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Brrunno [24]
3 years ago
8

Dividing Polynomials.. I need these 5 answered

Mathematics
2 answers:
natka813 [3]3 years ago
7 0
Simple...

you have: 

2.) \frac{( 3n^{4} + 3n^{3} +n^{2} )}{ 3n^{2} }

Simplify it-->>>

\frac{ 3n^{2}+3n+1 }{3}

4.)\frac{ v^{5}+ 2v^{4} + v^{3}  }{ v^{2} }

Simplify it-->>>

v^{3} + 2v^{2} +v

6.)\frac{ 2v^{3}+ 3v^{2} +2v }{4}

Factor it--->>>

\frac{v( 2v^{2}+3v+2 }{4}

8.)\frac{ 8m^{5}+ 2m^{4}+ 4m^{3}   }{4m^{3} }


Simplify it-->>

\frac{ 4m^{2}+m+2 }{2}

10.)\frac{6n^{4}+ 8n^{3}+ 2n^{2}   }{ 8n^{3} }

Simplify it-->>\frac{(3n+1)(n+1)}{4n}

Thus, your answer.
elixir [45]3 years ago
7 0
Simple: but i can't show working because of the browser am using 2.) is n^2+0.3 4.)is v^3+v2+v 8.) is 2m^2+0.5m
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3 years ago
What is the area of this figure, to the nearest unit?
Sauron [17]

Answer:

A, 95.

Step-by-step explanation:

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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
Using the Laws of exponents, simplify the following problem.
den301095 [7]
\frac{4 {}^{4} }{ {4}^{ - 8} ({4}^{ - 14}) } = \frac{4 {}^{4} }{ {4}^{ - 22 } }

(4^4)/(4^-22)
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7 0
3 years ago
Write the degree of the polynomial 2y – 13y11​
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Answer:

Assuming u meant 2y - 13y^(11), the degree is 11

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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