summer 2015 math practice problems solutions school of public policy and governance math practice...

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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) = + 4 = βˆ’ b) = 2 + 4 = βˆ’ c) = 3 + 5 = βˆ’ d) = 4 βˆ’ 8 = + e) = .25 βˆ’ 100 = + f) = βˆ’ ! ! βˆ’ 100 = βˆ’ βˆ’ g) = βˆ’.1 + 400 = βˆ’ + h) = 32 + ! ! = βˆ’ 2. SOLVING PAIRS OF SIMULTANEOUS EQUATIONS a) 3 + = 13 + 6 = βˆ’ 7 = βˆ’, = b) = + 4 = 2 + 4 = βˆ’, = βˆ’ c) 5 = 2 + 3 2 βˆ’ = 0 = , = d) + 2 = 8 βˆ’ 2 = 4 = , = e) 4 + = 9 βˆ’ = 1 = , = f) 2 + 3 = 28 + = 11 = , = g) 11 + 6 = 79 11 + 3 = 67 = , = h) ! ! + ! ! = 8 2 3 + 3 2 = 17 = , = 3. GRAPHING FROM AN EQUATION OR ITS INVERSE a) Graph = 3 b) Graph = + 3 c) Graph = ! ! + 1

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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) 𝑦 = π‘₯ + 4 Γ  𝒙 = π’š βˆ’ πŸ’

b) π‘₯ = 2𝑦 + 4 Γ   π’š = 𝟏

πŸπ’™ βˆ’ 𝟐

c) 𝑦 = 3π‘₯ + 5 Γ  𝒙 = 𝟏

πŸ‘π’š βˆ’ πŸ“

πŸ‘

d) 𝑦 = 4π‘₯ βˆ’  8 Γ  𝒙 = 𝟏

πŸ’π’š + 𝟐

e) 𝑃 = .25𝑄 βˆ’ 100 Γ  𝑸 = πŸ’π‘· + πŸ’πŸŽπŸŽ

f) 𝑦 = βˆ’ !!π‘₯ βˆ’  100 Γ  𝒙 = βˆ’πŸ’π’š βˆ’  πŸ’πŸŽπŸŽ

g) 𝑃 = βˆ’.1𝑄 +  400 Γ  𝑸 = βˆ’πŸπŸŽπ‘· +  πŸ’πŸŽπŸŽπŸŽ

h) 𝑦 = 32 + !

!π‘₯ Γ  𝒙 = πŸ“

πŸ–π’š βˆ’ 𝟐𝟎

2. SOLVING PAIRS OF SIMULTANEOUS EQUATIONS a) 3π‘₯ + 𝑦 = 13

π‘₯ + 6𝑦 = βˆ’  7 π’š =  βˆ’πŸ, 𝒙 = πŸ“

b) 𝑦 = π‘₯ + 4 π‘₯ = 2𝑦 + 4 π’š =  βˆ’πŸ–, 𝒙 = βˆ’πŸπŸ

c) 5π‘Ž = 2𝑏 + 3 2π‘Ž βˆ’ 𝑏 = 0 𝒂 =  πŸ‘,𝒃 = πŸ”

d) π‘₯ + 2𝑦 = 8 π‘₯ βˆ’ 2𝑦 = 4 π’š =  πŸ, 𝒙 = πŸ”

e) 4π‘₯ + 𝑦 = 9 π‘₯ βˆ’ 𝑦 = 1 π’š =  πŸ, 𝒙 = 𝟐

f) 2π‘₯ + 3𝑦 =  28 π‘₯ + 𝑦 =  11 π’š =  πŸ”, 𝒙 = πŸ“

g) 11π‘₯   +  6𝑦 = 79

11π‘₯ + 3𝑦 = 67 π’š =  πŸ’, 𝒙 = πŸ“

h) !!π‘₯   +  !

!𝑦 = 8

23π‘₯   +  

32𝑦 = 17

π’š =  πŸ”, 𝒙 = 𝟏𝟐

3. GRAPHING FROM AN EQUATION OR ITS INVERSE a) Graph 𝑦 = 3π‘₯ b) Graph 𝑦 = π‘₯ + 3 c) Graph 𝑦 = !

!π‘₯ + 1

d) Graph 𝑦 = !!π‘₯ + 8 e) Graph π‘₯ = 𝑦 + 3 f) Graph π‘₯ = !

!𝑦 + 1

g) Graph 𝑦 = 5π‘₯ and 𝑦 = βˆ’5π‘₯ + 50 h) Graph π‘₯ = !

!𝑦 βˆ’ 7.5 and 𝑦 = βˆ’ !

!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’š = βˆ’ 𝟐

πŸ‘π’™ + 𝟐𝟏 π’š = βˆ’ 𝟏

πŸπŸŽπ’™ + πŸ— π’š = πŸ‘π’™ + 𝟏𝟎

d) Find the equation for the line. e) Find the equation for the line. f) Find the equation for the line.

π’š = βˆ’ πŸπŸ“

πŸ—π’™ + πŸπŸ“ π’š = πŸ“

πŸπŸ•π’™ π’š = 𝟏

πŸ—π’™

a) Find the equation for the line. b) Find the equation for the line. c) Find the equation for the line.

π’š = βˆ’ 𝟐

πŸ—π’™ + πŸπŸ• π’š = πŸ“ 𝒙 = πŸπŸ“

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‰ˆ 14.88 in logarithmic form. Γ  Solution: ln 14.88 = 2.7 c) Express the equation e2 β‰ˆ 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 – ln 2 Γ  Solution: ln 3 (using the quotient property) f) Expand the following expression: ln 12π‘₯! Γ  Solution: π₯𝐧 𝟏𝟐 + πŸ’ π₯𝐧 𝒙 (using the product and power

properties) g) Expand the following expression: ln !!

!

!! Γ  Solution: ln 4 + 3 ln y – 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