Answer:
B = 146 deg
Step-by-step explanation:
firstly, for angle FEB, do this formula:
180-34=146
FEB would be alternate with B, making them equal, therefore:
B = 146
Yes
for a square
P=4s
s=legnth o side
since all sides are equal
20=4s
divide 4
5=s
each sie is 5 units
5/3 = 1.66
5/3 = 166/100 [cross multiplication]
5 × 100 = 3 × 166
500 ≠ 498
NO, the equation is not balanced.
______
Hope it helps ⚜
Given:
The area of the squares are given.
To find:
The exact side length or estimate side length of the square.
Solution:
We know that, the area of a square is
![A=a^2](https://tex.z-dn.net/?f=A%3Da%5E2)
Where, a is the side length of the square.
![a=\sqrt{A}](https://tex.z-dn.net/?f=a%3D%5Csqrt%7BA%7D)
Area of the square is 100 square units. So, the side length is:
![a=\sqrt{100}](https://tex.z-dn.net/?f=a%3D%5Csqrt%7B100%7D)
![a=10](https://tex.z-dn.net/?f=a%3D10)
Therefore, the side length is 10 units.
Area of the square is 95 square units. So, the side length is:
![a=\sqrt{95}](https://tex.z-dn.net/?f=a%3D%5Csqrt%7B95%7D)
It is not exact. We know that
.
Therefore, the side length is between 9 and 10.
Area of the square is 36 square units. So, the side length is:
![a=\sqrt{36}](https://tex.z-dn.net/?f=a%3D%5Csqrt%7B36%7D)
![a=6](https://tex.z-dn.net/?f=a%3D6)
Therefore, the side length is 6 units.
Area of the square is 30 square units. So, the side length is:
![a=\sqrt{30}](https://tex.z-dn.net/?f=a%3D%5Csqrt%7B30%7D)
It is not exact. We know that
.
Therefore, the side length is between 5 and 6.
This problem is a real bear. Whoever wrote it has a sense of humor, and expects
a lot from you ... probability, permutations, combinations, and geometry.
<u>First, let's talk about probability:</u>
The probability of something happening is (the number of outcomes that meet your description) divided by (the total number of all possible outcomes).
<u>#62:</u>
There are 4 points on the drawing. How many different ways could you pick
two of them ?
The first one you pick could be any one of 4 points.
For each of those, the other point could be any one of the remaining 3.
So the total number of ways to pick 2 points out of 4 is (4 x 3) = 12 ways.
But wait ! Whether you pick 'A' and then 'C', or pick 'C' and then 'A', you still wind up with the same two points. So, although there are 12 ways to pick them, there are only 6 different distinct pairs of points.
OK. How many of those pairs are collinear ? ANY two points lie on the same line, because a line can always be drawn between any two points. So out of the 6 different possible pairs of points, ALL 6 pairs are collinear. The probability of picking a pair that are collinear is 100% .
<u>#63:</u>
There are 4 points on the drawing. How many different ways could you pick
three of them ?
The first one you pick could be any one of 4 points. For each of those ...
The second point could be any one of the remaining 3. For each of those ...
The third point could be either one of the remaining 2.
So the total number of ways to pick 3 points out of 4 is (4 x 3 x 2) = 24 ways.
But
wait ! Whether you pick ABC, ACB, BAC, BCA, CAB, or CBA, you still wind up with the same three points. So, although there are 24
ways to pick them, there are only 4 different distinct sets of three points.
OK. How many sets of 3 points in this drawing are collinear ?
There is only one ! ONLY A, B, and C are collinear.
Probability = 1 out of 4 = 1/4 = 25 percent .
<u>#64:</u>
In the last problem, we saw that there are 4 distinct sets of three points.
How many of them are coplanar ?
They ALL are. A plane can be drawn through ANY three points.
So whichever three points you pick, they are coplanar.
The probability is 4 out of 4 = 100 percent.
Thank you for your 5 points. I shall cherish them.