we have the equation
![\frac{x}{3}+5=8](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B3%7D%2B5%3D8)
Solve for x
That means ----> isolate the variable x
step 1
Subtract 5 on both sides
![\begin{gathered} \frac{x}{3}+5-5=8-5 \\ \\ \frac{x}{3}=3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bx%7D%7B3%7D%2B5-5%3D8-5%20%5C%5C%20%20%5C%5C%20%5Cfrac%7Bx%7D%7B3%7D%3D3%20%5Cend%7Bgathered%7D)
step 3
Multiply by 3 on both sides
![\begin{gathered} \frac{x}{3}*3=3*3 \\ x=9 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7Bx%7D%7B3%7D%2A3%3D3%2A3%20%5C%5C%20x%3D9%20%5Cend%7Bgathered%7D)
<h2>The answer is x=9</h2>
Answer: 0.01111
Step-by-step explanation: find attached picture for explanation
The average rate of change of a function f(x) in an interval, a < x < b is given by
![\frac{f(b) - f(a)}{b - a}](https://tex.z-dn.net/?f=%5Cfrac%7Bf%28b%29%20-%20f%28a%29%7D%7Bb%20-%20a%7D)
Given q(x) = (x + 3)^2
1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by
![\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4](https://tex.z-dn.net/?f=%20%5Cfrac%7Bq%28-4%29-q%28-6%29%7D%7B-4-%28-6%29%7D%20%3D%20%5Cfrac%7B%28-4%2B3%29%5E2-%28-6%2B3%29%5E2%7D%7B-4%2B6%7D%20%3D%20%5Cfrac%7B1-9%7D%7B2%7D%20%3D%20%5Cfrac%7B-8%7D%7B2%7D%20%3D-4)
2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by
![\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3](https://tex.z-dn.net/?f=%20%5Cfrac%7Bq%280%29-q%28-3%29%7D%7B0-%28-3%29%7D%20%3D%20%5Cfrac%7B%280%2B3%29%5E2-%28-3%2B3%29%5E2%7D%7B0%2B3%7D%20%3D%20%5Cfrac%7B9-0%7D%7B3%7D%20%3D%20%5Cfrac%7B9%7D%7B3%7D%20%3D3)
3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by
![\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3](https://tex.z-dn.net/?f=%20%5Cfrac%7Bq%28-3%29-q%28-6%29%7D%7B-3-%28-6%29%7D%20%3D%20%5Cfrac%7B%28-3%2B3%29%5E2-%28-6%2B3%29%5E2%7D%7B-3%2B6%7D%20%3D%20%5Cfrac%7B0-9%7D%7B3%7D%20%3D%20%5Cfrac%7B-9%7D%7B3%7D%20%3D-3)
4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by
![\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bq%28-2%29-q%28-3%29%7D%7B-2-%28-3%29%7D%20%3D%20%5Cfrac%7B%28-2%2B3%29%5E2-%28-3%2B3%29%5E2%7D%7B-2%2B3%7D%20%3D%20%5Cfrac%7B1-0%7D%7B1%7D%20%3D%20%5Cfrac%7B1%7D%7B1%7D%20%3D1)
5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by
![\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bq%28-3%29-q%28-4%29%7D%7B-3-%28-4%29%7D%20%3D%20%5Cfrac%7B%28-3%2B3%29%5E2-%28-4%2B3%29%5E2%7D%7B-3%2B4%7D%20%3D%20%5Cfrac%7B0-1%7D%7B1%7D%20%3D%20%5Cfrac%7B-1%7D%7B1%7D%20%3D-1)
6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by
The answer is 5.2π in² = 16.3 in²
The area (A) of the circle with radius r is: A = r²π
It is known:
π = 3.14
![d=6- \frac{50}{t^{2}+10 }](https://tex.z-dn.net/?f=d%3D6-%20%5Cfrac%7B50%7D%7Bt%5E%7B2%7D%2B10%20%7D%20)
![r = \frac{d}{2}= \frac{6- \frac{50}{t^{2}+10 }}{2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7Bd%7D%7B2%7D%3D%20%5Cfrac%7B6-%20%5Cfrac%7B50%7D%7Bt%5E%7B2%7D%2B10%20%7D%7D%7B2%7D%20)
t = 5
⇒
![=\frac{ \frac{42-10}{7}}{2}=\frac{ \frac{32}{7}}{ \frac{2}{1} }= \frac{32}{7}* \frac{1}{2} = \frac{16}{7}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%20%5Cfrac%7B42-10%7D%7B7%7D%7D%7B2%7D%3D%5Cfrac%7B%20%5Cfrac%7B32%7D%7B7%7D%7D%7B%20%5Cfrac%7B2%7D%7B1%7D%20%7D%3D%20%5Cfrac%7B32%7D%7B7%7D%2A%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%5Cfrac%7B16%7D%7B7%7D%20)
A = r²π = (16/7)²π = (2.28)²π = 5.2π = 5.2 · 3.14 = 16.3 in²
Answer:
-29 and -12
Step-by-step explanation: