So, the equation has a leading term with a negative coefficient, and is a quadratic, meaning is a parabola opening downwards, so it goes up and up and up and then makes a U-turn then down down, so it has a maximum point, that is, at the vertex point

so.. where the dickens is the vertex at?
we're asked to make the vertex with a y-coordinate of 104, so the maximum value is 104

well
$\bf y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)\\\\
-----------------------------\\\\
\textit{so we want }\implies {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}=104
\\\\
\textit{let's take a peek at your equation}\quad 
\begin{array}{ccccllll}
f(x)=&-x^2&+bx&-17\\
f(x)=&-1x^2&+bx&-17\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\\\\
thus
\\\\
(-17)=\cfrac{b^2}{4(-1)}=104$

solve for "b"

5 0

3 years ago

Hii! please help asap. i’ll give brainliest

grin007 [14]

only the mean

Step-by-step explanation:

The mean which by where you add all of the data 6+7+9+10+11+13+14/7 Equals 10 which is the middle value, making it the measure of center

6 0

3 years ago