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Katyanochek1 [597]
3 years ago
6

Who can solve this in 1 min 5678÷78

Mathematics
1 answer:
andreev551 [17]3 years ago
5 0

It is 72.7948718

~Hope that helped!~

~Izzy <3

You might be interested in
Help me ASAP PLEASE I NEED THIS DONE BY 7:30 this is test graded
dimaraw [331]
I'm not quite sure but the first one I'm pretty sure....

+3,+5,+7,+9 so then +11,+13,+15,+17,+19

24+11=35 => 6th term

35+13=48 => 7th term

48+15=63 => 8th term

63+17=80 => 9th term

80+19=99 => 10th term

14,34,54,74,94

Each increases by 20...

94+20= 114

114+20= 134

134+20= 154

154+20= 174

174+20= 194

37,46,55,64,73

Each increasing by 9,

73+9= 82

82+9= 91

91+9= 100

100+9= 109

109+9= 118
5 0
3 years ago
Read 2 more answers
Communicate Mathematical Ideas Why do you subtract exponents
zysi [14]

Answer:

We subtract exponents when diving powers with the same base because they eventually get canceled out when written in expanded form.

Step-by-step explanation:

For example, if we expand the given fractions:

10^3/10^1 = 1000/10 = 100 = 10^2

Now, if we subtract the exponents we get the same result.

10^3-1 = 10^2

So, a^m/a^n = a^m-

n is one of the exponent rules which saves time and makes calculation easier.

5 0
2 years ago
A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e
My name is Ann [436]

Answer:

\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)}  \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

Step-by-step explanation:

If the system contains n atoms, we can arrange r quanta of energy in

                         \binom{n}{r} = \dfrac{n!}{r!(n-r)!}

ways.

\mathbf{a)}

In this case,

                                n  = 2, r=1.

Therefore,

                    \binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2

which means that we can arrange 1 quanta of energy in 2 ways.

\mathbf{b)}

In this case,

                                n  = 20, r=10.

Therefore,

                    \binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756

which means that we can arrange 10 quanta of energy in 184 756 ways.

\mathbf{c)}

In this case,

                                n = 2 \times 10^{23}, r = 10^{23}.

Therefore, we obtain that the number of ways is

                    \binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

3 0
3 years ago
Leah cut a 7 inch ribbon into pieces that are 1/2 an inch long. How many pieces of ribbon did she cut? ​
photoshop1234 [79]

14 ribbons.

7 divided by 1/2 is 14. dividing by a fraction or decimal is just like multiplying without the decimal point or numerator.

7 0
3 years ago
How many times is 3 in 43.999 than the value of the digit 3 in 42.103
Dennis_Churaev [7]

Answer:

1. 14.666 ¯ over the 3

2. 14.034 ¯ over the 3

Step-by-step explanation:

5 0
3 years ago
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