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rjkz [21]
2 years ago
10

Evaluate each expression 6! = 3! • 2= 31

Mathematics
1 answer:
Airida [17]2 years ago
6 0
<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.

_______________∞_______________

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A restaurant has 60 tables. The graph shows the average number of tables occupied during each day of the week. During which day
Naddika [18.5K]

Answer: Saturday

Step-by-step explanation:

Saturday is at the highest point meaning it has less empty tables

8 0
3 years ago
When 8 is subtracted from Ollie’s age, the result is the same as subtracting 16 from 3 times his age.
Scilla [17]
X - Ollie's age

8 is subtracted from Ollie's age = x-8
the result is the same as
subtracting 16 from 3 times his age = 3x-16

x-8=3x-16 \\&#10;x-3x=-16+8 \\&#10;-2x=-8 \\&#10;x=4

Ollie is 4 years old.
5 0
3 years ago
Read 2 more answers
Please see attachment
Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }  

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

                       f(x) = \frac{-x}{2x^{2} +1}     ...(i)

Differentiating equation (i) with respective to 'x'

                     f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2}  }   ...(ii)

                    f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  }

Equating Zero

                   f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                 \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                2 x^{2}-1 = 0

               2 x^{2} = 1

             x^{2}  = \frac{1}{2}

             x = \frac{-1}{\sqrt{2} }  , x = \frac{1}{\sqrt{2} }

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4}  }

put

      x = \frac{1}{\sqrt{2} }

f^{ll} (x) > 0

The absolute minimum value at   x = \frac{1}{\sqrt{2} }

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

                         f(x) = \frac{-x}{2x^{2} +1}

                       f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}

         on calculation we get

The value of absolute minimum value = - 0.3536      

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }    

3 0
3 years ago
Write the first four terms of the sequence:<br> a = 2<br> an = a(n-1) + 4
eduard

We have sequence equation a_n=a(n-1)+4.

In this case n is a natural number (1, 2, 3, ...).

So start inserting and computing value of a_n given that you know the value of n and a.

a_1=2(1-1)+4=4 (first term)

a_2=2(2-1)+4=6 \\a_3=2(3-1)+4=8 \\a_4=2(4-1)+4=10

Hope this helps.

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