In which graph do all the plotted points lie on the line y = x + 2?

(a) In order to solve this problem we will introduce new variables u and v, where u = 4t - 4 and v = -2t + 7.  Notice that as t gets very large, u = 4t - 4 looks like 4t, and v = -2t + 7 looks like -2t.  We can formalize this by saying that as t gets very large, v/u = (-2t + 7)/(4t - 4).  Divide both sides by t, to get v/u = (-2 + 7/t)/(4 - 4/t).  As t gets large, 7/t and 4/t are small, so u/v becomes (-2 + 7/t)/(4 - 4/t) = -1/2.  This tells us that we should look for a line of slope -1/2.

If y/x = (-2t + 7)/(4t - 4), then y(4t - 4) = x(-2t + 7), so 4yt - 4y = -2xt + 7x.  Then 4y + 7x = 4yt + 2xt = t(4y + 2x).  Since this is true for every value of t, we must have 4y + 7x = 0 and 4y + 2x = 0.  When we solve this system, we get that x = y = 0, so that tells us that our line passes through the origin.  And we know that the slope of the line is -1/2, so the equation of our line is y = -x/2.  But that is only the base line - we can take our line and shift it in the x-direction, or y-direction, until it matches what we want.  Here, the problem tells us that our line passes through the point (4,3), so we can shift it 3 units up, to get y = -x/2 + 3.  We can then shift it four units in the x-direction, to get y = -(x + 4)/2 + 3 = -x/2 + 5.  So this is the equation of the line.

(b) We can use the same argument as in part (a).  Let u = 4t - 4 and v = -2t + 7.  As we said in part (a), as t gets very large, u/v gets close to -1/2.  So we can find a value of t so that u/v is close to -1/2.  If y = -x/2 + 5, then y/x is close to -1/2.  So choose the value of t so that y/x = u/v.  Then y/x = (-2t + 7)/(4t - 4).  This leads to the same equation 4y + 7x = 4yt + 2xt = t(4y + 2x).  Then again, 4y + 7x = 0 and 4y + 2x = 0, so x = y = 0.  This gives us a base line of y/x = -1/2, or y = -x/2.  We can then shift five units up, because we want the line to pass through (0,5), to get the line y = -x/2 + 5.  So the line of y = -x/2 + 5 is the graph of G.

(a) In order to solve this problem we will introduce new variables u and v, where u = 4t - 4 and v = -2t + 7.  Notice that as t gets very large, u = 4t - 4 looks like 4t, and v = -2t + 7 looks like -2t.  We can formalize this by saying that as t gets very large, v/u = (-2t + 7)/(4t - 4).  Divide both sides by t, to get v/u = (-2 + 7/t)/(4 - 4/t).  As t gets large, 7/t and 4/t are small, so u/v becomes (-2 + 7/t)/(4 - 4/t) = -1/2.  This tells us that we should look for a line of slope -1/2.

If y/x = (-2t + 7)/(4t - 4), then y(4t - 4) = x(-2t + 7), so 4yt - 4y = -2xt + 7x.  Then 4y + 7x = 4yt + 2xt = t(4y + 2x).  Since this is true for every value of t, we must have 4y + 7x = 0 and 4y + 2x = 0.  When we solve this system, we get that x = y = 0, so that tells us that our line passes through the origin.  And we know that the slope of the line is -1/2, so the equation of our line is y = -x/2.  But that is only the base line - we can take our line and shift it in the x-direction, or y-direction, until it matches what we want.  Here, the problem tells us that our line passes through the point (4,3), so we can shift it 3 units up, to get y = -x/2 + 3.  We can then shift it four units in the x-direction, to get y = -(x + 4)/2 + 3 = -x/2 + 5.  So this is the equation of the line.

