If a polygon does not have six sides, then it is not a hexagon.
Answer:
4 √6
Step-by-step explanation:
We have a few right triangles. We know that a²+b²=c², with c being the side opposite the right angle. Representing the side without a value as z, we have:
m²+z² = (8+4)² = 12²
4²+n²=z²
8²+n²=m²
We have 3 equations with 3 unknown variables, so this should be solvable. One way to find a solution is to put everything in terms of m and go from there. First, we can take n out of the equations entirely, removing one variable. We can do this by solving for it in terms of z and plugging that into the third equation, removing a variable as well as an equation.
4²+n²=z²
subtract 4²=16 from both sides
z²-16 = n²
plug that into the third equation
64 + z² - 16 = m²
48 + z² = m²
subtract 48 from both sides to solve for z²
z² = m² - 48
plug that into the first equation
m² + m² - 48 = 144
2m² - 48 = 144
add 48 to both sides to isolate the m² and its coefficient
192 = 2m²
divide both sides by 2 to isolate the m²
96 = m²
square root both sides to solve for m
√96 = m
we know that 96 = 16 * 6, and 16 = 4², so
m = √96 = √(4²*6) = 4 √6
Answer:
87q-6
sorry if i am incorrect.
Step-by-step explanation:
Answer:
Gila Monster is 1.54 times that of Chuckwalla.
Step-by-step explanation:
Given:
Average Length of Gila Monster = 0.608 m
Average Length of Chuckwalla = 0.395 m
We need to find the number of times the Gila monster is as the Chuckwalla.
Solution:
Now we know that;
To find the number of times the Gila monster is as the Chuckwalla we will divide the Average Length of Gila Monster by Average Length of Chuckwalla.
framing in equation form we get;
number of times the Gila monster is as the Chuckwalla = 
Rounding to nearest hundredth's we get;
number of times the Gila monster is as the Chuckwalla = 1.54
Hence Gila Monster is 1.54 times that of Chuckwalla.
