Question

TEST II. Identify whether the expression is a polynomial function. If it is,
write YES and identify the type of polynomial and if it is not, write NO.
1. 0 _____________ 2. y + 4 __________________
3. 1/y __________________ 4. 3x + 2 ________________ 
5. z + 2/z _______________ 6. 8/3 x 9/3 _______________
7. 1 2x3 + 2x 3 __________ 8. 2 + 5x/5 _________________

PLEASE HELP!!!

Answer 1

Correct option is

Correct option isA

Correct option isA2

Correct option isA2 log(2x) is valid when x>0

Correct option isA2 log(2x) is valid when x>0log5(2x2+3x+2)41=log25(2x)

Correct option isA2 log(2x) is valid when x>0log5(2x2+3x+2)41=log25(2x)$$\Rightarrow \dfrac{1}{4}\log _{ 5 }{ { \left( { 2x }^{ 2 }+3x+2 \right)  } } =\dfrac{1}{2}\log _{ 5 }{ (2x) }  \quad [\because \log a^m=m\log a  \text&  \log_{a^m}b=\dfrac{1}{m}\log_ab]$$

Correct option isA2 log(2x) is valid when x>0log5(2x2+3x+2)41=log25(2x)$$\Rightarrow \dfrac{1}{4}\log _{ 5 }{ { \left( { 2x }^{ 2 }+3x+2 \right)  } } =\dfrac{1}{2}\log _{ 5 }{ (2x) }  \quad [\because \log a^m=m\log a  \text&  \log_{a^m}b=\dfrac{1}{m}\log_ab]$$⇒log5(2x2+3x+2)=log5(2x)2

Correct option isA2 log(2x) is valid when x>0log5(2x2+3x+2)41=log25(2x)$$\Rightarrow \dfrac{1}{4}\log _{ 5 }{ { \left( { 2x }^{ 2 }+3x+2 \right)  } } =\dfrac{1}{2}\log _{ 5 }{ (2x) }  \quad [\because \log a^m=m\log a  \text&  \log_{a^m}b=\dfrac{1}{m}\log_ab]$$⇒log5(2x2+3x+2)=log5(2x)2⇒2x2+3x+2=(2x)2

Correct option isA2 log(2x) is valid when x>0log5(2x2+3x+2)41=log25(2x)$$\Rightarrow \dfrac{1}{4}\log _{ 5 }{ { \left( { 2x }^{ 2 }+3x+2 \right)  } } =\dfrac{1}{2}\log _{ 5 }{ (2x) }  \quad [\because \log a^m=m\log a  \text&  \log_{a^m}b=\dfrac{1}{m}\log_ab]$$⇒log5(2x2+3x+2)=log5(2x)2⇒2x2+3x+2=(2x)2⇒−2x2+3x+2=0⇒x=2asx>0

Answer 2

Answer:

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