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Anvisha [2.4K]
2 years ago
7

whats the answer to this? I don't know how to do it but i need to factor it completely with whatever method gies with it.​

Mathematics
1 answer:
sesenic [268]2 years ago
4 0

Answer:

3x²y²(1 - 4xy²z)

Step-by-step explanation:

________________________________________
Factor
3 out of 12
Factor x² out of x³
Factor y² out of y⁴

you will get the GCF of 3x²y² and left with (1 - 4xy²z)

3x²y²(1 - 4xy²z)
________________________________________

Hoped this helped

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masya89 [10]

Answer:

10/49

Step-by-step explanation:

You just have to multiply the numerators and denominators from each other, then cancel out the common factors that is 3 only.

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8 0
3 years ago
Job cashed a check for $1,765. The teller gave him nine hundred-dollar bills, seven fifty-dollar bills, nineteen twenty-dollar b
Luden [163]

Answer:

x = 13

Step-by-step explanation:

1765

x = 100 (9) + (7) 50 + (19) 20 + (7) 10 - 1765

x = 900 + 350 + 380 + 70 - 1765

x = 1700 - 1765

x = -65

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x = 13

5 0
2 years ago
in a litter of 7 kittens ,each kitten weighs more than 3.5 ounces.Find all the possible combined weights of the kittens. please
Liono4ka [1.6K]
It will help me if you put each weight of the kittens or do they all weight the same?
4 0
3 years ago
A baseball manager has 11 players of the same ability. How many 9 player starting lineups can he
Lana71 [14]

Answer:

55

Step-by-step explanation:

Combination will be used here as teams are a group.

So using combination:

<em>No. of combination of 'n' things taken 'r' at a time</em>

<em>\limits^n {C} \,_r= \frac{n!}{r! (n-r)!}</em>

here n = 11 ,

r = 9

11<em>C</em>9 <em>=  \frac{11!}{9! (11-9)!} = \frac{11!}{9! (2!)}</em>

11<em>C</em>9<em> =</em><em> 55</em>

<em>Which are the possible 9 player starting lineups</em>

4 0
3 years ago
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John paid 18% of his earnings in entertainment expenses.
3 0
3 years ago
Read 2 more answers
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