Answer:
The amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.
Step-by-step explanation:
Consider the provided information.
A chemical flows into a storage tank at a rate of (180+3t) liters per minute,
Let 
is the amount of chemical in the take at <em>t </em>time.
Now find the rate of change of chemical flow during the first 20 minutes.
![\int\limits^{20}_{0} {c'(t)} \, dt =\left[180t+\dfrac{3}{2}t^2\right]^{20}_0](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B20%7D_%7B0%7D%20%7Bc%27%28t%29%7D%20%5C%2C%20dt%20%3D%5Cleft%5B180t%2B%5Cdfrac%7B3%7D%7B2%7Dt%5E2%5Cright%5D%5E%7B20%7D_0)
So, the amount of the chemical flows into the tank during the firs 20 minutes is 4200 liters.
Answer:
Step-by-step explanation:
BDC=95
Answer:
X= 5
Explanation (Steps):
<u>Step 1: Simplify both sides of the equation.</u>
−3.4(x−2)=9.8−4x
(−3.4)(x)+(−3.4)(−2)=9.8+−4x(Distribute)
−3.4x+6.8=9.8+−4x
−3.4x+6.8=−4x+9.8
<u>Step 2: Add 4x to both sides.</u>
−3.4x+6.8+4x=−4x+9.8+4x
0.6x+6.8=9.8
<u>Step 3: Subtract 6.8 from both sides.</u>
0.6x+6.8−6.8=9.8−6.8
0.6x=3
<u>Step 4: Divide both sides by 0.6.</u>
0.6x0.6=3/0.6
x=5
(5/8)+(2/5)=
so you need a common den so in this case its 40
so we multiply to get 40 like
(5/5)(5/8)+ (2/5)(8/8)=
25/40 + 16/ 40 = 41/40
