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.
Take DEBC a parralelogram
Parralel sides are DE and BC
THEREFORE,
1/2*(a+b)*h
1/2*(6+15)*h
1/2*(21)*h=0
10.5*h=0
h=-10.5
The answer is 4x>_-12. the underscore is the line under the “>”
B how many ways can they register for the same section
Answer:
±sqrt( H *f•c)= L
Step-by-step explanation:
H=L^2/f•c
Multiply each side by fc
H *fc=L^2/f•c * fc
H *f•c=L^2
Take the square root of each side
±sqrt( H *f•c)= sqrt(L^2)
±sqrt( H *f•c)= L