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3 years ago

2/5, -2/5, 5/6, -5/6. Which list of numbers is ordered from least to greatest

Mathematics

1 answer:

g100num [7]3 years ago

6 0

Answer:

-5/6, -2/5, 2/5, 5/6

Step-by-step explanation:

put the fractions in your calculator and then click the fraction to decimal button and press enter.

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Maria and Nina each ordered a small pizza. Maria ate 3/8 of her pizza. Nina ate 3/6 of her pizza. Who ate more pizza?

Verdich [7]

First we have to get the LCM of 8 and 6, which is 24. so the fractions with be Maria 9/24 and Nina 12/24. therefore Nina are more

8 0

3 years ago

Read 2 more answers

Lin rode her bike 2 miles in 8 minutes. She rode at a constant speed. How

hodyreva [135]

Answer: 4

Step-by-step explanation:

For this question, we have to divide the number 8 by 2.

8 ÷ 2 = 4

Now, we have to take the quotient, 4, and multiply it by 1.

1 x 4 = 4

The answer is 4.

6 0

3 years ago

I forgot how to do this!! i'd like help please!!

user100 [1]

Answer:

9) 28

10) 22

Step-by-step explanation:

collinear means on the same line

9 ) AB+BC=AC

16+12=28

10) 13+9=22

3 0

3 years ago

One-third of 2c subtracted frome 5 times nine

lilavasa [31]

-44.8 is the answer.

4 0

3 years ago

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A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following

kiruha [24]

Answer:

After 47 days she will have more than 90 trillion pennies.

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

$1,2,4,8,16,32,.....$

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

$S_n=\dfrac{a(r^n-1)}{r-1}$

where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

$\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}$

$\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}$

$\Rightarrow 90\times 10^{12}\le 2^n-1$

$\Rightarrow 2^n\ge 90\times 10^{12}+1$

$\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)$

$\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)$

$\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}$

$\Rightarrow n \ge 46.4$

$\Rightarrow n\approx 47$

8 0

3 years ago