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djyliett [7]
3 years ago
9

2/3 (5 x 4) ÷ 3 + 6/7 I divided 3/2 is 1.5*(20)=30/3=10+1.16=11.16 is that correct not so good in math

Mathematics
2 answers:
Assoli18 [71]3 years ago
8 0
\frac{2}{3}(5*4):3+ \frac{6}{7} = \\ \\  \frac{2}{3}*20:3+ \frac{6}{7} = \\ \\   \frac{40}{3}:3+ \frac{6}{7} = \\ \\\frac{40}{9}+ \frac{6}{7}= \\ \\  \frac{40*7+6*9}{7*9}= \\ \\\frac{280+54}{63}= \\ \\  \frac{334}{63} = \\ \\  5 \frac{19}{63} \ \ \  [as \ a \  mixed \ fraction] \\ \\ \approx 5.3 \ \ \  [as \ a \  decimal\ fraction \ to \ the \ nearest \ tenth]
kramer3 years ago
5 0

2/3 (5 × 4) ÷ 3 + 6/7 = 334/63

<h3>Further explanation</h3>

Order of Operations in Mathematics follow this following rule :

  1. Parentheses
  2. Exponents
  3. Multiplication and Division
  4. Addition and Subtraction

This rule is known as the PEMDAS method.

In working on a mathematical problem, we first calculate operation that is in parentheses, follow by exponentiation, then multiplication or division, and finally addition or subtraction.

Let us tackle the problem.

Let's write down the problem that is wanted to be solved:

\frac{2}{3} ( 5 \times 4 ) \div 3 + \frac{6}{7}

\frac{2}{3} ( 20 ) \div 3 + \frac{6}{7} → Parentheses First

\frac{40}{3} \div 3 + \frac{6}{7} → Multiplication

\frac{40}{9} + \frac{6}{7} → Division

\frac{280}{63} + \frac{54}{63} → Match the denominator of the fraction

\frac{334}{63} → Final Answer

From the results above, it can be concluded that the results of this operation 2/3 (5 × 4) ÷ 3 + 6/7 is 334/63

<h3>Learn more</h3>
  • Infinite Number of Solutions : brainly.com/question/5450548
  • System of Equations : brainly.com/question/1995493
  • System of Linear equations : brainly.com/question/3291576

<h3>Answer details</h3>

Grade: Middle School

Subject: Mathematics

Chapter: Order of Operations

Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point , Multiplication , Division , Exponent , PEMDAS

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Answer: f(n)=3^{n-1}

Step-by-step explanation:

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We can write it as : f(n)=3^{n-1}

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8 0
3 years ago
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djyliett [7]
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</span>
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vodomira [7]
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