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aalyn [17]
2 years ago
15

Newton and descartes each have 42 glow in the dark stickers newton divides his into 6 equal groups descartes divides his into 7

equal groups who has more stickers in each group
Mathematics
2 answers:
LekaFEV [45]2 years ago
6 0

Newton has more stickers in each group if Newton and Descartes each have 42 glow-in-the-dark stickers newton divides his into 6 equal groups Descartes divides his into 7 equal groups.

<h3>What is a fraction?</h3>

Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

It is given that:

Newton and Descartes each have 42 glow-in-the-dark stickers newton divides his into 6 equal groups Descartes divides his into 7 equal groups.

For Newton:

= 42/6

= 7

For Descartes:

= 42/7

= 6

Thus, Newton has more stickers in each group if Newton and Descartes each have 42 glow-in-the-dark stickers newton divides his into 6 equal groups Descartes divides his into 7 equal groups.

Learn more about the fraction here:

brainly.com/question/1301963

#SPJ2

Elodia [21]2 years ago
5 0

Answer:

newton

Step-by-step explanation:

first you have to divide Newtons stickers and get 7, then you divide Descartes stickers and get 6, so therefore Newton has more

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What is the answer to 2(x-3)-4(x+3)
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Step-by-step explanation:

2 (x-3)-4 (x+3)

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It is:

40+1+0.3+0.02=41.32



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If 32 teams participate in a standard single elimination tournament, how many games must be played to decide a champion?
SCORPION-xisa [38]
<h3>Answer: 31 games</h3>

======================================================

Explanation:

  • 32/2 = 16 games will happen in round 1. Afterward, 16 teams are left.
  • 16/2 = 8 games will happen in round 2. Afterward, 8 teams are left.
  • 8/2 = 4 games happen in round 3.
  • 4/2 = 2 games in round 4.
  • 2/2 = 1 game as the final championship.

Count the number of times you divide by two and we have five occurrences of this. So there five rounds overall.

To get the total number of games played, we add up the quotients

16+8+4+2+1 = 31

----------------

Or as a shortcut we can simply subtract off 1 since 1+2+4+...+2^n = 2^(n+1)-1

We can write that rule as

\displaystyle \sum_{k=1}^{n}2^k = 2^{n+1}-1

which is equivalent to

\displaystyle \sum_{k=1}^{n}2^{k-1} = 2^n-1

3 0
4 years ago
What are the roots to<br> the equation<br> 2x² + 6x-1=0?
lys-0071 [83]

Answer:

The roots of  equations are as m =  \frac{-3}{2} + \frac{\sqrt{11} }{2}  And n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}    

Step-by-step explanation:

The given quadratic equation is 2 x² + 6 x - 1 = 0

This equation is in form of a x² + b x + c = 0

Let the roots of the equation are ( m , n )

Now , sum of roots = \frac{ - b}{a}

And products of roots = \frac{c}{a}

So, m + n = \frac{ - 6}{2} = - 3

And m × n =  \frac{ - 1}{2}

Or, (m - n)² = (m + n)² - 4mn

Or, (m - n)² = (-3)² - 4 (\frac{ - 1}{2})

Or, (m - n)² = 9 + 2 = 11

I.e m - n = \sqrt{11}

Again m + n = - 3    And m - n = \sqrt{11}

Solving this two equation

(m + n) + ( m - n) = - 3 + \sqrt{11}

I.e 2 m =  - 3 + \sqrt{11}

Or, m = \frac{-3}{2} + \frac{\sqrt{11} }{2}

Similarly n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}      

Hence the roots of  equations are as m =  \frac{-3}{2} + \frac{\sqrt{11} }{2}  And n =  \frac{-3}{2} - \frac{\sqrt{11} }{2}      Answer

6 0
3 years ago
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