Answer:
The radius of the spherical tank is 12 meters
Step-by-step explanation:
In this question, we are tasked with calculating the radius of the spherical tank given the volume of air it can hold
Mathematically, the volume V of a sphere is 4/3 * pi * r^3
Thus, 7234.56 = 4/3 * 3.14 * r^3
7,234.56 = 4.19 * r^3
r^3 = 7234.56/4.19
r^3 = 1,728
r = cube root of 1,728
r = 12 meters
<h2>Answer-Average rate of change(A(x)) of f(x) over a interval [a,b] is given by:</h2><h2 /><h2>A(x) = \frac{f(b)-f(a)}{b-a}A(x)= </h2><h2>b−a</h2><h2>f(b)−f(a)</h2><h2> </h2><h2> </h2><h2 /><h2>Given the function:</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^xf(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>x</h2><h2> </h2><h2 /><h2>We have to find the average rate of change from x = 1 to x= 2</h2><h2 /><h2>At x = 1</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^1 = 5f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>1</h2><h2> =5</h2><h2 /><h2>At x = 2</h2><h2 /><h2>then;</h2><h2 /><h2>f(x) = 20 \cdot(\frac{1}{4})^2=20 \cdot \frac{1}{16} = 1.25f(x)=20⋅( </h2><h2>4</h2><h2>1</h2><h2> </h2><h2> ) </h2><h2>2</h2><h2> =20⋅ </h2><h2>16</h2><h2>1</h2><h2> </h2><h2> =1.25</h2><h2 /><h2>Substitute these in above formula we have;</h2><h2 /><h2>A(x) = \frac{f(2)-f(1)}{2-1}A(x)= </h2><h2>2−1</h2><h2>f(2)−f(1)</h2><h2> </h2><h2> </h2><h2 /><h2>⇒A(x) = \frac{1.25-5}{1}=-3.75A(x)= </h2><h2>1</h2><h2>1.25−5</h2><h2> </h2><h2> =−3.75</h2><h2 /><h2>therefore, average rate of change of the function f(x) from x = 1 to x = 2 is, -3.75</h2>
<h2>Please Mark me as brainlist. </h2>
Option C is the right answer
½(12) = 6
½(64) = 32
½(30) = 15
½(78) = 39
Answer:
27p4-27p2-12p+12/p
Step-by-step explanation:
Step-by-step explanation:
Plug x in :
-8(7-3) = -32
Distribute :
-56 - (-24) = -32
Subtract :
-32 = -32
