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velikii [3]
3 years ago
11

Let f(x) = 8x3 - 28x + 61 and g(x) = 2x + 5. Find f(x) / g(x)

Mathematics
2 answers:
LiRa [457]3 years ago
8 0
This is a division of polynomials

   8x^3              - 28x + 61    | 2x + 5
                                             -----------------------
- 8x^3 - 20x^2                       4x^2 - 10x + 11x
---------------------------------
0         - 20x^2 - 28x + 61
           +20x^2 +50x
           ------------------------
                 0    +22x  +61
                       -22x   -55
                      ----------------
                                    6

Then, the division is not exact and the result is 4x^2 - 10x + 11x with remainder 6, or:

4x^2 - 10x + 11x + [ 6/ ( 2x + 5) 



Leona [35]3 years ago
8 0

Your answer would be 4x2 − 10x + 11+ 6 over quantity of 2 x plus 5.


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A beaded necklace is 2.5 feet long. Each bead is .25 inches wide & each pair of beads is separated by 1.25 inches of chain.
adelina 88 [10]

Answer:

There are <u>20 beads</u> on the necklace.

Step-by-step explanation:

Given:

A beaded necklace is 2.5 feet long. Each bead is .25 inches wide & each pair of beads is separated by 1.25 inches of chain.

Now, to find the number of beads on the necklace.

The length of necklace = 2.5 feet.

So, we convert feet into inches by using conversion factor:

<em>1 foot = 12 inches.</em>

<em>2.5 feet = 12 × 2.5 inches.</em>

<em>2.5 feet = 30 inches.</em>

<u>Thus, the length of necklace is 30 inches.</u>

Let the number of beads be x.

The width of each bead = 0.25 inches.

The each pair of beads is separated by 1.25 inches.

Now, we put an equation to get the number of beads:

(0.25+1.25)x=30

0.25x+1.25x=30

1.50x=30

<em>Dividing both sides by 1.50 we get:</em>

x=20.

<u><em>The number of beads = 20.</em></u>

Therefore, there are 20 beads on the necklace.

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3 years ago
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Orlov [11]

Answer:

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8+2m^3 is the answer. Hope it help!
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Svetlanka [38]
First do the ones in the parentheses. So, 2x3=6. 26-6=20. 20^2=400. Now you have 27-4x400. 4x400=1600. 27-1600=-1573.
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