Answer:
![\frac{51}{10} or \: 5 \times \frac{1}{10}](https://tex.z-dn.net/?f=%20%5Cfrac%7B51%7D%7B10%7D%20or%20%5C%3A%205%20%5Ctimes%20%5Cfrac%7B1%7D%7B10%7D%20)
Step-by-step explanation:
The solution is in the image
The correct answer is Choice D.
The equation y = kx is a proportional graph. It means that you multiply x by a number and you have k.
The point (0, 0) will always be on the graph, because no matter what K is, if you are multiplying by 0 for x, you will get 0 for y.
Answer: Arrows are used to show the feeding relationship between the animals.
Step-by-step explanation: The arrow points from the organism being eaten to the organism that eats it.
Answer:
So the coordinates are ![(1+ \sqrt{31}, 1-\sqrt{31}),(1-\sqrt{31} , 1+\sqrt{31})](https://tex.z-dn.net/?f=%20%281%2B%20%5Csqrt%7B31%7D%2C%201-%5Csqrt%7B31%7D%29%2C%281-%5Csqrt%7B31%7D%20%2C%201%2B%5Csqrt%7B31%7D%29)
Step-by-step explanation:
For this case we assume that the line is given by this formula ![y = 2-x](https://tex.z-dn.net/?f=%20y%20%3D%202-x)
We know that the general equation for a circle is given by:
![(x-h)^2 +(y-k)^2 = r^2](https://tex.z-dn.net/?f=%20%28x-h%29%5E2%20%2B%28y-k%29%5E2%20%3D%20r%5E2%20)
Since the circle for this case is centered at the origin then h,k =0 and we have this:
(1)
For this case we can replace the formula for y from the line into equation (1) and we got:
![x^2 + (2-x)^2= 64](https://tex.z-dn.net/?f=%20x%5E2%20%2B%20%282-x%29%5E2%3D%2064)
We know from algebra that
if we use this concept we got:
![x^2 +4 -4x + x^2 = 64](https://tex.z-dn.net/?f=%20x%5E2%20%2B4%20-4x%20%2B%20x%5E2%20%3D%2064)
We can subtract 64 on both sides and we got:
![2x^2 -4x -60 =0](https://tex.z-dn.net/?f=%202x%5E2%20-4x%20-60%20%3D0)
Now we can divide both sides by 2 and we got:
![x^2 -2x -30 = 0](https://tex.z-dn.net/?f=%20x%5E2%20-2x%20-30%20%3D%200)
And we can use the quadratic formula to solve this:
![x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2%20-4ac%7D%7D%7B2a%7D)
For this case
and if we replace we got:
![x = \frac{2 \pm \sqrt{2-4*(1)*(-30)}}{2}](https://tex.z-dn.net/?f=%20x%20%3D%20%5Cfrac%7B2%20%5Cpm%20%5Csqrt%7B2-4%2A%281%29%2A%28-30%29%7D%7D%7B2%7D)
And we got ![x_1= 1+\sqrt{31} , x_2 = 1-\sqrt{31}](https://tex.z-dn.net/?f=%20x_1%3D%201%2B%5Csqrt%7B31%7D%20%2C%20x_2%20%3D%201-%5Csqrt%7B31%7D)
Now we just need to replace into the original equation for the line and we get the y coordinates like this:
So the coordinates are ![(1+ \sqrt{31}, 1-\sqrt{31}),(1-\sqrt{31} , 1+\sqrt{31})](https://tex.z-dn.net/?f=%20%281%2B%20%5Csqrt%7B31%7D%2C%201-%5Csqrt%7B31%7D%29%2C%281-%5Csqrt%7B31%7D%20%2C%201%2B%5Csqrt%7B31%7D%29)