Download MATH 1010-2 Exam #1 Solutions: Validity, Simplifying, Solving Equations, Graphing and more Exams Algebra in PDF only on Docsity!
MATH 1010-2: PRACTICE EXAM #1 SOLUTIONS
- (24 points – 3 points each) Determine if each of the following assertions is valid. Indicate you answer by clearly circling either TRUE or FALSE. (a) The domain of f (x) = √x^1 +1 consists of all real numbers x greater than or equal to −1.
TRUE FALSE
(b) The slope of the line given by 3y + 6x − 10 = 0 is − 12.
TRUE FALSE
(c) There are real numbers which are not fractions.
TRUE FALSE
(d) The following
is the graph of f (x) = −x^2 + 2.
TRUE FALSE
(e) The following
is the graph of f (x) = x^3.
TRUE FALSE
(f) (−1)^10 = 1.
TRUE FALSE
(g) If m is the slope of a line ℓ, then −m is the slope of any line perpendicular to ℓ. TRUE FALSE
(h) No point on the line y = x + 2 lies in the third quadrant.
TRUE FALSE
- Simplify the following expression:
3
[
(x − 1)^2 + 2x(2x + 1) − x^3
]
Solution. According to the rules of operations, we should first treat the exponent in (x − 1)^2. Since (x − 1)^2 = (x − 1)(x − 1) = x^2 − 2 x + 1, we get
3
[
x^2 − 2 x + 1 + 2x(2x + 1) − x^3
]
The next operation we perform is the multiplication 2x(2x + 1) = 4x^2 + 2x. So now we have
3
[
x^2 − 2 x + 1 + 4x^2 + 2x − x^3
]
Combining like terms inside the parenthesis gives
3
[
5 x^2 − x^3 + 1
]
Finally multiplying through by 3 we get 15 x^2 − 3 x^3 + 3.
- Solve the following equation for x
|x + 5| = 2x + 3.
Solution. This is really two linear equations, namely
x + 5 = 2x + 3 or − (x + 5) = 2x + 3.
The first gives x = 2, the second gives x = − 83. So our answer is x = 2 or x = − 83.
- Ticket sales for a play total $2200. There are three times as many adult tickets sold as children’s tickets. The price of an adult ticket is $6 and the price of a child’s ticket is $4. Find the number of children’s tickets which were sold.
Solution. Let x be the number of children’s tickets sold. Since they cost $4 each, they contribute 4x dollars to the total ticket sales. Meanwhile the number of adult tickets sold is three times the number of children’s tickets, namely 3x, and since the cost of each is now $6, the adult tickets contribute 6 · 3 x = 18x dollars to the total sales. Thus 4 x + 18x = 2200
or 22 x = 2200
or finally x = 100.
So there were 100 children’s tickets sold.
