the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
Answer:
=204.13 inches.
Step-by-step explanation:
Using the side x, we can use sine to find the hypotenuse of the triangle with the angle marked 30°.
Sin 30 =x/hypotenuse.
sin 30 = 27/hyp
hyp= 27/sin 30
=54 inches
We can also find the adjacent as follows.
Cos 30 = adjacent/ 54
Adjacent= 54 cos 30
=46.77 inches
Using the angles marked 45 we can find the hypotenuse of the isosceles triangle.
sin 45= x/hypotenuse
sin 45 =27/hypotenuse
hypotenuse = 27/sin 45
=38.18 inches
The hypotenuse of both the triangles making the isosceles triangle are 38.18 inches long.
Perimeter = 54+ 46.77+27+ 38.18+38.18
=204.13 inches.
Answer:
Step-by-step explanation:
this is a set problem (mathematical set)
30 students played soccer
9 students played cricket and soccer
x students played neither cricket nor soccer
3x students played cricket only.
we can see that in the first set 21 appears and in the middle of the set 9 because it says 30 students played soccer adding x and 3x we have the following formula
30 + x + 3x = 50 where x is equal to 5 then ask us Determine the number of students who played cricket. then the value would be 3x + 9 = (3x5) + 9 = 24
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,
Here, 
,Perpendicular AB = 4000
Solving for BC, we get,
(approx)
(approx)
Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft
The answer is 10...............
