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EVANS SCHOOL OF PUBLIC POLICY AND GOVERNANCE Math practice problems for incoming MPA students Summer 2015
SOLUTIONS
1. INVERTING EQUATIONS a) π¦ = π₯ + 4 Γ π = π β π
b) π₯ = 2π¦ + 4 Γ π = π
ππ β π
c) π¦ = 3π₯ + 5 Γ π = π
ππ β π
π
d) π¦ = 4π₯ β 8 Γ π = π
ππ + π
e) π = .25π β 100 Γ πΈ = ππ· + πππ
f) π¦ = β !!π₯ β 100 Γ π = βππ β πππ
g) π = β.1π + 400 Γ πΈ = βπππ· + ππππ
h) π¦ = 32 + !
!π₯ Γ π = π
ππ β ππ
2. SOLVING PAIRS OF SIMULTANEOUS EQUATIONS a) 3π₯ + π¦ = 13
π₯ + 6π¦ = β 7 π = βπ, π = π
b) π¦ = π₯ + 4 π₯ = 2π¦ + 4 π = βπ, π = βππ
c) 5π = 2π + 3 2π β π = 0 π = π,π = π
d) π₯ + 2π¦ = 8 π₯ β 2π¦ = 4 π = π, π = π
e) 4π₯ + π¦ = 9 π₯ β π¦ = 1 π = π, π = π
f) 2π₯ + 3π¦ = 28 π₯ + π¦ = 11 π = π, π = π
g) 11π₯ + 6π¦ = 79
11π₯ + 3π¦ = 67 π = π, π = π
h) !!π₯ + !
!π¦ = 8
23π₯ +
32π¦ = 17
π = π, π = ππ
3. GRAPHING FROM AN EQUATION OR ITS INVERSE a) Graph π¦ = 3π₯ b) Graph π¦ = π₯ + 3 c) Graph π¦ = !
!π₯ + 1
d) Graph π¦ = !!π₯ + 8 e) Graph π₯ = π¦ + 3 f) Graph π₯ = !
!π¦ + 1
g) Graph π¦ = 5π₯ and π¦ = β5π₯ + 50 h) Graph π₯ = !
!π¦ β 7.5 and π¦ = β !
!π₯ + 25
4. RECOVERING THE EQUATION OF A LINE FROM A GRAPH a) Find the equation for the line. b) Find the equation for the line. c) Find the equation for the line.
π = β π
ππ + ππ π = β π
πππ + π π = ππ + ππ
d) Find the equation for the line. e) Find the equation for the line. f) Find the equation for the line.
π = β ππ
ππ + ππ π = π
πππ π = π
ππ
a) Find the equation for the line. b) Find the equation for the line. c) Find the equation for the line.
π = β π
ππ + ππ π = π π = ππ
5. CALCULATING AREAS UNDER STRAIGHT LINE CURVES
a) Find the area under the line. b) Find the area under the line. c) Find the area under the line.
(21 X 4) / 2 = 42 (35 X 2) / 2 = 35 (15 X 27) / 2 = 202.5
d) Find the area under the line. e) Find the area under the line. f) Find the area under the line.
((21-6) X 6) / 2 + (6 X 6) = 81 ((40-15) X 9) / 2 + (15 X 9) = 247.5 ((27-6) X 6) / 2 + (6X6) = 99
g) Find the area under the line. h) Find the area under the line. i) Find the area under the line.
(12 X 4) / 2 + ((21-12) X (10-4)) / 2 (30-5)X6 / 2 + (9-6)X5) / 2 ((12-6)X12)/2 + ((15-12) X (24-12)) + (12 X (10-4)) = 123 + (5 X 6) = 112.5 / 2 + (6X12) + (24-12) X 12 = 270
6. NATURAL LOGS AND EXPONENTIAL FUNCTIONS
a) Express ln 2.7183 = 1 in exponential form. Γ Solution: 2.7183 = e1 b) Write the equation e2.7 β 14.88 in logarithmic form. Γ Solution: ln 14.88 = 2.7 c) Express the equation e2 β 7.39 in logarithmic form. Γ Solution: ln 7.39 = 2 d) Write as a single logarithm: ln 3 + ln 7 Γ Solution: ln 21 (using the product property) e) Write as a single logarithm: ln 6 β ln 2 Γ Solution: ln 3 (using the quotient property) f) Expand the following expression: ln 12π₯! Γ Solution: π₯π§ ππ + π π₯π§ π (using the product and power
properties) g) Expand the following expression: ln !!
!
!! Γ Solution: ln 4 + 3 ln y β 5 ln x
h) Solve for x: ln x = 24 Γ Solution: x = e24 i) Solve for x: e4x+2 = 50 Γ Solution: x = .478 j) Solve 1 + 2e1-3z = 15. Γ z = -.315
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