Answer:
Step-by-step explanation:
Answer:
2.33 = BC
Step-by-step explanation:
Since this is a right triangle, we can use trig functions
tan theta = opp / adj
tan B = AC / BC
tan 25 = AC / 5
Multiply each side by 5
5 tan 25 = BC
2.331538291 = BC
To the nearest hundredth
2.33 = BC
Answer:
Step-by-step explanation:
volume = area of base * height = area of base * 9 = 72
-> area of base = 72/9 = 8
let a,b,c be the length of sides of isosceles right triangle base
we have a = b and a*b/2 = 8 ft2
-> a*a/2 = 8ft2
-> a*a = 16
-> a=4ft=b
have c^2 = a^2 + b^2 = 16 +16 = 32
-> c = 
= 5.656 ft
Since i dont know that is DF, i calculate both a,b and c
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.
Multiplication by 10ⁿ → move decimal n places to the right.
Multiplication by 10⁻ⁿ → move decimal n places to the left.
For n ∈ N
