Answer:
Boiling point of jet fuel = 165 degrees Celsius
Step-by-step explanation:
Given:
Boiling point of jet fuel = 329°F
Find:
Boiling point of jet fuel in degrees Celsius
Computation:
C = [5/9][F – 32]
C = [5/9][329 – 32]
C = 165
Boiling point of jet fuel = 165 degrees Celsius
Answer:
B. The slopes of the graphs are the same and the y-intercepts of the graphs are different
Step-by-step explanation:
The lines will also be parallel
Do you mean the following:
log(5^(4/7))=around 0.4
log(5)^(4/7)=around 0.815
log(5^4)/7= same thing as top one
log((5^4)/7)=around 1.95
When you rotate the ef counter lock 180 degrees, the new line e'f' will have the same length as ef.
Answer:
B
Step:
The general equation for linear is y=mx+b
So all is linear except B
