3 packages so there would be at least 21 invitiations
<span>We define
day--------------------------X
hockey cards away ------------f(X)
</span><span>He gives 2 cards away ----------------on day 1----------f(X)=2 X=1
</span>He gives 4 cards away ----------------on day 2----------f(X)=4 X=2
He gives 8 cards away ----------------on day 3----------f(X)=8 X=3
the sequence of cards away 2,4,8,16,32,64,128,256,512,1024
f(x) = 2**(x)
The answers are
D <span>Each day corresponds to a unique amount of hockey cards in this sequence.</span>
And B <span>The graph of this sequence would pass the vertical line test.
Each element of the Domain (X = day), relates to a single value of f(x) = number of hockey cards
That is, every value of X must have only one corresponding value of f(x)
(One value of y, for each input X)
This can also be checked by using the vertical line test.
If we plot f(x) and we draw vertical lines in the curve, we can verify that at every input the line intersects only one point of f(x)</span>
It's blurry try retaking it
Answer:
80 and 100 both qualify.
Step-by-step explanation:
Answer:
the greatest common multiple of 14 and 21 is 7.
