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3 years ago

15

What is the greatest common factor of 6 and 20

2 answers:

Alenkinab [10]3 years ago

5 0

Answer:

There is no common factor between the two.

Step-by-step explanation:

Nothing divided by 6 equals 20 and nothing divided by 20 equals 6. I could b wrong so my apologies. I hope this helped :)

xeze [42]3 years ago

4 0

Answer:

2

Step-by-step explanation:

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Can somebody please help I don't understand how to do this<br>

boyakko [2]

Answer:

hbsbipynoibnrioudjtoudjtuodjoudk

Step-by-step explanation:

linylnutdbitdiltcicyrkdvutodubtuodjtohtsuodhuthuodtjutodhodutohutdhusotohdtidgidgtidgtigdtuhothuodtohudtuodthhoudthoudtuohtd

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2 years ago

Can you please help me? Keep it a little basic. (no decimals) Thanks for the help!

Delvig [45]

When you are dividing fractions you just flip the second fraction upside down and multiply the two fractions
So the first is
½ x 6/1 = 6/2 = 3

For the second you write them as improper fractions then do the same
So it is
3/2 / 1/8 = 3/2 x 8/1 = 24/2 = 12

So the answer to the first question is 3 and the second is 12

4 0

3 years ago

Read 2 more answers

Please help me with this problem <br> Evaluate (-½·3)³

-Dominant- [34]

Answer: -27/8

Step-by-step explanation:

$\left(-\frac{1}{2}\cdot \:3\right)^3$

$\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=-a^n,\:\quad \mathrm{if\:}n\mathrm{\:is\:odd}$

$\left(-\frac{1}{2}\cdot \:3\right)^3=-\left(\frac{1}{2}\cdot \:3\right)^3$

$=-\left(\frac{1}{2}\cdot \:3\right)^3$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n$

$\left(\frac{1}{2}\cdot \:3\right)^3=\left(\frac{1}{2}\right)^3\cdot \:3^3$

$=-\left(\frac{1}{2}\right)^3\cdot \:3^3$

$=-\frac{1}{8}\cdot \:3^3$

$=-\frac{1}{8}\cdot \:27$

$=-\frac{27}{8}$

3 0

2 years ago

Jerry has two same size circles divided into the same number of equal parts one Circle has 3/4 shaded in the other 2/3 shaded hi

solmaris [256]

Yes Jerry's sister is right

5 0

2 years ago

An elementary school is offering 3 language classes: one in Spanish, one in French,and one in German. The classes are open to an

Alona [7]

Answer:

A. 0.5

B. 0.32

C. 0.75

Step-by-step explanation:

There are

28 students in the Spanish class,
26 in the French class,
16 in the German class,
12 students that are in both Spanish and French,
4 that are in both Spanish and German,
6 that are in both French and German,
2 students taking all 3 classes.

So,

2 students taking all 3 classes,
6 - 2 = 4 students are in French and German, bu are not in Spanish,
4 - 2 = 2 students are in Spanish and German, but are not in French,
12 - 2 = 10 students are in Spanish and French but are not in German,
16 - 2 - 4 - 2 = 8 students are only in German,
26 - 2 - 4 - 10 = 10 students are only in French,
28 - 2 - 2 - 10 = 14 students are only in Spanish.

In total, there are

2 + 4 + 2 + 10 + 8 + 10 +14 = 50 students.

The classes are open to any of the 100 students in the school, so

100 - 50 = 50 students are not in any of the languages classes.

A. If a student is chosen randomly, the probability that he or she is not in any of the language classes is

$\dfrac{50}{100} =0.5$

B. If a student is chosen randomly, the probability that he or she is taking exactly one language class is

$\dfrac{8+10+14}{100}=0.32$

C. If 2 students are chosen randomly, the probability that both are not taking any language classes is

$0.5\cdot 0.5=0.25$

So, the probability that at least 1 is taking a language class is

$1-0.25=0.75$

3 0

3 years ago