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Paha777 [63]
3 years ago
11

ormula1" title="3x - 2 \div 6x ^{2} + 11 ^{2} - 4x - 4" alt="3x - 2 \div 6x ^{2} + 11 ^{2} - 4x - 4" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
RoseWind [281]3 years ago
4 0

3 - 2 \div 6 \times 2 + 11 \times 2 - 4 - 4
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Solve 4.62 + (−12.3).<br><br> 56.83<br> 5.85<br> −7.68<br> −16.92
ikadub [295]

The sum of the given decimal numbers is -7.68

<h3>Sum of decimal numbers</h3>

Decimal numbers are numbers that contains decimal points. Given the expression below;

4.62 + (−12.3).

Convert to fraction to have:

4.62 + (−12.3) = 462/100 - 123/10

Find the LCM

4.62 + (−12.3) = 462-1230/100

4.62 + (−12.3) = -768/100

4.62 + (−12.3) =-7.68

Hence the sum of the given decimal numbers is -7.68

Learn more on sum of numbers here; brainly.com/question/25734188

#SPJ1

4 0
1 year ago
(2ײy²)⁰ I need help with exponents ​
balandron [24]

Answer:

1.

Step-by-step explanation:

Any real value to the power of 0 is 1.

7 0
3 years ago
Read 2 more answers
Given that<br> 5x:9=7:3<br> Calculate the value of <br> x
morpeh [17]

Answer:

  • x = 4.2

Step-by-step explanation:

<h3>Given  </h3>
  • 5x:9=7:3
<h3>To find</h3>
  • The value of  x
<h3>Solution</h3>
  • 5x:9=7:3
  • 5x = 7*9/3
  • 5x = 7*3
  • 5x = 21
  • x = 21/5
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7 0
3 years ago
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Does it make sense? In my data set of 10 exam scores, the mean turned out to be the score of the person with the third highest g
san4es73 [151]

Answer:

Yes

Step-by-step explanation:

Given that  In my data set of 10 exam scores, the mean turned out to be the score of the person with the third highest grade.

No two people got the same score.

Let the scores be a,b,c,d,e in ascending order where no two scores are equal

If a+b=d+e =2c then we can have c as the mean of the scores of 5 persons

This is because

Total sum = a+b+c+d+e = (a+b)+c+(d+e)

= 2c+c+2c=5c

Average= 5c/5 = c

It makes sense and there are chances as long as the above condition is satisfied.

8 0
3 years ago
Question 4 of 17
Vlada [557]

Answer:

I think it is A OR D

4 0
2 years ago
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