precalculus

[ashley14]

i need help with this problem please, will you explain step by step instructions, stumped on these type problems

k(t)=-2t+1;find k(-2/3)




f(a)=a3-a2
g(a)=3a+3
find (f/g)(a)

thank you so much

 

[BigGlenntheHeavy]

$\displaystyle k(t) \ = \ 2t+1; \ find \ k(-2/3)$

$\displaystyle k(-2/3) \ = \ 2(-2/3) +1 \ = \ (-4/3)+1 \ = \ -1/3.$

$\displaystyle By \ definition \ (f/g)(a) \ = \ \frac{f(a)}{g(a)}, \ g(a) \ doesn't \ equal \ zero.$

$\displaystyle f(a) \ = \ a^{3}-a^{2} \ and \ g(a) \ = \ 3a+3.$

$\displaystyle Hence, \ (f/g)(a) \ = \ \frac{f(a)}{g(a)} \ = \ \frac{a^{3}-a^{2}}{3a+3}.$