Answer:
Yes! You can
Step-by-step explanation:
1. If this is rows and columns it is doable, however if you are speaking columns by rows it isn't
2. You multiply them like normal except when you reach the end on your 2x2 matrix you repeat the cycle
3. http://www.calcul.com/show/calculator/matrix-multiplication_;2;2;2;3 there's a link to check your answer after you've done it
4
Set y equal to 0 to find x intercept
set x equal to 0 to find y intercept
The answer is C
all of them plugged in as x work
You could simplify this work by factoring "3" out of all four terms, as follows:
3(x^2 + 2x - 3) =3(0) = 0
Hold the 3 for later re-insertion. Focus on "completing the square" of x^2 + 2x - 3.
1. Take the coefficient (2) of x and halve it: 2 divided by 2 is 1
2. Square this result: 1^2 = 1
3. Add this result (1) to x^2 + 2x, holding the "-3" for later:
x^2 +2x
4 Subtract (1) from x^2 + 2x + 1: x^2 + 2x + 1 -3 -1 = 0,
or x^2 + 2x + 1 - 4 = 0
5. Simplify, remembering that x^2 + 2x + 1 is a perfect square:
(x+1)^2 - 4 = 0
We have "completed the square." We can stop here. or, we could solve for x: one way would be to factor the left side:
[(x+1)-2][(x+1)+2]=0 The solutions would then be:
x+1-2=0=> x-1=0, or x=1, and
x+1 +2 = 0 => x+3=0, or x=-3. (you were not asked to do this).
Answer:
you would need 12 rooms
Step-by-step explanation:
168/12=11.2 so you would have 11 full rooms and then a room for the leftover two
