L1: 2x+4y-3=0 ..........(1)
P: (2,0)
The point on the line L1 closest to the given point P is at the intersection of L1 with L2, which is the perpendicular passing through P.
Slope of L1=-2/4=-1/2
Slope of L2=-1/(-1/2)=2
Since it passes throug P(2,0), we can use the point-slope formula:
(y-0)=2(x-2) =>
L2: 2x-y-4=0.............(2)
Solve for x & y using (1) and (2) to get intersection point required:
(1)-(2)
2x-2x + 4y-(-y) -3 -(-4) =0
5y=-1, y=-1/5
Substitute y=1/5 in equation (1)
2x+4(-1/5)-3=0 =>
2x-19/5=0
x=19/10
=> the point on L1 closest to (2,0) is (19/10, -1/5)
The angles ∠ABC and ∠CBE together should equal 180º.
∠ABC is 155º.
180 - 155 = 25
So, ∠CBE = 25º.
Hope this helps!
9514 1404 393
Answer:
x^2/1024 -y^2/7744 = 1
Step-by-step explanation:
The parent hyperbola relation is ...
x^2 -y^2 = 1
This has asymptotes of y = ±x and x-intercepts of ±1.
For the given hyperbola, we want to scale x by a factor of 32, and y by a factor that is 2.75 times that, or 88. Then the equation could be written as ...
(x/32)^2 -(y/88)^2 = 1
More conventionally, the denominator is shown at full value:
x^2/1024 -y^2/7744 = 1
Answer:
✔ a/f
✔ cf
✔ c/b
✔ cf + ec
✔ c
Step-by-step explanation:
just did it on edg
Median is in the middle witch would be 479 and mode is 113
