Answer:
three thousand five hundred seventy nine
Step-by-step explanation:
Answer:
The graph is here, in the photo.
Step-by-step explanation:
-4.46cm
Step-by-step explanation:
The horizontal midpoint of the page is;
The horizontal midpoint of the image is given as;
For the image to be centered, these to midpoints must be on the same point in the axis.
If we take the leftmost edge of the paper to be point zero on the axis, then, the distance accommodated by the paper is 10.795cm. The distance that goes beyond this leftmost edge is computed as;
On the axis, this is on the negative side. Therefore, the horizontal position for the left edge of the image that Joelle should use to center the image is -4.46cm.
The answer is 0.07
have a good day
Answer:
![t = t_1 +t_2 = 4 sec+6 sec = 10 sec](https://tex.z-dn.net/?f=%20t%20%3D%20t_1%20%2Bt_2%20%3D%204%20sec%2B6%20sec%20%3D%2010%20sec)
Step-by-step explanation:
Let's call the initial point
and the second point
we can find the distance between these two points with the following formula:
![d_1 = \sqrt{(x_r -x_i)^2 +(y_r -y_i)^2}](https://tex.z-dn.net/?f=%20d_1%20%3D%20%5Csqrt%7B%28x_r%20-x_i%29%5E2%20%2B%28y_r%20-y_i%29%5E2%7D)
And if we replace we got:
![d_1 = \sqrt{(16 -8)^2 +(20 -10)^2} =2\sqrt{41}](https://tex.z-dn.net/?f=%20d_1%20%3D%20%5Csqrt%7B%2816%20-8%29%5E2%20%2B%2820%20-10%29%5E2%7D%20%3D2%5Csqrt%7B41%7D)
So since we know the time in order to reach the point R we can find the velocity like this:
![v=\frac{d_1}{t_1} = \frac{2\sqrt{41}}{4}=\frac{\sqrt{41}}{2}](https://tex.z-dn.net/?f=%20v%3D%5Cfrac%7Bd_1%7D%7Bt_1%7D%20%3D%20%5Cfrac%7B2%5Csqrt%7B41%7D%7D%7B4%7D%3D%5Cfrac%7B%5Csqrt%7B41%7D%7D%7B2%7D)
And this velocity is constant along all the displacement.
Let's call the final point ![F =(x_f = 28, y_f = 35)](https://tex.z-dn.net/?f=%20F%20%3D%28x_f%20%3D%2028%2C%20y_f%20%3D%2035%29)
And we can find the distance between the point R and F
like this:
![d_2 = \sqrt{(x_f -x_r)^2 +(y_f -y_r)^2}](https://tex.z-dn.net/?f=%20d_2%20%3D%20%5Csqrt%7B%28x_f%20-x_r%29%5E2%20%2B%28y_f%20-y_r%29%5E2%7D)
And if we replace we got:
![d_2 = \sqrt{(28 -16)^2 +(35 -20)^2}=3\sqrt{41}](https://tex.z-dn.net/?f=%20d_2%20%3D%20%5Csqrt%7B%2828%20-16%29%5E2%20%2B%2835%20-20%29%5E2%7D%3D3%5Csqrt%7B41%7D)
Since the velocity is constant we can find the time between point R and F like this:
![t_2 = \frac{3\sqrt{41}}{\frac{\sqrt{41}}{2}}=6 sec](https://tex.z-dn.net/?f=%20t_2%20%3D%20%5Cfrac%7B3%5Csqrt%7B41%7D%7D%7B%5Cfrac%7B%5Csqrt%7B41%7D%7D%7B2%7D%7D%3D6%20sec)
And we are interested on "When will it pick up the ball located at position (28,35)?" And then the total time would be:
![t = t_1 +t_2 = 4 sec+6 sec = 10 sec](https://tex.z-dn.net/?f=%20t%20%3D%20t_1%20%2Bt_2%20%3D%204%20sec%2B6%20sec%20%3D%2010%20sec)
The figure attached illustrate the problem.