Mth 113
Review Sheet for Exam 2
Our Exam will take place on Thursday, 11/3. The exam will be closed book though you have free rein with your
calculator. The exam will cover the material we have discussed in class through Thursday, 10/27.
Expect writing/explaining problems, T/F problems, Fill-in-the-Blank problems, computational problems and
problems involving graphic interpretation.
Topics to be covered:
 Anything and everything from your previous math classes and Previous Exam(s)

Solving Single Variable Equations- Algebraically and Graphically

Graphing: Lines ↔ Eqns, Quadratics ↔ Eqns, Circles ↔ Eqns

Vector arithmetic: u ± v , u • v , u × v , k v , v /k, - v , v̂ , || v ||

Graphical Interpretation of Vectors and Vector Relationships

Vector Applications
A Few Practice Problems for Midterm 2
Notes: Be sure you write neatly using complete sentences and correct grammar. Diagrams should be included
where appropriate. Neatness and organization are necessary for full credit. Consider every one of these
questions as a possible test question. If you cannot figure out how to do the problem on your own come by for
hints and help before the last minute.
1)
Outline the procedure for algebraically solving linear equations.
2)
Outline the procedure for algebraically solving quadratic equations (include both QF and factoring).
3)
Outline the procedure for algebraically solving radical equations.
4)
Outline the procedure for graphically solving general single variable equations (include both cases f(x) = 0
and f(x) = g(x)).
5)
Give an interpretation of each parameter in y ± k = a f(b(x ± h))
6)
Algebraically solve for x (check your answer graphically):
(a)
2x + 1
5 − 2 = 2x + 3
(b) 1+
x
x = 3x
1−1+x
x2 + 5x + 1 − 2 = 2x − 3
7)
Solve for x algebraically and check your solution graphically:
8)
2y + 1
Solve for y: (a) x − 1 = y + 1
2y + 3x
(b) 2x − 4 + 1 = y + x
9)
2y + 1
Solve for x: (a) x − 1 = y + 1
2y + 3x
(b) 2x − 4 + 1 = y + x
10)
Solve for x: (a)
11)
Solve for x: (a) (x − a)(x − b) = (x − c)(x − d)
1
1
(b) a + x − b = c + x − d
12)
Solve for x & y: (a) 3x + 7y = -1 & 2y − 4x = 24
(b) ax + by = 1 & x − y = 1
13)
Outline the procedure for algebraically finding the equation of a line through two points (a, b) & (c, d).
14)
Find the equation of the line through (5, -7) and (-2, 3).
15)
Find the equation of the line through (a, f(a)) and (b, f(b)).
ax2 + 4x + 1 − 2 = x
(b) 3ax2 + 5bx + 6c = 3x2 + 2x − 5
16)
Find the equation of the line through (-3, 8) and parallel to 2x + 5y = 10.
17)
Find the equation of the line through (-3, 8) and perpendicular to 2x + 5y = 10.
18)
Algebraically find the intersection(s) between 2x + 3y = 12 and y = -2x2 + 5x + 7. Then graph the
functions and check your answer.
19)
Find a parabola with y-intercept of 4 and vertex at (4, 2)
20)
Find a parabola with a maximum value of 4 and having roots at x = -3 and x = 7.
21)
Find the center and radius of the circle given by x2 + 4x + y2 − 6x = 12
22)
Simplify to all positive exponents: (a)
23)
For f(x) = 3x2 − 4x + 5, compute and simplify:
x2 x5 y3
y5 x−3
