F(1) = -5(1)^2+2(1)+9 = 6
f(2) = -5(2)^2+2(2)+9 = -9

So, because f(1) = 6 is positive and f(2) = -9 is negative, f(1) > f(2).

f(1) is greater than f(2)!

6 0

3 years ago

Read 2 more answers

If a = {6 6 4 7 -4 4 -6 9 -3} and B= {1 5 -8 9 -5 -8 9 -5 2}, find 3a +9B Matrix operations

V125BC [204]

Answer:

$3A+9B=\begin{bmatrix} 27 & 63 & -60 \\\\ 102 & -57 & -60\\\\ 63 & -18 & 9 \end{bmatrix}$

Step-by-step explanation

$A=\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix}\: and \: B=\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}$

$\rightarrow 3A=3\begin{bmatrix} 6 & 6 & 4\\\\7 & -4 & 4\\\\ -6 & 9 & -3\end{bmatrix},\: \: 9B =9\begin{bmatrix}1 & 5 & -8\\\\ 9 & -5 & -8\\\\ 9 & -5 & 2\end{bmatrix}$

$\rightarrow 3A=\begin{bmatrix} 3*6 & 3*6 & 3*4\\\\3*7 & 3(-4) & 3*4\\\\ 3(-6) & 3*9 & 3(-3)\end{bmatrix},\:\: 9B =\begin{bmatrix}9*1 &9* 5 &9( -8)\\\\ 9*9 & 9(-5) & 9(-8)\\\\ 9*9 & 9(-5) & 9*2\end{bmatrix}$

$\rightarrow 3A=\begin{bmatrix} 18 & 18 & 12 \\\\ 21 & -12 & 12\\\\ -18 & 27 & -9\end{bmatrix},\:\: 9B =\begin{bmatrix} 9 & 45 & -72 \\\\ 81 & -45 & -72\\\\ 81 & -45 & 18\end{bmatrix}$

$\rightarrow 3A+9B=\begin{bmatrix}18+9 & 18+45 & 12+(-72) \\\\ 21+81 & -12 +(-45) & 12+(-72)\\\\ -18+81 & 27+(-45) & -9+18 \end{bmatrix}$

$\rightarrow 3A+9B=\begin{bmatrix} 18+9 & 18+45 & 12-72 \\\\ 21+81 & -12 -45 & 12-72\\\\ -18+81 & 27-45 & -9+18 \end{bmatrix}$

$\rightarrow \purple{\bold{3A+9B=\begin{bmatrix} 27 & 63 & -60 \\\\ 102 & -57 & -60\\\\ 63 & -18 & 9 \end{bmatrix}}}$

6 0

2 years ago