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disa [49]
2 years ago
7

Can you guys tell me what 6!/3!

Mathematics
2 answers:
NeTakaya2 years ago
7 0

Answer:

  • \boxed{\sf{120}}

Step-by-step explanation:

Use the factorials formula.

<u>Factorials formula:</u>

\Longrightarrow: \sf{\dfrac{n!}{(n-m)!}=n*(n-1)* (n-m+1),\:\quad \:n > \:m}

→ 6!/3!=6*5*4

⇒ 6*5*4

<u>Solve a order of operations.</u>

<u>PEMDAS stands for:</u>

  • <u>Parenthesis</u>
  • <u>Exponents</u>
  • <u>Multiply</u>
  • <u>Divide</u>
  • <u>Add</u>
  • <u>Subtract</u>

<u>First, multiply.</u>

→ 6*5*4

⇒ 6*5=30

⇒ 30*4=120

<u>= 120</u>

  • <u>Therefore, the correct answer is 120.</u>

I hope this helps you! Let me know if my answer is wrong or not.

Likurg_2 [28]2 years ago
4 0

\cfrac{6!}{3!}\implies \cfrac{6\cdot 5\cdot 4\cdot \qquad 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}\implies \cfrac{6\cdot 5\cdot 4\cdot 3!}{3!}\implies 6\cdot 5\cdot 4\implies 120

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Which of the binomials below is a factor of this Torinomial? X2 + 12x + 32
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F $1 \le a \le 10$ and $1 \le b \le 36$, for how many ordered pairs of integers $(a, b)$ is $\sqrt{a + \sqrt{b}}$ an integer?
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You have such entry data: 
1\ \textless \ a\ \textless \ 10 \\ 1\ \textless \ b\ \textless \ 36

Consider expression \sqrt{a+ \sqrt{b} }. If this expression becomes an integer, then b=4,9,16,25, because then \sqrt{b} = 2,3,4,5, respectively. In other cases \sqrt{b} is not integer and thus the expression \sqrt{a+ \sqrt{b} } also is not integer.

1. b=4, then \sqrt{a+ \sqrt{b} }=\sqrt{a+2}. Here a=2,7 (in other cases \sqrt{a+2} is not integer). When a=2, \sqrt{a+2}=\sqrt{2+2}=2 and when a=7, \sqrt{a+2}=\sqrt{7+2}=3.

2. b=9, then a=6 and \sqrt{a+ \sqrt{b} }= \sqrt{a+3}=\sqrt{6+3}=3.

3. b=16, then a=5 and \sqrt{a+ \sqrt{b} }= \sqrt{a+4}=\sqrt{5+4}=3.

4. b=25, then a=4 and \sqrt{a+ \sqrt{b} }=\sqrt{a+5}=\sqrt{4+5}=3.

Answer: (2,4), (7,4), (6,9), (5,16), (4,25).







3 0
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