(b)
x-2 (x5 y3)4
y5 x3
f(x + h) − f(x)
h
24) Given: u = (6, -5), v = (-8, 6), w = (10, 12)
(a) Draw and label: u , v , u + v + w , u − v + w , û
(e) Find compv u
(b) Compute: = u + v , u − v , v̂ , || u ||, u • v , u × v
(f) Find proju v
(c) Compute: 2 u • 3 v , 7.69263434468 w × 3,452,875 w
(g) Find p such that u + v + w + p = 0
(d) Show that u  w but u is not perpendicular to v
(h) Find q̂ such that q̂  v
25) One bulldozer pulls on a skid shack as shown. How should the other
bulldozer pull to move the shack exactly due east?
26) A 1,000# weight (W) is attached as shown. a = 50° and b = 20°.
What is the tension in cable A and cable B? What tension should be
applied to cables L and R to support and balance the system?
lbs
00
2,0
Shack
a
A
B
b
L
R
W
27)
A smuggler is headed straight for the landing strip at 120 mph on a heading of N 40° E. At exactly 12
midnight he is 20 miles from the drop zone. At that point a strong wind begins that has the effect of
moving him at 30 mph in direction S 10° E. How close will he get to the landing zone?
M
28) Andy (A) and Betty (B) are fighting over a toy when Mom (M) steps in to the fray.
Andy is pulling due SW with 40# of force and Betty is pulling due East with 50# of
force. What must Mom's vector be if the toy is to be kept stationary?
B
29) A swimmer heads across a river at 2 mph. Normally the swimmer would land
straight across on the opposite bank but the river is flowing at 3 mph so he lands
downstream of his intended position directly across from his start. How far down
stream does he land if the river is 200 ft across?
30) Find the minimum force necessary to keep the 2 ton SUV from rolling back.
A
5°
1)
An Efficient Scheme for Solving Linear Equations
(0)
rewrite subtractions as additions e.g. 5 – 3(8x + 2) → 5 + (-3)(8x + 2)
(1)
Remove all fractions (multiply both sides (all terms) by LCD then cancel)
Multiply EVERY TERM, ONE LCD PER TERM
(2)
Remove all parentheses (apply Distributive Property)
(3)
Combine like terms (repeat (1) , (2) & (3) as necessary)
(4)
Shift variable term(s) to one side (Additive Property)
(5)
Shift everything else to the other side (Additive Property)
(6)
If necessary write variable term(s) as (coefficient) × (variable). (Distributive Law)
(7)
(8)
i.e. 2 x + 4x = ( 2 + 4)x or π x + ax = (π + a)x
Divide both sides by the variable's coefficient (Multiplicative property)
Check the answer
2)
An Efficient Scheme for Solving Quadratic Equations
(0)
rewrite subtractions as additions e.g. 5 – 3(8x + 2) → 5 + (-3)(8x + 2)
(1)
Remove all fractions (multiply both sides (all terms) by LCD then cancel)
Multiply EVERY TERM, ONE LCD PER TERM
(2)
Remove all parentheses (apply Distributive Property)
(3)
Combine like terms (repeat (1) , (2) & (3) as necessary)
(4)
Shift ALL term(s) to one side (Additive Property)
(5)
Write in standard form ax2 + bx + c = 0
(6)
Factor or use the Quadratic formula to solve for x
(7)
Check the answer
3)
An Efficient Scheme for Solving Radical Equations
(0)
rewrite subtractions as additions e.g. 5 – 3(8x + 2) → 5 + (-3)(8x + 2)
(1)
