Question

Classify the following as linear, quadratic and cubic polynomials:

(i) 4x + 5 (ii) 2x

2 + 5 (iii) 5x

3 + 6x

^2

(iv) 2y + 4 (v) u^

2 – 5u/2 (vi) 2t

^3 + 4t^2

+ 5t + 7 (vii) t +root3 (viii) 5y

2 + 3y + 7 (ix) 5 – y^

2 + y

^3

(x) 4x^3

​

Answer 1

Answer:

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = { \huge{ \red{ \boxed{1}}}}$

Answer 2

Step-by-step explanation:

${\mathcal{\fcolorbox{darkblack}{pink}{߷√~verified~answer}}}$

${\mathcal{\fcolorbox{brown}{yellow}{♧brainleiest~answer}}}$

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = {\tt{\underline {\underbrace { \blue{1}}}}}$