Answer:
Step-by-step explanation:
Algebraic expressions tend to have atleast one unknown variable in it. In order to solve the expression we need to assign a fixed value to that variable or isolate it if the expression is equal to another value. For example, in the following expression we have the variable x, if we give it a value of 5 we simply solve like a regular expression...
5x + 3
5(5) + 3
25 + 3
28
Therefore, if we swap the variable x for the value of 5 we would get a value of 28 in this algebraic expression.
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It takes 6 seconds for it to hit the ground.
0 = -5x²+20x+60
We can solve this by factoring. First factor out the GCF, -5:
0 = -5(x²-4x-12)
Now we want factors of -12 that sum to -4. -6(2) = -12 an -6+2 = -4:
0 = -5(x-6)(x+2)
Using the zero product property, we know that either x-6=0 or x+2=0; this gives us the answers x=6 or x=-2. Since we cannot have negative time, x=6.
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.