Assignment1

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Math 121
Assignment #1
Mr. Jacobs
Page 14
45)
Simplify
(3π‘š4 )3 (2π‘š−5 )4 → (33 π‘š12 )(24 π‘š−20 ) → (27)(16)(π‘š−8 ) → 432⁄ 8
π‘š
50)
Simplify
(
79)
125𝑝12 π‘ž−14 π‘Ÿ 22
25𝑝8 π‘ž 6 π‘Ÿ −15
−4
)
25𝑝8 π‘ž6 π‘Ÿ −15
4
π‘ž 20
4
→ (125𝑝12 π‘ž−14 π‘Ÿ 22 ) → (5𝑝4 π‘Ÿ 37) →
π‘ž 80⁄
625𝑝16 π‘Ÿ 148
The distance from the Earth to the Sun is defined a 1 Astronomical Unit or AU. It is
about 93 million miles. The average distance from Earth to Pluto is 39 AU’s. Find this
distance in miles.
Set up as a Ratio
39 π΄π‘ˆ ′ 𝑠
1 π΄π‘ˆ
=
93,000,000 π‘šπ‘–π‘™π‘’π‘ 
π‘₯ π‘šπ‘–π‘™π‘’π‘ 
→ πΆπ‘Ÿπ‘œπ‘ π‘  𝑀𝑒𝑙𝑑𝑖𝑝𝑙𝑦 → 1π‘₯ = 39 × 93,000,000
3,627,000,000 π‘šπ‘–π‘™π‘’π‘  → 3.627 × 109 π‘šπ‘–π‘™π‘’π‘ 
80)
One Parsec is about 3.26 Light Years and 1 Light Year is about 5.88 x 1012 miles. Find
the number of miles in 1 Parsec.
Set up as a Ratio
1 πΏπ‘Œ
3.26 πΏπ‘Œ
5.88×1012 π‘šπ‘–π‘™π‘’π‘ 
= π‘₯ π‘šπ‘–π‘™π‘’π‘  → πΆπ‘Ÿπ‘œπ‘ π‘  𝑀𝑒𝑙𝑑𝑖𝑝𝑙𝑦 → 1π‘₯ = 1.91688 × 1013 π‘šπ‘–π‘™π‘’π‘ 
1 π‘ƒπ‘Žπ‘Ÿπ‘ π‘’π‘ = 1.91688 × 1013 π‘šπ‘–π‘™π‘’π‘ 
93)
Suppose that $2,125.00 is invested at 6.2%, compounded semiannually. How much is in
the account at the end of 5 years?
P = 2,125
r = 6.2% = 0.062
n = 2 (for Semiannual)
t=5
π‘Ÿ
𝐴 = 𝑃(1 + 𝑛)𝑛𝑑 = (2125)(1 +
0.062 (2)(5)
)
2
There is $2,883.67 in the account.
= 2,883.67
Math 121
96)
Assignment #1
Mr. Jacobs
Suppose that $4,875.00 is invested at 5.8%, compounded quarterly. How much is in the
account at the end of 9 years?
P = 4,875
r = 5.8% = 0.058
n = 4 (for Quarterly)
t=9
π‘Ÿ
𝐴 = 𝑃(1 + 𝑛)𝑛𝑑 = (4875)(1 +
0.058 (4)(9)
)
4
= 8,185.56
There will be $8,185.56 in the account.
Page 20
7)
Perform the operations indicated.
(2π‘₯ + 3𝑦 + 𝑧 − 7) + (4π‘₯ − 2𝑦 − 𝑧 + 8) + (−3π‘₯ + 𝑦 − 2𝑧 − 4) →
2π‘₯ + 4π‘₯ − 3π‘₯ + 3𝑦 − 2𝑦 + 𝑦 + 𝑧 − 𝑧 − 2𝑧 − 7 + 8 − 4 → πŸ‘π’™ + πŸπ’š − πŸπ’› − πŸ‘
13)
Perform the operations indicated.
(π‘Ž − 𝑏)(2π‘Ž3 − π‘Žπ‘ + 3𝑏 2 ) →
π‘Ž(2π‘Ž3 ) + π‘Ž(−π‘Žπ‘) + π‘Ž(3𝑏 2 ) − 𝑏(2π‘Ž3 ) − 𝑏(−π‘Žπ‘) − 𝑏(3𝑏 2 ) →
2π‘Ž3 − π‘Ž2 𝑏 + 3π‘Žπ‘ 2 − 2π‘Ž3 𝑏 + π‘Žπ‘ 2 − 3𝑏 3 → πŸπ’‚πŸ‘ − πŸπ’‚πŸ‘ 𝒃 − π’‚πŸ 𝒃 + πŸ’π’‚π’ƒπŸ − πŸ‘π’ƒπŸ‘
32)
Perform the operations indicated.
(4π‘₯ 2 − 5𝑦)2 → (4π‘₯ 2 − 5𝑦)(4π‘₯ 2 − 5𝑦) →
4π‘₯ 2 (4π‘₯ 2 ) + 4π‘₯ 2 (−5𝑦) − 5𝑦(4π‘₯ 2 ) − 5𝑦(−5𝑦) → 16π‘₯ 4 − 20π‘₯ 2 𝑦 − 20π‘₯ 2 𝑦 + 25𝑦 2 →
16π‘₯ 4 − 40π‘₯ 2 𝑦 + 25𝑦 2
40)
Perform the operations indicated.
(5π‘₯ + 2𝑦 + 3)(5π‘₯ + 2𝑦 − 3)
5π‘₯(5π‘₯) + 5π‘₯(2𝑦) + 5π‘₯(−3) + 2𝑦(5π‘₯) + 2𝑦(2𝑦) + 2𝑦(−3) + 3(5π‘₯) + 3(2𝑦) − 9
25π‘₯ 2 + 10π‘₯𝑦 − 15π‘₯ + 10π‘₯𝑦 + 4𝑦 2 − 6𝑦 + 15π‘₯ + 6𝑦 − 9
25π‘₯ 2 + 4𝑦 2 + 20π‘₯𝑦 − 9
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