Remove all fractions (multiply both sides (all terms) by LCD then cancel)
Multiply EVERY TERM, ONE LCD PER TERM
(2)
Remove all parentheses (apply Distributive Property)
(3)
Combine like terms (repeat (1) , (2) & (3) as necessary)
(4)
Shift ALL radical term(s) to one side (Additive Property)
(4)
Shift remaining term(s) to other side (Additive Property)
(5)
Square both sides.
(6)
Continue with procedure corresponding to new form of the equation
4a)
An Efficient Scheme for Solving f(x) = g(x)
(1)
Enter f(x) into Y1, Enter g(x) into Y2. Be sure to use 'X' as the variable.
(2)
Graph
(3)
Find all intersections; the x-value is the solution(s) to the original equation.
4b)
An Efficient Scheme for Solving f(x) = 0
(1)
Enter f(x) into Y1. Be sure to use 'X' as the variable.
(2)
Graph
(3)
Find all roots (zeros); the x-value is the solution(s) to the original equation.
y ± k = a f(b(x ± h)) In this form:
a > 1 = y-stretch, a < 1 = y-compression, a < 0 flip around x-axis
b > 1 = x-compression, b < 1 = x- stretch, b < 0 flip around y-axis
+h = shift left, −h = shift right, +k shift down, −k shift up
5)
6)
(a) x = -3; (b) x2 − 2x + 1 = 0 → x = 1
7)
3x2 – 9x = 0 → x = 0, 3
8)
x–2
2x2 – 9x + 2xy + 4
(a) y = 3 – x ; (b) y =
6 – 2x
9)
x–2
2x2 – 9x + 2xy + 4
(a) y = 3 – x ; (b) y =
6 – 2x
10)
(a) (a – 1)x2 – 3 = 0 → x = ±
3
a–1;
(b) (3a – 3)x2 + (5b – 2)x + (6c + 5) = 0 → x =
11)
ab – cd
(a) x = a + b – c – d ; (b) (a – c)x2 + (-ad – ab + cd + bc)x + (b – d + abd – bcd) = 0 →
x=
12)
13)
-(5b – 2) ± (5b – 2)2 − 4(3a – 3)(6c + 5)
2(3a – 3)
-(-ad – ab + cd + bc) ± (-ad – ab + cd + bc)2 − 4(a – c)( b – d + abd – bcd)
2(a – c)
b+1
a–1
(a) (x,y) = (-5,2); (b) x = a + b , y = a – b
An Efficient Scheme for finding the equation of a line through two points
b–d
(1)
Find the slope: m = a – c
(2)
b–d
Find the y-intercept: b = y – mx = b − a – c a
(3)
Write equation as y = mx + b.
14)
y = -10x/7 + 1/7
15)
y – f(a) =
16)
2x + 5y = 34
17)
5x − 5y = -31
18)
6x2 – 17x – 9 = 0 → (x,y) ≈ (3.29, 1.81), ≈ (-0.456, 4.30)
19)
y = (⅛)(x – 4)2 + 2
20)
y = -(4/25)(x + 3)(x – 7)
21)
(x + 2)2 + (y – 3)2 = 52 → Ctr = (-2,3), r = 5
22)
x7.5
(a) y3.5 ; (b) x16.5 y9.5
23)
3(x + h)2 – 4(x + h) + 5 – [3x2 – 4x + 5]
= 6x + 6h – 4
h
24a)
f(b) – f(a)
b – a (x – a)
(b) (-2,1), (14,-11),
(c)
(d)
(e)
(f)
(g)
(h)
(-8, 6)
10 = (-0.8, 0.6), 61 , -78, (0, 0, -4)
-468, (0, 0, 0)
u • w = 0, u • v ≠ 0
-7.8
-0.78(-8, 6)
(-8, -13)
k(-3, 4)
25)
(2000 sin (30°)/sin (20°), -2000 sin (20°))
26)
Ta cos 50° = Tb cos 70°, Ta sin 40° + Tb sin 70° = 1000 → Ta = 415#, Tb = 780#, L = 636#, R = 534#
27)
bad question
27a) A smuggler is headed straight for the landing strip at 120 mph on a heading of N 40° E. At exactly 12 midnight he is
20 miles from the drop zone. The coast line is running dead N-S. At that point a strong cross-wind begins to blow at
30 mph in direction S 10° E. What will be his adjusted speed? What will be his adjusted direction? How close will he
get to the landing zone?
speed = 103 mph, direction θ = 37°, 5.6 mi S of landing zone.
28)
(-21.7, 28.3)
29)
300 ft
30)
f = 4000 sin 5°