Keep adding 20
Ex. 20. 40. 60.....
.
Answer:
The difference between the two possible lengths is 3.2 in
Explanation:
Assume that the sides of the triangle are:
a = 5
b = 8
c = x
First, we will assume that the third side is the hypotenuse:
For the triangle to be right angled:
c^2 = a^2 + b^2
Substitute with the values of a, b and c in the equation and solve for x as follows:
c^2 = a^2 + b^2
x^2 = (5)^2 + (8)^2
x^2 = 89
either x = 9.433 in
or x = -9.433 in (refused as no length is in negative)
So, first possible value of the third side is 9.433 in
Second, we will assume that the 8 in is the hypotenuse of the triangle.
For the triangle to be right angled:
b^2 = a^2 + c^2
(8)^2 = (5)^2 + x^2
64 = 25 + x^2
x^2 = 64 - 25 = 39
either x = 6.245 in
or x = -6.245 in (refused as no length is in negative)
Therefore, the second possible value of the third side is 6.245 in
Finally, we will get the difference between the two length as follows:
difference = 9.433 - 6.245 = 3.188 which is approximately 3.2 in
Note:
We cannot assume that the 5 in is the hypotenuse because in the right-angled triangle the hypotenuse is the longest side. We are given that one side = 8 in therefore, it is impossible for the 5 in to be the hypotenuse.
Hope this helps :)
Answer:
Step-by-step explanation:
The diameter is 70 ml
The radius is worth half the diameter, which means that the radius is 35 ml
A = pi r squared
A = 3.14 x 1225
A = 3846.45
Answer:
a = 6, b = 3, and c = 2
1. ab1 = 6 × 3 × 1 = 18
2. c + 42 = 2 + 42 = 44
3. 18 = 18
4. a − b4 = 6 - 3(4) = 6 - 12 = - 6
5. 2c3 = 2 × 2 × 3 = 12
6. b ÷ 35 = 3÷35 = 0.086
7. a − 1 6 = 6 - 16 = -10
8. 6 + c8 = 6 + 2 (8) = 6 + 16 = 22
Step-by-step explanation:
I got 75582.
Explanation:
First, group the 40 identical candies into 20 pairs. It doesn't matter how since the candies are identical. This grouping will ensure that any assigment will contain at least two candies.
Then think of the 20 groups a 20 beads on a string. We are looking to place 11 separators between them to obtain 12 segments, each with a varying number of beads between them. How many ways are there to place 11 separators to 19 potential spaces between beads? The asnwer is ![n = {19\choose{11}}=75582](https://tex.z-dn.net/?f=n%20%3D%20%7B19%5Cchoose%7B11%7D%7D%3D75582)