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ELEN [110]
3 years ago
12

Find the quotient of 1/4 divided by 3/8​

Mathematics
2 answers:
kap26 [50]3 years ago
5 0

Answer:

1/4 divided by 3/8= 6

Step-by-step explanation:

hope this helps!! :)

gogolik [260]3 years ago
3 0

Answer: ⅔

Steps:  ¼ ÷ ⅜ = ⅔

Apply the fraction rule:  ¼ × ⁸⁄₃ = ⅔

Plz mark brainliest:)

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A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row. All the lights start out off. The first
Gnom [1K]

<u>Solution-</u>

A school has 1800 students and 1800 light bulbs, each with a pull cord and all in a row.

As all the lights start out off, in the first pass all bulbs will be turned on.

In the second pass all the multiples of 2 will be off and rest will be turned on.

In the third pass all the multiples of 3 will be off, but the common multiple of 2 and 3 will be on along with the rest. i.e all the multiples of 6 will be turned on along with the rest.

In the fourth pass 4th light bulb will be turned on and so does all the multiples of 4.

But, in the sixth pass the 6th light bulb will be turned off as it was on after the third pass.

This pattern can observed that when a number has odd number of factors then only it can stay on till the last pass.

1 = 1

2 = 1, 2

3 = 1, 3

<u>4 = 1, 2, 4</u>

5 = 1, 5

6 = 1, 2, 3, 6

7 = 1, 7

8 = 1, 2, 4, 8

9 = 1, 3, 9

10 = 1, 2, 5, 10

11 = 1, 11

12 = 1, 2, 3, 4, 6, 12

13 = 1, 13

14 = 1, 2, 7, 14

15 = 1, 3, 5, 15

16 = 1, 2, 4, 8, 16

so on.....

The numbers who have odd number of factors are the perfect squares.

So calculating the number of perfect squares upto 1800 will give the number of light bulbs that will stay on.

As,  \sqrt{1800} =42.42  , so 42 perfect squared numbers are there which are less than 1800.

∴ 42 light bulbs will end up in the on position. And there position is given in the attached table.

7 0
3 years ago
The average number of stars in a galaxy is about 5x10^11 and the total number of galaxies is about 5.2x10^9. what is the total n
Arisa [49]

Answer:

2.6 x 10^21

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the  10   If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

4 0
2 years ago
In a baseball tournament, teams get 5 points for a win, 3 points for a tie and 1 point for a loss Nathan’s team has 29 points ho
puteri [66]

Answer:

37 different combinations

Step-by-step explanation:

First of all, we will count the possible combinations that add up to 29.

-Losses only;

One possibility: 29 losses

-Ties & losses;

Nine possibilities: 1 tie and 26 losses; 2 ties and 23 losses; 3 ties and 20 losses;4 ties and 17 losses; 5 ties and 14 losses; 6 ties and 11 losses; 7 ties and 8 losses; 8 ties and 5 losses; 9 ties and 2 losses

-Wins & losses

Five possibilities: 1 win and 24 losses; 2 wins and 19 losses; 3 wins and 14 losses; 4 wins and 9 losses; 5 wins and 4 losses

Now, what we want to find from the question is number of possibilities for wins, ties, and losses all together. So, we will count up the ties and losses in the remainder for each case of a given number of wins.

Thus;

For 0 Wins: 0 ties and 29 losses; 1 tie and 26 losses; 2 ties and 23 losses; 3 ties and 20 losses;4 ties and 17 losses; 5 ties and 14 losses; 6 ties and 11 losses; 7 ties and 8 losses; 8 ties and 5 losses; 9 ties and 2 losses.

Which sums up to 10 possibilities

For 1 win; 0 ties and 24 losses; 1 tie and 21 losses........8 ties and 0 losses.

Which sums up to 9 possibilities

For 2 wins; 0 ties and 19 losses; 1 tie & 16 losses............ 6 ties and 1 loss.

Which sums up to 7 possibilities

For 3 wins; 0 ties and 14 losses; 1 tie and 11 losses ....... 4 ties and 2 losses.

Which sums up to 5 possibilities

For 4 wins; 0 ties & 9 losses; 1 tie and 6 losses....... 3 ties and 0 losses

Which sums up to 4 possibilities

For 5 wins; 0 ties and 4 losses; 1 tie and 1 loss

Which sums up to 2 possibilities

Thus;

Total number of possibilities of combinations of wins, ties and losses = 10 + 9 + 7 + 5 + 4 + 2 = 37

8 0
3 years ago
Ex 3.6<br> 6. find the area enclosed between the curve y= -2x²-5x+3 and the x-axis
Xelga [282]
When y=0,

-2{ x }^{ 2 }-5x+3=0\\ \\ 2{ x }^{ 2 }+5x-3=0\\ \\ \left( 2x-1 \right) \left( x+3 \right) =0

\\ \\ \therefore \quad x=\frac { 1 }{ 2 } \\ \\ \therefore \quad x=-3

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

\int _{ -3 }^{ \frac { 1 }{ 2 }  }{ -2{ x }^{ 2 } } -5x+3dx

\\ \\ ={ \left[ -\frac { { 2x }^{ 2+1 } }{ 2+1 } -\frac { 5{ x }^{ 1+1 } }{ 1+1 } +3x \right]  }_{ -3 }^{ \frac { 1 }{ 2 }  }

\\ \\ ={ \left[ -\frac { 2{ x }^{ 3 } }{ 3 } -\frac { 5{ x }^{ 2 } }{ 2 } +3x \right]  }_{ -3 }^{ \frac { 1 }{ 2 }  }

\\ \\ \\ =\left\{ -\frac { 2 }{ 3 } { \left( \frac { 1 }{ 2 }  \right)  }^{ 3 }-\frac { 5 }{ 2 } { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }+3\left( \frac { 1 }{ 2 }  \right)  \right\} -\left\{ -\frac { 2 }{ 3 } { \left( -3 \right)  }^{ 3 }-\frac { 5 }{ 2 } { \left( -3 \right)  }^{ 2 }+3\left( -3 \right)  \right\}

\\ \\ \\ =-\frac { 2 }{ 3 } \cdot \frac { 1 }{ 8 } -\frac { 5 }{ 2 } \cdot \frac { 1 }{ 4 } +\frac { 3 }{ 2 } -\left\{ -\frac { 2 }{ 3 } \left( -27 \right) -\frac { 5 }{ 2 } \cdot 9-9 \right\}

\\ \\ =-\frac { 2 }{ 24 } -\frac { 5 }{ 8 } +\frac { 3 }{ 2 } -\left\{ \frac { 54 }{ 3 } -\frac { 45 }{ 2 } -9 \right\}

\\ \\ =-\frac { 2 }{ 24 } -\frac { 15 }{ 24 } +\frac { 36 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +9

\\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 45 }{ 2 } +\frac { 18 }{ 2 } \\ \\ =\frac { 19 }{ 24 } -\frac { 54 }{ 3 } +\frac { 63 }{ 2 }

\\ \\ =\frac { 343 }{ 24 }

Answer: 343/24 units squared.
6 0
3 years ago
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