<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
Answer:
I believe it's 1.
Step-by-step explanation:
Answer:
d= -16
Step-by-step explanation:
To answer this question you'll have to first simplify the equation to
0.2d+-1.2=0.3d+5-3+0.1d
then combining like terms you get
0.2d-1.2=(0.3d+0.1d)+(5+-3)
0.2d-1.2=0.4d+2
then subtract 0.4d from both sides leaving you with
-0.2d-1.2=2
then add 1.2 to both sides
-0.2d=3.2
after you divide both sides by -0.2
leaving you with the answer
D= -16
You can get a vertical asymptote at x=1 using y = 1/(x-1)
You can generate a hole at x=3 by multiplying by (x - 3/(x - 3) which is undefined at x=3 but otherwise equals 1
You can move the horizontal asymptote up to y=2 by adding 2
y = (x - 3)/((x - 1)(x - 3)) + 2
Answer:
N= -8
Step-by-step explanation:
a
