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4 years ago

Find two consecutive numbers whose sum is 117

2 answers:

6 0

So,

x + y = 117
y = x + 1
x + x + 1 = 117

Collect Like Terms
2x + 1 = 117

Subtract 1 from both sides
2x = 116

Divide both sides by 2
x = 58

y = x + 1

y = 58 + 1

y = 59

The 2 consecutive integers are 58 and 59.

nika2105 [10]4 years ago

4 0

$n+n+1=117\\
2n=116\\
n=58\\
n+1=59$

These numbers are 58 and 59.

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a_sh-v [17]

Answer:

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Step-by-step explanation:

One dozen means 12. If you ask for 1/12 of 12, you multiply 12 and 1/12. You should get 1 as your answer.

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Alik [6]

Answer:

32, 16, 8, 4, 2, 1

Explanation:

The geometric mean can be represented by $\sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}$ .

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

$a_{n} = a_{1} • r^{n-1}$

Where $a_{n}$ is the nth term, $a_{1}$ is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, $a_{1}$ will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

$\sqrt[5]{\frac{a_{6}}{a_{1}}}$ =

$\sqrt[5]{\frac{1}{32}}$ =

$\frac{1}{2}$ .

Therefore: $r = \frac{1}{2}$ .

Using all the information we have, we can find the explicit rule:

$a_{n} = a_{1} • r^{n-1}$

$a_{1}$ = 32
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$a_{n} = a_{1} • r^{n-1}$ →

$\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}$

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We can test that this works by substituting the number location of the term you want to find.

For instance:

$a_{1} = 32 • (\frac{1}{2})^{1-1}$

$a_{1} = 32 • (\frac{1}{2})^{0}$

$a_{1} = 32 • 1$

$a_{1} = 32$

$a_{6} = 32 • (\frac{1}{2})^{6-1}$

$a_{6} = 32 • (\frac{1}{2})^{5}$

$a_{6} = 32 • \frac{1}{32}$

$a_{6} = 1$

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