6 A radar device has an antenna that revolves at a constant rate. The graph shows the
number of revolutions the device will make over time.
X RADAR DEVICE ANTENNA
Number of Revolutions
100
90
80
70
60
50
40
30
20
10
0 1 2 3 4 5 6 7 8 9 10
Time (minutes)
Which table shows the data for an antenna that revolves at exactly twice the rate
of the antenna described in the graph?
ANTENNA #1
ANTENNA #3
Time Number of
Time Number of
А
(minutes) Revolutions
С
(minutes) Revolutions
15
315
20
40
30
660
25
50
ANTENNA #2
ANTENNA #4
Time
(minutes)
Number of
Revolutions
Time
(minutes)
Number of
Revolutions
B
D
18
450
22
660
36
900
24
720

for any of you who need the translation here it is

step-by-step explanation:

find the angular size of the equatorial radius of the earth, if the horizontal parallax of the sun

75%

step-by-step explanation:

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.