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3 years ago

What is 12 divided by 1/2?o

Mathematics

2 answers:

Sedaia [141]3 years ago

4 0

6, because 12 divided by 1/2 is the same as 12 divided by 2

saw5 [17]3 years ago

3 0

<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ You are finding how many halves fit into 12

As there are 2 halves in 1, you would just need to double the value

Multiply by the denominator:

12 x 2 = 24

It is 24

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

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↬ ʜᴀɴɴᴀʜ ♡

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Exercise<br>1<br>Solve the inequality<br>of these equation<br>pretty please

Ad libitum [116K]

Answer:

y<6

Step-by-step explanation:

1 /3 y + 1/4 > 3 /8y

1 /3 y + 1/4 − 3/ 8 y > 3/ 8 y − 3/ 8 y

− 1 /24 y + 1 4 > 0

−1 /24 y + 1 /4 − 1/ 4 > 0 − 1 /4

−1 /24 y > −1/ 4

Multiply both sides by 24/(-1).

y<6

<u><em>Please mark as brainliest </em></u>

Have a great day, be safe and healthy

Thank u

6 0

2 years ago

Flying against the wind, an airplane travels 4000 kilometers in 5 hours. Flying with the wind, the same plane travels 5040 kilom

andriy [413]

Answer:the plane in still air is 904 km/h

the rate of the wind is 104 km/h

Step-by-step explanation:

Let x represent the rate or speed of the plane in still air.

Let y represent the rate or speed of the wind.

Flying against the wind, an airplane travels 4000 kilometers in 5 hours. This means that the total speed of the airplane would be (x - y) km/h

Distance = speed × time

Therefore

4000 = 5(x - y)

4000 = 5x - 5y - - - - - - - - -1

Flying with the wind, the same plane travels 5040 kilometers in 4 hours. This means that the total speed of the airplane would be (x + y) km/h. Therefore,

5040 = 4(x + y)

5040 = 4x + 4y - - - - - - - - - 2

Multiplying equation 1 by 4 and equation 2 by 5, it becomes

16000 = 20x - 20y

20160 = 20x + 20y

Subtracting, it becomes

- 4160 = - 40y

y = - 4160/-40

y = 104

Substituting y = 104 into equation 1, it becomes

4000 = 5x - 5 × 104

4000 = 5x - 520

5x = 4000 + 520

5x = 4520

x = 4520/5 = 904

7 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

2 years ago

What property is 6+0=6

liraira [26]

Additive Identity Property

3 0

3 years ago

To which graph does the point (−1, 3) belong? y ≥ 2x + 6 y ≥ 3x − 1 y ≥ −x + 3 y ≥ −2x + 5

Anna35 [415]

Well first find the slope of the point
y=mx+b y=-1x+3
slope = -1
then put the problem in point slope form y - y1 = m (x-x1)
y + 3 = -1 (x+1) then solve for y=mx+b again
so y> 2x + 6 is the answer

4 0

3 years ago

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