Hey all, just going over some work and these are some questionsI 'm getting that I've gotten wrong. Looking back on them and still can't figure out what I'm doing wrong on some of them.
The first question deals with geometric sequences. Always had trouble with this for some reason =(
I will add questions as they come up, thanks all for your help over the last few days, it really is appreciated. I have my exam tommorow
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1) Determine the indicated term of the geomtric sequence.
a) 1,2,4,...t6
Answer:
Use tn=ar^(n-1)
a = 1
r = 2
n-1 = 5
tn=1(2)^5 = 32
t6 = 32
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2) Prezwalski's wild horse is the only known species of wild horse left in the world. In 1956, there was only 36 of them. They were taken into protective captivity and there was an international effort towards registered breeding. By 1976, there were 250 wild horses. Copy and complete the table. Assume the populations are terms of a geomtric sequence.
Year - Population
1956-36
1961- ?
1966- ?
1971- ?
1976- 250
1981- ?
1986- ?
1991- ?
1996- ?
Answer:
t5 = 36 * r ^(n-1)
t5 = 36 * r^(4) = 250
r^(4) = 250/36
r=(250/36)^(1/4)
r= 1.62333958
so
1956 - t1 = 36
1961 - t2 = 58
1966 - t3 = 95
1971 - t4 = 154
1976 - t5 = 250
1981 - t6 = 406
1986 - t7 = 659
1991 - t8 = 1069
1996 - t9 = 1736
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Let *root* represent your square root sign, let ^# (replace # with value) represent your exponents
The first question deals with geometric sequences. Always had trouble with this for some reason =(
I will add questions as they come up, thanks all for your help over the last few days, it really is appreciated. I have my exam tommorow
-----------
1) Determine the indicated term of the geomtric sequence.
a) 1,2,4,...t6
Answer:
Use tn=ar^(n-1)
a = 1
r = 2
n-1 = 5
tn=1(2)^5 = 32
t6 = 32
------------
2) Prezwalski's wild horse is the only known species of wild horse left in the world. In 1956, there was only 36 of them. They were taken into protective captivity and there was an international effort towards registered breeding. By 1976, there were 250 wild horses. Copy and complete the table. Assume the populations are terms of a geomtric sequence.
Year - Population
1956-36
1961- ?
1966- ?
1971- ?
1976- 250
1981- ?
1986- ?
1991- ?
1996- ?
Answer:
t5 = 36 * r ^(n-1)
t5 = 36 * r^(4) = 250
r^(4) = 250/36
r=(250/36)^(1/4)
r= 1.62333958
so
1956 - t1 = 36
1961 - t2 = 58
1966 - t3 = 95
1971 - t4 = 154
1976 - t5 = 250
1981 - t6 = 406
1986 - t7 = 659
1991 - t8 = 1069
1996 - t9 = 1736
------------
Let *root* represent your square root sign, let ^# (replace # with value) represent your exponents
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