Answer:
Step-by-step explanation:
It's linear because the general form of a linear function is:
f(x) = ax + b
In your example a = -10, b = 10
A) | 3 - t | <68
This is because anytime you seeless than, the number will go in front. From makes me think of = so I hope that helped!
= ![\frac{5}{7}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B7%7D%20%20)
Cross multiply. (7)(x) = (5)(x - 2)
7x = 5x - 10
Subtract 5x from both sides.
2x = -10
Divide both sides by 2.
x = -5
1000 is the answer that you are looking for
Answer:
![\frac{(x-1)^2}{81}+\frac{(y-6)^2}{72}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-1%29%5E2%7D%7B81%7D%2B%5Cfrac%7B%28y-6%29%5E2%7D%7B72%7D%3D1)
Step-by-step explanation:
We have been given that the center of an ellipse is (1,6). One focus of the ellipse is (-2,6). One vertex of the ellipse is (10,6).
We know that standard equation of an ellipse centered at (h,k) is in form:
, where,
a = Horizontal radius
b = Vertical radius.
Since center of the ellipse is at point (1,6) and one vertex is (10,6), so the horizontal radius would be 9 (10-1) units.
Now, we will find vertical radius using formula
, where c represents focal length.
![\text{Focal length}=1-(-2)](https://tex.z-dn.net/?f=%5Ctext%7BFocal%20length%7D%3D1-%28-2%29)
![\text{Focal length}=1+2](https://tex.z-dn.net/?f=%5Ctext%7BFocal%20length%7D%3D1%2B2)
![\text{Focal length}=3](https://tex.z-dn.net/?f=%5Ctext%7BFocal%20length%7D%3D3)
Substituting given values:
![b^2=9^2-3^2](https://tex.z-dn.net/?f=b%5E2%3D9%5E2-3%5E2)
![b^2=81-9](https://tex.z-dn.net/?f=b%5E2%3D81-9)
![b^2=72](https://tex.z-dn.net/?f=b%5E2%3D72)
Upon substituting
,
and
in standard form of ellipse, we will get:
![\frac{(x-1)^2}{9^2}+\frac{(y-6)^2}{72}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-1%29%5E2%7D%7B9%5E2%7D%2B%5Cfrac%7B%28y-6%29%5E2%7D%7B72%7D%3D1)
![\frac{(x-1)^2}{81}+\frac{(y-6)^2}{72}=1](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-1%29%5E2%7D%7B81%7D%2B%5Cfrac%7B%28y-6%29%5E2%7D%7B72%7D%3D1)
Therefore, the required equation of the ellipse in standard form would be
.