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ArbitrLikvidat [17]
3 years ago
12

I need help asapppppp

Mathematics
2 answers:
aivan3 [116]3 years ago
4 0

Answer:

A

Step-by-step explanation:

Supplementary angles are angles that add up to 180 degrees. Essentially, you can make a straight line with them.

1 and 7 are supplementary

1 and 4 are verticle, so no.

2 and 6 are complementary, so no.

3 and 6 are complementary, so no.

Nataly [62]3 years ago
3 0
The answer Is A I need help for algebra if you know algebra please help me
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Which expression is equivalent to this polynomial? 4x^2 + 27
ch4aika [34]
Lemme guess i would try 2x+9.29 is the equipment of the answer of the 2929
6 0
3 years ago
What property is 74.5*0=0
Nadya [2.5K]
<span>74.5 x 0 = 0 belongs to the multiplication property of zero.
Because this property has a rule that any number multiplied by zero, the answer is zero.
It will not matter how big or small the number is, if it will be multiplied by zero, the answer will always be zero.
Another example:
=> 1000 x 0 = 0
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7 0
3 years ago
Choosing an odd number 123468957111210
dezoksy [38]

Answer:

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probably 3 an including the other

Step-by-step explanation:

>3

5 0
3 years ago
PLZ HELPPP!!!!!!!!!! ASAP
salantis [7]

Answer:

They are consecutive terms in the Fibonacci Sequence.

Step-by-step explanation:

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8 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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