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user100 [1]
3 years ago
7

Find the sixth term of this geometric sequence 3584,896,224

Mathematics
1 answer:
Alex3 years ago
7 0

Answer:

3.5

Step-by-step explanation:

these are being decided by 4 everytime, so you can just continue the sequence. 3584,896,224,56,14,3.5

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Write the inequality represented by the graph below.
Ipatiy [6.2K]

Answer:

y > -x + 3

Step-by-step explanation:

find the slope and y-intercept so you can create a linear equation:

y = mx + b ;   m = slope and  b = y-intercept

slope (m) = (3-0) / (0-3)

m = 3/-3 or -1

we can see the y-intercept by looking at the graph, it is 3

y > -x + 3

6 0
2 years ago
I need help plzzzzz this is really confusing
Anuta_ua [19.1K]

Answer:

B. <u>x</u><u><</u><u> </u>-3

hope it helps.

<h3>stay safe healthy and happy.</h3>
5 0
3 years ago
Read 2 more answers
What is The difference of 3 and a number X
slamgirl [31]

Answer:

"the difference of" means to subtract

Therefore, "the difference of 3 and a number" would be:

3-x

Step-by-step explanation:

7 0
3 years ago
The perimeter of the picture frame is 38 inches. The width of the frame is 7 inches, what is the length?
o-na [289]

Answer:

12 inches

Step-by-step explanation:

So you know that the perimeter of a picture frame (a rectangle) is 2 widths plus 2 lengths.

Let x= lengths and w= widths

From the question we already know that w=7, so we can set up an equation that says:

2x + 2w = 38

2x + 2(7) = 38

2x+14 = 38

2x +14 -14 = 38-14

2x = 24

2x/2 = 24/2

x= 12

Since x= lengths =12

the length of the frame is 12 inches.

7 0
3 years ago
How many permutations are there of the letters in the words (a) TRISKAIDEKAPHOBIA (fear of the number 13)? (b) FLOCCINAUCINIHILI
zubka84 [21]

Answer: Hello!

Let's start with the word TRISKAIDEKAPHOBIA wich has 17 letters (some of them repeat, but it does not matter in this problem)

We want to know how many permutations we can do with 17 letters: then think this way, Lets compose a word. The first letter of this word has 17 options, the second letter of the word has 16 options (you already took one of the set) the third letter of the word has 15 options, and so on.

The total number of permutations is the product of the number of options that you have for each letter, this is:

17*16*15*14*....*3*2*1 = 17! = 3.6e+14

(b) FLOCCINAUCINIHILIPILIFICATION now we have 30 letters in total, using the same reasoning as before, here we have 30! permutations; this is

30! = 2.65e+32

(c) PNEUMONOULTRAMICROSCOPICSILICOVOLCANOCONIOSIS: now there are 47 letters.

then P = 47! = 2.59e+59

(d) DERMATOGLYPHICS: here are 18 letters, then:

p = 18! = 6.4e+15

6 0
3 years ago
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