Answer:
r = 8
Step-by-step explanation:
The circumference of a circle is given by
C = 2 * pi *r
16 pi = 2 * pi r
Divide each side by 2 pi
16 pi /2 pi = 2 pi r / 2 pi
8 = r
It’s 30 because they are vertical angles which means both sides are congruent
a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have
Evaluate the integral to solve for y :
Use the other known value, f(2) = 18, to solve for k :
Then the curve C has equation
b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:
The slope of the given tangent line 
is 1. Solve for a :
so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).
Decide which of these points is correct:
So, the point of contact between the tangent line and C is (-1, -3).
Answer:
4.9=4.2
Step-by-step explanation:
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.
