Factor a²-a-12 Please

Mathematics

1 answer:

Paha777 [63]2 years ago

7 0

Please find attached photograph for your answer.

You might be interested in

Prove the identity cos(A+B+C)=cosAcosBcosC-cosAsinBsinC-sinAcosBcosC-sinAsinBcosC

blagie [28]

Hello !

cos (a+b) = cos a cos b - sin a sin b
sin (a+b) = sin a cos b + sin b cos a

cos (a+b+c) = cos (a+(b+c))
cos (a+b+c) = cos a cos (b+c) - sin a sin (b+c)
cos (a+b+c) = cos a (cos b cos c - sin b sin c) - sin a (sin b cos c + sin c cos b)
cos (a+b+c)=cos a cos b cos c - cos a sin b sin c - sin a sin b cos c - sin a cos b sin c

4 0

3 years ago

If the coefficient of determination for a data set is 0.75 and the SSE for the data set is 9, what is the SST for the data set?

OverLord2011 [107]

0.1873 aaaaaaaaaaaaaaaaaaaaaaaa

6 0

3 years ago

Read 2 more answers

I NEED HELP PLEASE, THANKS! :)

Alenkasestr [34]

Answer:

Option A

Step-by-step explanation:

This is a great question!

The first thing we want to do here is to graph the system of " inequalities " -

$\begin{bmatrix}2x+9y\le \:100\\ 9x+y\le \:54\end{bmatrix}$

As x is ≥ 0, respectively y ≥ 0, it makes things a lot simpler, as it kind of restricts us to quadrant 1, but not entirely.

First change each inequality to " slope - intercept " form, as such -

$2x + 10y \leq 100,\\10y \leq 100 - 2x,\\y \leq - 1 / 5x + 10\\----------\\9x + y \leq 54,\\y \leq - 9x + 54$

The graphed solutions would be in the attachment. I have colored the region with which the two intersect, and the fact that we are limited to quadrant 1. In this colored region there are 3 major points, ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ). We have to determine which of these are are maximum points, as the minimum are the same. Substitute these values ( ( 5, 9 ), ( 0, 10 ), and ( 6, 0 ) ) into the equation f( x, y ) = 10x + 4y,

$f ( x, y ) = 10( 5 ) + 4( 9 ),\\f ( x, y ) = 86 -\\f ( x, y ) = 10( 0 ) + 4( 10 ),\\f ( x, y ) = 40 -\\f ( x, y ) = 10( 6 ) + 4( 0 ),\\f ( x, y ) = 60$