For #4, we know V = lwh
We have the values
5184 = 2(18)h
Solve for h
5184 = 36h
144 = h.
for # 3:
we know A = pi r^2
We have
A = 3.14 * 6^2
A = 3.14 * 36
A = 113.04
for #2:
We know C = 2pi *r
C = 2(3.14)(14)
C = 28(3.14)
C = 87.92
for #1:
C = 2pi * r
C = 2(3.14)(26.3/2)
C = 2(3.14)(13.15)
C = 26.3(3.14)
C = 82.582
Answer:
<h2>45, 46, 47</h2>
Step-by-step explanation:
n, n + 1, n + 2 - three consecutive integers
The equation:
138 - n = (n + 1) + (n + 2)
138 - n = n + 1 + n + 2 <em>combile like terms</em>
138 - n = 2n + 3 <em>subtract 138 from both sides</em>
-n = 2n - 135 <em>subtract 2n from both sides</em>
-3n = -135 <em>divide both sides by (-3)</em>
n = 45
n + 1 = 46
n + 2 = 47
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:
10
10
Step-by-step explanation:
One fifth of fifty is 10.
Answer:
the solutions of a function are the points where for some value of x the function becomes zero
thus the solns for this graph would be
<h3>-3 , 2</h3>
that's option 1.