(b) We can use the same argument as in part (a).  Let u = 4t - 4 and v = -2t + 7.  As we said in part (a), as t gets very large, u/v gets close to -1/2.  So we can find a value of t so that u/v is close to -1/2.  If y = -x/2 + 5, then y/x is close to -1/2.  So choose the value of t so that y/x = u/v.  Then y/x = (-2t + 7)/(4t - 4).  This leads to the same equation 4y + 7x = 4yt + 2xt = t(4y + 2x).  Then again, 4y + 7x = 0 and 4y + 2x = 0, so x = y = 0.  This gives us a base line of y/x = -1/2, or y = -x/2.  We can then shift five units up, because we want the line to pass through (0,5), to get the line y = -x/2 + 5.  So the line of y = -x/2 + 5 is the graph of G.

The maxima of a differential equation can be obtained by getting the 1st derivate dx/dy and equating it to 0.

Given the equation h = - 2 t^2 + 12 t       , taking the 1st derivative result in:

dh = - 4 t dt + 12 dt

dh / dt = 0 = - 4 t + 12   calculating for t:

t = -12 / - 4

t = 3 s

Therefore the maximum height obtained is calculated by plugging in the value of t in the given equation.

h = -2 (3)^2 + 12 (3)

h = 18 m

This problem can also be solved graphically by plotting t (x-axis) against h (y-axis). Then assigning values to t and calculate for h and plot it in the graph to see the point in which the peak is obtained. Therefore the answer to this is:

The ball reaches a maximum height of 18 meters. The maximum of h(t) can be found both graphically or algebraically, and lies at (3,18). The x-coordinate, 3, is the time in seconds it takes the ball to reach maximum height, and the y-coordinate, 18, is the max height in meters.

The maxima of a differential equation can be obtained by getting the 1st derivate dx/dy and equating it to 0.

Given the equation h = - 2 t^2 + 12 t       , taking the 1st derivative result in:

dh = - 4 t dt + 12 dt

dh / dt = 0 = - 4 t + 12   calculating for t:

t = -12 / - 4

t = 3 s

Therefore the maximum height obtained is calculated by plugging in the value of t in the given equation.

h = -2 (3)^2 + 12 (3)

h = 18 m

This problem can also be solved graphically by plotting t (x-axis) against h (y-axis). Then assigning values to t and calculate for h and plot it in the graph to see the point in which the peak is obtained. Therefore the answer to this is:

The ball reaches a maximum height of 18 meters. The maximum of h(t) can be found both graphically or algebraically, and lies at (3,18). The x-coordinate, 3, is the time in seconds it takes the ball to reach maximum height, and the y-coordinate, 18, is the max height in meters.

Step-by-step explanation:

y intercept is -6

the slope is already in it's fraction form but if you want it back to whole number it's y=1.33x-6

the slope is positive

Some plots you can put are (6, 2) and (12, 10)

Just start at (0, -6) and counts 4 up and 3 right

Step-by-step explanation:

We need to find the scatter plot that is closest to y = x.

First of all, you must know the behaviour of y = x. That equation represents a straight line that passes through the origin of the coordinate system.

So, the right scatter plot must have the majority of points on this line that passes through the origin.

Notice that the Scatter plot A has this beahivour, if you draw a straight line through the origin and the points, you'll observe that the line best fits.

On the other hand, the other scatter plots are not following this linear behaviour.

Therefore, the right answer is A.

Step-by-step explanation:

y intercept is -6

the slope is already in it's fraction form but if you want it back to whole number it's y=1.33x-6

the slope is positive

Some plots you can put are (6, 2) and (12, 10)

Just start at (0, -6) and counts 4 up and 3 right

Step-by-step explanation:

We need to find the scatter plot that is closest to y = x.

First of all, you must know the behaviour of y = x. That equation represents a straight line that passes through the origin of the coordinate system.

So, the right scatter plot must have the majority of points on this line that passes through the origin.

Notice that the Scatter plot A has this beahivour, if you draw a straight line through the origin and the points, you'll observe that the line best fits.

On the other hand, the other scatter plots are not following this linear behaviour.

Therefore, the right answer is A.