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.
The area of a rectangle is obtained through the equation,
A = L x W
The width of the yard is 4 ft less than the length and may be expressed as L - 4. Length may be solved through the following steps,
A = (L)(L-4) ; 96 = L(L - 4) ; L = 12 ft
The length and width are 12 ft and 8 ft, respectively. Perimeter may be solved through the equation,
P = 2 x (L + W)
Substituting the values of L and W
P = 2 x ( 12 ft + 8 ft) = 40 ft
Therefore, the perimeter of the yard is 40 ft.
Answer:
answer B 
Step-by-step explanation:
f(8)=-2*8+5=-11
g(8)=9*8*8+4=580
f(8)+g(8)= -11+580=569
Answer:
is 23,13 usjebiddje
Step-by-step explanation:
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Answer:
yes
Step-by-step explanation: