So I believe you go
![(x_{1}- x_{2)}](https://tex.z-dn.net/?f=%20%28x_%7B1%7D-%20x_%7B2%29%7D%20)
and substitute x an y like this
(1-7) (7-1)
and get a slope of -9/6
Sorry If I got it wrong,but I hope I helped
Answer:
Step-by-step explanation:
15.5884573 i
(7y² + 6xy) - (-2xy + 3)
⇒ (-1 * -2xy) + (-1 * 3) = 2xy - 3
7y² + 6xy + 2xy - 3
7y² + 8xy - 3 Choice B.
Answer:
542 minutes
Step-by-step explanation:
currently our mean is =530
i got that by adding all the number out which equal too =2,650
and divide it by how many number there are which is 5
2,650/5=530
After some trial and error i found it
so you add 494+690+502+478+486+542=3,192
3,192/6=532
and that was our goal
so the 6th month she talked on the phone for 542 minutes
pls mark me brainliest
Answer:
The horizontal distance from the plane to the person on the runway is 20408.16 ft.
Step-by-step explanation:
Consider the figure below,
Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°
BC is the horizontal distance from the plane to the person on the runway.
We have to find distance BC,
Using trigonometric ratio,
![\tan\theta=\frac{Perpendicular}{base}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cfrac%7BPerpendicular%7D%7Bbase%7D)
Here,
,Perpendicular AB = 4000
![\tan\theta=\frac{AB}{BC}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cfrac%7BAB%7D%7BBC%7D)
![\tan 11.1^{\circ} =\frac{4000}{BC}](https://tex.z-dn.net/?f=%5Ctan%2011.1%5E%7B%5Ccirc%7D%20%3D%5Cfrac%7B4000%7D%7BBC%7D)
Solving for BC, we get,
![BC=\frac{4000}{\tan 11.1^{\circ} }](https://tex.z-dn.net/?f=BC%3D%5Cfrac%7B4000%7D%7B%5Ctan%2011.1%5E%7B%5Ccirc%7D%20%7D)
(approx)
(approx)
Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft