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mrs_skeptik [129]
3 years ago
8

There are four basic operations: addition, subtraction, multiplication, and division. Do you think these four operations can be

performed on polynomials? What would it look like to perform these operations on polynomials? Which operation do you think would be the simplest? Which do you think will be difficult?
Mathematics
1 answer:
geniusboy [140]3 years ago
7 0
What is the greatest common factor of 91, 59, and 41?
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Ryan made 4 identical necklaces, each having beads and a pendant. the total cost of the beads and pendants for all 4 necklaces w
Monica [59]
Write an equation to show all elements:

Cost of 4 lots of b (beads) + cost of 4p (pendants) = $18.80

Put values in;
9.29 + 4p = 18.80
4p = 18.80 - 9.20
4p = 9.60
p = 2.40 cost of each pendant
5 0
3 years ago
FONTS
Wewaii [24]

Answer:

x=5

Step-by-step explanation:

9x^2 - 2x + 25 = 8x^2 + 8x

9x^2-8x^2-2x-8x+25=0

x^2-6x+25=0 factorize

(x-5)(x-5)=0

x-5=0 then x=5

8 0
3 years ago
How do I do any of this?!
evablogger [386]
U should get the app "photomath" it will help u out alot!!!
7 0
3 years ago
I need help! who ever get this right and explain nicely then I make you brainlist.
Lelechka [254]

Answer:

1. 64

2. 72 square feet

3. 31.5 square feet

Step-by-step explanation:

For the first one use the equation A=bh.

Plug in your values: A=12.8*5.

Solve: A=64

For the second one multiply 8*3 which gives you 24. Next multiply that by 3 to give you 72 square feet.

For the third one multiply 9 by 3.5 to get 31.5 square feet.

5 0
2 years ago
7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many
BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>
5 0
2 years ago
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