I believe the correct answer is a hope this helps
Option A ( 2 , 4 , 6 , 8) because domain are x values
Answer:
The approximate distance is 15416 miles....
Step-by-step explanation:
We have given:
A satellite is 19,000 miles from the horizon of earth.
The radius is 4,000 miles.
Lets say that BC =x
AO = OB = 4,000 miles
AC = 19,000 miles
The tangent from the external point forms right angle with the radius of the circle.
So in ΔABC
(OC)² = (AC)²+(OA)²
where OC = x+4000
AC = 19,000
OA = 4000
Therefore,
(x+4000)² = (19,000)² +(4,000)²
Take square root at both sides:
√(x+4000)² = √(19,000)² +(4,000)²
x+4000 =√361000000+16000000
x+4000 = √377000000
x+4000 = 19416.48
x= 19416.48 - 4000
x = 15416.48
Therefore the approximate distance is 15416 miles....
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS = = 57.5°
Now, tan(57.5°) =
⇒ 1.5697 =
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) =
⇒ tan65° =
⇒ QM =
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = () × (ST)
= () × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²
4/2 * 4/5 = (4*4) ^ (2*5) = 1 3/5 = 1.6