Answer:
90% sure it's the third option
Step-by-step explanation:
Answer:
x = 1
x = -5/8 + i* root( 39) / 8
or x = -5/8 - i* root( 39) / 8
Step-by-step explanation:
(x+x)^2 × -(x×(-1))=4+x-x^2
solve for x, show work.
(x+x)^2 × -(x×(-1)) = 4+x-x^2
Simplify:
(2x)^2 * (-(-x)) = 4 + x - x^2
4x*x * (-(-x)) = 4 + x - x*x
4*x*x*x = 4 + x - x*x
4*x*x*x - 4 = x *(1 - x)
4* (xxx - 1) = x * (1- x)
4* (x - 1)*( x*x +x + 1) = x (1 - x)
4( xx + x + 1) = -x
x = 1 is a solution.
4xx + 4x + 4 = -x
4xx + 5x + 4 = 0
x = -5/8 + root(25 - 4*4*4) / 2*4
x = -5/8 + i* root( 39) / 8
or x = -5/8 - i* root( 39) / 8
also x = 1
Hello,
Radius= √((4-0)²+(5-0)²)=√41
Equation of the circle : x²+y²=41
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.
