frac{the \space length \space of \space the \space side \space that \space is \space opposite \space that \space angle}{the \space length \space of \space the \space longest \space side \space of \space the \space triangle}=\frac{4}{5}\)

7- C
\(-7y=-6x-12⇒y=\frac{-6}{-7}x- \frac{12}{-7}⇒y=\frac{6}{7}x + \frac{12}{7}\)

8- D
\( E = 4 + A\)
\(A = S – 3\)

9- C
\(5x + y = 9 \)
\(10x-7y= -18\)
Multiplication \((–2)\) in the first equation
\(-10x- 2y = -18 \)
\(10x-7y= -18\)
Add two equations together \(⇒ –9y = –36 ⇒ y = 4\) then: \(x = 1\)

10- B
\((x – h)^2 + (y – k)^2 = r^2\) ⇒ center: (h,k) and radius: \(r\)
\((x – 3)^2 + (y + 6)^2 = 12\) ⇒ center: \((3,-6)\) and radius:\(2\sqrt 3\)

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