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Mars2501 [29]
3 years ago
9

HELP ILL GIVE BRAINLIST IF YOU HELP MEEEEEEEEEEE

Mathematics
2 answers:
slavikrds [6]3 years ago
8 0

Answer:

  • 2 groups of 6
  • 6 groups of 2
  • 4 groups of 3
  • 3 groups of 4
  • 1 group of 12
  • 12 groips of 1

Step-by-step explanation:

To figure out the possible groups find the factors of 12; in other words what two numbers would multiply to 12. These are: 1, 2, 3, 4, 6, and 12

  • 1 × 12 = 12
  • 2 × 6 = 12
  • 3 × 4 = 12
  • 4 × 3 = 12
  • 6 × 2 = 12
  • 12 × 1 = 12

So, your possible groupings are;

  • 2 groups of 6
  • 6 groups of 2
  • 4 groups of 3
  • 3 groups of 4
  • 1 group of 12
  • 12 groips of 1
astraxan [27]3 years ago
4 0

Answer:

the answer is 4 * 9

Step-by-step explanation:

np

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Can someone please help me evaluate this expression? 18÷2+7*3
Hatshy [7]

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6 0
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Read 2 more answers
A Rectangle with sides 100m and 50m. Person B and Person A start walking in the opposite direction at the same point with veloci
Maurinko [17]

Answer:

It takes takes them 31.67 secs to meet after Person A starts walking

OR

34.67 secs after Person B starts walking

Step-by-step explanation:

First, we will find the perimeter of the rectangle

From,

Perimeter of rectangle = 2 (l + b)

Where l is the length

and b is the breadth

From the question, l = 100m

and b = 50m

Hence,

Perimeter of the rectangle = 2 (100+50)

= 2(150) = 300m

Hence, the perimeter of the rectangle is 300m

This is the total distance that will be covered by both persons by the time they meet.

Let the distance covered by person A be S_{1}

and the distance covered by person B be S_{2}

We can write that

S_{1} + S_{2} = 300m

Velocity of person A, V_{A} = 4 m/s

and velocity of person B, V_{B} = 5 m/s

From the question, Person A starts walking 3 seconds after person B,

This means, if person A spends t secs before they meet, then person B would spend (3 + t) secs.

For Person A,

Velocity = 4 m/s

Time = t secs

Distance = S_{1}

From,

Velocity = \frac{Distance}{Time}

Then,

Distance = Velocity \times time

S_{1} = 4 \times t

S_{1} = 4t ...... (1)

For Person B

Velocity = 5 m/s

Time = (t + 3) secs

Distance = S_{2}

Also, from

Distance = Velocity \times time

S_{2} = 5 \times (3+t)

S_{2} = 5(3+t) ...... (2)

Recall that, S_{1} + S_{2} = 300m

Then, S_{2} = 300m - S_{1}

We can then write that,

300m - S_{1} = 5(3+t)

Then,

S_{1} = 300 - 5(3+t) ..... (3)

Equating equations (1) and (3), we get

300 - 5(3+t) = 4t

300 - 15 -5t = 4t\\9t = 285\\t = \frac{285}{9}\\

t = 31.67 secs

This is the time spent by Person A

Hence, it takes takes them 31.67 secs to meet after Person A starts walking OR

34.67 secs after Person B starts walking

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