Solve for r. 3r − 2r =
1 answer:
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Given:
The table for a geometric sequence.
To find:
The formula for the given sequence and the 10th term of the sequence.
Solution:
In the given geometric sequence, the first term is 1120 and the common ratio is:
The nth term of a geometric sequence is:
Where a is the first term and r is the common ratio.
Putting , we get
Therefore, the required formula for the given sequence is .
We need to find the 10th term of the given sequence. So, substituting in the above formula.
Therefore, the 10th term of the given sequence is 2.1875.
1. Remarks:
f(x) to f(x)-3 is the whole graph of f(x), shifted 3 units down.
f(x) to -2f(x):
The effect of "multiplication by -" is that the whole graph is reflected with respect to the x axis, so it is turned upside down.
The effect of "multiplication by 2" is that every point is "stretched vertically by a factor of 2" . So for example the point (-1, -4) in the original function, becomes (-1, -8) in the second one. Or (2, 5) would become (2, 10).
The only points that do not change (are not streched vertically) are the roots. For example if (4,0) is an x-intercept (a root) in the original function, (4,0) is still a root in the second one because 2 times 0 is still 0.
2. Consider the polynomial function of degree n:
a. Y-intercept
The y - intercept is the value of the polynomial function at x=0.
So it is f(0)=
, that is, the constant term of f(x)
in f(x)-3 the y intercept is shifted 3 units down as any other point, so it becomes
In -2f(x), the y-intercept becomes 
b. Regions of f decreasing or increasing
f(x)-3 is f(x) just shifted down 3 units, so they are both increasing and decreasing in the same intervals of x
-2f(x) is f(x) turned upside down, so -2f(x) is increasing in all intervals f(x) is decreasing and it is decreasing in all intervals f(x) is increasing.
c. End behaviors
By now it is clear that end behaviors of f(x) and f(x)-3 are same, and f(x) with -2f(x) are opposite
d. Evenness, oddness
If f(x) is even, then f(x)=f(-x)
Let g(x)=f(x)-3
g(x)=f(x)-3=f(-x)-3=g(-3), so in this case f(x)-3 is even
If f(x) is odd, then f(-x)=-f(x)
g(x)=f(x)-3=-f(-x)-3,
so -g(x)=f(-x)+3
g(-x)=f(-x)-3,
so g(-x) is not equal to -g(x). Which means f(x)-3 is not odd if f(x) is
Consider f(x)=-2f(x)
If f(x) is even, f(x)=f(-x)
g(x)=-2f(x)=-2f(-x)
g(-x)=-2f(-x)
So g(x)=g(-x), which means -2f(x) is even if f(x) is even
If f(x) is odd, f(x)=-f(-x)
let g(x)=-2f(x)=-2(-f(-x))=2f(-x)
g(-x)=-2f(-x)=-2(-f(x))=2f(x)
so g(-x) is not equal to -g(x), thus -2f(x) is not odd if f(x) is odd.
The conclusions about oddness and evenness can be also derived from the discussions about the graphs.
f(x) to f(x)-3 is the whole graph of f(x), shifted 3 units down.
f(x) to -2f(x):
The effect of "multiplication by -" is that the whole graph is reflected with respect to the x axis, so it is turned upside down.
The effect of "multiplication by 2" is that every point is "stretched vertically by a factor of 2" . So for example the point (-1, -4) in the original function, becomes (-1, -8) in the second one. Or (2, 5) would become (2, 10).
The only points that do not change (are not streched vertically) are the roots. For example if (4,0) is an x-intercept (a root) in the original function, (4,0) is still a root in the second one because 2 times 0 is still 0.
2. Consider the polynomial function of degree n:
a. Y-intercept
The y - intercept is the value of the polynomial function at x=0.
So it is f(0)=
in f(x)-3 the y intercept is shifted 3 units down as any other point, so it becomes
In -2f(x), the y-intercept
b. Regions of f decreasing or increasing
f(x)-3 is f(x) just shifted down 3 units, so they are both increasing and decreasing in the same intervals of x
-2f(x) is f(x) turned upside down, so -2f(x) is increasing in all intervals f(x) is decreasing and it is decreasing in all intervals f(x) is increasing.
c. End behaviors
By now it is clear that end behaviors of f(x) and f(x)-3 are same, and f(x) with -2f(x) are opposite
d. Evenness, oddness
If f(x) is even, then f(x)=f(-x)
Let g(x)=f(x)-3
g(x)=f(x)-3=f(-x)-3=g(-3), so in this case f(x)-3 is even
If f(x) is odd, then f(-x)=-f(x)
g(x)=f(x)-3=-f(-x)-3,
so -g(x)=f(-x)+3
g(-x)=f(-x)-3,
so g(-x) is not equal to -g(x). Which means f(x)-3 is not odd if f(x) is
Consider f(x)=-2f(x)
If f(x) is even, f(x)=f(-x)
g(x)=-2f(x)=-2f(-x)
g(-x)=-2f(-x)
So g(x)=g(-x), which means -2f(x) is even if f(x) is even
If f(x) is odd, f(x)=-f(-x)
let g(x)=-2f(x)=-2(-f(-x))=2f(-x)
g(-x)=-2f(-x)=-2(-f(x))=2f(x)
so g(-x) is not equal to -g(x), thus -2f(x) is not odd if f(x) is odd.
The conclusions about oddness and evenness can be also derived from the discussions about the graphs.