$y=ax^2+bx+c\\\\y=a(x-h)^2+k\ where\ h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}[tex]y=-3x^2+7x-5\\\\a=-3;\ b=7;\ c=-5\\\\h=\dfrac{-7}{2\cdot(-3)}=\dfrac{7}{6}\\\\k=\dfrac{-(7^2-4\cdot(-3)\cdot(-5)}{4\cdot(-3)}=\dfrac{-49+60}{-12}=-\dfrac{11}{12}\\\\y=-3\left(x-\dfrac{7}{6}\right)^2-\dfrac{11}{12}$

5 0

3 years ago

Question 26 OT 40

masha68 [24]

I think the answer is D

6 0

3 years ago

You currently have a 67% in class. You have 2 more assignments a quiz which is 10% and a project worth 45%. Lets say I get a 100

finlep [7]

Answer:

82% I think I hope this helped

6 0

3 years ago

Simplify each expression. Write your answers in exponential notation.

GenaCL600 [577]

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad
a^n\implies \cfrac{1}{a^{-n}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\left( \cfrac{36b^3}{9a^5} \right)^3\implies \left( \cfrac{4b^3}{a^5} \right)^3\implies \stackrel{\textit{distributing the exponent}}{\left(\cfrac{4^3b^{3\cdot 3}}{a^{5\cdot 3}}\right)}\implies \cfrac{64b^9}{a^{15}}$

3 0

3 years ago