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patriot [66]
3 years ago
10

The first number is the LCM of 2 ,7 and 10​

Mathematics
1 answer:
kati45 [8]3 years ago
8 0
<h2><em>Answer</em>:</h2>

The least common multiple of 2, 7 and 10 would be <em>70</em>.

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What the circumference of 4.5 cm
irina [24]
If 4.5 was the diameter then it would be

4.5 times 3.14 = 14.13
4 0
3 years ago
Evaluate each expression<br> 6! =<br> 3! • 2=<br> 31
Airida [17]
<h2><u>N-FACTORIAL!</u></h2>

<h3>\mathbb{  \color{brown} \: PROBLEM  }  \:   \:  \bold{{ \color{brown}{1}}} :</h3>

\mathcal{SOLUTION: }

  • \:   \bold{ \red{6!}= 6×5×4×3×2×1= \boxed{ \bold{ \blue{720}}}}

Therefore, <u>the value of 6! is 720</u>.

━┈─────────────────────┈━

<h3>\mathbb{  \color{brown} \: PROBLEM  }  \:   \:  \bold{{ \color{brown}{2}}} :</h3>

\mathcal{SOLUTION: }

  • \bold{3!  \cdot 2! }\implies  \bold{(3×2×1)  \cdot( 2×1 )}

  • \:   \: \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \implies \: \bold{ 6  \times 2} =  \boxed{ \bold{ \blue{12}}}

Therefore, <u>the value of the given expression is 12.</u>

━┈─────────────────────┈━

<h3><u>EXPLANATION</u><u>:</u></h3>
  • The mathematical symbol n! is read as "n factorial". The exclamation point "!" is read as factorial. And please remember or take note that 0! is equal to 1, and 1! is also equal to 1.

_______________∞_______________

6 0
2 years ago
Hwllppooppppoopppppppp
babunello [35]

Answer:

c

Step-by-step explanation:

because they are in large sizes but small amount of time

5 0
2 years ago
Read 2 more answers
Please hurry! Thanks &lt;3
Gwar [14]

Answer:

they match with each other

A -1

B-2

C-3

Step-by-step explanation:

6 0
2 years ago
The following is a linear programming formulation of a labor planning problem. There are four overlapping shifts, and management
dexar [7]

Answer:

d. 15

Step-by-step explanation:

Putting the values in the shift 2 function

X1 + X2 ≥ 15

where x1=  13, and x2=2

13+12≥ 15

15≥ 15

At least 15 workers must be assigned to the shift 2.

The LP model questions require that the constraints are satisfied.

The constraint for the shift 2 is that the  number of workers must be equal or greater than 15

This can be solved using other constraint functions e.g

Putting  X4= 0 in

X1 + X4 ≥ 12

gives

X1 ≥ 12

Now Putting the value X1 ≥ 12  in shift 2 constraint

X1 + X2 ≥ 15

12+ 2≥ 15

14 ≥ 15

this does not satisfy the condition so this is wrong.

Now from

X2 + X3 ≥ 16

Putting X3= 14

X2 + 14 ≥ 16

gives

X2  ≥ 2

Putting these in the shift 2

X1 + X2 ≥ 15

13+2 ≥ 15

15 ≥ 15

Which gives the same result as above.

6 0
3 years ago
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