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STALIN [3.7K]
3 years ago
9

Who plays among us?

Mathematics
2 answers:
Anna71 [15]3 years ago
7 0
I play among us. Let’s play
Andrei [34K]3 years ago
3 0

Answer:

Easy to guess with my profile picture

Step-by-step explanation:

PLS mark me brainliest it means a lot and I need them.

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Mark and Jen me at the beach yesterday July 31. Might go to the beach every third day in Django ceviche before three. How many t
Savatey [412]

Answer:

The number times both Mark and Jen will be at the beach on the same day over the next 30 days is 10 times.

Step-by-step explanation:

Every third day implies that they both go the beach and then have a two day break, and then go the beach on the day after.

Put simply, every third day implies one day in three days.

Based on the above, the number times both Mark and Jen will be at the beach on the same day over the next 30 days can be calculated by dividing 30 by 3 as follows:

Number times both Mark and Jen will be at the beach = 30 / 3 = 10

Therefore, the number times both Mark and Jen will be at the beach on the same day over the next 30 days is 10 times.

5 0
3 years ago
2(3+4y)=46<br> please help asap
sleet_krkn [62]

Answer:

y=5

Step-by-step explanation:

esa es la respuesta

3 0
3 years ago
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A 2-row table with 9 columns. The first row is labeled x with entries negative 4, negative 3, negative 2, negative 1, 0, 1, 2, 3
Margarita [4]

Answer:

x    -4    -3  -2  -1    0    1   2   3    4

y  -54 -20  -4   0   -2  -4  0   16   50

at 4 it is  maximum .

maximum value=50

hope that helps

7 0
3 years ago
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Please help me as soon as possible
Elena-2011 [213]
Answer: 3x^2 - 15 + 12

7 0
3 years ago
Lavinia has 9 glasses and 6 mugs . she would like to set them out in identical groups with none left over in preparation for a p
sergejj [24]
You must find the factors of each number...
9 = 1, 3, 3, 9
6 = 1, 2, 3, 6

Next, determine the common multiple...
9 = 1, 3, 3, 9
6 = 1, 2, 3, 6

Therefore, the most groups Lavinia can make without any glasses or mugs left to spare is 3.
3 0
3 years ago
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