Answer:
yes
Step-by-step explanation:
So the circumfrehce of a circle is basically the perimiter of the circle. Let me help with number 1. We need to find the radius. In order to do tuis we divide 6 in half giving us 3. Then we plug in 3 into the circumfrence formula (Pi times the radius multiplied by 2). So it would be 3.14*3*2. The answer for number 1 is 18.85
<span>The graph you plotted is the graph of f ' (x) and NOT f(x) itself. </span>
Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get
f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24
So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5
<span>f ' (x) = (x-4)(6-2x)
</span><span>f ' (3.5) = (3.5-4)(6-2*3.5)
</span>f ' (3.5) = 0.5
The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.
So the final answer is C) 3
<h2>
Answer: (x = 1)</h2>
Step-by-step explanation:
We are given the equation <em>6(5x - 3) = 1/3(24x + 12)</em>, and we must find the value of <em>x</em>.
The first step is to distribute. Distribute the <em>6 </em>to <em>5x </em>and <em>-3</em>, and distribute<em> 1/3 </em>to <em>24x </em>and <em>12</em>. This give us the equation <em>30x - 18 = 8x + 4</em> to work with.
The next step is to get <em>x </em>by itself. In order to do this, subtract 18 from each side of the equation. This gives get's rid of the <em>-18</em> and get's <em>x </em>by itself on one side of the equation and gives us the equation <em>30x = 8x +22</em>.
Now that x is by itself on one side of the equation, get 22 by itself on the other side of the equation. The way to do this is to subtract 8x from each side. This gives us the equation <em>22x = 22</em>.
Divide both sides by 22 and get the equation <em>x = 1</em>






<em>Hope that this helps and clears it up! Have a great day!</em>