If you have a graphing calculator just put in the equation in 'y=' (not the i equation), and then go to 2nd trace and see where the y=0, those numbers under the x column are the zeros. For the first one, the zeros are: -1, .5, and 2.8. For the second question the zeros are: -3 and about 1.9. The zeros with a decimal are estimations.
So we have 299790000, we want to have only one digit in front of the decimal, 2.9979x10^9
hope this helps :)
D. y=mx+b. m being slope and b being the y intercept. so the equation for the line is y=3x+2 and to be parallel your m must be the same number so D. 3x
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.
Answer:
Step-by-step explanation:
The equation of a line in the point-slope form:
We have:
Substitute:
