Answer:
On average, cars enter the highway during the first half hour of rush hour at a rate 97 per minute.
Step-by-step explanation:
Given that, the rate R(t) at which cars enter the highway is given the formula
The average rate of car enter the highway during first half hour of rush hour is the average value of R(t) from t=0 to t=30.
![=[100(t-0.0001\frac{t^3}{3})]_0^{30}](https://tex.z-dn.net/?f=%3D%5B100%28t-0.0001%5Cfrac%7Bt%5E3%7D%7B3%7D%29%5D_0%5E%7B30%7D)
![=100[(30-0.0001\frac{30^3}{3})-(0-0.0001\frac{0^3}{3})]](https://tex.z-dn.net/?f=%3D100%5B%2830-0.0001%5Cfrac%7B30%5E3%7D%7B3%7D%29-%280-0.0001%5Cfrac%7B0%5E3%7D%7B3%7D%29%5D)
=2901
The average rate of car is 
=97
On average, cars enter the highway during the first half hour of rush hour at a rate 97 per minute.
<h2>
Answer:</h2>
The value of y is 8.4375
<h2>
Step-by-step explanation:</h2><h3>Known :</h3>
<h3>Asked :</h3>
- The value of y when x is 2.25
<h3>Solution :</h3>
2/7.5 = 2.25/y
Do a cross multiplication,
2/7.5 = 2.25/y
=> 20/75 = 2.25/y
=> 75 . 2.25 = 20y
=> 168.75 = 20y
Reverse the equation,
168.75 = 20y
=> 20y = 168.75
Find the value of y,
20y = 168.75
=> y = 8.4375
<h3>Conclusion :</h3>
y = 8.4375
<span>f(x) = 1.5 x (1 - x)
f(0.8) = 1.5(0.8)(1 - 0.8) = 0.24
f(0.24) = 1.5(0.24)(1 - 0.24) ~= 0.274
f(0.274) = 1.5(0.274)(1 - 0.274) ~= 0.298
f(0.298) = 1.5(0.298)(1 - 0.298) ~= 0.314
f(0.314) = 1.5(0.314)(1 - 0.314) ~= 0.323
</span>So the answer is D : 0.24, 0.274, 0.298, 0.314, 0.323
Answer:
6 1/3 divided by 2 1/2 = x
Step-by-step explanation:
Hope this helps.
Answer:
45p-9p+8 simplified is 36p+8
