Answer:
C
Step-by-step explanation:
the distance is 15•6 = 18•5 = 90 feet
so, time needed = 90/30 = 3 second
Answer: I think that the cost now from hotel 5 would be $15.00.
Step-by-step explanation:
5 hours = $5.00
10 hours = $10.00
15 hours = $15.00
20 hours = $20.00
25 hours = $25.00
Answer:
An alternative definition for the acceleration ax that can be written in terms of
and
is 
Step-by-step explanation:
We know that :

Now we are supposed to find an alternative definition for the acceleration ax that can be written in terms of
and 
So, We will use chain rule over here :
![a_x=\frac{dv_x}{dt}\\a_x=\frac{dv_x}{dt} \times \frac{dx}{dx}\\a_x=\frac{dv_x}{dx} \times \frac{dx}{dt}\\a_x=\frac{dv_x}{dx} \times \frac{dx}{dt} [\frac{dx}{dt}=v_x]\\a_x=\frac{dv_x}{dx} \times v_x\\a_x=v_x\frac{dv_x}{dx}](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7Bdv_x%7D%7Bdt%7D%5C%5Ca_x%3D%5Cfrac%7Bdv_x%7D%7Bdt%7D%20%5Ctimes%20%5Cfrac%7Bdx%7D%7Bdx%7D%5C%5Ca_x%3D%5Cfrac%7Bdv_x%7D%7Bdx%7D%20%5Ctimes%20%5Cfrac%7Bdx%7D%7Bdt%7D%5C%5Ca_x%3D%5Cfrac%7Bdv_x%7D%7Bdx%7D%20%5Ctimes%20%5Cfrac%7Bdx%7D%7Bdt%7D%20%20%5B%5Cfrac%7Bdx%7D%7Bdt%7D%3Dv_x%5D%5C%5Ca_x%3D%5Cfrac%7Bdv_x%7D%7Bdx%7D%20%5Ctimes%20v_x%5C%5Ca_x%3Dv_x%5Cfrac%7Bdv_x%7D%7Bdx%7D)
Hence an alternative definition for the acceleration ax that can be written in terms of
and
is 
Answer:
The lateral area is equal to

Step-by-step explanation:
In this problem the lateral area is equal to the area of one equilateral triangle multiplied by 
To find the area of one equilateral triangle calculate the height
The area of the triangle is equal to

we have

Applying the Pythagoras theorem

The area of one triangle is equal to

so
The lateral area is equal to

Answer:
8.1 hours
Step-by-step explanation:
A model of the fraction remaining can be ...
f = (1/2)^(t/37) . . . . t in hours
So, for the fraction remaining being 86%, we can solve for t using ...
0.86 = 0.5^(t/37)
log(0.86) = (t/37)log(0.5)
t = 37·log(0.86)/log(0.5) ≈ 8.0509 ≈ 8.1 . . . hours
It takes about 8.1 hours to decay to 86% of the original concentration.