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Rzqust [24]
2 years ago
13

-2 x -2 x -2 = ? (-2) cubed

Mathematics
1 answer:
Lunna [17]2 years ago
5 0

Answer:

-8

Step-by-step explanation:

(-2)(-2)(-2)

4(-2)

-8

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a pharmacist has 10 and 90 iodine solutions on hand. how many liters of each iodine solution will be required to produce 8 liter
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1 litre of 10% solution contains 100mL of iodine
1 litre of 90% solution contains 900mL of iodine
Mixing them will produce 2 litres of solution containing 1000mL of iodine which is 50% iodine.
To produce 8 litres of 50% iodine they can mix 4 litres of 10% and 4 litres of 90%
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1 x 0 = 0?
Alexxandr [17]
Both

1 times a=a
and 
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Im very confused. I need help. Asap.
masya89 [10]
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7 0
3 years ago
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Plz..Help Me Solve This ...
puteri [66]
R = recycling and reuse industry jobs
w = waste management jobs

we know, those two jobs combined are 1,275,000
or
r + w = 1,275,000

we also know that, whatever the "w" jobs are, are really less than "r"
jobs by 1,025,000

"r" has 1,025,000 more than  "w"
or
r = w + 1,025,000
or
r-w = 1,025,000

so... there, just solve by elimination, simple, notice the "w",
is just waiting to be eliminated :)

\bf \begin{cases}
r+\quad  w=1,275,000\\
r-\quad w=1,025,000\\
\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\textendash\\
\boxed{?}+\boxed{?}=\boxed{?}
\end{cases}
7 0
3 years ago
Determine the commutators of the operators a and a+,where a = (x + ip)/2 ^1/2 and a+ = (x - ip)/ 2 ^1/2
Vsevolod [243]

Answer:

Given that:

a = (\frac{x+ip}{2})^{\frac{1}{2}} and a+= (\frac{x-ip}{2})^{\frac{1}{2}}

if a , a+ commutator, it obeys aa^+ = a^+a

First find:

aa^+ = (\frac{x+ip}{2})^{\frac{1}{2}} (\frac{x-ip}{2})^{\frac{1}{2}}

                = (\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}

Now;

a^+a =(\frac{x-ip}{2})^{\frac{1}{2}} (\frac{x+ip}{2})^{\frac{1}{2}} = (\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}

              =(\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}

therefore, aa^+ = a^+a which implies the operators a and a+ are commutators.    


7 0
3 years ago
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