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Colt1911 [192]
3 years ago
15

6. A store orders 8 boxes of notebooks.

Mathematics
1 answer:
Tamiku [17]3 years ago
5 0

Answer: 1,152 notebooks

Step-by-step explanation:

You just do 144 x 8. I’m not really sure what they mean when they say use breaking apart method. Maybe break them up into 8 group of 144. Then just add 144 together 8 times instead of multiplying hope this helps. :)

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9/10 divide by -3/15 as a mixed number
Ket [755]

-4 1/2 is the answer.



8 0
3 years ago
Read 2 more answers
How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

\displaystyle\sum_{n=1}^N (21n-20) = 2332

Expand the sum on the left as

\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

and recall the formulas,

\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2

So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that

21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332

Solve for <em>N</em> :

21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664

21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

4 0
2 years ago
The school orchestra has 25 woodwinds, 20 percussionists, 15 strings and 40 brass instruments.
tankabanditka [31]

Answer: select all what that apply?

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Help me please I need the answers
3241004551 [841]
1- 3/4
2- 64
3- 18
4- 36
5- 10
6- 3/4
7- $4.70
8- <
9- =
10- =
4 0
3 years ago
An easier question all wrong or unhelpful or no-work answers will be reported.
ipn [44]
Elimination: 

3x - 9y = 3
6x - 3y = -24

3x - 9y = 3
18x - 9y = -72
(subtract)
-15x = 75
÷ -15
x = -5

(3 × -5) - 9y = 3
-15 - 9y = 3
+ 15
-9y = 18
÷ -9
y = -2

Substitution:

6x - 3y = -24
+ 3y
6x = -24 + 3y

÷ 6
x = 4 + 0.5y

3(4 + 0.5y) - 9y = 3
12 + 1.5y - 9y = 3
12 - 7.5y = 3
- 12
-7.5y = -9
÷ -7.5
y = 1.2

x = 4 + (0.5 × 1.2)
x = 4 + 0.6
x = 4.6

So this one didn't fail as much, but I got different numbers. If you have to give in values, I'd give in the values from the elimination because I don't trust myself when it comes to the substitution
4 0
3 years ago
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