Answer:
You need to attach the pictures and ask again.
Step-by-step explanation:
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A = p + prt
The above equation can be solved for “r” in such a
way.
Subtract “p” from both sides;
A – p = p + prt – p
A – p = prt
Divide both sides by “p” and “t”;
A/pt – p/pt = prt/pt
A/pt – 1/t = r
Or
<span>r = A/pt – 1/t</span>
We want to create a linear equation to model the given situation.
A) c(r) = $6.00 + $1.50*r
B) 19 rides.
We know that the carnival charges $6.00 for entry plus $1.50 for each ride.
A) With the given information we can see that if you ride for r rides, then the cost equation will be:
c(r) = $6.00 + $1.50*r
Where c(r) is the cost for going to the carnival and doing r rides.
B) If you have $35.00, then we can solve:
c(r) = $35.00 = $6.00 + $1.50*r
Now we can solve the equation for r.
$35.00 = $6.00 + $1.50*r
$35.00 - $6.00 = $1.50*r
$29.00 = $1.50*r
$29.00/$1.50 = r = 19.33
Rounding to the next whole number we get: r = 19
This means that with $35.00, Dennis could go to 19 rides.
If you want to learn more, you can read:
brainly.com/question/13738061
One meaning of a 'linear' equation is that if you draw the graph
of the equation, the graph will be a straight line.
That's an easy way to test the equation . . . find 3 points on the
graph, and see whether they're all in a straight line.
This equation is y = 4 / x .
To find a point on the graph, just pick any number for 'x',
and figure out the value of 'y' that goes with it.
Do that 3 times, and you've got 3 points on the graph.
Here ... I'll do 3 quick points:
Point-A: x = 1 y = 4 / 1 = 4
Point-B: x = 2 y = 4 / 2 = 2
Point-C: x = 4 y = 4 / 4 = 1
Look at this:
Slope of the line from point-A to point-B
= (change in 'y') / (change in 'x') = -2 .
Slope of the line from point-B to point-C
= (change in 'y') / (change in 'x') = -1/2 .
The two pieces of line from A-B and from B-C don't even have
the same slope, so they're not pieces of the same straight line !
So my points A, B, and C are NOT in a straight line.
So the equation is NOT linear.
Try it again with three points of your own.