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Alex_Xolod [135]
3 years ago
6

Find the sum of the first 10 terms for the sequence {21, 63, 189, . . . }.

Mathematics
2 answers:
marishachu [46]3 years ago
8 0
It’ll be 620,004
Each number is multiplied by 3 so the 10 numbers would be
21; 63; 189; 567; 1,701; 5,103; 15,309; 45,927; 137,781; 413,343
Adding those numbers equals 620,004
antiseptic1488 [7]3 years ago
3 0
The sum of n terms of a geometric sequence with first term "a" and common ratio "r" is given by
S_{n}=a\dfrac{r^{n}-1}{r-1}
You have a=21, r=3, so the sum of 10 terms is
S_{10}=21\cdot \dfrac{3^{10}-1}{3-1}=620004

The appropriate choice is the 2nd one:
   620,004
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Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?
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On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

f(x) = \dfrac{2}{x - 1} + 4

From the given function, we have;

When x = 1, the denominator of the fraction, \dfrac{2}{x - 1}, which is (x - 1) = 0, and the function becomes, \dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as <em>x</em> increases, the fraction \dfrac{2}{x - 1} tends to 0, therefore as x increases, we have;

\lim_  {x \to \infty}  \dfrac{2}{(x - 1)} \to 0, and \  \dfrac{2}{(x - 1)}  + 4 \to 4

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, \dfrac{2}{0 - 1} + 4 = 2. When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

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Can any one help me i really need it
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