Without trying to list all of them, I just now thought of a way to figure out the number of different possibilities:
The total can be made from:
-- zero, 1, 2, or 3 quarters . . . 4 choices
-- zero, 1, or 2 dimes . . . . . . . 3 choices
-- zero or 1 nickel . . . . . . . . . . 2 choices
and
-- zero, 1, or 2 pennies . . . . . . 3 choices
So there are (4 x 3 x 2 x 3) = 72 different possible combinations of coins
Almost all of the possible combinations appear to be unique. I do
see one possible duplication: 1qtr is the same thing as (2dim + 1nkl).
That reduces the number somewhat, but I don't really know how to handle it.
So the number of different amounts of change is a few less than 72 .
I hope this answer is worth 5 points.
Answer:
D - The slope from point O to point A is three times the slope of the line from point A to point B.
Responder:
<em>a) 1
</em>
<em>b) 12
</em>
<em>c) 3
</em>
<em>d) -1 y 3
</em>
Explicación paso a paso:
Dadas las siguientes ecuaciones;
a) x²-x =0
x² = 0+x
x² = x
x = 1
b) -x²+12x = 0
-x² = -12x
x² = 12x
x = 12
c) 3x² - 9x = 0
Suma 9x a ambos lados
3x²-9x+9x = 0+9x
3x² = 9x
3x = 9
x = 9/3
x = 3
d) x²-2x-3 = 0
x²-3x+x-3 = 0
x(x-3)+1(x-3) = 0
(x+1)(x-3) = 0
x+1 = 0 y x-3 = 0
x = -1 y 3
For either square root to exist, you require that
. This is true for all
, since
is always non-negative. This means the domain of
as a function of
is all real numbers, or
or
.
Now, because
is non-negative, and the smallest value it can take on is 7, it follows that the minimum value for the positive square root must be
, while the maximum value of the negative root must be
. This means the range is
, or
, or
![(-\infty,-\sqrt7]\cup[\sqrt7,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C-%5Csqrt7%5D%5Ccup%5B%5Csqrt7%2C%5Cinfty%29)
.
6/3 = 6 divided by 3
= 2
12/6= 12 divided by 6
= 2
