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Shtirlitz [24]
3 years ago
8

Namid works for a florist and is making flower arrangements for the tables at a reception. He has 36 roses, 48 tulips, and 72 ca

rnations. Each case must contain the same number of each type of flower. How many cases will Namid need? How many of each type of flower will he put in one case?
Mathematics
2 answers:
julia-pushkina [17]3 years ago
8 0
3 roses, 4 tulips, and 6 carnations
Tatiana [17]3 years ago
7 0
The LCM is 12
So Namid will need 12 types of flowers. He will put 3 roses, 4 tulips, and 6 carnations
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eduard

Answer:

x - 5.

Step-by-step explanation:

x(y + 6) – 5(y + 6) = ( )(y + 6)

The missing factor  consists of the 2 terms outside  the parentheses.

That is x - 5.

6 0
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3 years ago
Can someone help me please?
lakkis [162]
Choice A is correct
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3 years ago
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How do I solve this, please help
Bas_tet [7]

9514 1404 393

Answer:

  • EF = DE = 44
  • FG = DG = 36
  • FH = DF = 31

Step-by-step explanation:

Since EH is the perpendicular bisector of DF, ∆DEF is isosceles and sides DE and EF have the same length.

  DE = EF

  (9x -1) = (7x +9)

  2x = 10 . . . . . . . add 1-7x

  x = 5 . . . . . . . . . divide by 2

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Similarly, marked sides GD and GF are the same length, so ...

  GD = GF

  (10y -4) = (7y +8)

  3y = 12 . . . . . . . . . . add 4-7y

  y = 4 . . . . . . . . . divide by 3

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Now, we have what we need to calculate the side lengths.

  EF = 7x+9 = 7·5 +9 = 44

  DE = 9x-1 = 9·4 -1 = 44

  FG = 7y+8 = 7·4 +8 = 36

  DG = 10y-4 = 10·4 -4 = 36

  FH = 3x+4y = 3·5 +4·4 = 31

  DF = FH = 31

6 0
3 years ago
HELP NEEDED ASAP... 20 POINTS. WILL MARK BRAINLIEST!
ira [324]
1 is C
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