If 1 in.= 2.54 cm, Then !9in. multiplied by 2.54 cm would be b. 48.26 cm
I believe it is ; for the number
Answer:
21.9°
Step-by-step explanation:
In ΔHIJ, the measure of ∠J=90°, HI = 8 feet, and JH = 1.9 feet. Find the measure of ∠H to the nearest tenth of a degree.
We solve this above question using the Sine rule
a/sin A = b/sin B
In ΔHIJ, the measure of ∠J=90°, HI = 8 feet, and JH = 1.9 feet.
Hence:
HI/∠J = JH/∠H
= 8/sin 90° = 1.9/sin ∠H
Cross Multiply
∠H = arc sin(sin 1.9 × 90/8)
∠H = 21.9°
Answer:
6
Step-by-step explanation:
36/6
6, 12,18,24,30,36
1 2. 3. 4. 5. 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.
