Answer: (0,-2)
hope that answered your question
Problem 1)
The base of the exponential is 12 which is also the base of the log as well. The only answer choice that has this is choice B.
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Problem 2)
log(x) + log(y) - 2log(z)
log(x) + log(y) - log(z^2)
log(x*y) - log(z^2)
log[(x*y)/(z^2)]
Answer is choice D
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Problem 3)
log[21/(x^2)]
log(21) - log(x^2)
log(21) - 2*log(x)
This matches with choice B
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Problem 4)
Ln(63) = Ln(z) + Ln(7)
Ln(63)-Ln(7) = Ln(z)
Ln(63/7) = Ln(z)
Ln(9) = Ln(z)
z = 9
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Problem 5)
Ln(5x-3) = 2
5x-3 = e^2
5x = e^2+3
x = (e^2+3)/5
This means choice A is the answer
Answer:is A
Step-by-step explanation: because if we add all of those numbers together we get 49vthen divided by 6 gives us 8.16
Take an example: The salary per hour:
Assume x (variable) is the number of hours worked and assume THAT THE RATE PER HOUR IS $10, then:
If you work x hours at the hourly rate of$10, you will generate an income,
say, y that is equal the number of hours worked by the rate per hour, that is:
y = 10.x. The more hour(x) you work, the more income(y) you get.
Hence we created a relationship between y and x , or in English, between the earning and the number of hours worked.
Moreover y depends on x, if there is no x, there is no y and if x increases, y increases. However x is an INDEPENDENT variable because you can
give to x any number & y will follow
Find the next two terms in the given sequence, then write it in recursive form. A.) {7,12,17,22,27,...} B.) { 3,7,15,31,63,...}
iren [92.7K]
Answer:
A) a_n = 5n + 2
B) a_n = (2^(n + 1)) - 1
Step-by-step explanation:
A) The sequence is given as;
{7,12,17,22,27,...}
The differences are:
5,5,5,5.
This is an arithmetic sequence following the formula;
a_n = a_1 + (n - 1)d
d is 5
Thus;
a_n = a_1 + (n - 1)5
Now, a_1 = 7. Thus;
a_n = 7 + 5n - 5
a_n = 5n + 2
B) The sequence is given as;
{ 3,7,15,31,63,...}
Now, let's write out the differences of this sequence:
Differences are:
4, 8, 16, 32
This shows that it is a geometric sequence with a common ratio of 2.
In the given sequence, a_1 = 3 and a_2 = 7 and a_3 = 15
Thus, a_2 = 2a_1 + 1
Also, a_(2 + 1) = 2a_2 + 1
Combining both equations, we can deduce that: a_(n + 1) = 2a_n + 1
Thus; a_n can be expressed as:
a_n = (2^(n + 1)) - 1