Answer:
D
Step-by-step explanation:
I don’t believe there is any work to show for this question
Let t be time and r be rate, then if time varies inversely with the rate, the equation is

. If it takes 5 hours to drive a fixed distance at a rate of 80, we can sub those values in to solve for the constant of variation, k.

. Solve for k by multiplying 5 and 80 to get that k = 400. Now let's find a new time t when r is a rate of 70. We will use that k value to do this:

and find that it will take 20 hours to drive the distance at 70 mph when it takes 5 hours to drive the distance at 80 mph. Makes sense that it takes longer to drive a fixed distance when you are going slower.
Answer:
x=½
Step-by-step explanation:
<em>3</em><em>x</em><em>-</em><em>9</em><em>=</em><em>x-8</em><em>(</em><em>Group</em><em> </em><em>like</em><em> </em><em>terms</em><em>)</em>
<em>3</em><em>x</em><em>-</em><em>x</em><em>=</em><em>-</em><em>8</em><em>+</em><em>9</em><em>(</em><em>You </em><em>will </em><em>subtract</em><em> </em><em>x</em><em> </em><em>which</em><em> </em><em>is </em><em>also</em><em> </em><em>1</em><em>x</em><em> </em><em>from </em><em>3</em><em>x</em><em> </em><em>which</em><em> </em><em>is</em><em> </em><em>going</em><em> </em><em>to</em><em> </em><em>be </em><em>2</em><em>x</em><em> </em><em>Then </em><em>you </em><em>will</em><em> </em><em>add</em><em> </em><em>-</em><em>8</em><em> </em><em>to</em><em> </em><em>9</em><em> </em><em>making </em><em>it</em><em> </em><em>1</em><em>(</em><em>positive</em><em>)</em><em>)</em><em>.</em>
<em>2</em><em>x</em><em>=</em><em>1</em><em>(</em><em>Divide</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>2</em><em>)</em>
<em>x</em><em>=</em><em>½</em>
Distribute the 2 through the parentheses
on the right side of the equation.
2(y) is 2y and 2(2) is 4.
So the problem now reads 6 = 2y + 4.
Now we can subtract 4 from both
sides of the equation to get 2 = 2y.
Now divide both sides of the
equation by 2 and <em>1 = y</em>.
Answer:
The graph increases if the line goes upwards. (The end of the segment is above the start of the segment)
The graph decreases if the line goes downwards. (The end of the segment is below the start of the segment)
The graph is constant if the line is horizontal. (The end of the segment is at the same height as the start of the segment)
Then:
The graph of the function is increasing in the segments c and e.
The graph of the function is decreasing in the segments a and f.
The graph of the function is constant in the segments b and d.