at x = 0, y = 100, y-intercept
meaning at 0 years, at the very beginning of functioning, the satellite had 100 units of power.
at y = 0, x = 10, x-intercept
meaning, at year 10, the satellite had 0 units of power, already out of power.
6a^2x-10 beccause u have to multiply
Answer:
432 pages.
Step-by-step explanation:
We know Madison reads 24pgs/ per day. She reads for nine days, so she reads 24*9 pages, or 216.
We know also that this is halfway through her book, so 2*216= the whole book, so the book has 432 pages
Answer:
1/4.
Step-by-step explanation:
That is 1 - (1/4 + 2/5 + 1/10).
The Lowest common multiple 4 5 and 10 is 20 so we have
20/20 - (5/20 + 8/20 + 2/20)
= 20/20 - 15/20
= 5/20
= 1/4 answer.
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.
