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....
Replace x = 3 into <span>6/x + 2x^2
</span>6/x + 2x^2
=6/3 + 2 (3)^2
= 2 + 2(9)
= 2 + 18
= 20
The digital time for quarter past three is 3:15, it is 15 minutes past 3 o'clock.
A. 72÷9÷2 = 8÷2 = 4
b. (18 ÷ 6) ÷ 3 = 3 ÷ 3 = 1
c. 45 ÷ 5 ÷ 3 = 9 ÷ 3 = 3
d. 144 ÷ (12 ÷ 2) = 144 ÷6 = 24
Undistribute z

divide both sides by (v-k^5)