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2 years ago

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How many different sets of initials can be formed if every person has one surname (i.e., last name) and either one or two given

names (i.e., a first name, or a first and middle name).

1 answer:

Dima020 [189]2 years ago

4 0

The different sets of initials can be formed if every person has one surname (i.e., last name) and either one or two given names (i.e., a first name, or a first and middle name) is 15600.

There are 26 possible letters, each set of initials would have 3 letters in it, order doesnt matter, i.e. ABC does not equal ACB. So type in 26P3 on your calculator (P=permutation).

15600.

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Pls help!! 25 points!!

ankoles [38]

Answer:

(6, 840) (5, 700) (3, 420) (4, 560)

Step-by-step explanation:

Use the one point that is completely filled out, (6, 840). Divide 840 by 6 and you get 140 so that is the slope. 5x140= what the y intercept will be, 700. 420/140= what the x intercept will be, 3. Now you can see a pattern which helps check your work. They are all decreasing pretty much, not really ordered much but. And finally the last point 4x140=560. Hope that helps! :)

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2 years ago

Then each week thereafter, you add $5 to the account but no interest is earned. Suppose you receive $100 for a graduation presen

irga5000 [103]

100 + 5x = y

100 +5x = 310
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3 years ago

Read 2 more answers

Xelga [282]

Answer:

b-m b-c b-r

Step-by-step explanation:

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2 years ago

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The first, third and thirteenth terms of an arithmetic sequence are the first 3 terms of a geometric sequence. If the first term

Salsk061 [2.6K]

Answer:

The first three terms of the geometry sequence would be $1$ , $5$ , and $25$ .

The sum of the first seven terms of the geometric sequence would be $127$ .

Step-by-step explanation:

<h3>1.</h3>

Let $d$ denote the common difference of the arithmetic sequence.

Let $a_1$ denote the first term of the arithmetic sequence. The expression for the $n$ th term of this sequence (where $n\!$ is a positive whole number) would be $(a_1 + (n - 1)\, d)$ .

The question states that the first term of this arithmetic sequence is $a_1 = 1$ . Hence:

The third term of this arithmetic sequence would be $a_1 + (3 - 1)\, d = 1 + 2\, d$ .
The thirteenth term of would be $a_1 + (13 - 1)\, d = 1 + 12\, d$ .

The common ratio of a geometric sequence is ratio between consecutive terms of that sequence. Let $r$ denote the ratio of the geometric sequence in this question.

Ratio between the second term and the first term of the geometric sequence:

$\displaystyle r = \frac{1 + 2\, d}{1} = 1 + 2\, d$ .

Ratio between the third term and the second term of the geometric sequence:

$\displaystyle r = \frac{1 + 12\, d}{1 + 2\, d}$ .

Both $(1 + 2\, d)$ and $\left(\displaystyle \frac{1 + 12\, d}{1 + 2\, d}\right)$ are expressions for $r$ , the common ratio of this geometric sequence. Hence, equate these two expressions and solve for $d$ , the common difference of this arithmetic sequence.

$\displaystyle 1 + 2\, d = \frac{1 + 12\, d}{1 + 2\, d}$ .

$(1 + 2\, d)^{2} = 1 + 12\, d$ .

$d = 2$ .

Hence, the first term, the third term, and the thirteenth term of the arithmetic sequence would be $1$ , $(1 + (3 - 1) \times 2) = 5$ , and $(1 + (13 - 1) \times 2) = 25$ , respectively.

These three terms ( $1$ , $5$ , and $25$ , respectively) would correspond to the first three terms of the geometric sequence. Hence, the common ratio of this geometric sequence would be $r = 25 /5 = 5$ .

<h3>2.</h3>

Let $a_1$ and $r$ denote the first term and the common ratio of a geometric sequence. The sum of the first $n$ terms would be:

$\displaystyle \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}$ .

For the geometric sequence in this question, $a_1 = 1$ and $r = 25 / 5 = 5$ .

Hence, the sum of the first $n = 7$ terms of this geometric sequence would be:

$\begin{aligned} & \frac{a_1 \, \left(1 - r^{n}\right)}{1 - r}\\ &= \frac{1 \times \left(1 - 2^{7}\right)}{1 - 2} \\ &= \frac{(1 - 128)}{(-1)} = 127 \end{aligned}$ .

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2 years ago

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aksik [14]

Answer:

y=16(x)+7(y)

y=16 + 7

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im not sure bc is nt info but i think i this

Step-by-step explanation:

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